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a/publication/publication.ptx b/publication/publication.ptx index 6fcbe811..f7e7aa7d 100644 --- a/publication/publication.ptx +++ b/publication/publication.ptx @@ -19,8 +19,8 @@ - - + + diff --git a/source/activities/act-changing-aroc-graphs.xml b/source/activities/act-changing-aroc-graphs.xml index cc8c872d..8569821f 100755 --- a/source/activities/act-changing-aroc-graphs.xml +++ b/source/activities/act-changing-aroc-graphs.xml @@ -21,7 +21,7 @@ -

f is a function defined on [-1,7] such that f(1) = 4 and AV_{[1,3]} = -2. +

f is a function defined on [-1,7] such that f(1) = 4 and AV_{[1,3]} = -2.

ADD ALT TEXT TO THIS IMAGE

 @@ -30,7 +30,20 @@ - + +

+ Since AV_{[1,3]} = -2 and f(1) = 4, we need + \frac{f(3)-4}{3-1} = -2, so f(3) = 0. Any graph passing through + (1,4) and (3,0) satisfies the conditions. Two examples: + the line f(x) = -2x+6, or a curve such as f(x) = -(x-1)^2 + 4 + (which also passes through both required points). +

+ + + + + + @@ -43,7 +56,20 @@ - + +

+ Since AV_{[0,4]} = 0.5 and g(4) = 3, we need + \frac{3 - g(0)}{4} = 0.5, so g(0) = 1. Any graph passing through + (0,1) and (4,3) that is not always increasing on (0,4) + satisfies the conditions. For example, a curve that dips below the secant line + between x=0 and x=4 before returning to (4,3). +

+ + + + + + @@ -56,7 +82,15 @@ - + +

+ These conditions are impossible to satisfy simultaneously. Given h(2) = 5 + and h(4) = 3, the average rate of change is determined: + AV_{[2,4]} = \frac{h(4) - h(2)}{4-2} = \frac{3-5}{2} = -1 \ne -2. + No function can have both the specified output values and the specified average + rate of change on [2,4]. +

+

diff --git a/source/activities/act-changing-aroc-population.xml b/source/activities/act-changing-aroc-population.xml index 0b76d780..be78a4c2 100755 --- a/source/activities/act-changing-aroc-population.xml +++ b/source/activities/act-changing-aroc-population.xml @@ -79,7 +79,24 @@ - + +

+ For Kent County, the population in 1990 was 500,631 and in 2010 was 602,622, so + + AV_{[1990,2010]} &= \frac{602{,}622 - 500{,}631}{2010 - 1990} + &= \frac{101{,}991}{20} + &\approx 5099.55 \text{ people per year.} + +

+

+ For Ottawa County, the population in 1990 was 187,768 and in 2010 was 263,801, so + + AV_{[1990,2010]} &= \frac{263{,}801 - 187{,}768}{2010 - 1990} + &= \frac{76{,}033}{20} + &\approx 3801.65 \text{ people per year.} + +

+ @@ -89,7 +106,11 @@ - + +

+ The units on AV_{[1990,2010]} for both counties are people per year. +

+ @@ -99,7 +120,12 @@ - + +

+ Between 1990 and 2010, the population of Ottawa County grew by an average of + approximately 3801.65 people per year. +

+ @@ -110,7 +136,19 @@ - + +

+ For Kent County on [2000,2010]: + AV_{[2000,2010]} = \frac{602{,}622 - 574{,}336}{10} = \frac{28{,}286}{10} \approx 2828.6 \text{ people per year.} + For Ottawa County on [2000,2010]: + AV_{[2000,2010]} = \frac{263{,}801 - 238{,}313}{10} = \frac{25{,}488}{10} = 2548.8 \text{ people per year.} + Kent County had a greater average rate of change during [2000,2010]. + In both counties, the population increased in every decade from 1960 to 2010, + so the average rate of change was positive on every decade interval for both counties. +

+ + + @@ -120,7 +158,14 @@ - + +

+ Using the most recent decade's rate of change as a guide, Ottawa County grew at + 2548.8 people per year from 2000 to 2010. Projecting this rate forward + eight years from 2010: + 263{,}801 + 8 \cdot 2548.8 \approx 284{,}191 \text{ people.} +

+

diff --git a/source/activities/act-changing-aroc-trends.xml b/source/activities/act-changing-aroc-trends.xml index 2a546740..0b1d1eb3 100755 --- a/source/activities/act-changing-aroc-trends.xml +++ b/source/activities/act-changing-aroc-trends.xml @@ -30,7 +30,16 @@ - + +

+ Computing the function values: q(0)=0, q(1)=3, q(2)=4, + q(3)=3, q(4)=0. Thus: + AV_{[0,1]} = \frac{3-0}{1} = 3, \quad AV_{[1,2]} = \frac{4-3}{1} = 1, + AV_{[2,3]} = \frac{3-4}{1} = -1, \quad AV_{[3,4]} = \frac{0-3}{1} = -3. + Since AV_{[2,3]} and AV_{[3,4]} are both negative, q is + decreasing on [2,4]. +

+ @@ -43,7 +52,18 @@ - + +

+ Computing function values: h(-1) = 3 - 2(0.5)^{-1} = 3 - 4 = -1, + h(1) = 3 - 2(0.5) = 2, h(3) = 3 - 2(0.5)^3 = \frac{11}{4} \approx 2.75, + h(5) = 3 - 2(0.5)^5 = \frac{47}{16} \approx 2.9375. Thus: + AV_{[-1,1]} = \frac{2-(-1)}{2} = \frac{3}{2} = 1.5, + AV_{[1,3]} = \frac{\frac{11}{4} - 2}{2} = \frac{3}{8} \approx 0.375, + AV_{[3,5]} = \frac{\frac{47}{16} - \frac{11}{4}}{2} = \frac{3}{32} \approx 0.094. + All three average rates of change are positive but decreasing, so h is + increasing on [-1,5], but at a decreasing rate. +

+ @@ -63,7 +83,20 @@ - + +

On the graph of q, draw four line segments: one connecting (0,0) and (1,3) (slope 3), one connecting (1,3) and (2,4) (slope 1), one connecting (2,4) and (3,3) (slope -1), and one connecting (3,3) and (4,0) (slope -3). On the graph of h, draw three line segments: one connecting (-1,-1) and (1,2) (slope 1.5), one connecting (1,2) and (3,2.75) (slope 0.375), and one connecting (3,2.75) and (5,2.9375) (slope 0.094). The decreasing slopes on the left graph reflect that q increases then decreases, while the positive but decreasing slopes on the right reflect that h is increasing at a decreasing rate.

+ + + + + + + + + + + + @@ -73,7 +106,14 @@ - + +

+ False. Although AV_{[0,3]} = \frac{q(3)-q(0)}{3} = \frac{3-0}{3} = 1 > 0, + a positive average rate of change over an interval does not mean the function is + increasing everywhere on that interval. From part (a), AV_{[2,3]} = -1 < 0, + so q is actually decreasing on [2,3]. +

+ @@ -86,13 +126,19 @@ - + +

+ Any linear function has constant average rate of change. For example, + f(x) = x has AV_{[a,b]} = \frac{b-a}{b-a} = 1 for every interval + [a,b]. +

+

-

+

See solutions to individual tasks above.

diff --git a/source/activities/act-changing-combining-arithmetic.xml b/source/activities/act-changing-combining-arithmetic.xml index fc4829ff..e978222e 100755 --- a/source/activities/act-changing-combining-arithmetic.xml +++ b/source/activities/act-changing-combining-arithmetic.xml @@ -37,7 +37,12 @@ - + +

+ From the graphs, f(0) = \frac{5}{2} and g(0) = 1, so + (f+g)(0) = \frac{5}{2} + 1 = \frac{7}{2}. +

+ @@ -47,7 +52,12 @@ - + +

+ From the graphs, f(1) = 4 and g(1) = 3, so + (g-f)(1) = 3 - 4 = -1. +

+ @@ -57,7 +67,12 @@ - + +

+ From the graphs, f(-1) = 3 and g(-1) = 3, so + (f \cdot g)(-1) = 3 \cdot 3 = 9. +

+ @@ -67,7 +82,12 @@ - + +

+ \left(\frac{f}{g}\right)(x) is undefined wherever g(x) = 0. + From the graph, g(x) = 0 at x = -2 and x = 3. +

+ @@ -77,7 +97,14 @@ - + +

+ (f \cdot g)(x) = 0 wherever f(x) = 0 or g(x) = 0. + From the graphs, f(x) = 0 at x = \frac{7}{3}, and + g(x) = 0 at x = -2 and x = 3. + So (f \cdot g)(x) = 0 at x \in \left\{-2,\, \frac{7}{3},\, 3\right\}. +

+ @@ -87,7 +114,19 @@ - + +

+ Yes: (f-g)(x) = 0 when f(x) = g(x). + One intersection is visible from the graphs at x = -1. + On the interval -1 \lt x \le 1, the formulas are f(x) = \frac{5}{2} + x + and g(x) = -x^2 + 4; setting them equal gives: + + x^2 + x - \frac{3}{2} &= 0 + x &= \frac{-1 + \sqrt{7}}{2}. + + So (f-g)(x) = 0 at x = -1 and x = \dfrac{\sqrt{7}-1}{2}. +

+

diff --git a/source/activities/act-changing-combining-context.xml b/source/activities/act-changing-combining-context.xml index f6370f69..67a4d1d9 100755 --- a/source/activities/act-changing-combining-context.xml +++ b/source/activities/act-changing-combining-context.xml @@ -34,7 +34,12 @@ - + +

+ At a speed of 60 miles per hour, the car consumes 0.04 gallons of fuel + for each mile traveled. +

+ @@ -47,7 +52,13 @@ - + +

+ g(60) = \frac{1}{f(60)} = \frac{1}{0.04} = 25 miles per gallon. + The function g measures the car's fuel economy in miles per gallon + at a given speed. +

+ @@ -60,7 +71,14 @@ - + +

+ h(60) = 60 \cdot f(60) = 60 \cdot 0.04 = 2.4 gallons per hour. + The units follow from: miles/hour \times gallons/mile = gallons/hour. + The function h measures the rate at which the car consumes fuel + (in gallons per hour) at a given speed. +

+ @@ -70,7 +88,14 @@ - + +

+ All three convey information about fuel consumption at 60 mph, but with + different units and perspectives: f(60) = 0.04 gives gallons per mile, + g(60) = 25 gives miles per gallon, and h(60) = 2.4 gives + gallons per hour. They describe the same phenomenon from different angles. +

+ @@ -84,7 +109,14 @@ - + +

+ AV_{[60,70]} = \frac{f(70)-f(60)}{70-60} = \frac{0.045 - 0.04}{10} = 0.0005 + \ \frac{\text{gal/mi}}{\text{mph}}. + This measures how much the car's fuel consumption rate (in gallons per mile) + increases per additional mile per hour of speed. +

+

diff --git a/source/activities/act-changing-combining-piecewise.xml b/source/activities/act-changing-combining-piecewise.xml index 6933598b..db411fce 100755 --- a/source/activities/act-changing-combining-piecewise.xml +++ b/source/activities/act-changing-combining-piecewise.xml @@ -37,7 +37,17 @@ - + +

+ For x \lt 0, use p(x) = -(x+2)^2 + 2: + p(-4) = -(-2)^2 + 2 = -2; + p(-2) = -(0)^2 + 2 = 2. + For x \ge 0, use p(x) = \frac{1}{2}(x-2)^2 + 1: + p(0) = \frac{1}{2}(4) + 1 = 3; + p(2) = 0 + 1 = 1; + p(4) = \frac{1}{2}(4) + 1 = 3. +

+ @@ -47,7 +57,14 @@ - + +

+ The left-side parabola -(x+2)^2 + 2 (valid for x \lt 0) opens + downward and has vertex (-2, 2). + The right-side parabola \frac{1}{2}(x-2)^2 + 1 (valid for x \ge 0) + opens upward and has vertex (2, 1). +

+ @@ -57,7 +74,16 @@ - + +

+ For x \lt 0, setting -(x+2)^2 + 2 = 0 gives (x+2)^2 = 2, + so x = -2 \pm \sqrt{2}. Both values x = -2 + \sqrt{2} \approx -0.586 + and x = -2 - \sqrt{2} \approx -3.414 lie in x \lt 0, so both are + zeros of p. The right-side parabola \frac{1}{2}(x-2)^2+1 \ge 1 \gt 0 + for all x \ge 0, so it contributes no zeros. + The y-intercept is p(0) = 3. +

+ @@ -77,7 +103,11 @@ - + +

The graph of y = p(x) consists of two parabolic pieces. For x \lt 0, the piece -(x+2)^2+2 is a downward-opening parabola with vertex (-2,2) that passes through (-4,-2), (-2+\sqrt{2},0), (-2-\sqrt{2},0), and approaches (0,3) from the left (open endpoint). For x \ge 0, the piece \frac{1}{2}(x-2)^2+1 is an upward-opening parabola with vertex (2,1) that starts at (0,3) (closed endpoint) and passes through (4,3). The graph has a jump at x = 0 but the values approach 3 from both sides, so the graph is actually continuous there.

+ + + @@ -87,13 +117,15 @@ - + +

Reading from the graph of f, the function appears piecewise linear on three intervals. For -2.5 \lt x \le -1, the line passes through (-2,3.5) and (-1,3), giving slope -\frac{1}{2} and formula f(x) = 3 - \frac{1}{2}(x+1). For -1 \lt x \le 1, the line passes through (0,2.5) and (1,4), giving slope \frac{3}{2} and formula f(x) = 4 + \frac{3}{2}(x-1). For 1 \lt x \lt 3.5 (excluding x = 2), the line passes through (1,1) and (3,-0.5), giving slope -\frac{3}{4} and formula f(x) = 1 - \frac{3}{4}(x-1). Thus f(x) = \begin{cases} 3 - \frac{1}{2}(x+1), \amp -2.5 \lt x \le -1 \\ 4 + \frac{3}{2}(x-1), \amp -1 \lt x \le 1 \\ 1 - \frac{3}{4}(x-1), \amp 1 \lt x \lt 3.5, \ x \ne 2 \end{cases}

+

-

+

See solutions to individual tasks above.

diff --git a/source/activities/act-changing-composite-aroc.xml b/source/activities/act-changing-composite-aroc.xml index ac60b27f..90841075 100755 --- a/source/activities/act-changing-composite-aroc.xml +++ b/source/activities/act-changing-composite-aroc.xml @@ -27,7 +27,15 @@ - + +

+ + f(1+h) &= 2(1+h)^2 - 3(1+h) + 1 + &= 2(1 + 2h + h^2) - 3 - 3h + 1 + &= 2h^2 + h. + +

+ @@ -38,7 +46,12 @@ - + +

+ Since f(1) = 2 - 3 + 1 = 0 and f(1+h) = 2h^2 + h: + AV_{[1,1+h]} = \frac{f(1+h) - f(1)}{h} = \frac{2h^2 + h}{h} = 2h + 1. +

+ @@ -49,7 +62,12 @@ - + +

+ g(1+h) = \frac{5}{1+h}. + This expression cannot be simplified further. +

+ @@ -60,7 +78,15 @@ - + +

+ Since g(1) = 5: + + AV_{[1,1+h]} &= \frac{g(1+h) - g(1)}{h} = \frac{\frac{5}{1+h} - 5}{h} + &= \frac{5 - 5(1+h)}{(1+h)h} = \frac{-5h}{(1+h)h} = \frac{-5}{1+h}. + +

+

diff --git a/source/activities/act-changing-composite-crickets-celsius.xml b/source/activities/act-changing-composite-crickets-celsius.xml index ca7ada63..ae7a3b91 100755 --- a/source/activities/act-changing-composite-crickets-celsius.xml +++ b/source/activities/act-changing-composite-crickets-celsius.xml @@ -27,7 +27,11 @@ - + +

+ H(N) = G(D(N)) = \frac{5}{9}((40 + 0.25N) - 32) = \frac{5}{9}(8 + 0.25N). +

+ @@ -37,7 +41,12 @@ - + +

+ The function H converts the number of cricket chirps per minute directly + into a temperature in degrees Celsius. +

+ @@ -57,7 +66,14 @@ - + +

Both Dolbear's function D and H = (G \circ D) are linear functions of N, so their graphs have the same shape. Dolbear's function has slope \frac{1}{4} (degrees Fahrenheit per chirp per minute), while H(N) = \frac{5}{9}(0.25N+8) has a smaller slope of \frac{5}{36} (degrees Celsius per chirp per minute). The vertical scale on the plot of H should reflect the Celsius range: H(40) = 10 (corresponding to 50^\circ F) and H(180) = \frac{265}{9} \approx 29.4 (corresponding to 85^\circ F). The plot of H is an increasing line on [40, 180], similar to Dolbear's function but with a shallower slope and Celsius units on the vertical axis.

+ + + + + + @@ -67,13 +83,20 @@ - + +

+ As an abstract mathematical function, both the domain and range are all real + numbers. In the context of Dolbear's model (domain [40, 180] chirps per + minute), the domain of H is [40, 180] and the corresponding range + is approximately [10, 29.4] degrees Celsius. +

+

-

+

See solutions to individual tasks above.

diff --git a/source/activities/act-changing-composite-tables-graphs.xml b/source/activities/act-changing-composite-tables-graphs.xml index ecbddb72..7857ecd8 100755 --- a/source/activities/act-changing-composite-tables-graphs.xml +++ b/source/activities/act-changing-composite-tables-graphs.xml @@ -71,7 +71,11 @@ - + +

+ From the graph, q(0) = 2 and p(2) = 1, so p(q(0)) = 1. +

+ @@ -81,7 +85,12 @@ - + +

+ From the graph, p(0) = -\frac{1}{2} and q\!\left(-\frac{1}{2}\right) = 2, + so q(p(0)) = 2. +

+ @@ -89,7 +98,11 @@ - + +

+ From the graph, p(-1) = -1, so (p \circ p)(-1) = p(p(-1)) = p(-1) = -1. +

+ @@ -99,7 +112,11 @@ - + +

+ From the table, g(2) = 0 and f(0) = 6, so (f \circ g)(2) = 6. +

+ @@ -109,7 +126,11 @@ - + +

+ From the table, f(3) = 4 and g(4) = 2, so (g \circ f)(3) = 2. +

+ @@ -119,7 +140,12 @@ - + +

+ From the table, f(0) = 6, but g(6) is not defined in the table, + so g(f(0)) is undefined. +

+ @@ -129,7 +155,14 @@ - + +

+ We need f(g(x)) = 4, which requires g(x) = 1 or g(x) = 3 + (since f(1)=4 and f(3)=4 from the table). From the table, + g(0) = 1 and g(1) = 3, so f(g(x)) = 4 for x = 0 + and x = 1. +

+ @@ -139,7 +172,15 @@ - + +

+ We need q(p(x)) = 1, which (from the graph of q) requires + p(x) = -\frac{4}{3} or p(x) = 2. From the graph of p, + p(x) = 2 at x = 2 and at x \approx -2.5; there are no + x-values for which p(x) = -\frac{4}{3}. So q(p(x)) = 1 + for x = 2 and x \approx -2.5. +

+

diff --git a/source/activities/act-changing-functions-is-it.xml b/source/activities/act-changing-functions-is-it.xml index 7b4ba376..00ad7f1a 100755 --- a/source/activities/act-changing-functions-is-it.xml +++ b/source/activities/act-changing-functions-is-it.xml @@ -37,7 +37,17 @@ - + +

+ The circle is not a function of x, because some values of x + correspond to more than one value of y. For example, when x = 0, + both y = 4 and y = -4 lie on the circle. +

+

+ The curve in the righthand figure is a function of x, because + each value of x in the domain corresponds to exactly one value of y. +

+ @@ -47,7 +57,14 @@ - + +

+ The closing value of the S&P500 can be expressed as a function of the day of + the year, provided the domain is restricted to trading days (excluding weekends and + market holidays). With this restriction, each day in the domain corresponds to + exactly one closing value. +

+ @@ -57,7 +74,13 @@ - + +

+ The odometer reading cannot be viewed as a function of the car's velocity. + The same speed can occur at many different odometer readings throughout a trip, + so a single velocity value can correspond to multiple odometer values. +

+ @@ -112,7 +135,12 @@ - + +

+ The table does not represent a function, because the input values + x = 1 and x = 2 each correspond to more than one output value. +

+

diff --git a/source/activities/act-changing-functions-spherical-tank-draining.xml b/source/activities/act-changing-functions-spherical-tank-draining.xml index e3b7e09b..24c66467 100755 --- a/source/activities/act-changing-functions-spherical-tank-draining.xml +++ b/source/activities/act-changing-functions-spherical-tank-draining.xml @@ -34,7 +34,20 @@ - + +

+ At t = 0, the tank is full and the height of the water is 8 m. + At t = 1 minute, 0.5 m of water has drained and the height is 7.5 m. + At t = 2 minutes, 1 m of water has drained and the height is 7 m. +

+

+ The tank has 8 m of water (depth), which is 16 half-meters, with one + half-meter draining each minute. Thus it will take 16 minutes for the tank + to drain completely. The linear height model + h = q(t) = 8 - 0.5t + gives q(t) = 0 when t = 16. +

+ @@ -44,7 +57,14 @@ - + +

+ The domain of each model is determined by the time it takes the tank to drain. + Assuming the tank begins draining at t = 0 minutes and is empty at + t = 16 minutes, the domain of both models is 0 \le t \le 16, + or equivalently [0, 16]. +

+ @@ -54,7 +74,19 @@ - + +

+ When the tank is full, the depth of the water equals the diameter of the tank, + 2 \times 4 = 8 m. The volume of water in the full tank equals the volume + of the sphere: + \frac{4}{3}\pi (4)^3 = \frac{256\pi}{3} \approx 268 \text{ m}^3. +

+

+ The range of the volume model is 0 \le V \le \frac{256\pi}{3}, going from + completely empty to completely full. The range of the height model is likewise + 0 \le h \le 8. +

+ @@ -66,7 +98,11 @@ - + +

The graph of V = p(t) = \frac{256\pi}{3} - \frac{\pi}{24}t^2(24-t) on [0,16] is a curve that decreases from V(0) = \frac{256\pi}{3} \approx 268 cubic meters down to V(16) = 0. The curve is concave down for small t (volume decreasing rapidly as the full, wide sphere loses water) and concave up near t = 16 (volume decreasing more slowly as the narrow bottom empties). The domain is [0, 16] minutes and the range is \left[0, \frac{256\pi}{3}\right] cubic meters.

+ + + @@ -76,7 +112,14 @@ - + +

+ The model V = p(t) differs from the abstract function y = r(x) + in its domain and range. The model's domain is restricted to [0, 16] + and its range to \left[0, \frac{256\pi}{3}\right] by the physical context. + The abstract function has an unrestricted domain and an unrestricted range. +

+ @@ -86,7 +129,13 @@ - + +

+ The height function should be linear and decreasing, because the height decreases + by the same amount (0.5 m) every minute. The formula is + q(t) = 8 - 0.5t. +

+

diff --git a/source/activities/act-changing-functions-spherical-tank.xml b/source/activities/act-changing-functions-spherical-tank.xml index f32d3374..12a68ab4 100755 --- a/source/activities/act-changing-functions-spherical-tank.xml +++ b/source/activities/act-changing-functions-spherical-tank.xml @@ -33,7 +33,14 @@ - + +

+ Since depth can't be negative and can't exceed the tank's diameter of 8 m, + the values of h that make sense are 0 \le h \le 8. + The corresponding values of V range from 0 (empty tank) to + f(8) = \frac{256\pi}{3} (full tank), so 0 \le V \le \frac{256\pi}{3}. +

+ @@ -43,7 +50,16 @@ - + +

+ The domain of f in context is [0, 8], since the water depth + can range from 0 m (empty) to 8 m (full diameter). + The range is \left[0, \frac{256\pi}{3}\right]. + Since volumes are non-negative, the codomain can be taken as [0, \infty), + or more broadly as all real numbers if we consider the abstract formula + without physical constraints. +

+ @@ -54,7 +70,19 @@ - + +

+ + f(2) &= \frac{\pi}{3}(4)(10) = \frac{40\pi}{3} \approx 41.9 \text{ m}^3 + f(4) &= \frac{\pi}{3}(16)(8) = \frac{128\pi}{3} \approx 134.0 \text{ m}^3 + f(8) &= \frac{\pi}{3}(64)(4) = \frac{256\pi}{3} \approx 268.1 \text{ m}^3 + + These give the volume of water when the depth is 2, 4, and 8 + meters respectively. The value f(8) = \frac{256\pi}{3} equals the volume + of a sphere of radius 4, confirming that the tank is completely full when + h = 8. +

+ @@ -64,7 +92,15 @@ - + +

+ Neither claim is valid. The domain of f is [0, 8], so + h = 9 is outside the domain — the water cannot be 9 m deep in a + tank whose diameter is only 8 m. Similarly, f(13) falls outside the + domain, and the negative value it produces is further evidence that the formula + should not be applied beyond h = 8. +

+ @@ -74,7 +110,13 @@ - + +

+ No. The maximum value of f on its domain [0, 8] is + f(8) = \frac{256\pi}{3} \approx 268.1 m^3, which is less than + 300. The tank simply cannot hold 300 cubic meters of water. +

+

diff --git a/source/activities/act-changing-inverse-Dolbear.xml b/source/activities/act-changing-inverse-Dolbear.xml index 65c5fe6e..61a0ae57 100755 --- a/source/activities/act-changing-inverse-Dolbear.xml +++ b/source/activities/act-changing-inverse-Dolbear.xml @@ -27,7 +27,12 @@ - + +

+ Subtracting 40 and multiplying by 4: + E(F) = N = 4(F - 40). +

+ @@ -37,7 +42,12 @@ - + +

+ The function E takes a temperature in degrees Fahrenheit as its input and + outputs the corresponding number of snowy tree cricket chirps per minute. +

+ @@ -47,7 +57,15 @@ - + +

+ j(N) = E(D(N)) = 4\!\left(\left(40 + \frac{1}{4}N\right) - 40\right) = 4 \cdot \frac{N}{4} = N. + k(F) = D(E(F)) = 40 + \frac{1}{4}(4(F-40)) = 40 + (F-40) = F. + Both j and k are the identity: composing D and E + in either order returns the original input, confirming that E and D + are inverse functions. +

+ @@ -57,7 +75,14 @@ - + +

+ They express the same relationship between F and N, just solved + for different variables. Each equation can be obtained from the other by + algebraic manipulation, so they describe the same connection between temperature + and chirp rate. +

+

diff --git a/source/activities/act-changing-inverse-does-it.xml b/source/activities/act-changing-inverse-does-it.xml index fd637c8c..42b460ed 100755 --- a/source/activities/act-changing-inverse-does-it.xml +++ b/source/activities/act-changing-inverse-does-it.xml @@ -52,7 +52,12 @@ - + +

+ The function f does not have an inverse, because f(1) = 2 = f(4), + so f^{-1}(2) would need to equal both 1 and 4 — not a function. +

+ @@ -87,7 +92,12 @@ - + +

+ The function g does have an inverse, since all five output values + are distinct. For example, g^{-1}(4) = 0 and g^{-1}(0) = 1. +

+ @@ -97,7 +107,14 @@ - + +

+ The function p(t) = 7 - \frac{3}{5}t is linear and one-to-one, so it + does have an inverse. Solving y = 7 - \frac{3}{5}t for t: + p^{-1}(y) = \frac{5}{3}(7-y). + For example, p^{-1}(7) = 0 and p^{-1}(4) = 5. +

+ @@ -107,7 +124,13 @@ - + +

+ The function q(t) = 7 - \frac{3}{5}t^4 does not have an inverse. + Since q(1) = q(-1) = 7 - \frac{3}{5}, the value q^{-1}\!\left(\frac{32}{5}\right) + could be either 1 or -1. +

+ @@ -127,7 +150,17 @@ - + +

+ The function r(t) passes the horizontal line test, so it does + have an inverse. For example, r^{-1}(2) = -1 and r^{-1}(0) = 0. +

+

+ The function s(t) does not have an inverse, because it fails the + horizontal line test: a horizontal line between y=1 and y=2 crosses + the graph in more than one place. +

+

diff --git a/source/activities/act-changing-inverse-rainfall.xml b/source/activities/act-changing-inverse-rainfall.xml index e9a1cb62..13a73d30 100755 --- a/source/activities/act-changing-inverse-rainfall.xml +++ b/source/activities/act-changing-inverse-rainfall.xml @@ -30,7 +30,13 @@ - + +

+ g(3) = \frac{4}{3+2} + 1 = \frac{4}{5} + 1 = \frac{9}{5} = 1.8 \text{ cm/hr.} + At t = 3 hours into the storm, the rain is falling at a rate of 1.8 + centimeters per hour. +

+ @@ -40,7 +46,15 @@ - + +

+ Using g(3) = \frac{9}{5} and g(5) = \frac{4}{7} + 1 = \frac{11}{7}: + AV_{[3,5]} = \frac{\frac{11}{7} - \frac{9}{5}}{2} = \frac{\frac{55-63}{35}}{2} = \frac{-8/35}{2} = -\frac{4}{35} \approx -0.114 \text{ cm/hr}^2. + Between hours 3 and 5, the rate of rainfall decreased on average by about + 0.114 cm/hr per hour. Since g is a decreasing function, we expect + the rainfall rate to continue decreasing throughout the storm. +

+ @@ -50,7 +64,17 @@ - + +

+ At the endpoints: g(0) = \frac{4}{2}+1 = 3 and + g(10) = \frac{4}{12}+1 = \frac{4}{3}. Since g is strictly + decreasing on [0,10], the range is \left[\frac{4}{3}, 3\right]. + The function has an inverse because it is strictly decreasing (one-to-one) on + this domain — it passes the horizontal line test. +

+ + + @@ -60,7 +84,17 @@ - + +

+ Setting \frac{9}{5} = \frac{4}{t+2} + 1: + + \frac{4}{5} &= \frac{4}{t+2} + t + 2 &= 5 \implies t = 3. + + So g^{-1}\!\left(\frac{9}{5}\right) = 3: the rainfall rate equals + 1.8 cm/hr at exactly 3 hours into the storm. +

+ @@ -70,7 +104,15 @@ - + +

+ No. Setting g(t) = 1 gives \frac{4}{t+2} + 1 = 1, so + \frac{4}{t+2} = 0, which has no solution. Since \frac{4}{t+2} > 0 + for all t in [0,10], we have g(t) > 1 throughout the storm. + (The range of g is \left[\frac{4}{3}, 3\right], and + 1 \notin \left[\frac{4}{3}, 3\right].) +

+

diff --git a/source/activities/act-changing-linear-Kilimanjaro.xml b/source/activities/act-changing-linear-Kilimanjaro.xml index ebb32d6f..26838d17 100755 --- a/source/activities/act-changing-linear-Kilimanjaro.xml +++ b/source/activities/act-changing-linear-Kilimanjaro.xml @@ -35,7 +35,14 @@ - + +

+ Using the points (0, 1951) and (7, 1555), the slope is + \frac{1555-1951}{7-0} = \frac{-396}{7} \approx -56.6 m^2/year. + Since t=0 gives A = 1951, the model is: + f(t) = 1951 - \frac{396}{7} t \approx 1951 - 56.6t. +

+ @@ -45,7 +52,13 @@ - + +

+ The slope \approx -56.6 m^2/year means the ice cover decreases by + about 56.6 square meters per year. The A-intercept of 1951 m^2 + represents the ice cover in the year 2000 (t=0). +

+ @@ -57,7 +70,13 @@ - + +

+ f(17) \approx 1951 - 56.6(17) \approx 989 \text{ m}^2. + In the year 2017, the model predicts that the ice field near the summit of + Mt. Kilimanjaro will have an area of approximately 989 square meters. +

+ @@ -68,7 +87,16 @@ - + +

+ Setting f(t) = 0: + + 0 &\approx 1951 - 56.6t + t &\approx \frac{1951}{56.6} \approx 34.5. + + The model predicts the ice will vanish around the year 2034. +

+ @@ -80,7 +108,12 @@ - + +

+ A reasonable domain is 0 \le t \le 34.5 (from the year 2000 until the ice + disappears). The corresponding range is 0 \le A \le 1951. +

+

diff --git a/source/activities/act-changing-linear-finding-eqs.xml b/source/activities/act-changing-linear-finding-eqs.xml index 75ad7cd6..42322d66 100755 --- a/source/activities/act-changing-linear-finding-eqs.xml +++ b/source/activities/act-changing-linear-finding-eqs.xml @@ -28,7 +28,11 @@ - + +

+ y - (-17) = \frac{3}{7}(x - (-11)), \quad \text{i.e.,} \quad y + 17 = \frac{3}{7}(x+11). +

+ @@ -38,7 +42,13 @@ - + +

+ The slope is m = \frac{-1-5}{3-(-2)} = -\frac{6}{5}. Using the point + (3,-1): + y + 1 = -\frac{6}{5}(x-3). +

+ @@ -48,7 +58,13 @@ - + +

+ Solving 2x - 3y = 5 for y gives y = \frac{2}{3}x - \frac{5}{3}, + so the slope is m = \frac{2}{3}. Using the point (4,9): + y - 9 = \frac{2}{3}(x-4). +

+ @@ -99,7 +115,17 @@ - + +

+ The function appears to be linear because the rate of change is constant: + for each increase of 1 in x, f(x) decreases by 2, + giving slope m = -2. Using the point (1,7): + + y - 7 &= -2(x-1) + f(x) &= 9 - 2x. + +

+ @@ -110,11 +136,19 @@ Plot of a linear function h. -->

ADD ALT TEXT TO THIS IMAGE

 + --> - + +

+ The graph has slope m = -\frac{1}{3} and passes through (4,1), so: + + y - 1 &= -\frac{1}{3}(x-4) + h(x) &= \frac{7}{3} - \frac{1}{3}x. + +

+

diff --git a/source/activities/act-changing-linear-in-context.xml b/source/activities/act-changing-linear-in-context.xml index f0466841..4c1acaf5 100755 --- a/source/activities/act-changing-linear-in-context.xml +++ b/source/activities/act-changing-linear-in-context.xml @@ -28,7 +28,11 @@ - + +

+ f(t) = 28750 + 825t. +

+ @@ -38,7 +42,12 @@ - + +

+ The slope is -465 people per year. This means the town's population + decreases by 465 people each year. +

+ @@ -48,7 +57,13 @@ - + +

+ p(t) = \frac{32\pi}{3} - 1.2t. + The tank empties when p(t) = 0, i.e., t = \frac{32\pi}{3 \cdot 1.2} \approx 27.9 minutes. + A reasonable domain is [0, 27.9]. +

+ @@ -58,7 +73,12 @@ - + +

+ Since q(t) = 0.65t is a linear function with slope 0.65, the water + level rises at a constant rate of 0.65 feet per minute. +

+ @@ -68,7 +88,18 @@ - + +

+ The slope is \frac{4600 - 10200}{10-5} = \frac{-5600}{5} = -1120 dollars per year. + Using the point (5, 10200): + + C - 10200 &= -1120(t-5) + L(t) &= 15800 - 1120t. + + The car reaches zero value when t \approx 14.1, so a reasonable domain is + [0, 14.1]. The slope means the car loses \$1120 in value each year. +

+

diff --git a/source/activities/act-changing-quadratic-falling-ball.xml b/source/activities/act-changing-quadratic-falling-ball.xml index f10d09cf..30b1e954 100755 --- a/source/activities/act-changing-quadratic-falling-ball.xml +++ b/source/activities/act-changing-quadratic-falling-ball.xml @@ -28,7 +28,13 @@ - + +

+ Using the standard model s(t) = -16t^2 + v_0 t + s_0 with initial height + s_0 = 37 ft and initial velocity v_0 = 41 ft/s: + s(t) = -16t^2 + 41t + 37. +

+ @@ -39,7 +45,11 @@ - + +

Plotting s(t) = -16t^2 + 41t + 37 in an appropriate window (for example, 0 \le t \le 4 and 0 \le s \le 70) shows an upside-down parabola that rises from s(0) = 37, reaches a maximum near t \approx 1.28 seconds and s \approx 63.3 feet, and then falls back to ground level near t \approx 3.3 seconds. Answers will vary on the exact sketch.

+ + + @@ -49,7 +59,11 @@ - + +

+ From the graph, the balloon appears to land at approximately t \approx 3.3 seconds. +

+ @@ -60,7 +74,18 @@ - + +

+ Setting s(t) = 0 and applying the quadratic formula with a=-16, + b=41, c=37: + + t &= \frac{-41 \pm \sqrt{41^2 - 4(-16)(37)}}{2(-16)} + &= \frac{-41 \pm \sqrt{1681 + 2368}}{-32} + &= \frac{-41 \pm \sqrt{4049}}{-32}. + + Taking the root that gives t > 0: t = \frac{-41 - \sqrt{4049}}{-32} \approx 3.27 seconds. +

+ @@ -70,7 +95,13 @@ - + +

+ The vertex occurs at t = -\frac{b}{2a} = -\frac{41}{2(-16)} = \frac{41}{32} \approx 1.28 seconds. + The maximum height is + s\!\left(\frac{41}{32}\right) = -16 \cdot \frac{1681}{1024} + 41 \cdot \frac{41}{32} + 37 = \frac{4049}{64} \approx 63.3 \text{ feet.} +

+ @@ -83,7 +114,18 @@ - + +

+ Using s(1.5) = 62.5, s(2) = 55, s(2.5) = 39.5, s(3) = 16: + AV_{[1.5,2]} = \frac{55-62.5}{0.5} = -15 \text{ ft/s}, + AV_{[2,2.5]} = \frac{39.5-55}{0.5} = -31 \text{ ft/s}, + AV_{[2.5,3]} = \frac{16-39.5}{0.5} = -47 \text{ ft/s.} + These values represent the average downward velocity of the balloon over each + half-second interval; the balloon is falling faster and faster as it descends. +

+ + +

diff --git a/source/activities/act-changing-quadratic-parameters.xml b/source/activities/act-changing-quadratic-parameters.xml index 2aa55eed..2d35eb25 100755 --- a/source/activities/act-changing-quadratic-parameters.xml +++ b/source/activities/act-changing-quadratic-parameters.xml @@ -31,7 +31,13 @@ - + +

+ If a > 0 the parabola opens upward; if a < 0 it opens downward. + The larger |a| is, the narrower the parabola; the closer |a| is to + zero, the wider the parabola. +

+ @@ -41,7 +47,13 @@ - + +

+ With a = 1 and c = 0, changing b shifts the vertex + diagonally: making b > 0 moves the vertex left and down, while + making b < 0 moves it right and down. +

+ @@ -51,7 +63,13 @@ - + +

+ With a = 1 and b = 0, changing c moves the vertex straight + up or down along the y-axis. The vertex is at (0, c), and c + is the y-intercept of the function. +

+ @@ -63,7 +81,15 @@ - + +

+ The parameter a has the most clearly understandable effect: it controls + whether the parabola opens up or down and how wide it is. The parameter c + has a simple effect as well (pure vertical translation). The parameter b + has the most complicated effect because changing it moves the vertex diagonally + rather than purely up, down, or in terms of width. +

+ @@ -74,7 +100,16 @@ - + +

+ Yes. Since (1,12) and (2,12) have the same output, the axis of + symmetry is x = 1.5 and the vertex has x-coordinate 1.5. + Since the parabola must pass through (0,8), which is below the values at + x=1 and x=2, the parabola opens downward. With c = q(0) = 8 + and using vertex form or solving the system, the formula is + q(x) = -2x^2 + 6x + 8. +

+

diff --git a/source/activities/act-changing-quadratic-properties.xml b/source/activities/act-changing-quadratic-properties.xml index 11df9e0b..ea52eed6 100755 --- a/source/activities/act-changing-quadratic-properties.xml +++ b/source/activities/act-changing-quadratic-properties.xml @@ -27,7 +27,15 @@ - + +

+ Three points determine a unique quadratic, so exactly one such function exists. + Using the factored form q(x) = a(x+5)(x-10) and the y-intercept + q(0) = -1: + a(5)(-10) = -50a = -1 \implies a = \frac{1}{50}. + Thus q(x) = \frac{1}{50}x^2 - \frac{1}{10}x - 1. +

+ @@ -37,7 +45,13 @@ - + +

+ We can guarantee exactly two x-intercepts. Since the vertex (-3,-4) + is below the x-axis and the parabola opens upward, both arms must cross the + x-axis. +

+ @@ -47,7 +61,16 @@ - + +

+ Yes. Using vertex form q(x) = a(x+3)^2 - 4 and the point (-1,-3): + + -3 &= a(-1+3)^2 - 4 = 4a - 4 + a &= \frac{1}{4}. + + Thus q(x) = \frac{1}{4}(x+3)^2 - 4. +

+ @@ -57,7 +80,17 @@ - + +

+ The vertex (-1,9) is above the x-axis and a=-3 < 0, so the + parabola opens downward and has two x-intercepts. Setting p(x)=0: + + -3(x+1)^2 + 9 &= 0 + (x+1)^2 &= 3 + x &= -1 \pm \sqrt{3}. + +

+ @@ -67,7 +100,15 @@ - + +

+ The discriminant is b^2 - 4ac = 10^2 - 4(-2)(-20) = 100 - 160 = -60 < 0, + so there are no real x-intercepts. Equivalently, the vertex is at + x = \frac{-10}{2(-2)} = 2.5 with w(2.5) = -7.5 < 0, and since + a = -2 < 0 the parabola opens downward and lies entirely below the + x-axis. +

+

diff --git a/source/activities/act-changing-tandem-conical-tank.xml b/source/activities/act-changing-tandem-conical-tank.xml index 840f7034..f331e089 100755 --- a/source/activities/act-changing-tandem-conical-tank.xml +++ b/source/activities/act-changing-tandem-conical-tank.xml @@ -27,7 +27,9 @@ - + +

Answers will vary. A sketch should show an inverted cone with labeled radius 2 ft and depth 4 ft, with water partially filling the bottom of the tank.

+ @@ -37,7 +39,9 @@ - + +

Quantities that are changing include the volume of water in the tank, the height of the water, the surface area of the top of the water, and time. Quantities that are not changing include the rate at which water enters, the height of the tank, and the radius of the tank.

+ @@ -116,7 +120,9 @@ - + +

Since water enters at a constant rate of 0.75 cubic feet per minute, V = 0.75t. The table values are: t = 0, V = 0; t = 1, V = 0.75; t = 2, V = 1.5; t = 3, V = 2.25; t = 4, V = 3.0; t = 5, V = 3.75. The graph of V versus t is a straight line through the origin with slope 0.75.

+ @@ -130,13 +136,15 @@ - + +

Since the radius of the cone increases as the water level rises, the height of the water increases rapidly at first but the rate of increase slows as the water level rises. Sketches should show a concave-down increasing curve approaching the maximum depth of 4 ft.

+

-

+

See solutions to individual tasks above.

diff --git a/source/activities/act-changing-tandem-spherical-tank.xml b/source/activities/act-changing-tandem-spherical-tank.xml index 4baa5189..f05d31ae 100755 --- a/source/activities/act-changing-tandem-spherical-tank.xml +++ b/source/activities/act-changing-tandem-spherical-tank.xml @@ -27,7 +27,9 @@ - + +

Answers will vary. A sketch should show a sphere with labeled radius 3 ft, with water partially filling the tank.

+ @@ -37,7 +39,9 @@ - + +

Quantities that are changing include the volume of water in the tank, the height of the water, the surface area of the top of the water, and time. Quantities that are not changing include the rate at which water is pumped out and the radius of the tank.

+ @@ -47,7 +51,9 @@ - + +

Since the radius is 3 feet, the full tank contains \frac{4}{3}\pi(3)^3 = 36\pi \approx 113.1 cubic feet of water.

+ @@ -57,7 +63,9 @@ - + +

Water is removed at 1.2 cubic feet per minute. The tank drains when 1.2t = 36\pi, so t = \frac{36\pi}{1.2} = 30\pi \approx 94.25 minutes.

+ @@ -122,7 +130,9 @@ - + +

Since water is removed at a constant rate, V = 36\pi - 1.2t. The approximate values are: t = 0, V \approx 113.1; t = 20, V \approx 89.1; t = 40, V \approx 65.1; t = 60, V \approx 41.1; t = 80, V \approx 17.1; t = 94.24, V \approx 0. The graph is linear because water is removed at a constant rate.

+ @@ -136,13 +146,15 @@ - + +

When t = 0 the tank is full, so the height of the water is the diameter h = 6 ft. When the tank is empty, h = 0. The height decreases quickly at first, then the rate of decrease slows as the widest part of the sphere is reached, then speeds up again as the tank nears empty. Sketches should show an S-shaped decreasing curve from h = 6 to h = 0.

+

-

+

See solutions to individual tasks above.

 diff --git a/source/activities/act-changing-transformations-combined.xml b/source/activities/act-changing-transformations-combined.xml index f82caf17..6ac8760a 100755 --- a/source/activities/act-changing-transformations-combined.xml +++ b/source/activities/act-changing-transformations-combined.xml @@ -37,7 +37,16 @@ - + +

+ The function p(x) = -\frac{1}{2}f(x-1)+2 is obtained from f by + shifting right 1 unit, applying a vertical compression by \frac{1}{2} and + a reflection across the x-axis, then shifting up 2 units. + The point (-2, 2) on f moves to (-1, 1) on p. +

+ + + @@ -47,7 +56,16 @@ - + +

+ The function q(x) = 2g(x+0.5) - 0.75 is obtained from g by + shifting left 0.5 units, stretching vertically by a factor of 2, then + shifting down 0.75 units. + The point (1.5, 1.5) on g moves to (1, 2.25) on q. +

+ + + @@ -57,7 +75,17 @@ - + +

+ Algebraically: + r(x) = \frac{1}{2}(-f(x-1)-4) = -\frac{1}{2}f(x-1) - 2. + Since p(x) = -\frac{1}{2}f(x-1)+2, we have r(x) \ne p(x) + — they differ by 4 in output. + As transformations, both shift f right 1 unit and apply a vertical + compression and reflection by -\frac{1}{2}, but p then shifts + the result up 2 units while r shifts it down 2 units. +

+ @@ -70,7 +98,14 @@ - + +

+ s(x) = -2.5\,g(x+1.25) + 1.75. + The point (1.5, 1.5) on g moves to (0.25, -2) on s. +

+ + +

diff --git a/source/activities/act-changing-transformations-translations.xml b/source/activities/act-changing-transformations-translations.xml index a8af2e3b..ea93f507 100755 --- a/source/activities/act-changing-transformations-translations.xml +++ b/source/activities/act-changing-transformations-translations.xml @@ -37,7 +37,18 @@ - + +

+ The graph of g(x) = r(x) + 2 is r shifted vertically up 2 units; + the point (-2, -1) moves to (-2, 1). + The graph of h(x) = r(x+1) is r shifted horizontally to the left + 1 unit; the point (-2, -1) moves to (-3, -1). + The graph of f(x) = r(x+1) + 2 combines both transformations — a shift + left 1 unit and up 2 units — moving (-2, -1) to (-3, 1). +

+ + + @@ -47,7 +58,16 @@ - + +

+ The graph of k(x) = s(x) - 1 is s shifted down 1 unit; + the point (-2, -3) moves to (-2, -4). + The graph of j(x) = s(x-2) is s shifted right 2 units; + the point (-2, -3) moves to (0, -3). + The graph of m(x) = s(x-2) - 1 combines both — a shift right 2 units + and down 1 unit — moving (-2, -3) to (0, -4). +

+ @@ -57,7 +77,19 @@ - + +

+ Substituting q(x) = x^2: + + p(x) &= q(x+3) - 4 = (x+3)^2 - 4 + &= x^2 + 6x + 9 - 4 = x^2 + 6x + 5. + + The function p is a translation of q three units to the left + and four units down. +

+ + +

diff --git a/source/activities/act-changing-transformations-vert-stretch.xml b/source/activities/act-changing-transformations-vert-stretch.xml index d6f88d94..c5b3a626 100755 --- a/source/activities/act-changing-transformations-vert-stretch.xml +++ b/source/activities/act-changing-transformations-vert-stretch.xml @@ -37,7 +37,17 @@ - + +

+ The graph of g(x) = 3r(x) is a vertical stretch of r by a factor + of 3; the point (-2, -1) moves to (-2, -3). + The graph of h(x) = \frac{1}{3}r(x) is a vertical compression of r + by a factor of \frac{1}{3}; the point (-2, -1) moves to + (-2, -\frac{1}{3}). +

+ + + @@ -47,7 +57,17 @@ - + +

+ The graph of k(x) = -s(x) is a reflection of s across the + x-axis; the point (-2, -3) moves to (-2, 3). + The graph of j(x) = -\frac{1}{2}s(x) is a vertical compression by + \frac{1}{2} combined with a reflection across the x-axis; + the point (-2, -3) moves to (-2, \frac{3}{2}). +

+ + + @@ -61,7 +81,19 @@ - + +

+ For m(x) = 2r(x+1) - 1: the point (-2, -1) on r moves + to (-3, -3) on m. + For n(x) = \frac{1}{2}s(x-2) + 2: the point (-2, -3) on s + moves to (0, \frac{1}{2}) on n. +

+ + + + + + @@ -71,7 +103,16 @@ - + +

+ The function m(x) = 2r(x+1) - 1 results from three elementary + transformations of r: a horizontal shift left 1 unit (replacing x + with x+1), a vertical stretch by a factor of 2 (multiplying by 2), and + a vertical shift down 1 unit (subtracting 1). + The vertical stretch and the horizontal shift may be applied in either order, + but both must be applied before the vertical shift. +

+

diff --git a/source/activities/act-circular-sine-cosine-computing.xml b/source/activities/act-circular-sine-cosine-computing.xml index 301ce1a1..c64e9062 100755 --- a/source/activities/act-circular-sine-cosine-computing.xml +++ b/source/activities/act-circular-sine-cosine-computing.xml @@ -27,7 +27,11 @@ - + +

+ Using x^2 + \left(-\frac{3}{4}\right)^2 = 1, we solve for x = -\frac{\sqrt{7}}{4} \approx -0.66143 (negative since in Quadrant III). +

+ @@ -37,7 +41,11 @@ - + +

+ \sin(2 \text{ rad}) \approx 0.90929. +

+ @@ -47,7 +55,11 @@ - + +

+ \cos(-3.05 \text{ rad}) \approx -0.99581. +

+ @@ -57,7 +69,11 @@ - + +

+ If \sin(t) = \frac{1}{2} and t is in Quadrant II, then t = \frac{5\pi}{6} and \cos(t) = -\frac{\sqrt{3}}{2}. +

+ @@ -67,7 +83,11 @@ - + +

+ Using \sin^2(t) + \cos^2(t) = 1, we get \sin(t) = -\sqrt{1 - 0.49} = -\sqrt{0.51} \approx -0.71414 (negative since in Quadrant III). +

+ @@ -77,7 +97,16 @@ - + +

+ + AV_{[0.1,0.2]} = \frac{\sin(0.2) - \sin(0.1)}{0.2 - 0.1} \approx \frac{0.19867 - 0.09983}{0.1} \approx 0.98836 + + + AV_{[0.8,0.9]} = \frac{\sin(0.9) - \sin(0.8)}{0.9 - 0.8} \approx \frac{0.78333 - 0.71736}{0.1} \approx 0.65971 + +

+ @@ -87,7 +116,16 @@ - + +

+ + AV_{[0.1,0.2]} = \frac{\cos(0.2) - \cos(0.1)}{0.2 - 0.1} \approx \frac{0.98007 - 0.99500}{0.1} \approx -0.14938 + + + AV_{[0.8,0.9]} = \frac{\cos(0.9) - \cos(0.8)}{0.9 - 0.8} \approx \frac{0.62161 - 0.69671}{0.1} \approx -0.75097 + +

+

diff --git a/source/activities/act-circular-sine-cosine-incr-CCU.xml b/source/activities/act-circular-sine-cosine-incr-CCU.xml index 9adce395..c8c5b2aa 100755 --- a/source/activities/act-circular-sine-cosine-incr-CCU.xml +++ b/source/activities/act-circular-sine-cosine-incr-CCU.xml @@ -28,7 +28,11 @@ - + +

+ The function f(t) = \sin(t) is decreasing on the interval \left[\frac{\pi}{2}, \frac{3\pi}{2}\right]. +

+ @@ -39,7 +43,11 @@ - + +

+ The function f(t) = \sin(t) is decreasing and concave down on the interval \left[\frac{\pi}{2}, \pi\right]. +

+ @@ -50,7 +58,11 @@ - + +

+ The function g(t) = \cos(t) is increasing on the interval [\pi, 2\pi]. +

+ @@ -61,7 +73,11 @@ - + +

+ The function g(t) = \cos(t) is increasing and concave up on the interval \left[\pi, \frac{3\pi}{2}\right]. +

+ @@ -71,7 +87,11 @@ - + +

+ The average rate of change of g(t) = \cos(t) is greater on the interval \left[\frac{3\pi}{2}, \frac{3\pi}{2} + 0.1\right] because g(t) has a steeper slope near \frac{3\pi}{2} (a zero crossing) than near \pi (a minimum). +

+ @@ -81,7 +101,11 @@ - + +

+ The most rapid average rates of change on both graphs occur near the zero crossing points (where the function crosses the midline). +

+ @@ -91,7 +115,11 @@ - + +

+ The function g(t) = \cos(t) is negative on quadrants II and III. +

+

diff --git a/source/activities/act-circular-sine-cosine.xml b/source/activities/act-circular-sine-cosine.xml index f394cd66..edf94b91 100755 --- a/source/activities/act-circular-sine-cosine.xml +++ b/source/activities/act-circular-sine-cosine.xml @@ -27,7 +27,11 @@ - + +

+ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}; \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}; \cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}. +

+ @@ -137,7 +141,59 @@ - + + + + t + 0 + \frac{\pi}{6} + \frac{\pi}{4} + \frac{\pi}{3} + \frac{\pi}{2} + \frac{2\pi}{3} + \frac{3\pi}{4} + \frac{5\pi}{6} + \pi + + + k + 1 + \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} + \frac{1}{2} + 0 + -\frac{1}{2} + -\frac{\sqrt{2}}{2} + -\frac{\sqrt{3}}{2} + -1 + + + + t + \pi + \frac{7\pi}{6} + \frac{5\pi}{4} + \frac{4\pi}{3} + \frac{3\pi}{2} + \frac{5\pi}{3} + \frac{7\pi}{4} + \frac{11\pi}{6} + 2\pi + + + k + -1 + -\frac{\sqrt{3}}{2} + -\frac{\sqrt{2}}{2} + -\frac{1}{2} + 0 + \frac{1}{2} + \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} + 1 + + + @@ -151,7 +207,9 @@ - + + + @@ -161,7 +219,11 @@ - + +

+ \cos\left(\frac{11\pi}{4}\right) = \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}; and \cos\left(\frac{14\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}. +

+ @@ -171,7 +233,11 @@ - + +

+ \cos(t) = -\frac{\sqrt{3}}{2} for t = \frac{5\pi}{6},\ \frac{7\pi}{6},\ \frac{17\pi}{6},\ \frac{19\pi}{6}. +

+ @@ -181,7 +247,11 @@ - + +

+ The graphs of k = \cos(t) and h = \sin(t) are actually identical, just with a horizontal shift of \pm\frac{\pi}{2}. +

+

diff --git a/source/activities/act-circular-sinusoidal-horiz-stretch.xml b/source/activities/act-circular-sinusoidal-horiz-stretch.xml index 9ce430c8..dfd9dc89 100755 --- a/source/activities/act-circular-sinusoidal-horiz-stretch.xml +++ b/source/activities/act-circular-sinusoidal-horiz-stretch.xml @@ -37,7 +37,12 @@ - + +

+ The function h(t) = f\!\left(\frac{1}{3}t\right) represents a horizontal scaling of f by a factor of 3 (stretching). The function j(t) = f(4t) represents a horizontal scaling by a factor of \frac{1}{4} (compression). +

+ + @@ -47,7 +52,12 @@ - + +

+ The function k(t) = g(2t) represents a horizontal scaling of g by a factor of \frac{1}{2} (compression). The function m(t) = g\!\left(\frac{1}{2}t\right) represents a horizontal scaling by a factor of 2 (stretching). +

+ + @@ -61,7 +71,12 @@ - + +

+ The function r(t) = 2f\!\left(\frac{1}{2}t\right) applies both a horizontal scaling by a factor of 2 and a vertical scaling by a factor of 2. The function s(t) = \frac{1}{2}g(2t) applies both a horizontal scaling by a factor of \frac{1}{2} and a vertical scaling by a factor of \frac{1}{2}. +

+ + @@ -71,7 +86,11 @@ - + +

+ The function r(t) = 2f\!\left(\frac{1}{2}t\right) is both a vertical and a horizontal scaling. The order does not matter: applying the horizontal stretch by 2 and then the vertical stretch by 2 produces the same result as applying them in the opposite order. +

+

diff --git a/source/activities/act-circular-sinusoidal-model.xml b/source/activities/act-circular-sinusoidal-model.xml index 3ba927ef..42a47e5b 100755 --- a/source/activities/act-circular-sinusoidal-model.xml +++ b/source/activities/act-circular-sinusoidal-model.xml @@ -31,7 +31,13 @@

-

+

+ The midline is c = \frac{1.5+4}{2} = 2.75 feet, the amplitude is a = \frac{4-1.5}{2} = 1.25 feet, the range is [1.5,4], and the anchor point is at approximately (0, 1.67). Since d(1.25) = 4 is the maximum and the period is 3, we have k = \frac{2\pi}{3} and b = 1.25. The formula is + + d(t) = 1.25\cos\!\left(\frac{2\pi}{3}(t - 1.25)\right) + 2.75. + +

+ diff --git a/source/activities/act-circular-sinusoidal-oscillator.xml b/source/activities/act-circular-sinusoidal-oscillator.xml index d4ae90a6..04d5232d 100755 --- a/source/activities/act-circular-sinusoidal-oscillator.xml +++ b/source/activities/act-circular-sinusoidal-oscillator.xml @@ -25,7 +25,9 @@

-

+

+ Note, the measurements of the system are physical, so exact values are not appropriate. The spring-mass system may be described in many ways. One formula is d(t) = -2.5\sin(t) + 4.5, and has an anchor point with coordinates (0, 4.5), corresponding to the starting position of the mass. Another formula with the same anchor point is d(t) = 2.5\cos\!\left(t + \frac{\pi}{2}\right) + 4.5. +

diff --git a/source/activities/act-circular-sinusoidal-period.xml b/source/activities/act-circular-sinusoidal-period.xml index b5ed366a..82c5eaf7 100755 --- a/source/activities/act-circular-sinusoidal-period.xml +++ b/source/activities/act-circular-sinusoidal-period.xml @@ -27,7 +27,11 @@ - + +

+ Period \frac{\pi}{5}, amplitude 1, midline y=2, range [1,3], no horizontal shift, anchor point (0,2). +

+ @@ -37,7 +41,11 @@ - + +

+ Period 8\pi, amplitude 3 (with reflection), midline y=-4, range [-7,-1], no horizontal shift, anchor point (0,-7). +

+ @@ -47,7 +55,11 @@ - + +

+ Period 8, amplitude 2, midline y=5, range [3,7], no horizontal shift, anchor point (0,5). +

+ @@ -57,7 +69,11 @@ - + +

+ Period 4, amplitude 2, midline y=5, range [3,7], horizontal shift 3 units to the right, anchor point (0,5). +

+ @@ -67,7 +83,11 @@ - + +

+ Note that u(x) = -0.25\sin(3(x-2)) + 5. Period \frac{2\pi}{3}, amplitude 0.25, midline y=5, range [4.75,5.25], horizontal shift 2 units to the right. The anchor point is at approximately (0, 4.93), since u(0) = -0.25\sin(-6)+5 is not a convenient exact value. +

+

diff --git a/source/activities/act-circular-traversing-2nd-ex.xml b/source/activities/act-circular-traversing-2nd-ex.xml index 0b6244b0..43330b23 100755 --- a/source/activities/act-circular-traversing-2nd-ex.xml +++ b/source/activities/act-circular-traversing-2nd-ex.xml @@ -37,7 +37,11 @@ - + +

+ The eight points on the circle are evenly spaced, and the circle has a circumference of 8 units. Thus, the distance between any two sequential points is 1 unit. +

+ @@ -47,7 +51,11 @@ - + +

+ The point P_2 is directly below the center, at coordinates (2, 2), and thus has a y-coordinate that is one radius less than 2. Using the formula for circumference, C = 2\pi r, the radius must be 8/(2\pi) = \frac{4}{\pi}. Therefore, the y-coordinate of point P_2 is 2 - \frac{4}{\pi} = \frac{2(\pi - 2)}{\pi}. At point P_4, the point is at the same height as the center, so the y-coordinate of P_4 is 2. +

+ @@ -174,7 +182,37 @@ - + +

+ The completed table is: +

+ + + d + 0123 + 4567 + 8 + + + h + 21.100.731.10 + 22.903.272.90 + 2 + + + + + d + 9101112 + 13141516 + + + h + 1.100.731.102 + 2.903.272.902 + + + @@ -184,7 +222,9 @@ - + + + @@ -194,7 +234,11 @@ - + +

+ The graph has a very similar shape to Figure 2.1.5. This graph is shifted up, shifted to the right, and is compressed horizontally compared to the textbook figure. +

+ @@ -204,7 +248,11 @@ - + +

+ To find h when d = 51, we use the periodicity of the function: 51 = 4 \cdot 8 + 3, so h(51) = h(3) \approx 1.10. Likewise, h(102) = h(6) \approx 3.27, because 102 = 12 \cdot 8 + 6. +

+

diff --git a/source/activities/act-circular-traversing-oscillator-aroc.xml b/source/activities/act-circular-traversing-oscillator-aroc.xml index 43ab841b..378917dd 100755 --- a/source/activities/act-circular-traversing-oscillator-aroc.xml +++ b/source/activities/act-circular-traversing-oscillator-aroc.xml @@ -179,7 +179,17 @@ - + +

+

    +
  • AV_{[2,2.25]} = \frac{9.913 - 8.00}{2.25 - 2} = \frac{1.913}{0.25} \approx 7.652 inches/sec
    • +
  • AV_{[2.25,2.5]} = \frac{11.536 - 9.913}{2.5 - 2.25} = \frac{1.623}{0.25} \approx 6.492 inches/sec
    • +
  • AV_{[2.5,2.75]} = \frac{12.619 - 11.536}{2.75 - 2.5} = \frac{1.083}{0.25} \approx 4.332 inches/sec
    • +
  • AV_{[2.75,3]} = \frac{13.000 - 12.619}{3.00 - 2.75} = \frac{0.381}{0.25} \approx 1.524 inches/sec
    • +
+ The weight is always moving in the same direction (away from the wall) on the interval [2,3], but is slowing down. +

+ @@ -189,7 +199,11 @@ - + +

+ The greatest (steepest) negative rates of change occur near the midline where the function is decreasing. Examples include intervals such as [0, 0.25], [3.75, 4.00], and [4.00, 4.25]. These occur near the midline where the function crosses from above to below (or vice versa), meaning the weight is moving most rapidly toward the equilibrium position. +

+ @@ -199,7 +213,11 @@ - + +

+ The function is decreasing from each peak to each trough. Examples of longest such intervals are [3,5], [7,9], and [11,13]. +

+ @@ -208,7 +226,11 @@ - + +

+ The intervals on which f is concave up include [0,2], [4,6], [8,10], and [12,14]. +

+ @@ -218,7 +240,11 @@ - + +

+ On intervals that are decreasing and concave up, the weight is moving toward the wall, but at a decreasing rate (slowing down). +

+ @@ -228,7 +254,11 @@ - + +

+ At the highest and lowest points, the function has the smallest rates of change. Near the midline, the function has its greatest rates of change. +

+

diff --git a/source/activities/act-circular-traversing-oscillator.xml b/source/activities/act-circular-traversing-oscillator.xml index 8b02c781..d2360730 100755 --- a/source/activities/act-circular-traversing-oscillator.xml +++ b/source/activities/act-circular-traversing-oscillator.xml @@ -181,7 +181,11 @@ - + +

+ The period of the motion is p = 4 seconds. The midline is y = 8 inches. The amplitude is a = 5 inches. +

+ @@ -191,7 +195,11 @@ - + +

+ The greatest displacement of the weight is equal to the maximum of the graph: 13 inches. The least displacement is the minimum: 3 inches. The range of f is [3, 13]. +

+ @@ -201,7 +209,19 @@ - + +

+ The average rate of change AV_{[4,4.25]} on the interval [4,4.25] is + + \frac{6.807 - 8.00}{4.25 - 4} = \frac{-1.193}{0.25} = -4.772 \text{ inches per second}. + + The average rate of change AV_{[4.75,5]} on the interval [4.75,5] is + + \frac{3.000 - 3.381}{5 - 4.75} = \frac{-0.381}{0.25} = -1.524 \text{ inches per second}. + + The average rate of change of this function represents the speed of the weight. Taken together, these two values indicate that the weight is moving more slowly on the interval [4.75,5] than on the interval [4,4.25], but that the weight is moving in the same direction on each of these intervals. +

+ @@ -211,7 +231,11 @@ - + +

+ The value of f(6.75) = f(2.75) = 12.619 inches. The value of f(11.25) = f(7.25) = f(3.25) = 12.619 inches. +

+

diff --git a/source/activities/act-circular-unit-circle-non-unit.xml b/source/activities/act-circular-unit-circle-non-unit.xml index e60a79ee..5669acb6 100755 --- a/source/activities/act-circular-unit-circle-non-unit.xml +++ b/source/activities/act-circular-unit-circle-non-unit.xml @@ -27,7 +27,11 @@ - + +

+ In a circle of radius 11, the arc length intercepted by a central angle of \frac{5\pi}{3} is 11 \cdot \frac{5\pi}{3} = \frac{55\pi}{3}. +

+ @@ -37,7 +41,11 @@ - + +

+ In a circle of radius 3, the central angle that intercepts an arc length \frac{\pi}{4} is \frac{\pi/4}{3} = \frac{\pi}{12}. +

+ @@ -48,7 +56,11 @@ - + +

+ The radius is \frac{\pi/2}{7\pi/6} = \frac{3}{7}. +

+ @@ -58,7 +70,11 @@ - + +

+ The angle corresponding to the arc length is \theta = \frac{25\pi/6}{5} = \frac{5\pi}{6}. But since \frac{25\pi}{6} = 4\pi + \frac{\pi}{6}, the effective angle is \frac{\pi}{6}. Using the unit circle coordinates at \frac{\pi}{6}, the point on the circle of radius 5 is \left( \frac{5\sqrt{3}}{2}, \frac{5}{2} \right). +

+

diff --git a/source/activities/act-circular-unit-circle-radians-degrees.xml b/source/activities/act-circular-unit-circle-radians-degrees.xml index e0a8565e..8cf334be 100755 --- a/source/activities/act-circular-unit-circle-radians-degrees.xml +++ b/source/activities/act-circular-unit-circle-radians-degrees.xml @@ -27,25 +27,37 @@ - + +

+ 30^\circ = \frac{\pi}{6} radians \approx 0.524 radians. +

+ -

\frac{2\pi}{3} radians +

\frac{2\pi}{3} radians

- + +

+ \frac{2\pi}{3} radians = 120^\circ. +

+ -

\frac{5\pi}{4} radians +

\frac{5\pi}{4} radians

- + +

+ \frac{5\pi}{4} radians = 225^\circ. +

+ @@ -55,7 +67,11 @@ - + +

+ 240^\circ = \frac{4\pi}{3} radians \approx 4.189 radians. +

+ @@ -65,7 +81,11 @@ - + +

+ 17^\circ = \frac{17\pi}{180} radians \approx 0.297 radians. +

+ @@ -74,7 +94,11 @@ - + +

+ 2 radians = \frac{360}{\pi}^\circ \approx 114^\circ. +

+

diff --git a/source/activities/act-circular-unit-circle-special-triangles.xml b/source/activities/act-circular-unit-circle-special-triangles.xml index 29b0ac94..1cf1c968 100755 --- a/source/activities/act-circular-unit-circle-special-triangles.xml +++ b/source/activities/act-circular-unit-circle-special-triangles.xml @@ -34,12 +34,16 @@

- For the 45^\circ-45^\circ-90^\circ triangle with legs of length x and y and hypotenuse of length 1, what does the fact that the triangle is isosceles tell us about the relationship between x and y? What are their exact values? + For the 45^\circ-45^\circ-90^\circ triangle with legs of length x and y and hypotenuse of length 1, what does the fact that the triangle is isosceles tell us about the relationship between x and y? What are their exact values?

- + +

+ Because a 45^\circ-45^\circ-90^\circ triangle is isosceles, the sides x and y must be equal. From the Pythagorean theorem, \sqrt{x^2 + y^2} = 1, which reduces to \sqrt{2x^2} = 1, giving x = y = \frac{\sqrt{2}}{2}. +

+ @@ -49,7 +53,11 @@ - + +

+ When the 30^\circ-60^\circ-90^\circ triangle is reflected across the x-axis, the outer perimeter of the resulting shape forms an equilateral triangle where all three interior angles are 60^\circ and all sides have the same length, 1 unit. This gives us the insight that side y must have a length equal to half of one unit: y = \frac{1}{2}. Side x is determined from the Pythagorean theorem: x = \frac{\sqrt{3}}{2}. +

+ @@ -59,7 +67,11 @@ - + +

+ When a 30^\circ-60^\circ-90^\circ triangle is constructed such that the angle between the hypotenuse and the x-side is 60^\circ, the sides are y = \frac{\sqrt{3}}{2} and x = \frac{1}{2}. +

+ @@ -86,7 +98,11 @@ - + +

+ The coordinates of the point at \theta = \frac{\pi}{6} on the unit circle are \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right). The coordinates at \theta = \frac{\pi}{4} are \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right). The coordinates at \theta = \frac{\pi}{3} are \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right). +

+

diff --git a/source/activities/act-exp-e-aroc-e.xml b/source/activities/act-exp-e-aroc-e.xml index cfe8ee05..5503b1e7 100755 --- a/source/activities/act-exp-e-aroc-e.xml +++ b/source/activities/act-exp-e-aroc-e.xml @@ -41,7 +41,11 @@ - + +

+ The function A evaluated at h = 0.5 gives the average rate of change of f(t) = e^t on the interval [1, 1.5]. +

+ @@ -51,7 +55,38 @@ - + + + + h + A(h) + + + 0.2 + 3.010 + + + 0.1 + 2.859 + + + 0.01 + 2.732 + + + 0.001 + 2.720 + + + 0.0001 + 2.718 + + + 0.00001 + 2.718 + + + @@ -61,7 +96,11 @@ - + +

+ As h gets smaller and smaller, the value of A(h) approaches the value of e. +

+ @@ -70,7 +109,11 @@ - + +

+ The statement makes sense because the rate of change on small intervals near t = 1 is close to the value of e^1 = e. +

+ @@ -84,7 +127,11 @@ - + +

+ The value of A(h) near h = 0 appears to be the same as the value of f(2) = e^2. +

+

diff --git a/source/activities/act-exp-e-graph-solve.xml b/source/activities/act-exp-e-graph-solve.xml index ab0d0174..0b70b5bb 100755 --- a/source/activities/act-exp-e-graph-solve.xml +++ b/source/activities/act-exp-e-graph-solve.xml @@ -27,7 +27,11 @@ - + +

+ For e^t = 2, t = \ln 2 \approx 0.6931. +

+ @@ -37,7 +41,11 @@ - + +

+ For e^{3t} = 5, t = \frac{\ln 5}{3} \approx 0.5365. +

+ @@ -47,7 +55,11 @@ - + +

+ The equation 2e^t - 4 = 7 simplifies to e^t = \frac{11}{2}, so t = \ln\!\left(\frac{11}{2}\right) \approx 1.7047. +

+ @@ -57,7 +69,11 @@ - + +

+ The equation 3e^{0.25t} + 2 = 6 simplifies to e^{0.25t} = \frac{4}{3}, so t = 4\ln\!\left(\frac{4}{3}\right) \approx 1.1507. +

+ @@ -67,7 +83,11 @@ - + +

+ The equation 4 - 2e^{-0.7t} = 3 simplifies to e^{-0.7t} = \frac{1}{2}, so t = \frac{\ln 2}{0.7} \approx 0.9902. +

+ @@ -77,13 +97,20 @@ - + +

+ The equation 2e^{1.2t} = 1.5e^{1.6t} simplifies to e^{-0.4t} = \frac{3}{4}, so t = \frac{-\ln(3/4)}{0.4} \approx 0.7192. +

+

-

+

+ A graph of f(t) = e^t showing the solutions to each equation is below. +

+ diff --git a/source/activities/act-exp-growth-a-b-t.xml b/source/activities/act-exp-growth-a-b-t.xml index 54c55a17..8c2e8fd3 100755 --- a/source/activities/act-exp-growth-a-b-t.xml +++ b/source/activities/act-exp-growth-a-b-t.xml @@ -27,7 +27,11 @@ - + +

+ The domain of g(t) = ab^t is all real numbers, (-\infty, \infty). +

+ @@ -37,7 +41,11 @@ - + +

+ Allowing for all possible values of a and b, the range of g(t) is all real numbers. If we restrict a and b to 0.1 \le a, b \le 10, the range of g(t) becomes (0, \infty). +

+ @@ -47,7 +55,11 @@ - + +

+ The y-intercept for g(t) is a. +

+ @@ -57,7 +69,11 @@ - + +

+ When b is less than 1, the graph approaches \infty as t approaches -\infty, and the graph approaches 0 when t approaches \infty. When b is greater than 1, the graph approaches 0 when t approaches -\infty and the graph approaches \infty when t approaches \infty. +

+ @@ -67,7 +83,11 @@ - + +

+ When b is greater than 1, the function describes positive growth (growing with time). When b is between 0 and 1, the function describes negative growth (decreasing with time). +

+ @@ -81,7 +101,11 @@ - + +

+ The function p(t) shows negative growth, so the value of b must be between 0 and 1. The function q(t) shows positive growth, so the value of d must be greater than 1. The function p has a y-intercept that is greater than the function q's y-intercept, thus a \gt c. +

+

diff --git a/source/activities/act-exp-growth-find-a-b.xml b/source/activities/act-exp-growth-find-a-b.xml index 4be850e7..145b0875 100755 --- a/source/activities/act-exp-growth-find-a-b.xml +++ b/source/activities/act-exp-growth-find-a-b.xml @@ -27,7 +27,11 @@ - + +

+ Given V(t) = ab^t, V(3) = \$12{,}500, and V(7) = \$6{,}500, the ratio of the two specified points gives b = \left( \dfrac{13}{25} \right)^{1/4} exactly. Solving either equation for a gives a = 12500 \left( \dfrac{25}{13} \right)^{3/4} = 6500 \left( \dfrac{25}{13} \right)^{7/4} exactly. In approximate form, V(t) \approx \$20{,}413 \times (0.8492)^t. +

+ @@ -37,7 +41,11 @@ - + +

+ Using the approximate model, the purchase value of the car is approximately $20{,}413. Solving 1000 = 20413 \times 0.8492^t using logarithms gives t \approx 18.45, so the car is worth less than $1000 after about 18 years. +

+ @@ -47,7 +55,11 @@ - + +

+ If the car's value is linear, then L(t) = b + mt. Using L(3) = 12500 and L(7) = 6500, we find m = (12500 - 6500)/(3 - 7) = -1500 dollars per year and b = \$17{,}000. Thus L(t) = 17000 - 1500t, and the car will be worth $1000 when t = \frac{32}{3} \approx 10.67 years. +

+ @@ -57,7 +69,11 @@ - + +

+ The exponential model seems more realistic because new cars decrease in value at a faster rate than older cars. +

+

diff --git a/source/activities/act-exp-growth-rates.xml b/source/activities/act-exp-growth-rates.xml index 781d623c..24fe5de2 100755 --- a/source/activities/act-exp-growth-rates.xml +++ b/source/activities/act-exp-growth-rates.xml @@ -27,7 +27,12 @@ - + +

+ If p is always decreasing at a constant rate, then p is a linear function of the form p(x) = mx + b, where m \lt 0. For example, p(x) = -4x + 2. +

+ + @@ -37,7 +42,12 @@ - + +

+ If q is always increasing and increases at an increasing rate, then q may be an exponential function. For example, q(x) = 2^x. +

+ + @@ -47,7 +57,16 @@ - + +

+ One option is the piecewise function + + r(t) = \begin{cases} -\left(\frac{1}{2}\right)^t & \text{if } t \lt 2 \\ \left(\frac{1}{2}\right)^t - \frac{1}{2} & \text{if } t \gt 2 \end{cases} + + whose rate of change is always decreasing with increasing t. +

+ + @@ -57,7 +76,12 @@ - + +

+ Let s(t) = -\left(\frac{1}{2}\right)^t, so that s(t) is always increasing with a decreasing rate. +

+ + @@ -67,7 +91,12 @@ - + +

+ The function u(t) = \left(\frac{1}{2}\right)^t is always decreasing at a decreasing rate. +

+ +

diff --git a/source/activities/act-exp-log-base-10.xml b/source/activities/act-exp-log-base-10.xml index 216b7c5a..ff5432bd 100755 --- a/source/activities/act-exp-log-base-10.xml +++ b/source/activities/act-exp-log-base-10.xml @@ -27,7 +27,11 @@ - + +

+ t = \log_{10}(0.00001) = -5, exactly. +

+ @@ -37,7 +41,11 @@ - + +

+ t = \log_{10}(1000000) = 6, exactly. +

+ @@ -47,7 +55,11 @@ - + +

+ t = \log_{10}(37) \approx 1.568. +

+ @@ -57,7 +69,11 @@ - + +

+ y = 10^{1.375} \approx 23.71. +

+ @@ -67,7 +83,11 @@ - + +

+ t = \log_{10}(0.04) \approx -1.398. +

+ @@ -77,7 +97,11 @@ - + +

+ 3 \cdot 10^t + 11 = 147 gives 10^t = \frac{136}{3}, so t = \log_{10}\!\left(\frac{136}{3}\right) \approx 1.656. +

+ @@ -87,7 +111,11 @@ - + +

+ 2\log_{10}(y) + 5 = 1 gives \log_{10}(y) = -2, so y = 10^{-2} = 0.01, exactly. +

+

diff --git a/source/activities/act-exp-log-equations.xml b/source/activities/act-exp-log-equations.xml index bd907701..6fc070ca 100755 --- a/source/activities/act-exp-log-equations.xml +++ b/source/activities/act-exp-log-equations.xml @@ -27,7 +27,11 @@ - + +

+ t = \ln\!\left(\frac{1}{10}\right) \approx -2.30 +

+ @@ -37,7 +41,11 @@ - + +

+ t = \ln\!\left(\frac{7}{5}\right) \approx 0.336 +

+ @@ -47,7 +55,11 @@ - + +

+ t = e^{-1/3} \approx 0.717 +

+ @@ -57,7 +69,11 @@ - + +

+ e^{1-3t} = 4 gives 1 - 3t = \ln 4, so t = \dfrac{1 - \ln 4}{3} \approx -0.129. +

+ @@ -67,7 +83,11 @@ - + +

+ 2\ln(t) + 1 = 4 gives \ln(t) = \frac{3}{2}, so t = e^{3/2} \approx 4.48. +

+ @@ -77,7 +97,11 @@ - + +

+ 4 - 3e^{2t} = 2 gives e^{2t} = \frac{2}{3}, so t = \frac{1}{2}\ln\!\left(\frac{2}{3}\right) \approx -0.203. +

+ @@ -87,7 +111,11 @@ - + +

+ There is no solution because the equation gives e^{2t} = -\frac{2}{3}, and there is no exponent for which e raised to that power is negative. +

+ @@ -97,7 +125,11 @@ - + +

+ \ln(5 - 6t) = -2 gives 5 - 6t = e^{-2}, so t = \frac{5 - e^{-2}}{6} \approx 0.811. +

+

diff --git a/source/activities/act-exp-log-exponential-equations.xml b/source/activities/act-exp-log-exponential-equations.xml index ee1c454f..9d0c4498 100755 --- a/source/activities/act-exp-log-exponential-equations.xml +++ b/source/activities/act-exp-log-exponential-equations.xml @@ -27,7 +27,11 @@ - + +

+ From 3^t = 5, we take the natural log of both sides to get t \ln 3 = \ln 5, so t = \dfrac{\ln 5}{\ln 3} \approx 1.46497. +

+ @@ -37,7 +41,11 @@ - + +

+ From 4 \cdot 2^t - 2 = 3, we get 2^t = \frac{5}{4}, so t \ln 2 = \ln\!\left(\frac{5}{4}\right) and t = \dfrac{\ln 5 - \ln 4}{\ln 2} \approx 0.32193. +

+ @@ -47,7 +55,11 @@ - + +

+ From 3.7 \cdot (0.9)^{0.3t} + 1.5 = 2.1, we get (0.9)^{0.3t} = \frac{0.6}{3.7}, so 0.3t \ln(0.9) = \ln\!\left(\frac{0.6}{3.7}\right) and t = \dfrac{\ln(0.6) - \ln(3.7)}{0.3\ln(0.9)} \approx 57.55. +

+ @@ -57,7 +69,11 @@ - + +

+ From 72 - 30(0.7)^{0.05t} = 60, we get (0.7)^{0.05t} = \frac{12}{30} = \frac{2}{5}, so 0.05t \ln(0.7) = \ln\!\left(\frac{2}{5}\right) and t = \dfrac{\ln(12) - \ln(30)}{0.05\ln(0.7)} \approx 51.38. +

+ @@ -67,7 +83,11 @@ - + +

+ From \ln(t) = -2, we get t = e^{-2} \approx 0.13534. +

+ @@ -77,7 +97,11 @@ - + +

+ From 3 + 2\log_{10}(t) = 3.5, we get \log_{10}(t) = 0.25, so t = 10^{0.25} \approx 1.77828. +

+

diff --git a/source/activities/act-exp-log-natural.xml b/source/activities/act-exp-log-natural.xml index 83ccb9b0..be0a98bf 100755 --- a/source/activities/act-exp-log-natural.xml +++ b/source/activities/act-exp-log-natural.xml @@ -27,7 +27,11 @@ - + +

+ The domain of E(t) = e^t is all real numbers, and its range is all positive real numbers. +

+ @@ -37,7 +41,11 @@ - + +

+ The domain of N(y) = \ln(y) is all positive real numbers, and its range is all real numbers. +

+ @@ -47,7 +55,11 @@ - + +

+ For every real number t, \ln(e^t) = t. +

+ @@ -57,7 +69,11 @@ - + +

+ For every positive real number y, e^{\ln(y)} = y. +

+ @@ -147,7 +163,53 @@ - + + + + t + -2 + -1 + 0 + 1 + 2 + + + E(t) = e^t (exact) + e^{-2} + e^{-1} + 1 + e + e^2 + + + E(t) = e^t (approx.) + 0.135 + 0.368 + 1 + 2.718 + 7.39 + + + + + y + e^{-2} + e^{-1} + 1 + e^1 + e^2 + + + N(y) = \ln(y) + -2 + -1 + 0 + 1 + 2 + + + + diff --git a/source/activities/act-exp-log-properties-exp-or-log.xml b/source/activities/act-exp-log-properties-exp-or-log.xml index 8b0dd7ea..f17470f2 100755 --- a/source/activities/act-exp-log-properties-exp-or-log.xml +++ b/source/activities/act-exp-log-properties-exp-or-log.xml @@ -27,7 +27,11 @@ - + +

+ Both f(t) = 1 - e^{-(t-1)} and g(t) = \ln(t) are always increasing and concave down, and both have a t-intercept at t = 1. The domain of f is all real numbers, while the domain of g is all positive real numbers. The range of f is the interval (-\infty, 1), while the range of g is all real numbers. The y-intercept of f is f(0) = 1 - e \approx -1.718, while g has no y-intercept since t = 0 is not in its domain. As t \to \infty, f(t) \to 1 (bounded), while g(t) \to \infty (unbounded). +

+ @@ -37,7 +41,11 @@ - + +

+ The function h(t) = a - be^{-k(t-c)} is a transformation of E(t) = e^t in which E(t) is reflected over the horizontal axis and vertically stretched by a factor of b, horizontally compressed by a factor of k (with a horizontal reflection), shifted horizontally to the right by c units, and shifted vertically up by a units. +

+ @@ -47,7 +55,11 @@ - + +

+ The function r(t) = a + b\ln(t - c) is a transformation of L(t) = \ln(t) that is vertically stretched by a factor of b, shifted horizontally to the right by c units, and shifted vertically up by a units. +

+ @@ -98,7 +110,11 @@ - + +

+ Experimenting with Desmos, the exponential model q(t) = m + ne^{-rt} provides a good fit to the data; for instance, q(t) = 20.3 - 20.8e^{-0.35t} fits well. A logarithmic model is not as effective because the data levels off at a horizontal asymptote, which is consistent with an exponential model of the form a + be^{-kt} but not with a logarithmic model whose values grow without bound. +

+

diff --git a/source/activities/act-exp-log-properties-find-k.xml b/source/activities/act-exp-log-properties-find-k.xml index 3994098f..486b83ab 100755 --- a/source/activities/act-exp-log-properties-find-k.xml +++ b/source/activities/act-exp-log-properties-find-k.xml @@ -27,7 +27,11 @@ - + +

+ From 41 = 50e^{-7k}, we get e^{-7k} = \frac{41}{50}, so -7k = \ln\!\left(\frac{41}{50}\right) and k = -\dfrac{\ln(41/50)}{7}. +

+ @@ -37,7 +41,11 @@ - + +

+ From 65 = 34 + 47e^{-45k}, we get 31 = 47e^{-45k}, so e^{-45k} = \frac{31}{47} and k = -\dfrac{\ln(31/47)}{45}. +

+ @@ -47,7 +55,11 @@ - + +

+ From 7e^{2k-1} + 4 = 32, we get e^{2k-1} = 4, so 2k - 1 = \ln 4 and k = \dfrac{\ln 4 + 1}{2}. +

+ @@ -57,7 +69,11 @@ - + +

+ From \dfrac{5}{1 + 2e^{-10k}} = 4, we get 5 = 4 + 8e^{-10k}, so e^{-10k} = \frac{1}{8} and k = -\dfrac{\ln(1/8)}{10} = \dfrac{\ln 8}{10}. +

+

diff --git a/source/activities/act-exp-modeling-behavior.xml b/source/activities/act-exp-modeling-behavior.xml index 8c09b602..52e174c6 100755 --- a/source/activities/act-exp-modeling-behavior.xml +++ b/source/activities/act-exp-modeling-behavior.xml @@ -44,7 +44,11 @@ - + +

+ The function p(t) = 4372(1.000235)^t + 92856 is always increasing, always concave up, and increasing without bound. The y-intercept is p(0) = 97{,}228 and the range is (92856, \infty). +

+ @@ -54,7 +58,11 @@ - + +

+ The function q(t) = 27931(0.97231)^t + 549786 is always decreasing, always concave up, and decreasing toward 549786. The y-intercept is q(0) = 577{,}717 and the range is (549786, \infty). +

+ @@ -64,7 +72,11 @@ - + +

+ The function r(t) = -17398(0.85234)^t is always increasing, always concave down, and increasing toward 0. The y-intercept is r(0) = -17398 and the range is (-\infty, 0). +

+ @@ -74,7 +86,11 @@ - + +

+ The function s(t) = -17398(0.85234)^t + 19411 is always increasing, always concave down, and increasing toward the value 19411. The y-intercept is s(0) = 2013 and the range is (-\infty, 19411). +

+ @@ -84,7 +100,11 @@ - + +

+ The function u(t) = -7522(1.03817)^t is always decreasing, always concave down, and decreasing without bound. The y-intercept is u(0) = -7522 and the range is (-\infty, 0). +

+ @@ -94,7 +114,11 @@ - + +

+ The function v(t) = -7522(1.03817)^t + 6731 is always decreasing, always concave down, and decreasing without bound. The y-intercept is v(0) = -791 and the range is (-\infty, 6731). +

+

diff --git a/source/activities/act-exp-modeling-potato.xml b/source/activities/act-exp-modeling-potato.xml index a2beb55a..20309a4c 100755 --- a/source/activities/act-exp-modeling-potato.xml +++ b/source/activities/act-exp-modeling-potato.xml @@ -31,7 +31,11 @@ - + +

+ The numerical value of F(0) is 68^\circ, because the potato begins at room temperature. Therefore, F(0) = a - b(0.98)^0 = a - b = 68. +

+ @@ -44,7 +48,11 @@ - + +

+ The long-range behavior of F(t) should reflect the potato approaching the oven temperature of 350^\circ. The expression (0.98)^t approaches zero as t gets large, so F(t) \to a. Therefore a = 350^\circ. +

+ @@ -54,7 +62,11 @@ - + +

+ The value of b is a - 68 = 350 - 68 = 282^\circ, since we established that a - b = 68 and a = 350. +

+ @@ -70,7 +82,12 @@ - + +

+ The function F(t) = 350 - 282(0.98)^t is shown below. The potato reaches 325 degrees after about 120 minutes of baking in the 350 degree oven. +

+ + @@ -80,7 +97,11 @@ - + +

+ We can view F(t) = a - b(0.98)^t as a transformation of the parent function f(t) = (0.98)^t by first stretching f vertically by a factor of b, then reflecting over the t-axis, and finally shifting vertically up by a units. +

+

diff --git a/source/activities/act-exp-modeling-soda.xml b/source/activities/act-exp-modeling-soda.xml index f057d5d4..d981498c 100755 --- a/source/activities/act-exp-modeling-soda.xml +++ b/source/activities/act-exp-modeling-soda.xml @@ -36,7 +36,12 @@ - + +

+ The graph of p(t) = (0.95)^t is always decreasing, always concave up, and decreasing toward zero. +

+ + @@ -47,7 +52,12 @@ - + +

+ The function r(t) = 30(0.95)^t is similar to p(t) but stretched vertically by a factor of 30. It is always decreasing, always concave up, and decreasing toward zero. +

+ + @@ -63,7 +73,12 @@ - + +

+ Since F(t) = 42 + 30(0.95)^t has a vertical shift of 42 up, it remains always decreasing, always concave up, but decreasing toward 42. +

+ + @@ -75,7 +90,11 @@ - + +

+ Because F(t) decreases toward 42, that is the temperature of the refrigerator. Because the function starts at 42 + 30 = 72, that is the room temperature of the surroundings outside the refrigerator. +

+ @@ -88,7 +107,16 @@ - + +

+ + AV_{[10,20]} &= \frac{30(0.95^{20} - 0.95^{10})}{20-10} \approx -0.721 \text{ degrees per minute} \\ + AV_{[20,30]} &= \frac{30(0.95^{30} - 0.95^{20})}{30-20} \approx -0.432 \text{ degrees per minute} \\ + AV_{[30,40]} &= \frac{30(0.95^{40} - 0.95^{30})}{40-30} \approx -0.258 \text{ degrees per minute} + + The average rate of change is always negative and is decreasing in magnitude as t increases. This is consistent with a function that is always decreasing and always concave up. In context, the soda is cooling more and more slowly as time goes on. +

+

diff --git a/source/activities/act-exp-temp-pop-NLOC1.xml b/source/activities/act-exp-temp-pop-NLOC1.xml index 9455a153..92b702a4 100755 --- a/source/activities/act-exp-temp-pop-NLOC1.xml +++ b/source/activities/act-exp-temp-pop-NLOC1.xml @@ -30,7 +30,11 @@ - + +

+ Since the soda cools toward the refrigerator temperature in the long run, and e^{-kt} \to 0 as t \to \infty, we have c = 37.7. At t = 0, F(0) = a + c = a + 37.7 = 72.3, so a = 34.6. Finally, using F(30) = 59.5: 34.6e^{-30k} + 37.7 = 59.5, so 34.6e^{-30k} = 21.8 and k = -\dfrac{1}{30}\ln\!\left(\dfrac{21.8}{34.6}\right) \approx 0.01540. +

+ @@ -41,7 +45,17 @@ - + +

+ Setting F(t) = 42.4: + + 34.6e^{-kt} + 37.7 &= 42.4 + 34.6e^{-kt} &= 4.7 + e^{-kt} &= \frac{4.7}{34.6} + t &= -\frac{1}{k}\ln\!\left(\frac{4.7}{34.6}\right) \approx 131 \text{ minutes.} + +

+ @@ -54,7 +68,11 @@ - + +

+ Using F(t) = 34.6e^{-0.01540t} + 37.7 in Desmos, the average rate of change on [25,30] is approximately \frac{F(30)-F(25)}{5} \approx -0.349 degrees Fahrenheit per minute. This means the soda is cooling at a rate of about 0.349^\circF per minute during the five-minute interval from t = 25 to t = 30. +

+ @@ -67,7 +85,11 @@ - + +

+ If F(0) = 65, then a = 65 - 37.7 = 27.3, which is smaller than the original a = 34.6. Using the same condition F(30) = 59.5: 27.3e^{-30k} = 21.8, giving k \approx 0.00751. This is smaller than the original k \approx 0.0154. With a smaller initial temperature difference between the soda and the refrigerator, the soda cools more slowly, so k is smaller. +

+

diff --git a/source/activities/act-exp-temp-pop-logistic-Desmos.xml b/source/activities/act-exp-temp-pop-logistic-Desmos.xml index 83a0a09d..bf5b1ca3 100755 --- a/source/activities/act-exp-temp-pop-logistic-Desmos.xml +++ b/source/activities/act-exp-temp-pop-logistic-Desmos.xml @@ -31,7 +31,12 @@ - + +

+ A typical graph of P(t) = \dfrac{A}{1 + Me^{-kt}} has an S-shape (sigmoidal curve) that rises from near 0 as t \to -\infty to the value A as t \to \infty. The parameter A sets the upper asymptote (the carrying capacity). The parameter M affects the y-intercept: since P(0) = \frac{A}{1+M}, larger M gives a lower initial value, producing an effect similar to a horizontal shift. The parameter k controls the steepness of the transition from the lower asymptote to the upper asymptote. +

+ + @@ -41,7 +46,11 @@ - + +

+ The population appears to grow most rapidly at the midpoint of the transition from near 0 to the carrying capacity A. This occurs when P(t) = \dfrac{A}{2}, that is, at half the carrying capacity. +

+ @@ -51,7 +60,11 @@ - + +

+ As t \to \infty, e^{-kt} \to 0 (since k > 0), so Me^{-kt} \to 0 and therefore 1 + Me^{-kt} \to 1. This causes P(t) = \frac{A}{1 + Me^{-kt}} \to \frac{A}{1} = A, which is why A is the carrying capacity. +

+ @@ -61,7 +74,11 @@ - + +

+ Since P(t) \to 9 as t \to \infty, we have A = 9. From P(0) = \frac{9}{1+M} = 2, we get M = \frac{7}{2} = 3.5. From P(2) = \frac{9}{1 + 3.5e^{-2k}} = 4, we get 1 + 3.5e^{-2k} = 2.25, so e^{-2k} = \frac{1.25}{3.5} = \frac{5}{14} and k = \dfrac{1}{2}\ln\!\left(\dfrac{14}{5}\right) \approx 0.516. +

+

diff --git a/source/activities/act-exp-temp-pop-logistic-exact.xml b/source/activities/act-exp-temp-pop-logistic-exact.xml index 7b542dab..e8999cf0 100755 --- a/source/activities/act-exp-temp-pop-logistic-exact.xml +++ b/source/activities/act-exp-temp-pop-logistic-exact.xml @@ -31,7 +31,17 @@ - + +

+ Since P(t) \to 11.7 as t \to \infty, A = 11.7. From P(0) = \frac{11.7}{1+M} = 2.45, we get 1+M = \frac{11.7}{2.45}, so M = \frac{11.7}{2.45} - 1 = \frac{9.25}{2.45} \approx 3.776. Using P(3) = 4.52: + + \frac{11.7}{1 + 3.776e^{-3k}} &= 4.52 + 1 + 3.776e^{-3k} &= \frac{11.7}{4.52} \approx 2.589 + e^{-3k} &= \frac{1.589}{3.776} \approx 0.421 + k &= -\frac{1}{3}\ln(0.421) \approx 0.289. + +

+ @@ -44,7 +54,16 @@ - + +

+ Using P(t) = \dfrac{11.7}{1 + 3.776e^{-0.289t}}, we compute: + P(0) \approx 2.45, P(2) \approx 3.75, P(4) \approx 5.34, P(6) \approx 7.02, P(8) \approx 8.51. + The average rates of change are: + AV_{[0,2]} \approx 0.650, AV_{[2,4]} \approx 0.796, AV_{[4,6]} \approx 0.836, AV_{[6,8]} \approx 0.747 thousand animals per year. + These represent the average rate at which the population grows on each interval. The population grows most rapidly on [4,6], which makes sense since the midpoint between the asymptotes (0 and 11.7) is 5.85, which falls in this interval. +

+ + @@ -54,7 +73,18 @@ - + +

+ Setting P(t) = 10: + + \frac{11.7}{1 + 3.776e^{-0.289t}} &= 10 + 1 + 3.776e^{-0.289t} &= 1.17 + 3.776e^{-0.289t} &= 0.17 + e^{-0.289t} &= \frac{0.17}{3.776} \approx 0.04503 + t &= -\frac{\ln(0.04503)}{0.289} \approx 10.74 \text{ years.} + +

+

diff --git a/source/activities/act-poly-infty-limit.xml b/source/activities/act-poly-infty-limit.xml index 8c36c68a..e6e88f76 100755 --- a/source/activities/act-poly-infty-limit.xml +++ b/source/activities/act-poly-infty-limit.xml @@ -151,7 +151,24 @@

-

+ + + f(x) + \lim_{x \to \infty} f(x) + \lim_{x \to -\infty} f(x) + + e^x\infty0 + e^{-x}0\infty + \ln(x)\inftynot defined + x\infty-\infty + x^2\infty\infty + x^3\infty-\infty + x^4\infty\infty + \frac{1}{x}00 + \frac{1}{x^2}00 + \sin(x)no limitno limit + +

Note: \ln(x) is undefined for x \le 0, so \lim_{x \to -\infty} \ln(x) does not exist.

diff --git a/source/activities/act-poly-infty-natural-powers.xml b/source/activities/act-poly-infty-natural-powers.xml index 56dc9c29..1a7c21b1 100755 --- a/source/activities/act-poly-infty-natural-powers.xml +++ b/source/activities/act-poly-infty-natural-powers.xml @@ -30,7 +30,11 @@ - + +

+ Two trends to observe: (1) As n increases, the graph becomes flatter and wider near x = 0 before rising steeply. (2) When n is even, both ends of the graph go to +\infty (symmetric about the y-axis); when n is odd, the left end goes to -\infty and the right end goes to +\infty. +

+ @@ -44,7 +48,11 @@ - + +

+ On 0 \lt x \lt 1, the graph of x^b stays closer to 0 than x^a when b \gt a. That is, x^a \gt x^b on (0,1) when a \lt b. +

+ @@ -58,7 +66,11 @@ - + +

+ Two trends to observe: (1) The graphs with even n are symmetric about the y-axis, while odd-n graphs pass through the origin from lower-left to upper-right. (2) For larger n, the function stays very close to 0 on (-1,1) and then rises very steeply outside that interval. +

+ @@ -72,7 +84,11 @@ - + +

+ On x \gt 1, the graph of x^b lies above the graph of x^a when b \gt a: larger exponents produce faster growth for x \gt 1. +

+

diff --git a/source/activities/act-poly-infty-negative-powers.xml b/source/activities/act-poly-infty-negative-powers.xml index 139a3e37..5d495d04 100755 --- a/source/activities/act-poly-infty-negative-powers.xml +++ b/source/activities/act-poly-infty-negative-powers.xml @@ -31,7 +31,11 @@ - + +

+ Two trends: (1) For even-n exponents (like x^{-2}, x^{-4}, \ldots), the graph is symmetric about the y-axis and stays positive; for odd exponents (like x^{-1}, x^{-3}, \ldots), the graph is negative for x \lt 0 and positive for x \gt 0. (2) As the exponent becomes more negative, the graph becomes more sharply L-shaped near x = 0. +

+ @@ -46,7 +50,11 @@ - + +

+ On x \gt 1, x^a \lt x^b when a \lt b (with both negative). For example, x^{-3} \lt x^{-2} for x \gt 1, since dividing by a higher power gives a smaller result. +

+ @@ -56,7 +64,11 @@ - + +

+ On 0 \lt x \lt 1, the situation reverses: x^a \gt x^b when a \lt b (i.e., more negative exponents give larger values on (0,1) since dividing by a small number raised to a high power gives a large result). +

+ @@ -70,7 +82,11 @@ - + +

+ Two trends with the wider window: (1) The graphs with even exponents are symmetric about the y-axis; odd-exponent graphs are antisymmetric. (2) As the exponent becomes more negative, the function rises more steeply near x = 0 (steeper asymptotic behavior). +

+ @@ -80,7 +96,11 @@ - + +

+ As x \to \infty, x^n \to \infty for any positive integer n, so \frac{1}{x^n} = x^{-n} \to 0. Dividing 1 by an ever-larger number drives the result to 0. +

+

diff --git a/source/activities/act-poly-polynomial-applications-Taylor.xml b/source/activities/act-poly-polynomial-applications-Taylor.xml index 6c9b7d00..3bbbc824 100755 --- a/source/activities/act-poly-polynomial-applications-Taylor.xml +++ b/source/activities/act-poly-polynomial-applications-Taylor.xml @@ -30,7 +30,9 @@ - + +

The polynomial T_1(x) = x. It appears that \sin(x) \approx T_1(x) for values of x near 0, roughly for |x| \lesssim 0.5.

+ @@ -40,7 +42,9 @@ - + +

The polynomial T_3(x) = x - \dfrac{x^3}{3!}. It appears that \sin(x) \approx T_3(x) for roughly |x| \lesssim 1.5.

+ @@ -50,7 +54,9 @@ - + +

The polynomial T_5(x) = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!}. It appears that \sin(x) \approx T_5(x) for roughly |x| \lesssim 2.5.

+ @@ -60,7 +66,9 @@ - + +

Each additional term in the polynomial extends the range of x-values over which the approximation is accurate. T_{19}(x) provides an excellent approximation to \sin(x) over a very large interval, essentially indistinguishable from \sin(x) on most standard viewing windows.

+ @@ -70,13 +78,23 @@ - + +

Following the pattern of alternating signs and even powers, the next polynomials are + + P_6(x) &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!}, + P_8(x) &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!}, + P_{18}(x) &= \sum_{j=0}^{9} \frac{(-1)^j x^{2j}}{(2j)!}. + + Polynomial approximations can approximate y = \cos(x) just as well as y = \sin(x); each added term extends the range of accuracy.

+

-

+

+ T_1(x)=x is good near x=0; each higher-degree polynomial extends the interval of accuracy. T_{19} approximates \sin(x) extremely well. The cosine analogues follow the pattern P_{2k}(x) = \sum_{j=0}^{k} \frac{(-1)^j x^{2j}}{(2j)!}. +

diff --git a/source/activities/act-poly-polynomial-applications-postal.xml b/source/activities/act-poly-polynomial-applications-postal.xml index 80d2c3ab..91b7463e 100755 --- a/source/activities/act-poly-polynomial-applications-postal.xml +++ b/source/activities/act-poly-polynomial-applications-postal.xml @@ -34,7 +34,9 @@ - + +

The square end has side length x inches and the length of the package is y inches.

+ @@ -44,7 +46,9 @@ - + +

The girth is the perimeter of the square end: 4x. The constraint is that girth plus length is at most 120 inches. To maximize volume we use the entire allowance, giving the constraint equation 4x + y = 120.

+ @@ -54,7 +58,9 @@ - + +

Solving for y: y = 120 - 4x.

+ @@ -64,7 +70,13 @@ - + +

The volume of the rectangular box with square end of side x and length y is V = x^2 y. Substituting y = 120 - 4x: + + V(x) = x^2(120 - 4x) = 4x^2(30 - x). + +

+ @@ -74,13 +86,17 @@ - + +

We need both x > 0 and y = 120 - 4x > 0, which gives x < 30. So the domain of V in context is (0, 30).

+

-

+

+ With a square end of side x and length y = 120 - 4x, the volume is V(x) = x^2(120 - 4x) = 4x^2(30-x) for x \in (0, 30). +

diff --git a/source/activities/act-poly-polynomial-applications-soup.xml b/source/activities/act-poly-polynomial-applications-soup.xml index 4f40268b..51c9856f 100755 --- a/source/activities/act-poly-polynomial-applications-soup.xml +++ b/source/activities/act-poly-polynomial-applications-soup.xml @@ -30,7 +30,13 @@ - + +

The surface area of a closed cylinder is 2\pi r^2 + 2\pi r h. Setting this equal to 60 square inches gives + + 2\pi r^2 + 2\pi r h = 60. + +

+ @@ -40,7 +46,13 @@ - + +

Solving for h: + + h = \frac{60 - 2\pi r^2}{2\pi r} = \frac{30}{\pi r} - r. + +

+ @@ -50,7 +62,13 @@ - + +

Substituting into V = \pi r^2 h: + + V(r) = \pi r^2 \left(\frac{30}{\pi r} - r\right) = 30r - \pi r^3. + +

+ @@ -60,13 +78,17 @@ - + +

We need r > 0 and h > 0. From h = \frac{30}{\pi r} - r > 0, we get \frac{30}{\pi r} > r, so r^2 < \frac{30}{\pi}, giving r < \sqrt{\frac{30}{\pi}}. The domain is \left(0,\ \sqrt{\frac{30}{\pi}}\right).

+

-

+

+ The surface area constraint 2\pi r^2 + 2\pi r h = 60 gives h = \frac{30}{\pi r} - r, so the volume is V(r) = 30r - \pi r^3 for r \in \left(0,\ \sqrt{\frac{30}{\pi}}\right). +

diff --git a/source/activities/act-poly-polynomials-find.xml b/source/activities/act-poly-polynomials-find.xml index 1f34489a..a9cbec75 100755 --- a/source/activities/act-poly-polynomials-find.xml +++ b/source/activities/act-poly-polynomials-find.xml @@ -27,7 +27,11 @@ - + +

+ One example: p(x) = 1 - x - 2x^2 + 4x^3 + 2x^4 - 2x^5. This degree-5 polynomial has exactly 3 real zeros, 4 turning points, and since the leading coefficient is -2 \lt 0 and the degree is odd: \lim_{x \to -\infty} p(x) = +\infty and \lim_{x \to \infty} p(x) = -\infty. +

+ @@ -37,7 +41,11 @@ - + +

+ No such polynomial exists. A degree-4 polynomial has even degree, so both limits as x \to \pm\infty must be either both +\infty or both -\infty. It is impossible for the limits at -\infty and +\infty to be opposite infinities. +

+ @@ -47,7 +55,11 @@ - + +

+ One example: p(x) = -1 + x - 2x^5 - x^6. This degree-6 polynomial has negative leading coefficient, so \lim_{x \to \pm\infty} p(x) = -\infty. It has 2 real zeros and 3 turning points. +

+ @@ -57,7 +69,11 @@ - + +

+ No such polynomial exists. A degree-5 polynomial with odd degree and a positive leading coefficient has \lim_{x \to -\infty} p(x) = -\infty and \lim_{x \to \infty} p(x) = +\infty, while a negative leading coefficient gives the reverse. It is impossible for a degree-5 polynomial to have \lim_{x \to -\infty} p(x) = +\infty and simultaneously 3 turning points (an odd number of turning points forces matching end behavior, but odd-degree polynomials have opposite end behaviors). +

+

diff --git a/source/activities/act-poly-polynomials-multiple-zeros.xml b/source/activities/act-poly-polynomials-multiple-zeros.xml index e1dbd464..d25b5e12 100755 --- a/source/activities/act-poly-polynomials-multiple-zeros.xml +++ b/source/activities/act-poly-polynomials-multiple-zeros.xml @@ -27,7 +27,11 @@ - + +

+ One formula: f(x) = -21\!\left(\dfrac{x+12}{12}\right)^3\!\left(\dfrac{x+9}{9}\right)^2\!\left(\dfrac{x-4}{4}\right)^4\!\left(\dfrac{x-10}{10}\right). We can verify: f(0) = -21\cdot 1 \cdot 1 \cdot 1 \cdot (-1) = 21 ✓. Since the degree is 10 (even) and the leading coefficient is negative, \lim_{x \to \pm\infty} f(x) = -\infty. +

+ @@ -40,7 +44,11 @@ - + +

+ From the graph (reading off the zeros and their behavior), one example with p(0)=-2 is: p(x) = -2\!\left(\dfrac{x+8}{8}\right)^2\!\left(\dfrac{x+4}{4}\right)^3(x-1)\!\left(\dfrac{x-5}{5}\right)^2\!\left(\dfrac{x-7.5}{7.5}\right), which has degree 9 and p(0)=-2. +

+ @@ -53,7 +61,11 @@ - + +

+ From the sign chart (positive, then negative, then positive, then negative, with the pattern showing even-multiplicity zeros where sign doesn't change and odd-multiplicity zeros where it does), one example: q(x) = -10\!\left(\dfrac{x+2}{2}\right)^4\!\left(\dfrac{x-3}{3}\right)\!\left(\dfrac{x-9}{9}\right)^3. This has degree 8, three distinct real zeros, and q(0) = -10 \cdot 1 \cdot (-1) \cdot (-1) = -10 ✓. +

+ @@ -63,7 +75,11 @@ - + +

+ It is not possible to have a degree 9 polynomial satisfying the sign chart from part (c) and q(0)=-10, if the sign chart shows the same behavior at both ends (both positive or both negative for large |x|). A degree-9 polynomial has odd degree, so its limits at +\infty and -\infty must be opposite. If the sign chart requires both ends to have the same sign, degree 9 is impossible. +

+ @@ -73,7 +89,11 @@ - + +

+ It is not possible to have a degree-11 polynomial matching the graph in part (b) if that graph shows the same long-term behavior at both ends. A degree-11 polynomial has odd degree, so its limits at +\infty and -\infty must have opposite signs, while the graph in (b) (a degree-9 example) already illustrates this odd-degree behavior. However, a degree-11 polynomial could match the graph if the overall end behavior is also opposite-going. +

+

diff --git a/source/activities/act-poly-polynomials-sign-chart.xml b/source/activities/act-poly-polynomials-sign-chart.xml index f4880773..69604295 100755 --- a/source/activities/act-poly-polynomials-sign-chart.xml +++ b/source/activities/act-poly-polynomials-sign-chart.xml @@ -30,7 +30,11 @@ - + +

+ The degree is 1 + 2 + 2 + 1 = 6. We add the exponents from each factor: (x+1520)^1(x^2+10000)^1(x-3471)^2(x-9738)^1 contributes degrees 1+2+2+1=6. +

+ @@ -40,7 +44,11 @@ - + +

+ The factor (x^2 + 10000) is always positive for all real x, since x^2 \ge 0 so x^2 + 10000 \ge 10000 \gt 0. +

+ @@ -50,7 +58,11 @@ - + +

+ The real zeros are x = -1520, x = 3471, and x = 9738. (The factor x^2+10000 has no real zeros since x^2 = -10000 has no real solution.) +

+ @@ -60,7 +72,16 @@ - + +

+ The constant 4692 \gt 0, and (x^2+10000) \gt 0 always, and (x-3471)^2 \ge 0 always (no sign change at x=3471). The sign is thus determined by (x+1520) and (x-9738): +

    +

  • x \lt -1520: (x+1520) \lt 0 and (x-9738) \lt 0, product is positive: p \gt 0.

    • +

  • -1520 \lt x \lt 9738 (and x \ne 3471): (x+1520) \gt 0 and (x-9738) \lt 0: p \lt 0. (At x=3471, p=0 and the sign does not change.)

    • +

  • x \gt 9738: both factors positive: p \gt 0.

    • +
+

+ @@ -70,7 +91,11 @@ - + +

+ The graph is positive (above the x-axis) for x \lt -1520, crosses zero at x=-1520 (passes through), touches zero at x=3471 (bounce, even multiplicity), remains negative until x=9738, then crosses zero and becomes positive. The leading term 4692 x^6 means both ends go to +\infty. +

+ @@ -80,7 +105,11 @@ - + +

+ Using a graphing utility, the zeros at x=-1520, 3471, and 9738 are spread over a large x-range, making it challenging to see all features at once. It is difficult to tell from the graph that x=3471 is a bounce point (touching zero without crossing) unless the scale is set appropriately. +

+

diff --git a/source/activities/act-poly-rational-application.xml b/source/activities/act-poly-rational-application.xml index 72e58458..8d299514 100755 --- a/source/activities/act-poly-rational-application.xml +++ b/source/activities/act-poly-rational-application.xml @@ -27,7 +27,9 @@ - + +

The box has a shorter base side of length x, a longer base side of length 2x, and height h.

+ @@ -37,7 +39,9 @@ - + +

The volume is V = x \cdot 2x \cdot h = 2x^2 h. Setting V = 15: 2x^2 h = 15, so h = \dfrac{15}{2x^2}.

+ @@ -47,7 +51,13 @@ - + +

The open box (no top) has: one rectangular base (2x^2), two short sides (2 \cdot xh = 2xh), and two long sides (2 \cdot 2xh = 4xh). So + + S = 2x^2 + 2xh + 4xh = 2x^2 + 6xh. + +

+ @@ -57,7 +67,13 @@ - + +

Substituting h = \dfrac{15}{2x^2}: + + S(x) = 2x^2 + 6x \cdot \frac{15}{2x^2} = 2x^2 + \frac{45}{x}. + +

+ @@ -67,7 +83,9 @@ - + +

S(x) = 2x^2 + \dfrac{45}{x} = \dfrac{2x^3 + 45}{x} is a rational function. In the physical context, x > 0 (the side length must be positive), so the domain is (0, \infty).

+ @@ -77,13 +95,17 @@ - + +

From a graph of S(x) = 2x^2 + \dfrac{45}{x} on (0, \infty), the minimum value of S occurs at approximately x \approx 2.24 feet, giving a minimum surface area of approximately S \approx 30.0 square feet. (The exact minimum occurs at x = \sqrt[3]{45/4}.)

+

-

+

+ With base sides x and 2x and volume constraint 2x^2 h = 15, we get h = 15/(2x^2) and surface area S(x) = 2x^2 + 45/x, a rational function on domain (0, \infty). Its minimum is approximately S \approx 30.0 sq ft at x \approx 2.24 ft. +

diff --git a/source/activities/act-poly-rational-domain.xml b/source/activities/act-poly-rational-domain.xml index 7220cd35..43843d11 100755 --- a/source/activities/act-poly-rational-domain.xml +++ b/source/activities/act-poly-rational-domain.xml @@ -27,7 +27,9 @@ - + +

The denominator x^2+1 is always positive, so it is never zero. The domain of f is all real numbers.

+ @@ -37,7 +39,9 @@ - + +

Factoring the denominator: x^2+3x-4 = (x-1)(x+4). The denominator is zero when x = 1 or x = -4. The domain of g is all real numbers except x = 1 and x = -4.

+ @@ -47,7 +51,9 @@ - + +

Each fraction is undefined when its denominator is zero: 1/x is undefined at x=0, 1/(x-1) is undefined at x=1, and 1/(x-2) is undefined at x=2. The domain of h is all real numbers except x = 0, x = 1, and x = 2.

+ @@ -57,7 +63,9 @@ - + +

The denominator is zero when x = -1, x = -3, or x = 5. The domain of j is all real numbers except x = -1, x = -3, and x = 5.

+ @@ -67,7 +75,9 @@ - + +

Factoring the denominator: 3x^3 - 12x = 3x(x^2 - 4) = 3x(x-2)(x+2). The denominator is zero when x = 0, x = 2, or x = -2. The domain of k is all real numbers except x = -2, x = 0, and x = 2.

+ @@ -77,13 +87,17 @@ - + +

The denominator is zero when x = 2, x = 3, or x = -1 (the factor x^2+9 has no real zeros since x^2 \geq 0). The domain of m is all real numbers except x = -1, x = 2, and x = 3.

+

-

+

+ Rational functions are undefined wherever the denominator is zero. Factor each denominator and exclude those x-values: f has domain \mathbb{R}; g excludes x \in \{-4,1\}; h excludes x \in \{0,1,2\}; j excludes x \in \{-3,-1,5\}; k excludes x \in \{-2,0,2\}; m excludes x \in \{-1,2,3\}. +

diff --git a/source/activities/act-poly-rational-features-ZAH.xml b/source/activities/act-poly-rational-features-ZAH.xml index 989d0a4e..40182218 100755 --- a/source/activities/act-poly-rational-features-ZAH.xml +++ b/source/activities/act-poly-rational-features-ZAH.xml @@ -27,7 +27,17 @@ - + +

Factor: f(x) = \dfrac{x(x-1)(x-5)}{(x+1)(x-1)}. The factor (x-1) cancels, giving the reduced form \dfrac{x(x-5)}{x+1}. +

    +

  • Domain: all reals except x = 1 and x = -1.

    • +

  • Zeros: x = 0 and x = 5 (where reduced numerator is zero and x \neq -1, 1).

    • +

  • Vertical asymptote: x = -1 (denominator zero, numerator nonzero in reduced form).

    • +

  • Hole: at x = 1; hole value \frac{1(1-5)}{1+1} = -2, so hole at (1, -2).

    • +

  • Horizontal asymptote: none (reduced form has degree 2 over degree 1; there is an oblique asymptote y = x - 6).

    • +
+

+ @@ -37,7 +47,17 @@ - + +

The numerator x^2+1 > 0 always; the only real zero of the numerator is x = 7. The denominator zeros: x = 1 from (x-1); x^2+4 > 0 always. No common factors. +

    +

  • Domain: all reals except x = 1.

    • +

  • Zero: x = 7.

    • +

  • Vertical asymptote: x = 1.

    • +

  • Holes: none.

    • +

  • Horizontal asymptote: y = \dfrac{11}{23} (degree 3 over degree 3, ratio of leading coefficients).

    • +
+

+ @@ -47,7 +67,17 @@ - + +

Factor: h(x) = \dfrac{(x-2)(x-6)}{(x-6)(x+3)}. The factor (x-6) cancels. +

    +

  • Domain: all reals except x = 6 and x = -3.

    • +

  • Zero: x = 2.

    • +

  • Vertical asymptote: x = -3.

    • +

  • Hole: at x = 6; hole value \dfrac{6-2}{6+3} = \dfrac{4}{9}, so hole at \left(6,\, \dfrac{4}{9}\right).

    • +

  • Horizontal asymptote: y = 1 (same degree, leading coefficients both 1).

    • +
+

+ @@ -57,7 +87,17 @@ - + +

Factor: q(x) = \dfrac{(x-2)(x-3)(x+3)}{(x-3)(x^2+4)}. The factor (x-3) cancels; x^2+4 > 0 always. +

    +

  • Domain: all reals except x = 3.

    • +

  • Zeros: x = 2 and x = -3.

    • +

  • Vertical asymptotes: none (the only denominator zero x=3 is a hole).

    • +

  • Hole: at x = 3; hole value \dfrac{(3-2)(3+3)}{3^2+4} = \dfrac{6}{13}, so hole at \left(3,\, \dfrac{6}{13}\right).

    • +

  • Horizontal asymptote: y = 1 (degree 3 over degree 3).

    • +
+

+ @@ -67,7 +107,17 @@ - + +

The factor (x+1) cancels. +

    +

  • Domain: all reals except x = -1, x = 4, and x = 5.

    • +

  • Zeros: x = 2 and x = 3.

    • +

  • Vertical asymptotes: x = 4 and x = 5.

    • +

  • Hole: at x = -1; hole value \dfrac{19(-1-2)(-1-3)^2}{17(-1-4)^2(-1-5)} = \dfrac{19(-3)(16)}{17(25)(-6)} = \dfrac{-912}{-2550} = \dfrac{152}{425}, so hole at \left(-1,\, \dfrac{152}{425}\right).

    • +

  • Horizontal asymptote: y = \dfrac{19}{17} (degree 4 over degree 4).

    • +
+

+ @@ -77,13 +127,25 @@ - + +

The numerator is always 1 \neq 0, and the denominator x^2+1 > 0 always. +

    +

  • Domain: all real numbers.

    • +

  • Zeros: none.

    • +

  • Vertical asymptotes: none.

    • +

  • Holes: none.

    • +

  • Horizontal asymptote: y = 0 (degree 0 over degree 2).

    • +
+

+

-

+

+ Factor numerator and denominator, cancel common factors, then identify: zeros (numerator zero in reduced form), vertical asymptotes (denominator zero in reduced form), holes (cancelled factors), horizontal asymptote (degree comparison of leading terms). +

diff --git a/source/activities/act-poly-rational-formula.xml b/source/activities/act-poly-rational-formula.xml index 4dda2143..50fb1f5d 100755 --- a/source/activities/act-poly-rational-formula.xml +++ b/source/activities/act-poly-rational-formula.xml @@ -27,7 +27,13 @@ - + +

We need: a factor (x+2) in the denominator for the vertical asymptote at x=-2; a factor (x-1) in the numerator for the zero at x=1; the factor (x-5) in both numerator and denominator for the hole at x=5; and a ratio of leading coefficients of -3 for the horizontal asymptote y=-3. One such function is + + r(x) = \frac{-3(x-1)(x-5)}{(x+2)(x-5)}. + +

+ @@ -37,7 +43,13 @@ - + +

We need: exactly one factor (x+4) in the denominator for the vertical asymptote; the other two denominator factors must introduce no new real zeros (e.g., x^2+1); and ratio of leading coefficients \frac{3}{7}. One such function is + + u(x) = \frac{3x(x^2+3)}{7(x+4)(x^2+1)}. + + Both numerator and denominator have degree 3, there is exactly one vertical asymptote at x=-4, and the horizontal asymptote is y = \frac{3}{7}.

+ @@ -49,7 +61,13 @@ - + +

From the graph, w appears to have zeros at x = -4 and at x = 5 (with even multiplicity, since w(x) > 0 for other x > 3), and vertical asymptotes at x = -1 and x = 3. One formula consistent with these features is + + w(x) = \frac{9(x+4)(x-5)^2}{50(x+1)^2(x-3)}. + +

+ @@ -62,7 +80,13 @@ - + +

From the sign chart, the function has zeros at x = -4 and x = 5, vertical asymptotes at x = -1 and x = 3, no horizontal asymptote, and the behavior shown in the chart requires the numerator degree to exceed the denominator degree. One formula is + + z(x) = -\frac{(x+4)(x-5)^2}{(x+1)(x-3)}. + +

+ @@ -72,13 +96,21 @@ - + +

We need: two factors that cancel for the holes; two denominator factors that don't cancel for the VAs; two remaining numerator factors for the zeros; and equal numerator/denominator degree for the HA. One formula is + + f(x) = \frac{(x-1)(x+1)(x+2)(x-2)}{(x-1)(x+1)(x-3)(x+3)}. + + After simplification, f(x) = \dfrac{(x+2)(x-2)}{(x-3)(x+3)}. This has holes at \left(-1,\dfrac{3}{8}\right) and \left(1,\dfrac{3}{8}\right), vertical asymptotes at x = 3 and x = -3, zeros at x = 2 and x = -2, and horizontal asymptote y = 1.

+

-

+

+ Sample answers: (a) r(x)=-3(x-1)(x-5)/[(x+2)(x-5)]; (b) u(x)=3x(x^2+3)/[7(x+4)(x^2+1)]; (c) based on graph features; (d) based on sign chart; (e) f(x)=(x-1)(x+1)(x+2)(x-2)/[(x-1)(x+1)(x-3)(x+3)]. +

diff --git a/source/activities/act-poly-rational-long-term-1.xml b/source/activities/act-poly-rational-long-term-1.xml index d8a763a9..043ba3a7 100755 --- a/source/activities/act-poly-rational-long-term-1.xml +++ b/source/activities/act-poly-rational-long-term-1.xml @@ -30,7 +30,14 @@ - + +

Multiplying numerator and denominator by \dfrac{1}{x^2}: + + r(x) &= \frac{3x^2 - 5x + 1}{7x^2 + 2x - 11} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}} + &= \frac{3 - \frac{5}{x} + \frac{1}{x^2}}{7 + \frac{2}{x} - \frac{11}{x^2}}. + +

+ @@ -45,7 +52,13 @@ - + +

As x \to \infty, the terms \frac{5}{x}, \frac{1}{x^2}, \frac{2}{x}, and \frac{11}{x^2} all approach 0. Therefore + + \lim_{x \to \infty} r(x) = \frac{3 - 0 + 0}{7 + 0 - 0} = \frac{3}{7}. + +

+ @@ -58,7 +71,13 @@ - + +

For the same reason, as x \to -\infty the terms with x in the denominator all approach 0, so + + \lim_{x \to -\infty} r(x) = \frac{3}{7}. + +

+ @@ -68,13 +87,17 @@ - + +

The graph of r approaches the horizontal line y = \dfrac{3}{7} as x \to \pm\infty. This line is called a horizontal asymptote of r. The limits found in (b) and (c) say that far to the left and far to the right of the origin, r(x) gets arbitrarily close to \dfrac{3}{7}.

+

-

+

+ Multiplying by \frac{1/x^2}{1/x^2} gives r(x) = \frac{3 - 5/x + 1/x^2}{7 + 2/x - 11/x^2}. Since the terms with x in the denominator vanish as x \to \pm\infty, both limits equal \frac{3}{7}, the ratio of the leading coefficients. The horizontal asymptote is y = \frac{3}{7}. +

diff --git a/source/activities/act-poly-rational-long-term-2.xml b/source/activities/act-poly-rational-long-term-2.xml index 3d203fec..a2eadc96 100755 --- a/source/activities/act-poly-rational-long-term-2.xml +++ b/source/activities/act-poly-rational-long-term-2.xml @@ -31,7 +31,13 @@ - + +

Multiplying s(x) = \dfrac{3x-5}{7x^2+2x-11} by \dfrac{1/x^2}{1/x^2}: + + s(x) = \frac{\frac{3}{x} - \frac{5}{x^2}}{7 + \frac{2}{x} - \frac{11}{x^2}}. + + As x \to \infty, the numerator \to 0 and the denominator \to 7, so \displaystyle\lim_{x \to \infty} s(x) = 0.

+ @@ -41,7 +47,9 @@ - + +

The graph of y = s(x) has a horizontal asymptote at y = 0. Far to the left and right, s(x) approaches the x-axis. This occurs because the denominator grows much faster than the numerator (degree 2 vs. degree 1).

+ @@ -55,7 +63,13 @@ - + +

Multiplying u(x) = \dfrac{3x^2-5x+1}{7x+2} by \dfrac{1/x^2}{1/x^2}: + + u(x) = \frac{3 - \frac{5}{x} + \frac{1}{x^2}}{\frac{7}{x} + \frac{2}{x^2}}. + + As x \to \infty, the numerator \to 3 and the denominator \to 0^+, so \displaystyle\lim_{x \to \infty} u(x) = +\infty. The numerator degree exceeds the denominator degree, so there is no horizontal asymptote.

+ @@ -65,13 +79,17 @@ - + +

The graph of y = u(x) increases without bound as x \to \infty; u has no horizontal asymptote. Instead, it has an oblique (slant) asymptote. The graph shows that u(x) grows like a linear function for large |x|.

+

-

+

+ For s(x) (degree 1 over degree 2): \lim_{x \to \infty} s(x) = 0, horizontal asymptote y=0. For u(x) (degree 2 over degree 1): \lim_{x \to \infty} u(x) = +\infty, no horizontal asymptote. +

diff --git a/source/activities/act-trig-finding-angles-baseball.xml b/source/activities/act-trig-finding-angles-baseball.xml index f688e09b..daada1cb 100755 --- a/source/activities/act-trig-finding-angles-baseball.xml +++ b/source/activities/act-trig-finding-angles-baseball.xml @@ -30,7 +30,15 @@

-

+

+ Place home plate at the origin, first base at (90, 0), second base at (90, 90), and third base at (0, 90). The third baseman is at (0, 80) (10 feet from third base toward home plate). +

+

+ Throwing to first base. The straight-line distance is \sqrt{90^2 + 80^2} = 10\sqrt{145} feet. The angle the throw makes with the first base line is \arctan\!\left(\dfrac{80}{90}\right) = \arctan\!\left(\dfrac{8}{9}\right) \approx 41.6^{\circ}, and with the third base line it is \arctan\!\left(\dfrac{90}{80}\right) = \arctan\!\left(\dfrac{9}{8}\right) \approx 48.4^{\circ}. +

+

+ Throwing to second base. The displacement from (0, 80) to second base at (90, 90) is (90, 10), so the distance is \sqrt{90^2 + 10^2} = 10\sqrt{82} feet. The angle with the first base line is \arctan\!\left(\dfrac{10}{90}\right) = \arctan\!\left(\dfrac{1}{9}\right) \approx 6.3^{\circ}, and with the third base line it is approximately 83.7^{\circ}. +

diff --git a/source/activities/act-trig-finding-angles-exactly.xml b/source/activities/act-trig-finding-angles-exactly.xml index 45ecace2..76b76b52 100755 --- a/source/activities/act-trig-finding-angles-exactly.xml +++ b/source/activities/act-trig-finding-angles-exactly.xml @@ -28,7 +28,11 @@ - + +

+ The hypotenuse has length \sqrt{11^2 + 13^2} = \sqrt{290}. The angle \alpha opposite the leg of length 11 satisfies \alpha = \arcsin\!\left(\dfrac{11}{\sqrt{290}}\right) = \arccos\!\left(\dfrac{13}{\sqrt{290}}\right) \approx 0.702 radians. The angle \beta opposite the leg of length 13 satisfies \beta = \arcsin\!\left(\dfrac{13}{\sqrt{290}}\right) = \arccos\!\left(\dfrac{11}{\sqrt{290}}\right) \approx 0.869 radians. +

+ @@ -38,7 +42,11 @@ - + +

+ Since \cos(\alpha) = -\frac{1}{2} and \alpha is in quadrant III, the reference angle is \arccos\!\left(\frac{1}{2}\right) = \frac{\pi}{3}, so \alpha = \pi + \frac{\pi}{3} = \frac{4\pi}{3}. In addition, \sin(\alpha) = -\dfrac{\sqrt{3}}{2}. +

+ @@ -50,7 +58,11 @@ - + +

+ Since \sin(\beta) = 0.1 and \beta is in quadrant II, we have \beta = \pi - \arcsin(0.1) \approx 3.041 radians. In addition, \cos(\beta) = -\cos(\arcsin(0.1)) = -\sqrt{1 - 0.01} = -\sqrt{0.99} \approx -0.9950. +

+

diff --git a/source/activities/act-trig-finding-angles-rocket.xml b/source/activities/act-trig-finding-angles-rocket.xml index 216006f1..32977012 100755 --- a/source/activities/act-trig-finding-angles-rocket.xml +++ b/source/activities/act-trig-finding-angles-rocket.xml @@ -36,7 +36,25 @@

-

+

+ When the rocket is 3000' off the ground, the right triangle has legs 3000 and 4000 and hypotenuse 5000. The camera angle is \theta_0 = \arctan\!\left(\dfrac{3000}{4000}\right) = \arccos\!\left(\dfrac{4}{5}\right) \approx 0.644 radians. +

+

+ For general height h, \theta(h) = \arctan\!\left(\dfrac{h}{4000}\right). Using the table of values below (with angles in radians): +

+ + + h (feet) + 300035005000550070007500 + + + \theta(h) + 0.6440.7190.8960.9421.0501.080 + + +

+ The average rates of change are: AV_{[3000,3500]} \approx \frac{0.075}{500} = 1.5 \times 10^{-4} rad/ft; AV_{[5000,5500]} \approx \frac{0.046}{500} = 9.2 \times 10^{-5} rad/ft; AV_{[7000,7500]} \approx \frac{0.030}{500} = 6 \times 10^{-5} rad/ft. The camera angle is increasing at a decreasing rate as the rocket climbs higher. +

diff --git a/source/activities/act-trig-finding-angles-roof.xml b/source/activities/act-trig-finding-angles-roof.xml index 906682fb..9040e79d 100755 --- a/source/activities/act-trig-finding-angles-roof.xml +++ b/source/activities/act-trig-finding-angles-roof.xml @@ -28,7 +28,9 @@

-

+

+ The angle of the roof with the horizontal is \arctan\!\left(\dfrac{7}{12}\right) \approx 30.26^{\circ} (exactly \arctan\!\left(\frac{7}{12}\right) radians). Alternatively, this angle equals \arcsin\!\left(\dfrac{7}{\sqrt{193}}\right) = \arccos\!\left(\dfrac{12}{\sqrt{193}}\right). The angle at the peak of the roof is formed by the two sloping surfaces; since each makes an angle of \arctan\!\left(\frac{7}{12}\right) with the horizontal, the ridge angle is 180^{\circ} - 2\arctan\!\left(\frac{7}{12}\right) \approx 119.5^{\circ}. +

diff --git a/source/activities/act-trig-inverse-arccos.xml b/source/activities/act-trig-inverse-arccos.xml index 65b66066..56d35a0b 100755 --- a/source/activities/act-trig-inverse-arccos.xml +++ b/source/activities/act-trig-inverse-arccos.xml @@ -27,7 +27,9 @@ - + +

\arccos\!\left(\frac{1}{2}\right) = \frac{\pi}{3}

+ @@ -37,7 +39,9 @@ - + +

\arccos\!\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}

+ @@ -47,7 +51,9 @@ - + +

\arccos\!\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}

+ @@ -57,7 +63,9 @@ - + +

\arccos\!\left(-\frac{1}{2}\right) = \frac{2\pi}{3}

+ @@ -67,7 +75,9 @@ - + +

\arccos\!\left(-\frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}

+ @@ -77,7 +87,9 @@ - + +

\arccos\!\left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6}

+ @@ -87,7 +99,9 @@ - + +

\arccos(-1) = \pi

+ @@ -97,7 +111,9 @@ - + +

\arccos(0) = \frac{\pi}{2}

+ @@ -107,7 +123,9 @@ - + +

\cos\!\left(\arccos\!\left(-\frac{1}{2}\right)\right) = -\frac{1}{2}, since cosine and arccosine are inverse functions.

+ @@ -117,7 +135,9 @@ - + +

\arccos\!\left(\cos\!\left(\frac{7\pi}{6}\right)\right) = \frac{5\pi}{6}, since \cos\!\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2} and \arccos\!\left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6}.

+

diff --git a/source/activities/act-trig-inverse-arcsin.xml b/source/activities/act-trig-inverse-arcsin.xml index a0db87e0..240751d6 100755 --- a/source/activities/act-trig-inverse-arcsin.xml +++ b/source/activities/act-trig-inverse-arcsin.xml @@ -27,7 +27,9 @@ - + +

The domain of \arcsin is [-1, 1] and the range is \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

+ @@ -37,7 +39,9 @@ - + +

\arcsin(-1) = -\frac{\pi}{2}, \arcsin\!\left(-\frac{\sqrt{2}}{2}\right) = -\frac{\pi}{4}, \arcsin(0) = 0, \arcsin\!\left(\frac{1}{2}\right) = \frac{\pi}{6}, and \arcsin\!\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}.

+ @@ -51,7 +55,9 @@ - + + + @@ -61,7 +67,9 @@ - + +

False. Since \sin(5\pi) = 0 and \arcsin(0) = 0, we have \arcsin(\sin(5\pi)) = 0 \ne 5\pi. The arcsine function only returns values in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], so \arcsin(\sin(t)) = t only when t is already in that interval.

+

diff --git a/source/activities/act-trig-inverse-arctan.xml b/source/activities/act-trig-inverse-arctan.xml index 13f4b05d..c6e2e05b 100755 --- a/source/activities/act-trig-inverse-arctan.xml +++ b/source/activities/act-trig-inverse-arctan.xml @@ -27,7 +27,11 @@ - + +

+ The domain of the arctangent function is all real numbers (-\infty, \infty), and the range is the open interval \left(-\frac{\pi}{2}, \frac{\pi}{2}\right). +

+ @@ -37,7 +41,11 @@ - + +

+ \arctan(-\sqrt{3}) = -\frac{\pi}{3}, \arctan(-1) = -\frac{\pi}{4}, \arctan(0) = 0, and \arctan\!\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}. +

+ @@ -51,7 +59,9 @@ - + + + @@ -61,7 +71,11 @@ - + +

+ As t increases without bound, \arctan(t) approaches \frac{\pi}{2}. +

+

diff --git a/source/activities/act-trig-other-aroc-sine.xml b/source/activities/act-trig-other-aroc-sine.xml index fd09c6ae..afb081e0 100755 --- a/source/activities/act-trig-other-aroc-sine.xml +++ b/source/activities/act-trig-other-aroc-sine.xml @@ -33,7 +33,11 @@ - + +

+ AV_{[a,a+h]} = \dfrac{\sin(a+h) - \sin(a)}{h} +

+ @@ -43,7 +47,11 @@ - + +

+ Using the sum identity: \sin(a+h) = \sin(a)\cos(h) + \cos(a)\sin(h). +

+ @@ -56,7 +64,17 @@ - + +

+ Substituting into the average rate of change: + + AV_{[a,a+h]} &= \frac{\sin(a)\cos(h) + \cos(a)\sin(h) - \sin(a)}{h} + &= \frac{\sin(a)(\cos(h)-1) + \cos(a)\sin(h)}{h} + &= \sin(a) \cdot \frac{\cos(h)-1}{h} + \cos(a) \cdot \frac{\sin(h)}{h} + + which matches the stated formula (noting the sign: the formula in the problem has \frac{\cos(h)-1}{h}, which equals the same as written). +

+ @@ -70,7 +88,11 @@ - + +

+ As h approaches 0, the expression \dfrac{\cos(h)-1}{h} approaches 0, and \dfrac{\sin(h)}{h} approaches 1. Therefore AV_{[a,a+h]} approaches \sin(a) \cdot 0 + \cos(a) \cdot 1 = \cos(a). This tells us that the instantaneous rate of change of the sine function at a is \cos(a). +

+

diff --git a/source/activities/act-trig-other-cot.xml b/source/activities/act-trig-other-cot.xml index 5d7f512d..c7fd8409 100755 --- a/source/activities/act-trig-other-cot.xml +++ b/source/activities/act-trig-other-cot.xml @@ -304,7 +304,11 @@ - + +

+ Since \cot(t) = \cos(t)/\sin(t), the cotangent values for Q1–Q2 are: u, \sqrt{3}, 1, \frac{1}{\sqrt{3}}, 0, -\frac{1}{\sqrt{3}}, -1, -\sqrt{3}, u; and for Q3–Q4: \sqrt{3}, 1, \frac{1}{\sqrt{3}}, 0, -\frac{1}{\sqrt{3}}, -1, -\sqrt{3}, u. +

+ @@ -314,7 +318,11 @@ - + +

+ The cotangent function is positive in quadrants I and III (where sine and cosine have the same sign) and negative in quadrants II and IV (where they have opposite signs). +

+ @@ -324,7 +332,11 @@ - + +

+ The cotangent function has vertical asymptotes wherever \sin(t) = 0, that is, at every integer multiple of \pi: t = 0, \pm\pi, \pm 2\pi, \ldots +

+ @@ -334,7 +346,11 @@ - + +

+ The domain of the cotangent function is all real numbers except integer multiples of \pi. The range is all real numbers (-\infty, \infty). +

+ @@ -348,7 +364,9 @@ - + + + @@ -358,7 +376,11 @@ - + +

+ The cotangent function is always decreasing on every interval where it is defined. +

+ @@ -368,7 +390,11 @@ - + +

+ The period of the cotangent function is \pi radians. +

+ @@ -378,7 +404,11 @@ - + +

+ The graph of r(t) = \cot(t) can be obtained from the graph of \tan(t) by reflecting over the y-axis and then shifting \frac{\pi}{2} to the right. Equivalently, \cot(t) = \tan\!\left(\frac{\pi}{2} - t\right). +

+

diff --git a/source/activities/act-trig-other-csc.xml b/source/activities/act-trig-other-csc.xml index d75646a5..ec43c9b5 100755 --- a/source/activities/act-trig-other-csc.xml +++ b/source/activities/act-trig-other-csc.xml @@ -224,7 +224,11 @@ - + +

+ The csc values are reciprocals of sin: Q1-Q2: u, 2, \sqrt{2}, \frac{2\sqrt{3}}{3}, 1, \frac{2\sqrt{3}}{3}, \sqrt{2}, 2, u; Q3-Q4: -2, -\sqrt{2}, -\frac{2\sqrt{3}}{3}, -1, -\frac{2\sqrt{3}}{3}, -\sqrt{2}, -2, u. +

+ @@ -234,7 +238,11 @@ - + +

+ Since \csc(t) = 1/\sin(t), the cosecant function is positive wherever sine is positive, that is, in quadrants I and II, and negative wherever sine is negative, that is, in quadrants III and IV. +

+ @@ -244,7 +252,11 @@ - + +

+ The cosecant function has vertical asymptotes wherever \sin(t) = 0, which occurs at every integer multiple of \pi: t = 0, \pm\pi, \pm 2\pi, \ldots At these values \csc(t) = 1/\sin(t) is undefined (division by zero). +

+ @@ -254,7 +266,11 @@ - + +

+ The domain of the cosecant function is all real numbers except integer multiples of \pi. The range is (-\infty, -1] \cup [1, \infty). +

+ @@ -264,7 +280,9 @@ - + + + @@ -274,7 +292,11 @@ - + +

+ The period of the cosecant function is 2\pi radians. +

+

diff --git a/source/activities/act-trig-right-SOH-CAH.xml b/source/activities/act-trig-right-SOH-CAH.xml index db263fa2..2851cf11 100755 --- a/source/activities/act-trig-right-SOH-CAH.xml +++ b/source/activities/act-trig-right-SOH-CAH.xml @@ -27,7 +27,11 @@ - + +

+ With hypotenuse 7 and non-right angle \theta = \frac{\pi}{7}, the adjacent leg is x = 7\cos\!\left(\frac{\pi}{7}\right) \approx 6.3, the opposite leg is y = 7\sin\!\left(\frac{\pi}{7}\right) \approx 3.0, and the other non-right angle is \phi = \frac{\pi}{2} - \frac{\pi}{7} = \frac{5\pi}{14}. +

+ @@ -37,7 +41,11 @@ - + +

+ We know \sin(\alpha) = \frac{3}{5}, so the opposite side and hypotenuse are in ratio 3:5, and by the Pythagorean theorem the adjacent side is in ratio 4:5. Without a specific side length, we can only determine that the triangle is similar to a 3-4-5 right triangle: sides are 3a, 4a, 5a for some a > 0. The angles are \alpha = \arcsin\!\left(\frac{3}{5}\right) \approx 0.6435 and \frac{\pi}{2} - \alpha \approx 0.9273 radians. +

+ @@ -47,7 +55,11 @@ - + +

+ With angle \beta = 1.2 radians and hypotenuse 2.7: adjacent leg x = 2.7\cos(1.2) \approx 0.98, opposite leg y = 2.7\sin(1.2) \approx 2.5, and other angle \alpha = \frac{\pi}{2} - 1.2 \approx 0.371 radians. +

+ @@ -57,7 +69,11 @@ - + +

+ With hypotenuse 13 and one leg 6.5, the other leg is \sqrt{13^2 - 6.5^2} = \sqrt{169 - 42.25} \approx 11.26. The angles are \arccos\!\left(\frac{6.5}{13}\right) = \arccos\!\left(\frac{1}{2}\right) = \frac{\pi}{3} \approx 1.047 and \arcsin\!\left(\frac{1}{2}\right) = \frac{\pi}{6} \approx 0.524 radians. +

+ @@ -67,7 +83,11 @@ - + +

+ With legs 5 and 12, the hypotenuse is \sqrt{5^2 + 12^2} = \sqrt{169} = 13. The angles are \arctan\!\left(\frac{5}{12}\right) \approx 0.395 and \arctan\!\left(\frac{12}{5}\right) \approx 1.176 radians. +

+ @@ -77,7 +97,11 @@ - + +

+ With angle \beta = \frac{\pi}{5} and opposite leg 4: \sin\!\left(\frac{\pi}{5}\right) = \frac{4}{h} where h is the hypotenuse, so h = \frac{4}{\sin(\pi/5)} \approx 6.805. The adjacent leg is \sqrt{h^2 - 16} \approx 5.505. The other angle is \frac{\pi}{2} - \frac{\pi}{5} = \frac{3\pi}{10}. +

+

diff --git a/source/activities/act-trig-right-similar.xml b/source/activities/act-trig-right-similar.xml index fba4acd2..df55ec06 100755 --- a/source/activities/act-trig-right-similar.xml +++ b/source/activities/act-trig-right-similar.xml @@ -31,7 +31,11 @@ - + +

+ Both triangles share the angle \theta at vertex O, and both have a right angle (\angle OPQ and \angle ONM). Since two pairs of angles are equal, the triangles are similar by AA similarity. +

+ @@ -41,7 +45,11 @@ - + +

+ The ratio \frac{OP}{OM} = \frac{r}{1} = r. Since the triangles are similar with scale factor r, every pair of corresponding sides has the same ratio, so \frac{OQ}{ON} = r and \frac{PQ}{MN} = r. +

+ @@ -51,7 +59,11 @@ - + +

+ In the smaller right triangle \triangle OMN with hypotenuse OM = 1 and angle \theta at O, we have ON = \cos(\theta) (adjacent side) and MN = \sin(\theta) (opposite side). +

+ @@ -61,7 +73,11 @@ - + +

+ Since \frac{OQ}{ON} = r, we get x = OQ = r \cdot ON = r\cos(\theta). Since \frac{PQ}{MN} = r, we get y = PQ = r \cdot MN = r\sin(\theta). +

+

diff --git a/source/activities/act-trig-tangent-mountain.xml b/source/activities/act-trig-tangent-mountain.xml index 2f2d31b0..77d1344a 100755 --- a/source/activities/act-trig-tangent-mountain.xml +++ b/source/activities/act-trig-tangent-mountain.xml @@ -31,7 +31,11 @@ - + +

+ Using the right triangle with the 25^\circ angle, the opposite side is h and the adjacent side is x, so \tan(25^\circ) = \dfrac{h}{x}. +

+ @@ -41,7 +45,11 @@ - + +

+ Using the right triangle with the 19^\circ angle, the opposite side is h and the adjacent side is x + 1000, so \tan(19^\circ) = \dfrac{h}{x+1000}. +

+ @@ -51,7 +59,17 @@ - + +

+ From (a), h = x\tan(25^\circ). From (b), h = (x+1000)\tan(19^\circ). Setting equal: + + x\tan(25^\circ) &= (x+1000)\tan(19^\circ) + x\tan(25^\circ) - x\tan(19^\circ) &= 1000\tan(19^\circ) + x(\tan(25^\circ) - \tan(19^\circ)) &= 1000\tan(19^\circ) + x &= \dfrac{1000\tan(19^\circ)}{\tan(25^\circ) - \tan(19^\circ)}. + +

+ @@ -61,7 +79,11 @@ - + +

+ The exact value is x = \dfrac{1000\tan(19^\circ)}{\tan(25^\circ) - \tan(19^\circ)}. Numerically, x \approx 3412.527 feet. +

+ @@ -71,7 +93,11 @@ - + +

+ Using h = x\tan(25^\circ) with x \approx 3412.527: h \approx 3412.527 \cdot \tan(25^\circ) \approx 1591.287 feet. +

+ @@ -81,7 +107,11 @@ - + +

+ If the initial measurements were taken from 78 feet above sea level, the peak of the hill is approximately 1591.287 + 78 \approx 1669 feet above sea level. +

+

diff --git a/source/exercises/ez-changing-aroc.xml b/source/exercises/ez-changing-aroc.xml index 94d18855..35dfb2ee 100755 --- a/source/exercises/ez-changing-aroc.xml +++ b/source/exercises/ez-changing-aroc.xml @@ -105,9 +105,9 @@

-

- Exercise Solution -

+

  1. AV_{[0,5]} = \frac{44.74 - 37.00}{5} = 1.548 F/min; AV_{[5,10]} = \frac{50.77 - 44.74}{5} = 1.206 F/min; AV_{[10,15]} = \frac{55.47 - 50.77}{5} = 0.940 F/min. For example, on the interval [0,5], the temperature of the soda is increasing on average by 1.548 degrees Fahrenheit for each additional minute.

  2. Change on [10,20]: 59.12 - 50.77 = 8.35 F. Change on [25,35]: 65.92 - 61.97 = 3.95 F. There is more total change on [10,20].

  3. The soda's temperature is changing most rapidly at the beginning (when t is smallest), and the rate of change decreases over time as the soda approaches room temperature.

  4. The rate of change on [30,35] is AV_{[30,35]} = \frac{65.92 - 64.19}{5} = 0.346 F/min. Extending this trend by 2 minutes: F(37) \approx 65.92 + 2(0.346) \approx 66.6 F.

+ + @@ -171,9 +171,7 @@

-

- Exercise Solution -

+

  1. Read from the graph: describe behavior on each interval. Where the graph rises steeply, the car moves quickly; where flat, the car is stopped; where the slope is gentler, the car moves more slowly.

  2. From the graph: AV_{[3,4]} = \frac{s(4)-s(3)}{1}, AV_{[4,6]} = \frac{s(6)-s(4)}{2}, AV_{[5,8]} = \frac{s(8)-s(5)}{3}, each in thousands of feet per minute.

  3. On the graph, sketch line segments connecting (3, s(3)) to (4, s(4)), (4, s(4)) to (6, s(6)), and (5, s(5)) to (8, s(8)); their slopes represent the respective average velocities.

  4. An average velocity of 5000 ft/min requires a slope of 5 on the graph. Examine whether any secant line connecting two points on the graph achieves this slope.

  5. If the position graph is always non-decreasing, the car never goes in reverse. Reverse motion would correspond to a portion of the graph with negative slope (decreasing s).

 @@ -215,9 +213,7 @@

-

- Exercise Solution -

+

  1. As the conical tank fills, the cone widens, so equal amounts of water produce smaller increases in height. Therefore h = g(t) is increasing but at a decreasing rate (concave down).

  2. Using g(t) = (t/\pi)^{1/3}: g(0)=0, g(2)=(2/\pi)^{1/3}\approx 0.860, g(4)=(4/\pi)^{1/3}\approx 1.084, g(6)=(6/\pi)^{1/3}\approx 1.240. Thus AV_{[0,2]}\approx 0.430 ft/min, AV_{[2,4]}\approx 0.112 ft/min, AV_{[4,6]}\approx 0.078 ft/min.

  3. Yes. For intervals [0,b] starting at t=0: AV_{[0,b]} = g(b)/b = (1/\pi)^{1/3} b^{-2/3}. Setting this equal to 2 gives b^{2/3} = (1/\pi)^{1/3}/2, so b = (1/(8\pi))^{1/2} = 1/(2\sqrt{\pi}\,) \approx 0.282 min. The interval [0, 1/(2\sqrt{\pi})] gives AV = 2 ft/min.

  4. Since V = f(t) = 0.75t is linear with constant slope 0.75, the average rate of change of volume is always 0.75 cu ft/min on every interval, including [0,2], [2,4], and [4,6].

 diff --git a/source/exercises/ez-changing-combining.xml b/source/exercises/ez-changing-combining.xml index 89c0e815..e9fc3a66 100755 --- a/source/exercises/ez-changing-combining.xml +++ b/source/exercises/ez-changing-combining.xml @@ -93,9 +93,7 @@

-

- Exercise Solution -

+

  1. f(t) = (r+s)(t) = (2t-3)+(5-3t) = 2-t

  2. g(t) = \left(\frac{s}{r}\right)(t) = \frac{5-3t}{2t-3} (undefined when t = \frac{3}{2})

  3. h(t) = (r\cdot s)(t) = (2t-3)(5-3t) = -6t^2+19t-15

  4. q(t) = (s\circ r)(t) = s(2t-3) = 5-3(2t-3) = 14-6t

  5. w(t) = r(t-4)+7 = 2(t-4)-3+7 = 2t-4

 @@ -170,9 +168,7 @@

-

- Exercise Solution -

+

  1. Read the graph of s to identify its behavior on each piece. Find the equations of each linear piece and write them in bracket notation with the corresponding domain intervals.

  2. Similarly, read the graph of g and write a piecewise formula.

  3. With formulas for s and g in hand: (i) (s\cdot g)(1) = s(1)\cdot g(1); (ii) (g-s)(3) = g(3)-s(3); (iii) (s\circ g)(1.5) = s(g(1.5)); (iv) (g\circ s)(-4) = g(s(-4)). For each, evaluate using the piecewise formulas.

 @@ -301,5 +297,17 @@

+ +

+ Exercise Answer +

+ + +

  1. The triathlete exits the water after the swim at t = 2.4/2.5 = 0.96 hours.

  2. The bike finishes at t = 0.96 + 0.05 + 112/21 \approx 6.34 hours; the race ends at t \approx 6.37 + 26.2/8.5 \approx 9.45 hours.

  3. Key points: exits water at (0.96, 2.4); starts bike at (1.01, 2.4); finishes bike at (6.34, 114.4); starts run at (6.37, 114.4); finishes at (9.45, 140.6).

  4. f(t) = \begin{cases} 2.5t, \amp 0 \le t \le 0.96 \\ 2.4, \amp 0.96 \lt t \le 1.01 \\ 2.4 + 21(t-1.01), \amp 1.01 \lt t \le 6.34 \\ 114.4, \amp 6.34 \lt t \le 6.37 \\ 114.4 + 8.5(t-6.37), \amp 6.37 \lt t \le 9.45 \end{cases}

  5. See graphs below.

+ + + + + diff --git a/source/exercises/ez-changing-composite.xml b/source/exercises/ez-changing-composite.xml index 6946671e..771b6554 100755 --- a/source/exercises/ez-changing-composite.xml +++ b/source/exercises/ez-changing-composite.xml @@ -162,9 +162,7 @@

-

- Exercise Solution -

+

  1. From the graphs: g(1) can be read, then f(g(1)); and f(-2) can be read, then g(f(-2)).

  2. g(f(x)) is undefined if f(x) falls outside the domain of g or if f(x) is itself undefined.

  3. From the table: (s\circ r)(3) = s(r(3)) = s(-1) = -8; (s\circ r)(-4) = s(r(-4)) = s(4) = 5; for example (s\circ r)(0) = s(r(0)) = s(0) = 0.

  4. Domain of r: \{-4,-3,-2,-1,0,1,2,3,4\}; range of r: \{-4,-3,-1,0,1,2,3,4\}. Domain of s: \{-4,...,4\}; range of s: \{-8,-7,-6,-5,0,5,6,7,8\}. For (r\circ s)(b), we need s(b) in the domain of r. The only value of s in \{-4,...,4\} is s(0)=0, so only b=0 works: (r\circ s)(0) = r(0) = 0.

  5. m(x^2) = x^6 + 4x^4 - 5x^2 + 1. m(2+h) = (2+h)^3 + 4(2+h)^2 - 5(2+h) + 1 = h^3 + 10h^2 + 23h + 15. m(a+h) = (a+h)^3 + 4(a+h)^2 - 5(a+h) + 1.

  6. F(x) = 4-3x-x^2. F(2) = -6 and F(2+h) = 4-3(2+h)-(2+h)^2 = -6-7h-h^2. Thus AV_{[2,2+h]} = \frac{-6-7h-h^2-(-6)}{h} = -7-h.

 @@ -220,9 +218,7 @@

-

- Exercise Solution -

+

  1. Solving F = 40 + \frac{1}{4}N for N: N = 4(F-40) = 4F - 160.

  2. g(F) = 4F - 160 converts temperature in Fahrenheit to expected chirps per minute.

  3. g(82) = 4(82) - 160 = 168 chirps per minute, written N = g(82) = 168.

  4. Solving C = \frac{5}{9}(F-32) for F: F = \frac{9}{5}C + 32 = p(C). This converts Celsius to Fahrenheit.

  5. Yes: N = (g\circ p)(C) = g(p(C)) = g\!\left(\frac{9C}{5}+32\right) = 4\!\left(\frac{9C}{5}+32\right) - 160 = \frac{36C}{5} - 32. This function converts Celsius temperature to expected chirps per minute.

 @@ -321,9 +317,12 @@

-

- Exercise Solution -

+

  1. Domain: 0 \le h \le 8 (diameter of sphere). Range: 0 \le V \le \frac{256\pi}{3}.

  2. Since the height increases at a constant rate, h = p(t) = 0.4t is a linear function.

  3. Domain: [0, 20] minutes (height reaches 8 m at t = 8/0.4 = 20 min). Range: [0, 8] m.

  4. V = q(t) = f(p(t)) = \frac{\pi}{3}(0.4t)^2(12 - 0.4t) = \frac{\pi}{3}(0.16t^2)(12-0.4t) = \frac{\pi}{3}\cdot\frac{0.064t^2(30-t)}{1} = \frac{8\pi}{375}t^2(30-t).

  5. Domain: [0, 20] minutes. Range: \left[0, \frac{256\pi}{3}\right] cubic meters.

  6. The graph of h = p(t) is a straight line (height rises at constant rate). The graph of V = q(t) is an S-shaped increasing cubic: slow at first (sphere is narrow near the bottom), then faster through the middle (widest part), then slower again as the sphere narrows near the top.

+ + + + + diff --git a/source/exercises/ez-changing-functions.xml b/source/exercises/ez-changing-functions.xml index ec17e7ea..870f5216 100644 --- a/source/exercises/ez-changing-functions.xml +++ b/source/exercises/ez-changing-functions.xml @@ -92,9 +92,13 @@

-

- Exercise Solution -

+

  1. The full volume is V = \frac{1}{3}\pi(3)^2(2) = 6\pi \approx 18.85 cubic feet. At a constant fill rate of 0.75 cu ft/min, the tank fills in \frac{6\pi}{0.75} = 8\pi \approx 25.1 minutes.

  2. The graph of V = f(t) is a straight line from (0,0) to (8\pi, 6\pi) (constant rate). The graph of h = g(t) is a concave-down increasing curve from (0,0) to (8\pi, 2): since the cone widens as it fills, equal volumes of water raise the height by smaller and smaller amounts.

  3. The domain of h = g(t) is [0, 8\pi] minutes and the range is [0, 2] feet.

  4. The model h = g(t) = \left(\frac{t}{\pi}\right)^{1/3} has domain [0, 8\pi] and range [0,2]. The abstract function y = p(t) = \left(\frac{t}{\pi}\right)^{1/3} has domain [0, \infty) and range [0, \infty). The model is the abstract function restricted to the physical scenario.

+ + + + + + @@ -141,9 +145,7 @@

-

- Exercise Solution -

+

  1. The values of v(2) and v(7) can be read directly from the graph at t = 2 and t = 7, respectively.

  2. Any time the graph of v crosses or touches the horizontal line v = 3 gives a time when the velocity equals exactly 3 ft/s. Read these times from the graph.

  3. On the interval [4,5], examine whether the graph is positive (walking forward), negative (walking backward), or zero (stationary), and whether the velocity is increasing or decreasing.

  4. The distance traveled is proportional to the area under the velocity curve (or its absolute value). Compare the areas under the graph on [1,3] and on [6,8] to determine which interval has greater total distance.

 @@ -216,9 +218,7 @@

-

- Exercise Solution -

+

  1. No, G cannot be viewed as a function of D. At D = 100, there are two recorded values (G = 1.5 before refueling and G = 10.0 after), and similarly at D = 300. A function must assign exactly one output to each input.

  2. The fuel economy appears to be constant. Between D = 0 and D = 100, the car used 4.5 - 1.5 = 3.0 gallons for 100 miles, giving 100/3 \approx 33.3 mpg. The same rate holds for subsequent segments.

  3. At D = 100, the driver added 10.0 - 1.5 = 8.5 gallons. At D = 300, the driver added 11.0 - 4.0 = 7.0 gallons. So the driver put the most gas in the tank at D = 100 miles.

 diff --git a/source/exercises/ez-changing-inverse.xml b/source/exercises/ez-changing-inverse.xml index 45ebef3b..f2ece523 100755 --- a/source/exercises/ez-changing-inverse.xml +++ b/source/exercises/ez-changing-inverse.xml @@ -76,9 +76,9 @@

-

- Exercise Solution -

+

  1. From the graph: p^{-1}(2.5) is the input to p that gives output 2.5, found by reading p(x) = 2.5 from the graph. The problem states p^{-1}(2.5) = -3.5. p^{-1}(-2) and p^{-1}(0) are read similarly; q^{-1}(2) requires reading q(x) = 2.

  2. Since the graph of p^{-1} is the reflection of the graph of p across the line y = x, any point (a, b) on the graph of p gives a point (b, a) on the graph of p^{-1}. Use this to find additional points.

  3. Plot the six known points and sketch the complete graph of y = p^{-1}(x). The graphs of p and p^{-1} are reflections of each other across the line y = x.

+ + @@ -152,9 +152,9 @@

-

- Exercise Solution -

+

  1. Since the water level rises at a constant rate of 0.25 m/min starting from 0, the height at time t is h = 0.25t, so f(t) = 0.25t.

  2. At h = 2.5: 0.25t = 2.5 \Rightarrow t = 10 min. Full (h=4): t = 4/0.25 = 16 min.

  3. Yes: as time increases, the volume strictly increases (more water is added continuously), so the function g is one-to-one and has an inverse.

  4. Substituting h = 0.25t: V = \frac{\pi}{12}(0.25t)^3 = \frac{\pi}{12} \cdot \frac{t^3}{64} = \frac{\pi t^3}{768}. The graph is an increasing cubic curve (concave up), since the cone widens as it fills.

  5. Solving V = \frac{\pi t^3}{768} for t: t^3 = \frac{768V}{\pi}, so t = g^{-1}(V) = \left(\frac{768V}{\pi}\right)^{1/3}. This gives the time (in minutes) for the volume in the tank to reach V cubic meters.

  6. At V = \frac{8\pi}{3}: t = \left(\frac{768 \cdot \frac{8\pi}{3}}{\pi}\right)^{1/3} = \left(2048\right)^{1/3} = 8\cdot 2^{2/3} = 8\sqrt[3]{4} minutes.

+ + diff --git a/source/exercises/ez-changing-linear.xml b/source/exercises/ez-changing-linear.xml index 9ed2dd19..999a4997 100755 --- a/source/exercises/ez-changing-linear.xml +++ b/source/exercises/ez-changing-linear.xml @@ -110,9 +110,7 @@

-

- Exercise Solution -

+

  1. Each \$50 increase in rent results in a decrease of exactly 7 occupied apartments (constant rate of change), which means the relationship is linear.

  2. Slope = \frac{196-203}{700-650} = \frac{-7}{50} = -0.14 apartments per dollar. For each dollar increase in rent, the expected number of occupied apartments decreases by 0.14.

  3. A = f(R) = 203 - 0.14(R - 650) = 294 - 0.14R. A reasonable domain is [650, 2100] (rents below \$650 are below current data, and above \$2100 all apartments would be empty).

  4. At R = 1000: A = 294 - 0.14(1000) = 154 apartments. Monthly revenue = 1000 \times 154 = \$154{,}000.

  5. The manager wants to find the rent that maximizes total revenue, R \cdot f(R) = 294R - 0.14R^2. This is maximized at R = 294/(2 \times 0.14) = \$1050, giving 147 apartments and \$154{,}350 monthly revenue.

 @@ -164,9 +162,7 @@

-

- Exercise Solution -

+

  1. Read the formulas for A = f(t) and D = g(t) from the graph by identifying slope and intercept for each linear piece.

  2. The slope of A = f(t) is its constant rate of change (acceleration in ft/s²). Write a sentence interpreting this value: e.g., "Alicia's velocity increases/decreases at a rate of [slope] feet per second per second."

  3. AV_{[4,8]} for D is computed from the graph values g(4) and g(8); its meaning is the average rate of change of Damon's velocity on that interval.

  4. Set f(t) = g(t) and solve for t. If a solution exists in [0,10], they walk at the same velocity at that time.

  5. The velocity functions do not give us location information (we would need initial positions), so it is generally not possible to determine when/if they are at the same location.

 @@ -219,9 +215,7 @@

-

- Exercise Solution -

+

  1. The full cone has volume V(0) = \frac{1}{3}\pi(2)^2(4) = \frac{16\pi}{3} \approx 16.76 cu ft.

  2. The volume decreases at a constant rate of 1.5 cu ft/min (given), so the rate of change of V with respect to t is constant: V is linear in t.

  3. V = f(t) = \frac{16\pi}{3} - 1.5t.

  4. Setting f(t) = 0: t = \frac{16\pi}{3 \times 1.5} = \frac{16\pi}{4.5} = \frac{32\pi}{9} \approx 11.17 minutes.

  5. Domain: \left[0, \frac{32\pi}{9}\right]. Range: \left[0, \frac{16\pi}{3}\right].

 diff --git a/source/exercises/ez-changing-quadratic.xml b/source/exercises/ez-changing-quadratic.xml index 2805e79c..e8e6569c 100755 --- a/source/exercises/ez-changing-quadratic.xml +++ b/source/exercises/ez-changing-quadratic.xml @@ -88,9 +88,9 @@

-

- Exercise Solution -

+

  1. For f: read the vertex (h,k) and one additional point from the graph. Write f(x) = a(x-h)^2 + k and solve for a using the additional point.

  2. For g: similarly read the vertex and one additional point and determine the formula g(x) = a(x-h)^2 + k.

  3. Set h(x) = f(x): 2x^2 - 8x + 6 = f(x). Solve the resulting equation (quadratic). If the discriminant is non-negative, there are intersection points; otherwise the graphs do not intersect.

  4. Set h(x) = g(x) and solve similarly. The discriminant determines whether intersection points exist.

+ + @@ -196,9 +196,7 @@

-

- Exercise Solution -

+

  1. Vertex at (2, 1) (from vertex form).

  2. The minimum value is f(2) = 1 \gt 0, so f has no x-intercepts.

  3. Function values: f(0) = 3, f(1) = \frac{3}{2}, f(2) = 1, f(3) = \frac{3}{2}, f(4) = 3, f(5) = \frac{11}{2}. Average rates of change: AV_{[0,1]} = -\frac{3}{2}, AV_{[1,2]} = -\frac{1}{2}, AV_{[2,3]} = \frac{1}{2}, AV_{[3,4]} = \frac{3}{2}, AV_{[4,5]} = \frac{5}{2}.

  4. The function values are symmetric about x = 2 (the vertex). The average rates of change form an arithmetic sequence increasing by 1 at each step, a characteristic pattern of quadratic functions.

 @@ -239,9 +237,7 @@

-

- Exercise Solution -

+

  1. The vertex occurs at t = 1.2. By symmetry of the parabola, t = 2.4 is symmetric to t = 0 about the vertex: s(2.4) = s(0) = 56.3 feet.

  2. At the vertex, the initial velocity is v_0 = 32 \times 1.2 = 38.4 ft/s (since vertex is at t = v_0/32). Maximum height: s(1.2) = -16(1.44) + 38.4(1.2) + 56.3 = -23.04 + 46.08 + 56.3 = 79.34 feet.

  3. Set s(t) = 0: -16t^2 + 38.4t + 56.3 = 0. Using the quadratic formula: t = \frac{38.4 + \sqrt{38.4^2 + 4(16)(56.3)}}{32} = \frac{38.4 + \sqrt{5077.76}}{32} \approx \frac{38.4 + 71.26}{32} \approx 3.43 seconds.

  4. The initial velocity is v_0 = 38.4 = \frac{192}{5} feet per second.

 diff --git a/source/exercises/ez-changing-tandem.xml b/source/exercises/ez-changing-tandem.xml index c2a1a214..ae350fb6 100755 --- a/source/exercises/ez-changing-tandem.xml +++ b/source/exercises/ez-changing-tandem.xml @@ -77,9 +77,7 @@

-

- Exercise Solution -

+

  1. At a fill rate of 1.25 cu ft/min, the time to fill is \frac{V_{\text{full}}}{1.25} = \frac{\pi(20 + \frac{38}{3}\sqrt{2})}{1.25} \approx \frac{119.11}{1.25} \approx 95.3 minutes.

  2. The graph of V vs t is a straight line with slope 1.25 from (0,0) to (95.3, 119.11). The graph of h vs t is an S-shaped increasing curve: initially h rises quickly as the narrow bottom of the sphere fills, then more slowly as the widest part of the sphere fills, then speeds up slightly as the narrow chimney fills. Key points include (0,0), the time when h = \sqrt{8} + 3 \approx 5.83 ft (sphere full, chimney starts), and the endpoint (95.3, \sqrt{8}+5).

  3. If the chimney were an inverted cone, its cross-section would shrink as it fills, so the height would increase more quickly in the chimney. The V-vs-t graph would not change (still linear), but the h-vs-t graph would rise more steeply once the water enters the cone-shaped chimney.

 @@ -129,9 +127,7 @@

-

- Exercise Solution -

+

  1. At a constant rate of 0.4 ft/min, the depth reaches 12 ft (the diameter) in 12/0.4 = 30 minutes. At that time, V = \frac{4}{3}\pi(6)^3 = 288\pi \approx 904.8 cu ft.

  2. Since the height rises linearly, h = 0.4t is a straight line. The volume V as a function of t starts slowly (the bottom of the sphere is narrow), increases rapidly through the middle (widest part of sphere), and slows again as the top narrows. The V-vs-t graph is S-shaped (concave up then concave down). Key points: (0,0) and (30, 288\pi).

  3. For the draining scenario: h starts at 12 ft and decreases linearly at 0.25 ft/min to 0 in 48 minutes. The h-vs-t graph is a decreasing line. The V-vs-t graph is an S-shaped decreasing curve (symmetric to the filling case), from (0, 288\pi) to (48, 0).

 @@ -187,9 +183,7 @@

-

- Exercise Solution -

+

  1. Describe the car based on the graph: on intervals where s is increasing, the car moves forward; where s is flat, the car is stopped; where s decreases, the car is in reverse. The slope (steepness) indicates speed.

  2. From the graph, s(2) and s(10) can be read. The distance traveled between t = 2 and t = 10 is |s(10) - s(2)| thousand feet.

  3. If the position graph is always non-decreasing (never moves backward), the car never goes in reverse. A graph that decreases (negative slope) over some interval would indicate reverse motion.

  4. For the described motion: t = 0: s = 0. At t = 5: s = 1000 \times 5 = 5000 ft (i.e., 5 thousand ft). From t = 5 to t = 7: car is stopped, so s = 5. At t = 7: U-turn, moving at -800 ft/min. At t = 12: s = 5000 - 800 \times 5 = 5000 - 4000 = 1000 ft (i.e., 1 thousand ft). Plot a line of slope 1 from (0,0) to (5,5), horizontal from (5,5) to (7,5), and a line of slope -0.8 from (7,5) to (12,1).

 diff --git a/source/exercises/ez-changing-transformations.xml b/source/exercises/ez-changing-transformations.xml index 62768814..8996b343 100755 --- a/source/exercises/ez-changing-transformations.xml +++ b/source/exercises/ez-changing-transformations.xml @@ -87,9 +87,7 @@

-

- Exercise Solution -

+

  1. For f(x)=x^2: AV_{[-3,-1]}=(1-9)/2=-4 and AV_{[2,5]}=(25-4)/3=7. For g(x)=x^2+5: AV_{[-3,-1]}=(6-14)/2=-4 and AV_{[2,5]}=(30-9)/3=7. The values are the same: vertical shifts do not change average rates of change.

  2. For h(x)=(x-2)^2: AV_{[-1,1]}=(1-9)/2=-4 (same as AV_{[-3,-1]} for f) and AV_{[4,7]}=(25-4)/3=7 (same as AV_{[2,5]} for f). A horizontal shift of 2 units right causes the average rate of change to match on intervals that are shifted right by 2.

  3. For k(x)=3x^2: AV_{[-3,-1]}=(3-27)/2=-12=3\cdot(-4) and AV_{[2,5]}=(75-12)/3=21=3\cdot 7. Multiplying the output by 3 multiplies all average rates of change by 3.

  4. Conjecture: AV_{[-1,1]} for m equals 3\cdot AV_{[-3,-1]} for f, which is 3(-4) = -12. Verification: m(x)=3(x-2)^2+5, m(-1)=3(9)+5=32, m(1)=3(1)+5=8, AV_{[-1,1]}=(8-32)/2=-12. \checkmark

 @@ -130,9 +128,7 @@

-

- Exercise Solution -

+

  1. The slope is -4. The most obvious point is (3, 5).

  2. Starting from f(x)=x: shift right by 3 gives f(x-3)=x-3; multiply by -4 gives -4(x-3) (vertical stretch by 4 and reflection); shift up by 5 gives -4(x-3)+5 = L(x). So 3 is the horizontal shift, -4 is the vertical stretch/reflection factor, and 5 is the vertical shift.

  3. P(x)=m(x-x_0)+y_0 is a horizontal shift of f by x_0, a vertical stretch by m, and a vertical shift by y_0.

  4. Horizontal shift left 7: f(x+7) = x+7. Reflection and vertical stretch by 3: -3(x+7). Vertical shift -11: -3(x+7)-11 = -3x-32.

 @@ -190,9 +186,12 @@

-

- Exercise Solution -

+

  1. With f(x)=(x-2)^2+1: g(x)=f(4x) has vertex at (\frac{1}{2}, 1), y-intercept 5, no x-intercepts. h(x)=f(2x) has vertex (1,1), y-intercept 5. k(x)=f(0.5x) has vertex (4,1), y-intercept 5. m(x)=f(0.25x) has vertex (8,1), y-intercept 5. All have the same minimum value and y-intercept; increasing a horizontally compresses the graph.

  2. The transformation y=f(ax) with a\gt 0 horizontally compresses the graph by factor a (when a\gt 1) or stretches it by factor 1/a (when 0\lt a\lt 1). Points on the y-axis are unchanged.

  3. r(x)=f(-x)=(-x-2)^2+1=(x+2)^2+1. This is the reflection of f across the y-axis: every point (a,b) on f maps to (-a,b) on r.

  4. s(x)=f(-2x) combines a horizontal compression by 2 with reflection across the y-axis. In general, y=f(ax) with a\lt 0 reflects across the y-axis and compresses horizontally by |a|.

+ + + + + diff --git a/source/exercises/ez-circular-sine-cosine.xml b/source/exercises/ez-circular-sine-cosine.xml index dc53e0e6..8c8aa81d 100755 --- a/source/exercises/ez-circular-sine-cosine.xml +++ b/source/exercises/ez-circular-sine-cosine.xml @@ -87,7 +87,7 @@

t in quadrant IV such that \sin(t) = -\frac{\sqrt{3}}{2}

- +

@@ -99,7 +99,15 @@

- Exercise Solution +

    +

  1. \sin\!\left(-\frac{11\pi}{4}\right) = \sin\!\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}

    2. +

  2. \cos\!\left(\frac{29\pi}{6}\right) = \cos\!\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}

    3. +

  3. \cos\!\left(\frac{17\pi}{3}\right) = \cos\!\left(\frac{5\pi}{3}\right) = \frac{1}{2}

    4. +

  4. \sin(47\pi) = 0

    5. +

  5. \cos(-113\pi) = \cos(113\pi) = (-1)^{113} = -1

    6. +

  6. t = \frac{7\pi}{6}

    7. +

  7. t = \frac{5\pi}{3}

    8. +

@@ -153,7 +161,38 @@

- Exercise Solution +

    +
  1. +

    + Yes, \cos(t+2\pi) = \cos(t) is true because the function has a periodicity of 2\pi. +

    +
    2. +
  2. +

    + Yes, \sin(t-\pi) = -\sin(t) is true because a shift of \pi around the unit circle is equivalent to making the argument negative, and \sin is an odd function. Also, graphically, \sin gives the y-coordinate of points on the unit circle, and shifting the argument by \pm\pi essentially reflects the point across the x-axis, changing the sign of the y-coordinate. +

    +
    3. +
  3. +

    + No, \cos\!\left(t - \frac{3\pi}{2}\right) \ne \sin(t). For example, \cos\!\left(\frac{\pi}{2} - \frac{3\pi}{2}\right) = \cos(-\pi) = -1 while \sin\!\left(\frac{\pi}{2}\right) = 1. +

    +
    4. +
  4. +

    + Yes, \sin^2(t) = 1 - \cos^2(t), because of the Fundamental Trigonometric Identity. +

    +
    5. +
  5. +

    + No, \sin(t) + \cos(t) \ne 1. For example, \sin\!\left(\frac{\pi}{3}\right) + \cos\!\left(\frac{\pi}{3}\right) = \frac{1 + \sqrt{3}}{2} \ne 1. +

    +
    6. +
  6. +

    + No, \sin(t) + \sin\!\left(\frac{\pi}{2}\right) \ne \cos(t). For example, \sin\!\left(\frac{\pi}{4}\right) + \sin\!\left(\frac{\pi}{2}\right) = \frac{\sqrt{2} + 2}{2} while \cos\!\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}. +

    +
    7. +

diff --git a/source/exercises/ez-circular-sinusoidal.xml b/source/exercises/ez-circular-sinusoidal.xml index b82bd4c6..4cfde29c 100755 --- a/source/exercises/ez-circular-sinusoidal.xml +++ b/source/exercises/ez-circular-sinusoidal.xml @@ -86,7 +86,10 @@

- Exercise Solution + There is more than one possible correct answer. If we take the points to indicate a period of 1.0 and midpoint at y = 2, with an anchor point at (0, 2), a formula is + + g(x) = \sin(2\pi x) + 2. +

@@ -108,7 +111,10 @@

- Exercise Solution + Assuming the period is 365 days, the amplitude is approximately 3.17545 hours, the midline is at y = 12.18405, and the maximum occurs at day t = 172 (with the minimum at day 355). Using the maximum as an anchor point, the duration of daylight is + + s(t) = 3.17545\cos\!\left(\frac{2\pi}{365}(t - 172)\right) + 12.18405. +

@@ -116,7 +122,7 @@

- We now understand the effects of the transformation h(t) = f(kt) where k \gt 0 for a given function f. Our goal is to understand what happens when k \lt 0. + We now understand the effects of the transformation h(t) = f(kt) where k \gt 0 for a given function f. Our goal is to understand what happens when k \lt 0.

@@ -161,7 +167,38 @@

- Exercise Solution +

    +
  1. +

    + The graphs of f(t) = 2t - 1 and g(t) = -2t - 1 have lines that are reflected across the y-axis. +

    +
    2. +
  2. +

    + For any function p, the function q(t) = p(-t) is a reflection of p across the y-axis. +

    +
    3. +
  3. +

    + The graph of y = \sin(-3t) is a graph of y = \sin(t) first horizontally scaled by a factor of \frac{1}{3} and then reflected across the y-axis. +

    +
    4. +
  4. +

    + The graph of y = \cos(-3t) is a graph of y = \cos(t) first horizontally scaled by a factor of \frac{1}{3} and then reflected across the y-axis. Since \cos is symmetric across the y-axis, y = \cos(-3t) appears identical to y = \cos(3t). +

    +
    5. +
  5. +

    + For any function p(t), the graph of q(t) = p(kt) with k \lt 0 is equivalent to the graph of p(t) scaled horizontally by a factor of \frac{1}{|k|} and then reflected across the y-axis. +

    +
    6. +
  6. +

    + The graphs \sin(-t) and \sin(t) are related such that \sin(-t) = -\sin(t), while \cos(-t) = \cos(t). +

    +
    7. +

diff --git a/source/exercises/ez-circular-traversing.xml b/source/exercises/ez-circular-traversing.xml index cf9a54c4..9d421604 100755 --- a/source/exercises/ez-circular-traversing.xml +++ b/source/exercises/ez-circular-traversing.xml @@ -144,7 +144,54 @@

- Exercise Solution +

    +
  1. +

    + The point P_0 is directly to the left of the center (2,2), so its x-coordinate is one radius less than 2. The radius is \frac{8}{2\pi} = \frac{4}{\pi}, so the exact x-coordinate is 2 - \frac{4}{\pi} = \frac{2(\pi-2)}{\pi} \approx 0.727. +

    +
    2. +
  2. + + + d + 0123 + 4567 + 8 + + + k + 0.731.1022.90 + 3.272.9021.10 + 0.73 + + + + + d + 9101112 + 13141516 + + + k + 1.1022.903.27 + 2.9021.100.73 + + +
    3. +
  3. + +
    4. +
  4. +

    + The graph is just a horizontal shift compared to the graph of h from . +

    +
    5. +
  5. +

    + When d = 51, k(51) = 2.90. When d = 102, k(102) = 2. +

    +
    6. +

@@ -200,7 +247,28 @@

- Exercise Solution +

    +
  1. +

    + The midline of each oscillation graph corresponds to the center y-coordinate, and the amplitude equals the radius. Function f has a maximum of 11 and a minimum of 3, so the radius (amplitude) is 4 and the midline is y = 7. Function g has a maximum of 12 and a minimum of 1, so the radius (amplitude) is 5.5 and the midline is y = 6.5. +

    +
    2. +
  2. +

    + Assuming counterclockwise motion: the circle for function f begins its traverse at the 9:00 position with coordinates (-4, 7). The circle for function g begins at the 12:00 position with coordinates (0, 12). +

    +
    3. +
  3. +

    + For function f, the period of the function is 8 units. For function g, the period is 8 units. For these graphs, one period is one complete traverse of the circle. The speed v of the point is equal to the circumference of the circle divided by the period T of one complete circle: v = \frac{2\pi r}{T}, so the horizontal scale corresponds to defining each period as \frac{2\pi r}{v}. +

    +
    4. +
  4. +

    + The changes to the graph depend in part on whether the speed stays the same. If the circle's radius gets one unit larger and the speed stays the same, then the period of the graph of function f increases to \frac{5}{4} \cdot 8 = 10 units; the amplitude of the graph changes to 5. For function g, a constant-speed scenario changes the period to \frac{7.5}{6.5} \cdot 8 \approx 9.23 units; the amplitude of g changes to 7.5. If the circle's radius gets one unit smaller and the speed stays the same, then the period of the graph of function f decreases to \frac{3}{4} \cdot 8 = 6 units; the amplitude of the graph changes to 3. For function g, a constant-speed scenario changes the period to \frac{5.5}{6.5} \cdot 8 \approx 6.77 units; the amplitude changes to 4.5. The amplitude of each graph will increase by 1 if the radius increases and decrease by 1 if the radius decreases, but there is no change to the midline. +

    +
    5. +

@@ -243,7 +311,26 @@

- Exercise Solution +

    +
  1. +

    + The center of the ferris wheel is at height \frac{92 - 7}{2} + 7 = 49.5 feet. +

    +
    2. +
  2. +

    + The radius is 42.5 feet, so the circumference is 2\pi \cdot 42.5 = 85\pi \approx 267 feet. +

    +
    3. +
  3. +

    + The circular function h = f(d) has amplitude 42.5 feet, midline y = 49.5 feet, and period 85\pi \approx 267 feet. +

    +
    4. +
  4. + +
    5. +

diff --git a/source/exercises/ez-circular-unit-circle.xml b/source/exercises/ez-circular-unit-circle.xml index 979095cf..26ba615c 100755 --- a/source/exercises/ez-circular-unit-circle.xml +++ b/source/exercises/ez-circular-unit-circle.xml @@ -77,7 +77,28 @@

- Exercise Solution +

    +
  1. +

    + Using x^2 + y^2 = 1 with x = -0.3 and y < 0: y = -\sqrt{1 - 0.09} = -\frac{\sqrt{91}}{10} \approx -0.954. +

    +
    2. +
  2. +

    + With x = -2y and x^2 + y^2 = 1: 4y^2 + y^2 = 1, so y = \frac{\sqrt{5}}{5} and x = -\frac{2\sqrt{5}}{5}. +

    +
    3. +
  3. +

    + Going \frac{29\pi}{6} clockwise corresponds to the angle -\frac{29\pi}{6} \equiv \frac{7\pi}{6} \pmod{2\pi}. The coordinates are \left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right). +

    +
    4. +
  4. +

    + Substituting y = \frac{1}{2}x + \frac{1}{2} into x^2 + y^2 = 1 and solving gives x = -1 or x = \frac{3}{5}. The intersection points are (-1, 0) and \left(\frac{3}{5}, \frac{4}{5}\right). +

    +
    5. +

@@ -125,10 +146,41 @@

- Exercise Solution +

    +
  1. +

    + A circle of radius r centered at (h,k) is simply the unit circle transformed with a horizontal translation h units to the right, a vertical translation k units up, and scaled both horizontally and vertically by a factor of r. Thus the parent equation x^2 + y^2 = 1 is transformed with x \to \frac{x-h}{r} and y \to \frac{y-k}{r}: + + \left(\frac{x-h}{r}\right)^2 + \left(\frac{y-k}{r}\right)^2 &= 1 \\ + \frac{(x-h)^2}{r^2} + \frac{(y-k)^2}{r^2} &= 1 \\ + (x-h)^2 + (y-k)^2 &= r^2. + +

    +
    2. +
  2. +

    + (x+3)^2 + (y-5)^2 = 4. +

    +
    3. +
  3. +

    + (x-4)^2 + (y+7)^2 = 25. +

    +
    4. +
  4. +

    + The circle has radius 4, so the arc length is 4 \cdot \frac{2\pi}{3} = \frac{8\pi}{3}. +

    +
    5. +
  5. +

    + The diameter has length \sqrt{(4-(-2))^2 + (2-(-1))^2} = \sqrt{45}, so the radius is \frac{\sqrt{45}}{2}. The center is the midpoint \left(1, \frac{1}{2}\right). The equation is (x-1)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{45}{4}. +

    +
    6. +

- + @@ -168,7 +220,26 @@

- Exercise Solution +

    +
  1. +

    + At \theta = \frac{\pi}{6}: a = \frac{5\sqrt{3}}{2}, b = \frac{5}{2}. +

    +
    2. +
  2. +

    + At \theta = \frac{\pi}{4}: a = b = \frac{5\sqrt{2}}{2}. At \theta = \frac{\pi}{3}: a = \frac{5}{2}, b = \frac{5\sqrt{3}}{2}. +

    +
    3. +
  3. +

    + One full revolution has distance equal to the circumference: 2\pi \cdot 5 = 10\pi. +

    +
    4. +
  4. + +
    5. +

diff --git a/source/exercises/ez-exp-e.xml b/source/exercises/ez-exp-e.xml index 66c449eb..f4985b53 100755 --- a/source/exercises/ez-exp-e.xml +++ b/source/exercises/ez-exp-e.xml @@ -97,7 +97,38 @@

- Exercise Solution +

    +
  1. +

    + With P = 100, r = 0.08, and n = 4: A(1) = 100\left(1 + \frac{0.08}{4}\right)^{4} = 100(1.02)^4 \approx \$108.24. +

    +
    2. +
  2. +

    + With n = 12: A(1) = 100\left(1 + \frac{0.08}{12}\right)^{12} \approx \$108.30. +

    +
    3. +
  3. +

    + With n = 52: A(1) = 100\left(1 + \frac{0.08}{52}\right)^{52} \approx \$108.32. +

    +
    4. +
  4. +

    + With n = 365: A(1) = 100\left(1 + \frac{0.08}{365}\right)^{365} \approx \$108.33. +

    +
    5. +
  5. +

    + With continuous compounding: A(1) = 100e^{0.08} \approx \$108.33. +

    +
    6. +
  6. +

    + In one year, continuously compounded interest yields approximately P \times 0.000855 more than quarterly compounding. Over 25 years, the difference grows to approximately P \times 0.144, with the greater amount coming from continuously compounded interest. +

    +
    7. +

@@ -140,7 +171,28 @@

- Exercise Solution +

    +
  1. +

    + When k \gt 0, g(t) is always increasing. When k \lt 0, g(t) is always decreasing. +

    +
    2. +
  2. +

    + The average rate of change of g(t) on [0,1] is greater (less negative) when k = -0.05, giving AV_{[0,1]} \approx -0.049, than when k = -0.1, giving AV_{[0,1]} \approx -0.095. However, the magnitude of the rate of change is greater for k = -0.1. +

    +
    3. +
  3. +

    + When k \lt 0, the long-term behavior of g(t) is that g(t) \to 0 as t \to \infty, since e^{-kt} \to \infty and e^{kt} = 1/e^{-kt} \to 0. +

    +
    4. +
  4. +

    + We need g(2) = e^{2k} = \frac{1}{2}, so k = -\frac{\ln 2}{2} \approx -0.3466. +

    +
    5. +

@@ -188,7 +240,28 @@

- Exercise Solution +

    +
  1. +

    + The long-term behavior of g(t) = e^{-0.05t} is that g(t) \to 0 as t \to \infty, since the exponent becomes increasingly negative. +

    +
    2. +
  2. +

    + The function F(t) \to 74.4 as t \to \infty. This means the soda approaches the room temperature of 74.4^\circ F as it warms up. +

    +
    3. +
  3. +

    + The temperature of the refrigerator is F(0) = 74.4 - 38.8 = 35.6^\circ F. +

    +
    4. +
  4. +

    + AV_{[10,20]} \approx 0.926 degrees per minute, AV_{[20,30]} \approx 0.562 degrees per minute, and AV_{[30,40]} \approx 0.341 degrees per minute. On the interval [10, 20], the soda is warming at an average rate of about 0.926 degrees per minute. The average rate of warming is decreasing over time, meaning the soda is warming more slowly as time goes on. +

    +
    5. +

diff --git a/source/exercises/ez-exp-growth.xml b/source/exercises/ez-exp-growth.xml index 02e9c35e..98f5e83e 100755 --- a/source/exercises/ez-exp-growth.xml +++ b/source/exercises/ez-exp-growth.xml @@ -100,7 +100,34 @@

- Exercise Solution +

    +
  1. +

    + The area of Grinnell Glacier is G(t) = 142(0.956)^t. +

    + +
    2. +
  2. +

    + In 1997 (t = -10), the model predicts 142(0.956)^{-10} \approx 223 acres. In 2012 (t = 5), the model predicts 142(0.956)^{5} \approx 113 acres. In 2022 (t = 15), the model predicts 142(0.956)^{15} \approx 72 acres. +

    +
    3. +
  3. +

    + According to the model, between 2007 and 2012 the glacier lost about 142 - 113 = 29 acres. +

    +
    4. +
  4. +

    + The average rate of change of G from 2007 to 2012 is approximately -29/5 \approx -5.8 acres per year. The average retreat of the glacier was about 5.8 acres per year from 2007 to 2012. This rate of change is expected to be larger in magnitude than the average rate of change from 2012 to 2017, because the rate of decrease is decreasing with time. +

    +
    5. +
  5. +

    + The Grinnell Glacier is retreating at a decreasing rate. +

    +
    6. +

@@ -150,7 +177,28 @@

- Exercise Solution +

    +
  1. +

    + Taking the ratio of the two given points: b = \left(\frac{1}{4}\right)^{1/5} \approx 0.7579, and substituting back gives a = 10 \cdot 2^{9/5} \approx 34.82. +

    +
    2. +
  2. +

    + AV_{[2,7]} = (5 - 20)/(7 - 2) = -3 and AV_{[7,12]} = (f(12) - 5)/(12 - 7) = -\frac{3}{4}. The greater average rate of change is on the interval [2,7]. +

    +
    3. +
  3. +

    + A linear equation passing through (2, 20) and (7, 5) is L = -3t + 26. +

    +
    4. +
  4. +

    + Given that the rate of decrease of the exponential function f is decreasing and the rate of change of the linear function L is constant, the average rate of change of f will be greater in magnitude than L on the interval [0, 2]. +

    +
    5. +

@@ -218,7 +266,28 @@

- Exercise Solution +

    +
  1. +

    + The function described by the data appears exponential because there is a consistent ratio of about 0.7408 between temperatures at sequential 2-minute intervals. +

    +
    2. +
  2. +

    + It is necessary to take the square root of the ratio between 2-minute measurements to find the 1-minute growth factor. A possible formula is F(t) = 175(0.7408)^{t/2} \approx 175(0.8607)^t. We can verify: F(0) = 175, F(2) \approx 129.64, F(4) \approx 96.04, and so on. +

    +
    3. +
  3. +

    + AV_{[4,6]} = (71.15 - 96.04)/(6 - 4) \approx -12.45 degrees per minute. The meaning of this rate of change is the average amount of cooling per minute on the interval [4,6]. +

    +
    4. +
  4. +

    + An insulating mug would reduce the rate of cooling, but the temperature would still follow an exponential model. The growth factor would be less than 1 but closer to 1 than 0.8607. +

    +
    5. +

@@ -271,7 +340,39 @@

- Exercise Solution +

    +
  1. +

    + The value of b is about 0.875, or \sqrt[3]{15.2/22.7}. The value of a is about 33.9 milligrams. +

    +
    2. +
  2. +

    + The initial dose is equal to the coefficient a, about 33.9 milligrams. +

    +
    3. +
  3. +

    + Eight hours after the initial dose, the remaining amount is 33.9(0.875)^8 \approx 11.6 milligrams. +

    +
    4. +
  4. +

    + There will be less than 1 mg in the body after about 26.4 hours. +

    +
    5. +
  5. +

    + AV_{[3,5]} \approx -2.7 mg/hour, AV_{[5,7]} \approx -2.0 mg/hour, and AV_{[7,9]} \approx -1.6 mg/hour. On the interval [3,5], the amount of drug in the patient is decreasing at an average rate of 2.7 mg per hour. The rate of decrease is decreasing over time. +

    +
    6. +
  6. +

    + A graph of the model is shown below. +

    + +
    7. +

diff --git a/source/exercises/ez-exp-log-properties.xml b/source/exercises/ez-exp-log-properties.xml index a72389a3..fef51bd6 100755 --- a/source/exercises/ez-exp-log-properties.xml +++ b/source/exercises/ez-exp-log-properties.xml @@ -82,7 +82,28 @@

- Exercise Solution +

    +
  1. +

    + Since the population initially has 100 members, A = 100. Since the population doubles in 3 years, we have P(3) = 200, so 200 = 100e^{3k}, giving e^{3k} = 2 and thus k = \frac{1}{3}\ln 2 \approx 0.2310. +

    +
    2. +
  2. +

    + Since P(4) = 250 and P(11) = 500, the population doubles every 11 - 4 = 7 years. So the doubling time is 7 years. The population will reach 2000 = 4 \cdot 500 members two doublings after t = 11, i.e., at t = 11 + 14 = 25 years. +

    +
    3. +
  3. +

    + If the doubling time is 21 years, then 2A = Ae^{21k}, so e^{21k} = 2 and k = \frac{1}{21}\ln 2. +

    +
    4. +
  4. +

    + If the doubling time is t_2, then k = \frac{1}{t_2}\ln 2 = \frac{\ln 2}{t_2}. +

    +
    5. +

@@ -125,10 +146,34 @@

- Exercise Solution +

    +
  1. +

    + At purchase (t = 0), V(0) = A = \$28000. Using V(1) = 23200, we get 23200 = 28000e^{-k}, so e^{-k} = \frac{23200}{28000} and k = -\ln\!\left(\frac{23200}{28000}\right) \approx 0.1880. +

    +
    2. +
  2. +

    + We solve 10000 = 28000e^{-kt}, giving e^{-kt} = \frac{10000}{28000}, so + + t = \frac{\ln(10000/28000)}{-k} = \frac{\ln(10000) - \ln(28000)}{\ln(23200) - \ln(28000)} \approx 5.48 \text{ years}. + +

    +
    3. +
  3. +

    + As t \to \infty, the model V(t) = Ae^{-kt} approaches 0, but if the car's value approaches \$500, adding a positive constant c = 500 gives V(t) = Ae^{-kt} + 500, whose long-term behavior approaches \$500 as desired. +

    +
    4. +
  4. +

    + With c = 500, from V(0) = 28000 we get A + 500 = 28000, so A = 27500. From V(1) = 23200 we get 27500e^{-k} + 500 = 23200, so e^{-k} = \frac{22700}{27500} and k = -\ln\!\left(\frac{22700}{27500}\right) \approx 0.1913. The values of A and k are different from part (a) because the model now incorporates the long-term value of \$500. +

    +
    5. +

- + @@ -180,9 +225,35 @@

- Exercise Solution +

    +
  1. +

    + The equation y = \log_b(x) is equivalent to b^y = x. +

    +
    2. +
  2. +

    + Taking the natural log of both sides of b^y = x gives y\ln(b) = \ln(x). +

    +
    3. +
  3. +

    + Dividing both sides by \ln(b) gives y = \dfrac{\ln(x)}{\ln(b)}. +

    +
    4. +
  4. +

    + Since y = \log_b(x) and we showed y = \dfrac{\ln(x)}{\ln(b)}, it must be true that \log_b(x) = \dfrac{\ln(x)}{\ln(b)}. +

    +
    5. +
  5. +

    + Since \log_b(x) = \dfrac{\ln(x)}{\ln(b)} = \dfrac{1}{\ln(b)} \cdot \ln(x), the value of k that gives a vertical stretch is k = \dfrac{1}{\ln(b)}. +

    +
    6. +

- + diff --git a/source/exercises/ez-exp-log.xml b/source/exercises/ez-exp-log.xml index 64835447..c51f66b2 100755 --- a/source/exercises/ez-exp-log.xml +++ b/source/exercises/ez-exp-log.xml @@ -95,7 +95,43 @@

- Exercise Solution +

    +
  1. +

    + The domain of y = g(x) = e^{-0.25x} is all real numbers and the range is all positive real numbers. Solving for x: x = g^{-1}(y) = \dfrac{\ln y}{-0.25}. The domain of the inverse is all positive real numbers and the range is all real numbers. +

    +
    2. +
  2. +

    + The domain of y = h(x) = 2e^x + 1 is all real numbers and the range is all real numbers greater than 1. The inverse is x = h^{-1}(y) = \ln\!\dfrac{y-1}{2}. The domain of the inverse is all real numbers greater than 1 and the range is all real numbers. +

    +
    3. +
  3. +

    + The domain of y = r(x) = 21 + 15e^{-0.1x} is all real numbers and the range is all real numbers greater than 21. The inverse is x = r^{-1}(y) = \dfrac{\ln\!\left(\frac{y-21}{15}\right)}{-0.1}. The domain of the inverse is all real numbers greater than 21 and the range is all real numbers. +

    +
    4. +
  4. +

    + The domain of y = s(x) = 72 - 40e^{-0.05x} is all real numbers and the range is all real numbers less than 72. The inverse is x = s^{-1}(y) = \dfrac{\ln\!\left(\frac{y-72}{-40}\right)}{-0.05}. The domain of the inverse is all real numbers less than 72 and the range is all real numbers. +

    +
    5. +
  5. +

    + The domain of y = u(x) = -5e^{3x-4} + 8 is all real numbers and the range is all real numbers less than 8. The inverse is x = u^{-1}(y) = \dfrac{\ln\!\left(\frac{y-8}{-5}\right) + 4}{3}. The domain of the inverse is all real numbers less than 8 and the range is all real numbers. +

    +
    6. +
  6. +

    + The domain of y = w(x) = 3\ln(x) + 4 is all positive real numbers and the range is all real numbers. The inverse is x = w^{-1}(y) = e^{(y-4)/3}. The domain of the inverse is all real numbers and the range is all positive real numbers. +

    +
    7. +
  7. +

    + The domain of y = z(x) = -0.2\ln(2x-5) + 1 is all real numbers greater than \frac{5}{2} and the range is all real numbers. The inverse is x = z^{-1}(y) = \dfrac{e^{(y-1)/(-0.2)} + 5}{2}. The domain of the inverse is all real numbers and the range is all real numbers greater than \frac{5}{2}. +

    +
    8. +

@@ -150,7 +186,28 @@

- Exercise Solution +

    +
  1. +

    + In Desmos, entering V(t) = k * ln(t) and f(t) = log_2(t), we can adjust the slider until the graphs coincide. The value k = \frac{1}{\ln 2} \approx 1.443 makes \log_2(t) = k\ln(t). +

    +
    2. +
  2. +

    + Similarly, the values of k for the other functions are: for g(t) = \log_3(t), k = \frac{1}{\ln 3} \approx 0.910; for h(t) = \log_5(t), k = \frac{1}{\ln 5} \approx 0.621; and for p(t) = \log_{1.25}(t), k = \frac{1}{\ln(1.25)} \approx 4.481. In each case, k = \frac{1}{\ln b}. +

    +
    3. +
  3. +

    + True. For any b \gt 1, the function \log_b(t) is a vertical scaling of \ln(t) by a factor of \frac{1}{\ln b}. +

    +
    4. +
  4. +

    + We compute \frac{1}{\ln 2} \approx 1.443, \frac{1}{\ln 3} \approx 0.910, \frac{1}{\ln 5} \approx 0.621, and \frac{1}{\ln(1.25)} \approx 4.481. These are exactly the values of k found in (a) and (b), confirming that \log_b(t) = \frac{1}{\ln b} \cdot \ln(t). +

    +
    5. +

@@ -198,7 +255,35 @@

- Exercise Solution +

    +
  1. +

    + We solve 74.4 - 38.8e^{-0.05t} = 50. Subtracting 74.4 and dividing by -38.8 gives e^{-0.05t} = \frac{50 - 74.4}{-38.8} = \frac{24.4}{38.8}. Taking the natural log, -0.05t = \ln\!\left(\frac{24.4}{38.8}\right), so + + t = \frac{\ln\!\left(\frac{24.4}{38.8}\right)}{-0.05} \approx 9.28 \text{ minutes}. + +

    +
    2. +
  2. +

    + At t = 0, F(0) = 74.4 - 38.8 = 35.6^\circ. Since F is an increasing function (the exponential term decreases as t increases, so F increases from its initial value), and F(0) = 35.6 \lt 36, yes, there is a time when the soda's temperature reaches 36^\circ. +

    +
    3. +
  3. +

    + Since e^{-0.05t} \to 0 as t \to \infty, the temperature approaches 74.4^\circ but never reaches it. At t = 0, F(0) = 35.6^\circ. Thus the range of F on the domain t \gt 0 is (35.6, 74.4). +

    +
    4. +
  4. +

    + Solving y = 74.4 - 38.8e^{-0.05t} for t: we get e^{-0.05t} = \frac{y - 74.4}{-38.8}, so + + t = F^{-1}(y) = \frac{\ln\!\left(\frac{y - 74.4}{-38.8}\right)}{-0.05}. + + This inverse function tells us the time (in minutes) at which the soda reaches a given temperature y (in degrees Fahrenheit). +

    +
    5. +

diff --git a/source/exercises/ez-exp-modeling.xml b/source/exercises/ez-exp-modeling.xml index c6bf9bc2..1168ea75 100755 --- a/source/exercises/ez-exp-modeling.xml +++ b/source/exercises/ez-exp-modeling.xml @@ -65,7 +65,28 @@

- Exercise Solution +

    +
  1. +

    + The soda's initial temperature is 41^\circ Fahrenheit. Since F(0) = ab^0 + c = a + c, we have a + c = 41. +

    +
    2. +
  2. +

    + The soda's long-term temperature is the room temperature, 72^\circ. Since 0 \lt b \lt 1, the term ab^t \to 0 as t \to \infty, so F(t) \to c. Thus c = 72. +

    +
    3. +
  3. +

    + Using a + c = 41 and c = 72, we find a = 41 - 72 = -31. +

    +
    4. +
  4. +

    + With b = 0.931, the soda's temperature after 10 minutes is F(10) = -31(0.931)^{10} + 72 \approx 56.8^\circ Fahrenheit. +

    +
    5. +

@@ -118,7 +139,7 @@

- Exercise Solution + Each function has the form ab^t + c, where b \gt 0. For the function p(t) (upper left graph), a \gt 0, 0 \lt b \lt 1, and c \lt 0. For the function q(t) (upper right graph), a \lt 0, 0 \lt b \lt 1, and c appears to be close to zero. For the function r(t) (lower left graph), a \gt 0, b \gt 1, and c \gt 0. Finally, for s(t) (lower right graph), a \lt 0, b \gt 1, and c \gt 0.

@@ -140,7 +161,10 @@

- Exercise Solution + Assuming a cooling law of the form C(t) = ab^t + c, we use C(0) = 95, C(10) = 80, and end behavior C(t) \to 0 as t \to \infty (so c = 0). From C(0) = a = 95 and C(10) = 95b^{10} = 80, we find b = (80/95)^{1/10} \approx 0.983. So C(t) = 95(0.983)^t. +

+

+ Converting to Fahrenheit: F(t) = \frac{9}{5}C(t) + 32 = \frac{9}{5}(95(0.983)^t) + 32 = 171(0.983)^t + 32. The cooling law has the same basic ab^t + c form in either temperature scale. The coefficient c happens to be zero in the Celsius scale and 32 in the Fahrenheit scale, reflecting the different zero points of the two scales.

diff --git a/source/exercises/ez-exp-temp-pop.xml b/source/exercises/ez-exp-temp-pop.xml index bdf1aa1b..d68f8b0a 100755 --- a/source/exercises/ez-exp-temp-pop.xml +++ b/source/exercises/ez-exp-temp-pop.xml @@ -90,7 +90,28 @@

- Exercise Solution +

    +
  1. +

    + The temperature is always increasing (the ice water warms toward room temperature), concave down, and approaches 71^\circF asymptotically in the long run. +

    +
    2. +
  2. +

    + The function F(t) = c - ae^{-kt} is a transformation of e^t that is reflected over the horizontal axis, vertically stretched by a, horizontally stretched by k, and shifted vertically by c. As t \to \infty, e^{-kt} \to 0 so F(t) \to c, giving the correct long-term behavior. The function increases at a decreasing rate (concave down) because -ae^{-kt} is increasing toward 0, and the rate of increase slows as the exponential term shrinks. +

    +
    3. +
  3. +

    + Since F(t) \to c = 71 as t \to \infty, we have c = 71. From F(0) = 71 - a = 34.2, we get a = 36.8. Using F(20) = 41.7: 71 - 36.8e^{-20k} = 41.7, so 36.8e^{-20k} = 29.3 and k = -\dfrac{1}{20}\ln\!\left(\dfrac{29.3}{36.8}\right) \approx 0.01140. +

    +
    4. +
  4. +

    + Setting F(t) = 60: 71 - 36.8e^{-kt} = 60, so 36.8e^{-kt} = 11 and t = -\dfrac{1}{k}\ln\!\left(\dfrac{11}{36.8}\right) \approx 106 minutes. +

    +
    5. +

@@ -136,7 +157,23 @@

- Exercise Solution +

    +
  1. +

    + Since all 5000 passengers will eventually get sick, A = 5000. From S(0) = \frac{5000}{1+M} = 5, we get M = 999. Using S(1) = 20: \frac{5000}{1 + 999e^{-k}} = 20, so 999e^{-k} = 249 and k = \ln\!\left(\frac{999}{249}\right) \approx 1.389. +

    +
    2. +
  2. +

    + Setting S(t) = 4000: \frac{5000}{1 + 999e^{-1.389t}} = 4000, so 999e^{-1.389t} = 0.25 and t = -\dfrac{\ln(0.25/999)}{1.389} \approx 5.97 days. +

    +
    3. +
  3. +

    + The average rates of change are approximately AV_{[1,2]} \approx 59.3, AV_{[3,4]} \approx 726.1, and AV_{[5,7]} \approx 1084.3 people per day. These represent the average number of people per day acquiring the virus on each time interval. The rate of spread is increasing through the first seven days, but must eventually decrease since the total number who can be infected is bounded by 5000. +

    +
    4. +

@@ -191,7 +228,28 @@

- Exercise Solution +

    +
  1. +

    + At t = 0: A(0) = -500e^{0} + 750 = -500 + 750 = 250 grams of salt. +

    +
    2. +
  2. +

    + As t \to \infty, e^{-0.25t} \to 0, so A(t) \to 750 grams. +

    +
    3. +
  3. +

    + Setting A(t) = 500: -500e^{-0.25t} + 750 = 500, so e^{-0.25t} = \frac{1}{2} and t = -\dfrac{\ln(1/2)}{0.25} = 4\ln 2 \approx 2.773 minutes. +

    +
    4. +
  4. +

    + Yes. In the long run, the tank contains 750 grams of salt in 100 liters of solution, giving a concentration of \frac{750}{100} = 7.5 grams per liter. Since the salt concentration in the tank approaches that of the inflow over time, the inflow concentration is 7.5 grams per liter. +

    +
    5. +

diff --git a/source/exercises/ez-poly-infty.xml b/source/exercises/ez-poly-infty.xml index cd44f5ce..267e8f10 100755 --- a/source/exercises/ez-poly-infty.xml +++ b/source/exercises/ez-poly-infty.xml @@ -57,7 +57,29 @@

- Exercise Solution +

    +
  1. +

    + f(10) = \ln(10) \approx 2.303, g(10) = 100, h(10) = e^{10} \approx 22026; + f(100) \approx 4.605, g(100) = 10000, h(100) \approx 2.69 \times 10^{43}. + The exponential h grows dramatically faster than the polynomial g, which in turn grows faster than the logarithm f. +

    +
    2. +
  2. +

    + f(x) = \ln(x) \ge 10^{10} requires x \ge e^{10^{10}} \approx 10^{4.3 \times 10^9} (an astronomically large value). + g(x) = x^2 \ge 10^{10} requires x \ge 10^5 = 100000. + h(x) = e^x \ge 10^{10} requires x \ge 10\ln(10) \approx 23.03. + The exponential reaches 10^{10} at a tiny input compared to the others, confirming that e^x grows fastest. +

    +
    3. +
  3. +

    + r(10) = \frac{100}{e^{10}} \approx 0.00454, r(100) \approx 3.7 \times 10^{-40}, r(1000) \approx 10^{-432}. + Despite both x^2 and e^x growing without bound, r(x) \to 0 because the exponential grows far faster than the polynomial, so the ratio shrinks toward zero. +

    +
    4. +

@@ -118,7 +140,33 @@

- Exercise Solution +

    +
  1. +

    + g is f shifted 1 unit to the left and 2 units down: g(x) = f(x+1) - 2. +

    +
    2. +
  2. +

    + g(x) = \dfrac{1}{x+1} - 2. The domain is all real numbers except x = -1. +

    +
    3. +
  3. +

    + As x \to \infty, \frac{1}{x+1} \to 0, so g(x) \to -2. As x \to -1^+, x+1 \to 0^+ so \frac{1}{x+1} \to +\infty, giving g(x) \to +\infty. +

    +
    4. +
  4. +

    + A possible formula is h(x) = \dfrac{1}{x-5} + 10. +

    +
    5. +
  5. +

    + r(x) = \frac{1}{x+35} - 27 has vertical asymptote x = -35, horizontal asymptote y = -27, and domain all real numbers except x = -35. The graph looks like y = 1/x shifted left 35 units and down 27 units. +

    +
    6. +

@@ -164,7 +212,28 @@

- Exercise Solution +

    +
  1. +

    + All three functions are positive for x \gt 0 and grow roughly like x^2. They differ for x \lt 0: f(x) = x^{2.4} is even-symmetric (defined and positive for all x); g(x) = x^{2.5} is only defined for x \ge 0; h(x) = x^{2.6} is odd-symmetric (defined for all x, negative for x \lt 0). +

    +
    2. +
  2. +

    + g(x) = x^{5/2} = \sqrt{x^5}, defined only for x \ge 0 since square roots are undefined for negative inputs. h(x) = x^{13/5} = \sqrt[5]{x^{13}}, defined for all real x. g has a different domain (only x \ge 0) because it involves an even root. +

    +
    3. +
  3. +

    + f and h resemble x^2 and x^3 respectively (since 2.4 and 2.6 are close to those integers). These are natural comparisons because they represent the even and odd symmetry templates for power functions with exponents near 2.4 and 2.6. +

    +
    4. +
  4. +

    + p(x) = x^{-2.4} = \sqrt[5]{x^{-12}}, defined for all x \ne 0, even-symmetric, approaching 0 as x \to \pm\infty. q(x) = x^{-2.5} = \frac{1}{\sqrt{x^5}}, defined only for x \gt 0. r(x) = x^{-2.6} = \sqrt[5]{x^{-13}}, defined for all x \ne 0, odd-symmetric. All approach 0 as x \to \pm\infty. +

    +
    5. +

diff --git a/source/exercises/ez-poly-polynomial-applications.xml b/source/exercises/ez-poly-polynomial-applications.xml index a060cb1d..dc62ccec 100755 --- a/source/exercises/ez-poly-polynomial-applications.xml +++ b/source/exercises/ez-poly-polynomial-applications.xml @@ -81,7 +81,49 @@

- Exercise Solution +

    +
  1. +

    + The area of an equilateral triangle with side length s is A_{\triangle} = \dfrac{\sqrt{3}}{4}s^2. +

    +
    2. +
  2. +

    + The volume of the trough is cross-sectional area times length: V = \dfrac{\sqrt{3}}{4}s^2 l. +

    +
    3. +
  3. +

    + The trough has two equilateral triangular ends and two rectangular side panels (the open top means there is no rectangular top panel). The surface area is + + A = 2 \cdot \frac{\sqrt{3}}{4}s^2 + 2sl = \frac{\sqrt{3}}{2}s^2 + 2sl. + +

    +
    4. +
  4. +

    + Setting the surface area equal to 100: \dfrac{\sqrt{3}}{2}s^2 + 2sl = 100, so + + l = \frac{100 - \frac{\sqrt{3}}{2}s^2}{2s} = \frac{50}{s} - \frac{\sqrt{3}}{4}s. + +

    +
    5. +
  5. +

    + Substituting into the volume formula: + + V(s) &= \frac{\sqrt{3}}{4}s^2\left(\frac{50}{s} - \frac{\sqrt{3}}{4}s\right) + &= \frac{50\sqrt{3}}{4}s - \frac{3}{16}s^3 + &= \frac{25\sqrt{3}}{2}s - \frac{3}{16}s^3. + +

    +
    6. +
  6. +

    + We need s > 0 and l > 0. From l = \dfrac{50}{s} - \dfrac{\sqrt{3}}{4}s > 0, we get s^2 < \dfrac{200}{\sqrt{3}} = \dfrac{200\sqrt{3}}{3}, so s < \sqrt{\dfrac{200\sqrt{3}}{3}}. The domain is \left(0,\ \sqrt{\dfrac{200\sqrt{3}}{3}}\right). +

    +
    7. +

@@ -99,7 +141,15 @@

- Exercise Solution + Let the box have width w, length 2w, and height h. The areas of the base and top are each 2w^2, and the combined area of the four sides is 2(wh) + 2(2wh) = 6wh. The cost constraint is + + 2(2w^2 + 2w^2) + 1.5(6wh) = 8w^2 + 9wh = 10. + + Solving for h: h = \dfrac{10 - 8w^2}{9w}. The volume is + + V(w) = w \cdot 2w \cdot h = 2w^2 \cdot \frac{10 - 8w^2}{9w} = \frac{20w - 16w^3}{9}. + + For the domain, we need h > 0, so 10 - 8w^2 > 0, giving w < \sqrt{\dfrac{10}{8}} = \dfrac{\sqrt{5}}{2}. With w > 0, the domain is \left(0,\ \dfrac{\sqrt{5}}{2}\right).

@@ -147,7 +197,36 @@

- Exercise Solution +

    +
  1. +

    + If the radius is very small, the cylinder must be very tall to hold 8 cubic feet; if the radius is large, the cylinder is short and flat. +

    +
    2. +
  2. +

    + The volume constraint \pi r^2 h = 8 gives h = \dfrac{8}{\pi r^2}. +

    +
    3. +
  3. +

    + Substituting into the surface area formula: + + A(r) = 2\pi r^2 + 2\pi r \cdot \frac{8}{\pi r^2} = 2\pi r^2 + \frac{16}{r}. + +

    +
    4. +
  4. +

    + The domain is r > 0, since the radius must be positive. +

    +
    5. +
  5. +

    + A(r) = 2\pi r^2 + \dfrac{16}{r} is not a polynomial because of the term \dfrac{16}{r} = 16r^{-1}, which has a negative exponent. Polynomial functions have only non-negative integer exponents. +

    +
    6. +

diff --git a/source/exercises/ez-poly-polynomials.xml b/source/exercises/ez-poly-polynomials.xml index 27fb3b20..bf9e928a 100755 --- a/source/exercises/ez-poly-polynomials.xml +++ b/source/exercises/ez-poly-polynomials.xml @@ -102,7 +102,33 @@

- Exercise Solution +

    +
  1. +

    + The degree of p is 3+2+1+2 = 8. +

    +
    2. +
  2. +

    + The real zeros are x=-21.7 (multiplicity 3), x=20.9 (multiplicity 2), and x=31.4 (multiplicity 1). (The factor x^2+100 has no real zeros.) +

    +
    3. +
  3. +

    + The sign of p is determined by (x+21.7)^3 and (x-31.4) (since the other factors are always non-negative and (x-20.9)^2 doesn't change sign): p \gt 0 for x \lt -21.7; p \lt 0 for -21.7 \lt x \lt 31.4 (with p=0 at x=20.9, a bounce); p \gt 0 for x \gt 31.4. +

    +
    4. +
  4. +

    + The zeros span a large range (-21.7 to 31.4), so a viewing window such as -40 \le x \le 40 and -3 \times 10^8 \le y \le 2 \times 10^8 is needed. Even then, the bounce at x=20.9 is subtle. +

    +
    5. +
  5. +

    + q(x) = -0.0005(x+21.7)^3(x-20.9)^2(x-31.4)(x^2+100)(x-92.3) has degree 9. Its real zeros are x=-21.7 (mult 3), x=20.9 (mult 2), x=31.4 (mult 1), and x=92.3 (mult 1). With such widely-spaced zeros, no single window shows all features clearly. +

    +
    6. +

@@ -154,7 +180,28 @@

- Exercise Solution +

    +
  1. +

    + Since e^{-x^2} \gt 0 always and x^2+1 \gt 0 always, the only real zeros come from (x-2)(x-3). The real zeros are x = 2 and x = 3. +

    +
    2. +
  2. +

    + The sign of r equals the sign of (x-2)(x-3): positive for x \lt 2, negative for 2 \lt x \lt 3, positive for x \gt 3. +

    +
    3. +
  3. +

    + In Desmos, the zeros at x=2 and x=3 are visible, but the graph decays very rapidly to 0 for large |x| due to the e^{-x^2} factor. +

    +
    4. +
  4. +

    + \lim_{x \to \infty} r(x) = 0. This is surprising because (x^2+1)(x-2)(x-3) \to \infty, but the factor e^{-x^2} decays to zero much faster than any polynomial grows, so the product is driven to zero. +

    +
    5. +

@@ -203,7 +250,28 @@

- Exercise Solution +

    +
  1. +

    + q(x) = -\dfrac{14}{27}(x+7)(x-11). The vertex is at x = \frac{-7+11}{2} = 2. Setting q(2) = -\frac{14}{27}(9)(-9) = 42 confirms the vertex value. +

    +
    2. +
  2. +

    + r(x) = \dfrac{28}{225}(x+3)^2(x-5)^2. We have r(0) = \frac{28}{225}(9)(25) = 28 ✓. +

    +
    3. +
  3. +

    + Since the stated multiplicities sum to 1+1+2+2+3=9 but the degree must be 11, we need two additional roots (a complex conjugate pair). One example: f(x) = -(x+3)(x-5)(x+2)^2(x-3)^2(x-1)^3(x^2+1), which has degree 11 and \lim_{x \to \infty} f(x) = -\infty. +

    +
    4. +
  4. +

    + Reading from the graph (which shows zeros at approximately x = -3, 1, 4 with the zero at x=-3 being a touch/bounce and those at x=1, 4 being crossings): one possible formula is g(x) = -\dfrac{27}{50}\!\left(\dfrac{x+3}{3}\right)^3(x-1)^2\!\left(\dfrac{x-4}{4}\right), a degree-6 polynomial. +

    +
    5. +

@@ -258,7 +326,33 @@

- Exercise Solution +

    +
  1. +

    + p has degree 3. +

    +
    2. +
  2. +

    + \lim_{x \to -\infty} p(x) = -\infty and \lim_{x \to \infty} p(x) = +\infty (since the leading term is x^3, positive coefficient, odd degree). +

    +
    3. +
  3. +

    + Factor: p(x) = x(x^2-a^2) = x(x-a)(x+a). The zeros are x = -a, 0, a. +

    +
    4. +
  4. +

    + Sign chart (with a \gt 0): p \lt 0 for x \lt -a; p \gt 0 for -a \lt x \lt 0; p \lt 0 for 0 \lt x \lt a; p \gt 0 for x \gt a. +

    +
    5. +
  5. +

    + As a increases, the zeros at \pm a spread further apart and the local maximum and minimum values grow in magnitude. The overall shape remains an S-curve, but wider and taller. +

    +
    6. +

diff --git a/source/exercises/ez-poly-rational-features.xml b/source/exercises/ez-poly-rational-features.xml index c9561f3e..3aa52630 100755 --- a/source/exercises/ez-poly-rational-features.xml +++ b/source/exercises/ez-poly-rational-features.xml @@ -71,7 +71,41 @@

- Exercise Solution +

    +
  1. +

    + r(x) = \dfrac{-19(x+11.3)^2(x-15.1)(x-17.3)}{41(x+5.7)(x+11.3)(x-8.4)(x-15.1)}. Both (x+11.3) and (x-15.1) appear in both numerator and denominator, so they cancel. The reduced form is \dfrac{-19(x+11.3)(x-17.3)}{41(x+5.7)(x-8.4)}. +

      +

    • Horizontal asymptote: y = -\dfrac{19}{41} (same degree after cancellation).

      • +

    • Vertical asymptotes: x = -5.7 and x = 8.4.

      • +

    • Zero: x = 17.3 (the factor x+11.3 in reduced numerator gives x=-11.3, but that is a hole).

      • +

    • Holes: at x = -11.3 (hole value 0, since x+11.3=0 makes reduced numerator zero) and at x = 15.1.

      • +
    +

    +
    2. +
  2. +

    + s(x) = \dfrac{-29(x^2-16)(x^2+99)(x-53)}{101(x^2-4)(x-13)^2(x+104)}. Factor: numerator = -29(x-4)(x+4)(x^2+99)(x-53); denominator = 101(x-2)(x+2)(x-13)^2(x+104). No common factors. +

      +

    • Horizontal asymptote: y = -\dfrac{29}{101} (degree 5 over degree 5).

      • +

    • Vertical asymptotes: x = 2, x = -2, x = 13, x = -104.

      • +

    • Zeros: x = 4, x = -4, x = 53 (note x^2+99 > 0 always).

      • +

    • Holes: none.

      • +
    +

    +
    3. +
  3. +

    + u(x) = \dfrac{-71(x^2-13x+36)(x-58.4)(x+78.2)}{83(x+58.4)(x-78.2)(x^2-12x+27)}. Factor: x^2-13x+36 = (x-4)(x-9) and x^2-12x+27 = (x-3)(x-9). The factor (x-9) cancels. +

      +

    • Horizontal asymptote: y = -\dfrac{71}{83} (degree 4 over degree 4).

      • +

    • Vertical asymptotes: x = -58.4, x = 78.2, x = 3.

      • +

    • Zeros: x = 4, x = 58.4, x = -78.2.

      • +

    • Hole: at x = 9.

      • +
    +

    +
    4. +

@@ -120,7 +154,38 @@

- Exercise Solution +

    +
  1. +

    + We need b = -\dfrac{3}{7} (horizontal asymptote) and a = \dfrac{5}{2} (vertical asymptote). Then r(0) = \dfrac{k}{0 - \frac{5}{2}} - \dfrac{3}{7} = -\dfrac{2k}{5} - \dfrac{3}{7} = 4, so -\dfrac{2k}{5} = \dfrac{31}{7} and k = -\dfrac{155}{14}. Thus r(x) = -\dfrac{155/14}{x - 5/2} - \dfrac{3}{7}. +

    +
    2. +
  2. +

    + We need: numerator degree greater than denominator degree (no horizontal asymptote); zeros at x=-5 and x=3; single vertical asymptote at x=-1; and the correct end behavior. The function + + s(x) = -\frac{(x+5)(x-3)}{x+1} + + satisfies all criteria: as x \to \infty, s(x) \to -\infty (since -(+)(+)/(+) \to -\infty); as x \to -\infty, s(x) \to +\infty. +

    +
    3. +
  3. +

    + Zeros only at x=-4 and x=-2, with x=-2 having even multiplicity (no sign change), VAs at x=1 and x=5, hole at x=4. One formula: + + u(x) = \frac{(x+4)(x+2)^2(x-4)}{(x-1)(x-5)(x-4)}. + +

    +
    4. +
  4. +

    + Reading from the graph: zeros near x=-1 and x=5, vertical asymptotes near x=3 and x=8, a hole near x=10, and a horizontal asymptote near y=0. One formula consistent with the graph is + + w(x) = -\frac{(x+2)(x-5)(x-10)}{5(x-3)(x-8)(x-10)}. + +

    +
    5. +

@@ -163,7 +228,40 @@

- Exercise Solution +

    +
  1. +

    + r(x) = -\dfrac{3}{x-4} + 5. Domain: x \neq 4. Range: r \neq 5. The graph passes the horizontal line test, so an inverse exists. Solving y = -\dfrac{3}{x-4}+5 for x: y-5 = -\dfrac{3}{x-4}, so x-4 = -\dfrac{3}{y-5} and x = 4 - \dfrac{3}{y-5}. Thus + + r^{-1}(x) = 4 - \frac{3}{x-5}. + + Domain of r^{-1}: x \neq 5. Range of r^{-1}: r^{-1} \neq 4. +

    +
    2. +
  2. +

    + s(x) = \dfrac{4-3x}{7x-2}. Domain: x \neq \dfrac{2}{7}. Range: s \neq -\dfrac{3}{7}. The graph passes the horizontal line test. Solving y = \dfrac{4-3x}{7x-2} for x: y(7x-2) = 4-3x, so 7xy + 3x = 4+2y, giving x = \dfrac{2y+4}{7y+3}. Thus + + s^{-1}(x) = \frac{2x+4}{7x+3}. + + Domain of s^{-1}: x \neq -\dfrac{3}{7}. Range of s^{-1}: s^{-1} \neq \dfrac{2}{7}. +

    +
    3. +
  3. +

    + u(x) = \dfrac{2x-1}{(x-1)^2}. Domain: x \neq 1. This function fails the horizontal line test (not monotone), so it has no inverse function. +

    +
    4. +
  4. +

    + w(x) = \dfrac{11}{(x+4)^3} - 7. Domain: x \neq -4. Range: w \neq -7. The graph passes the horizontal line test. Solving for x: y+7 = \dfrac{11}{(x+4)^3}, so (x+4)^3 = \dfrac{11}{y+7} and x = -4 + \sqrt[3]{\dfrac{11}{y+7}}. Thus + + w^{-1}(x) = -4 + \sqrt[3]{\frac{11}{x+7}}. + + Domain of w^{-1}: x \neq -7. Range of w^{-1}: w^{-1} \neq -4. +

    +
    5. +

@@ -211,7 +309,33 @@

- Exercise Solution +

    +
  1. +

    + r(x) = \dfrac{x^2-16}{x+4} = \dfrac{(x-4)(x+4)}{x+4}. At x = -4, both numerator and denominator are zero. After cancellation, r(x) = x-4 for x \neq -4. Near x = -4, the values approach -4-4 = -8 (e.g., r(-3.9) \approx -7.9). So the graph has a hole at (-4, -8). +

    +
    2. +
  2. +

    + s(x) = \dfrac{(x-2)^2(x+3)}{x^2-5x-6} = \dfrac{(x-2)^2(x+3)}{(x-6)(x+1)}. The denominator zeros are x = 6 and x = -1; neither equals a numerator zero. There are no holes. +

    +
    3. +
  3. +

    + u(x) = \dfrac{(x-2)^3(x+3)}{(x^2-5x-6)(x-7)} = \dfrac{(x-2)^3(x+3)}{(x-6)(x+1)(x-7)}. The denominator zeros x=6, x=-1, x=7 share no factors with the numerator. There are no holes. +

    +
    4. +
  4. +

    + w(x) = \dfrac{x^2+x-6}{(x^2+5x+6)(x+3)} = \dfrac{(x+3)(x-2)}{(x+2)(x+3)^2}. One factor of (x+3) cancels, leaving \dfrac{x-2}{(x+2)(x+3)}. At x=-3, the original is 0/0 and the reduced form still has (x+3) in the denominator, so x=-3 is a vertical asymptote, not a hole. There are no holes. +

    +
    5. +
  5. +

    + False. Part (d) provides a counterexample: at x=-3, both p(-3)=0 and q(-3)=0, yet the graph has a vertical asymptote rather than a hole. A hole occurs only when the factor cancels completely in the reduced form. +

    +
    6. +

@@ -264,7 +388,38 @@

- Exercise Solution +

    +
  1. +

    + A(x) = \dfrac{f(x)-f(1)}{x-1} = \dfrac{x^2-1}{x-1}, which is a ratio of polynomials and hence a rational function. +

    +
    2. +
  2. +

    + A is undefined at x = 1 (divides by zero). Domain: all real numbers except x = 1. +

    +
    3. +
  3. +

    + Factoring: A(x) = \dfrac{(x+1)(x-1)}{x-1} = x+1 for x \neq 1. Near x=1, A(x) \to 2. So A has a hole at (1, 2), not a vertical asymptote. +

    +
    4. +
  4. +

    + As x \to 1, A(x) = x+1 \to 2. The average rate of change of f(x)=x^2 on [1,x] approaches 2. +

    +
    5. +
  5. +

    + For g(x) = x^3: B(x) = \dfrac{x^3-1}{x-1} = \dfrac{(x-1)(x^2+x+1)}{x-1} = x^2+x+1 for x \neq 1. This is a rational function. Domain: x \neq 1. Hole at (1, 3) (since 1^2+1+1=3). As x \to 1, B(x) \to 3. +

    +
    6. +
  6. +

    + For h(x) = x^4: C(x) = \dfrac{x^4-1}{x-1} = \dfrac{(x-1)(x^3+x^2+x+1)}{x-1} = x^3+x^2+x+1 for x \neq 1. This is a rational function. Domain: x \neq 1. Hole at (1, 4). As x \to 1, C(x) \to 4. +

    +
    7. +

diff --git a/source/exercises/ez-poly-rational.xml b/source/exercises/ez-poly-rational.xml index 92a83818..f98c17af 100755 --- a/source/exercises/ez-poly-rational.xml +++ b/source/exercises/ez-poly-rational.xml @@ -83,7 +83,28 @@

- Exercise Solution +

    +
  1. +

    + f(x) = \dfrac{17x^2+34}{19x^2-76} = \dfrac{17(x^2+2)}{19(x-2)(x+2)}. Domain: all real numbers except x = \pm 2. Horizontal asymptote: y = \dfrac{17}{19} (ratio of leading coefficients, same degree). +

    +
    2. +
  2. +

    + g(x) = \dfrac{29}{53} + \dfrac{1}{x-2} is undefined at x = 2. Domain: all real numbers except x = 2. Horizontal asymptote: y = \dfrac{29}{53} (as x \to \pm\infty, the term \frac{1}{x-2} \to 0). +

    +
    3. +
  3. +

    + h(x) = \dfrac{4-31x}{11x-7} is undefined at x = \dfrac{7}{11}. Domain: all real numbers except x = \dfrac{7}{11}. Horizontal asymptote: y = \dfrac{-31}{11} (ratio of leading coefficients). +

    +
    4. +
  4. +

    + r(x) = \dfrac{151(x-4)(x+5)^2(x-2)}{537(x+5)(x+1)(x^2+1)(x-15)}. The factor (x+5) cancels, leaving a hole at x = -5. The reduced denominator zeros are x = -1 and x = 15 (note x^2+1 > 0 always). Domain: all real numbers except x = -5, x = -1, and x = 15. Since the numerator has degree 4 and denominator has degree 5, the horizontal asymptote is y = 0. +

    +
    5. +

@@ -136,7 +157,41 @@

- Exercise Solution +

    +
  1. +

    + The box has width x, length 1.5x, and height h. +

    +
    2. +
  2. +

    + The surface area is: base and top each 1.5x^2, two short sides each xh, two long sides each 1.5xh. So S = 3x^2 + 5xh. +

    +
    3. +
  3. +

    + Cost: C = 3.75(3x^2) + 2.50(5xh) = 11.25x^2 + 12.50xh. +

    +
    4. +
  4. +

    + Volume constraint: 1.5x^2 h = 8, so h = \dfrac{8}{1.5x^2} = \dfrac{16}{3x^2}. +

    +
    5. +
  5. +

    + Substituting: + + C(x) = 11.25x^2 + 12.50 \cdot x \cdot \frac{16}{3x^2} = 11.25x^2 + \frac{200}{3x}. + +

    +
    6. +
  6. +

    + The domain is x > 0. As x \to 0^+, C \to \infty; as x \to \infty, C \to \infty. A graph shows a minimum at x \approx 1.44 ft, suggesting the ideal box has approximate dimensions 1.44 \times 2.15 \times 2.58 feet. +

    +
    7. +

@@ -159,7 +214,13 @@

- Exercise Solution + With radius r and height h, the volume constraint \pi r^2 h = 16 gives h = \dfrac{16}{\pi r^2}. The cost is + + C(r) &= 0.11 \cdot 2\pi r^2 + 0.07 \cdot 2\pi r h + &= 0.22\pi r^2 + 0.14\pi r \cdot \frac{16}{\pi r^2} + &= 0.22\pi r^2 + \frac{2.24}{r}. + + Numerically, C(r) \approx 0.691 r^2 + \dfrac{2.240}{r}. The domain is r > 0 (radius must be positive). A graph shows a minimum at r \approx 1.48 inches, giving a height of about h \approx 2.33 inches and cost of about \$3.03.

diff --git a/source/exercises/ez-trig-finding-angles.xml b/source/exercises/ez-trig-finding-angles.xml index f38b3813..d347d6ba 100755 --- a/source/exercises/ez-trig-finding-angles.xml +++ b/source/exercises/ez-trig-finding-angles.xml @@ -90,7 +90,28 @@

- Exercise Solution +

    +
  1. +

    + The plane is at altitude 2000 feet and 7500 feet from the building (slant distance). The angle of depression is \arcsin\!\left(\dfrac{2000}{7500}\right) = \arcsin\!\left(\dfrac{4}{15}\right) \approx 0.2699 radians, or about 15.47^{\circ}. +

    +
    2. +
  2. +

    + When the slant distance is 6000 feet, the angle of depression is \arcsin\!\left(\dfrac{2000}{6000}\right) = \arcsin\!\left(\dfrac{1}{3}\right) \approx 0.3398 radians, or about 19.47^{\circ}. +

    +
    3. +
  3. +

    + The horizontal distance at 7500 feet is \sqrt{7500^2 - 2000^2} = 500\sqrt{209} \approx 7228 feet, and at 6000 feet it is \sqrt{6000^2 - 2000^2} = 4000\sqrt{2} \approx 5657 feet. The plane traveled 500\sqrt{209} - 4000\sqrt{2} = 500(\sqrt{209} - 8\sqrt{2}) \approx 1572 feet. +

    +
    4. +
  4. +

    + Traveling 500(\sqrt{209}-8\sqrt{2}) feet in 5 seconds gives a speed of 100(\sqrt{209}-8\sqrt{2}) \approx 314 feet per second, which is approximately 214 miles per hour. +

    +
    5. +

@@ -133,7 +154,28 @@

- Exercise Solution +

    +
  1. +

    + When the balloon is 200 feet off the ground, the angle of elevation is \arctan\!\left(\dfrac{200}{850}\right) \approx 0.2311 radians (about 13.24^{\circ}). +

    +
    2. +
  2. +

    + When the balloon is 275 feet off the ground, the angle of elevation is \arctan\!\left(\dfrac{275}{850}\right) \approx 0.3129 radians (about 17.93^{\circ}). +

    +
    3. +
  3. +

    + The angle of elevation as a function of height is \theta(h) = \arctan\!\left(\dfrac{h}{850}\right). +

    +
    4. +
  4. +

    + The average rate of change on [200,275] is AV_{[200,275]} = \dfrac{\theta(275) - \theta(200)}{75} \approx \dfrac{0.3129 - 0.2311}{75} \approx 0.001091 radians per foot. This means that as the balloon rises from 200 to 275 feet, the camera angle must increase at an average rate of about 0.001091 radians for each foot of altitude gained. +

    +
    5. +

@@ -186,7 +228,38 @@

- Exercise Solution +

    +
  1. +

    + The hypotenuse has length \sqrt{5^2 + 12^2} = 13. Since \alpha is opposite the leg of length 5, we have \cos(\alpha) = \dfrac{12}{13}. +

    +
    2. +
  2. +

    + \sin(\beta) = \dfrac{12}{13}. +

    +
    3. +
  3. +

    + \tan(\beta) = \dfrac{12}{5} and \tan(\alpha) = \dfrac{5}{12}. +

    +
    4. +
  4. +

    + \alpha = \arctan\!\left(\dfrac{5}{12}\right) \approx 0.3948 radians. +

    +
    5. +
  5. +

    + \beta = \arctan\!\left(\dfrac{12}{5}\right) \approx 1.176 radians. +

    +
    6. +
  6. +

    + True. If \theta + \gamma = \frac{\pi}{2}, then \theta and \gamma are complementary angles in a right triangle. Labeling the side opposite \theta as B and the side opposite \gamma as C, with hypotenuse H, we get \cos(\theta) = \frac{C}{H} = \sin(\gamma). +

    +
    7. +

diff --git a/source/exercises/ez-trig-inverse.xml b/source/exercises/ez-trig-inverse.xml index 703128bb..3a65aada 100755 --- a/source/exercises/ez-trig-inverse.xml +++ b/source/exercises/ez-trig-inverse.xml @@ -94,7 +94,18 @@

- Exercise Solution +

    +

  1. \arcsin\!\left(\frac{1}{2}\right) = \frac{\pi}{6}

    2. +

  2. \arctan(-1) = -\frac{\pi}{4}

    3. +

  3. \arcsin\!\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3}

    4. +

  4. \arctan\!\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}

    5. +

  5. \arccos\!\left(\sin\!\left(\frac{\pi}{3}\right)\right) = \frac{\pi}{6}

    6. +

  6. \cos\!\left(\arcsin\!\left(-\frac{\sqrt{3}}{2}\right)\right) = \frac{1}{2}

    7. +

  7. \tan\!\left(\arcsin\!\left(-\frac{\sqrt{2}}{2}\right)\right) = -1

    8. +

  8. \arctan\!\left(\sin\!\left(\frac{\pi}{2}\right)\right) = \frac{\pi}{4}

    9. +

  9. \sin\!\left(\arcsin\!\left(-\frac{1}{2}\right)\right) = -\frac{1}{2}

    10. +

  10. \arctan\!\left(\tan\!\left(\frac{7\pi}{4}\right)\right) = -\frac{\pi}{4}

    11. +

@@ -147,7 +158,38 @@

- Exercise Solution +

    +
  1. +

    + True. Sine is the inverse function of arcsine, and therefore \sin(\arcsin(y)) = y for any y with -1 \le y \le 1. +

    +
    2. +
  2. +

    + False. For example, \arcsin\!\left(\sin\!\left(\frac{3\pi}{4}\right)\right) = \arcsin\!\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4} \ne \frac{3\pi}{4}. +

    +
    3. +
  3. +

    + False. For example, \arccos\!\left(\cos\!\left(-\frac{\pi}{4}\right)\right) = \arccos\!\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4} \ne -\frac{\pi}{4}. +

    +
    4. +
  4. +

    + True. Cosine is the inverse function of arccosine, and therefore \cos(\arccos(y)) = y for any y with -1 \le y \le 1. +

    +
    5. +
  5. +

    + True. Tangent is the inverse function of arctangent, and therefore \tan(\arctan(y)) = y for any real number y. +

    +
    6. +
  6. +

    + False. For example, \arctan\!\left(\tan\!\left(\frac{3\pi}{4}\right)\right) = \arctan(-1) = -\frac{\pi}{4} \ne \frac{3\pi}{4}. +

    +
    7. +

@@ -207,7 +249,28 @@

- Exercise Solution +

    +
  1. +

    + The domain of h is [-1, 1] (the domain of arcsine). The range of h is [0, 1], since arcsin returns values in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] and cosine is non-negative on that interval. +

    +
    2. +
  2. +

    + With hypotenuse 1 and vertical leg x, the Pythagorean Theorem gives the horizontal leg length as \sqrt{1 - x^2}. +

    +
    3. +
  3. +

    + We have \cos(\theta) = \dfrac{\sqrt{1-x^2}}{1} = \sqrt{1-x^2}. Therefore h(x) = \cos(\arcsin(x)) = \sqrt{1 - x^2}. +

    +
    4. +
  4. +

    + Let \alpha = \arctan(x), so \tan(\alpha) = x. In the corresponding right triangle, the opposite leg has length x, the adjacent leg has length 1, and by the Pythagorean Theorem the hypotenuse has length \sqrt{1 + x^2}. Therefore p(x) = \cos(\arctan(x)) = \dfrac{1}{\sqrt{1+x^2}}. +

    +
    5. +

diff --git a/source/exercises/ez-trig-other.xml b/source/exercises/ez-trig-other.xml index dfa66d98..8d5bb934 100755 --- a/source/exercises/ez-trig-other.xml +++ b/source/exercises/ez-trig-other.xml @@ -42,7 +42,13 @@

- Exercise Solution + With \cos(\beta) = -\frac{12}{13} and \beta in quadrant II, the Pythagorean identity gives \sin(\beta) = \frac{5}{13} (positive in QII). The remaining values are: + + \sec(\beta) = -\frac{13}{12}, \quad \csc(\beta) = \frac{13}{5}, \quad \tan(\beta) = -\frac{5}{12}, \quad \cot(\beta) = -\frac{12}{5}. + +

+

+ If \beta lies in quadrant III, then \cos(\beta) is still -\frac{12}{13} but now \sin(\beta) = -\frac{5}{13} (negative in QIII). So \sec(\beta) = -\frac{13}{12} (unchanged), \csc(\beta) = -\frac{13}{5}, \tan(\beta) = \frac{5}{12}, and \cot(\beta) = \frac{12}{5}.

@@ -86,7 +92,28 @@

- Exercise Solution +

    +
  1. +

    + f(t) = 5\sec\!\left(t - \frac{\pi}{2}\right) + 3: The parent \sec(t) has period 2\pi, domain all reals except \frac{\pi}{2} + j\pi, range (-\infty,-1]\cup[1,\infty). Shifting by \frac{\pi}{2} moves the asymptotes to t = j\pi. The vertical stretch by 5 and shift by 3 give range (-\infty,-2]\cup[8,\infty). Period remains 2\pi. +

    +
    2. +
  2. +

    + g(t) = -\frac{1}{3}\csc(2t) - 4: The horizontal compression by 2 gives period \pi and moves asymptotes to all multiples of \frac{\pi}{2} (i.e., t = \frac{j\pi}{2} for any integer j), so the domain is all reals except multiples of \frac{\pi}{2}. The factor -\frac{1}{3} and vertical shift -4 give range (-\infty, -\frac{13}{3}]\cup[-\frac{11}{3},\infty). +

    +
    3. +
  3. +

    + h(t) = -7\tan\!\left(t + \frac{\pi}{4}\right) + 1: The parent \tan(t) has period \pi, asymptotes at \frac{\pi}{2}+j\pi, range all reals. The horizontal shift moves asymptotes to t = \frac{\pi}{4} + j\pi. Domain is all reals except \frac{\pi}{4}+j\pi. Range remains all reals. Period remains \pi. +

    +
    4. +
  4. +

    + j(t) = \frac{1}{2}\cot(4t) - 2: The horizontal compression by 4 gives period \frac{\pi}{4} and moves asymptotes to t = \frac{j\pi}{4}. Domain is all reals except \frac{j\pi}{4}. Range remains all reals. +

    +
    5. +

@@ -146,7 +173,15 @@

- Exercise Solution + In the right triangle, the horizontal leg has length \sqrt{1-x^2} by the Pythagorean Theorem. +

    +

  1. \sin(\theta) = x

    2. +

  2. \sec(\theta) = \dfrac{1}{\sqrt{1-x^2}}

    3. +

  3. \csc(\theta) = \dfrac{1}{x}

    4. +

  4. \tan(\theta) = \dfrac{x}{\sqrt{1-x^2}}

    5. +

  5. \cos(\arcsin(x)) = \cos(\theta) = \sqrt{1-x^2}

    6. +

  6. \cot(\arcsin(x)) = \cot(\theta) = \dfrac{\sqrt{1-x^2}}{x}

    7. +

diff --git a/source/exercises/ez-trig-right.xml b/source/exercises/ez-trig-right.xml index ef6e78ea..082fb7d2 100755 --- a/source/exercises/ez-trig-right.xml +++ b/source/exercises/ez-trig-right.xml @@ -45,7 +45,7 @@

- Exercise Solution + Let the angle of elevation be 27.5^\circ. The line from the streetlight to the tip of the shadow is the hypotenuse of a large right triangle with base 50 + 14 = 64 feet, so this hypotenuse is \frac{64}{\cos(27.5^\circ)} \approx 72.1 feet. The line from the person's head to the tip of the shadow is the hypotenuse of a smaller right triangle with base 14 feet, so it has length \frac{14}{\cos(27.5^\circ)} \approx 15.7 feet. The person's height is \sqrt{15.7^2 - 14^2} \approx 7.1 feet, and the streetlight's height is \sqrt{72.1^2 - 64^2} \approx 33.2 feet.

@@ -63,7 +63,7 @@

- Exercise Solution + With the range-finder elevated at 17.4^\circ and at distance 1650 meters from the rocket, the height of the rocket is 1650\sin(17.4^\circ) \approx 493.4 meters. The horizontal distance from the range-finder to the launch point is 1650\cos(17.4^\circ) \approx 1574 meters.

@@ -92,7 +92,10 @@

- Exercise Solution + Each side panel is a 2' \times 24' strip folded at angle \theta from the base. The depth of the trough is \sin(\theta) feet, and the two slanted sides each add \cos(\theta) feet to the top width, giving a total top width of 2 + 2\cos(\theta) feet. The cross-sectional area of the trapezoidal cross-section is \frac{1}{2}(2 + 2 + 2\cos(\theta))\sin(\theta) = \sin(\theta)(2 + \cos(\theta)) square feet. Multiplying by the length of 24 feet gives + + V(\theta) = 24\sin(\theta)(2 + \cos(\theta)). +

diff --git a/source/exercises/ez-trig-tangent.xml b/source/exercises/ez-trig-tangent.xml index da2056ef..9d007df3 100755 --- a/source/exercises/ez-trig-tangent.xml +++ b/source/exercises/ez-trig-tangent.xml @@ -63,7 +63,7 @@

- Exercise Solution + The ramp makes a 4^\circ angle with the ground and rises 3 feet. Its length is \frac{3}{\sin(4^\circ)} \approx 43.0 feet. It meets the sidewalk approximately \frac{3}{\tan(4^\circ)} \approx 42.9 feet from the porch. The slope is \frac{3}{42.9} \approx 0.0699.

@@ -105,7 +105,23 @@

- Exercise Solution +

    +
  1. +

    + With the kite at 170 feet elevation and elevation angle 40^\circ, the horizontal distance is \frac{170}{\tan(40^\circ)} \approx 202.6 feet. +

    +
    2. +
  2. +

    + The string length is \frac{170}{\sin(40^\circ)} \approx 264.5 feet. +

    +
    3. +
  3. +

    + With the same string length and elevation angle 50^\circ, the kite height is 264.5\sin(50^\circ) \approx 202.6 feet. +

    +
    4. +

@@ -127,7 +143,7 @@

- Exercise Solution + At angle 36^\circ, the plane's horizontal distance is \frac{2400}{\tan(36^\circ)} \approx 3301 feet. Two seconds later at 41^\circ, the distance is \frac{2400}{\tan(41^\circ)} \approx 2759 feet. The plane traveled 3301 - 2759 \approx 542 feet in 2 seconds, giving a speed of approximately 271 feet per second, or about 185 miles per hour.

diff --git a/source/previews/PA-changing-aroc.xml b/source/previews/PA-changing-aroc.xml index ba44e289..e593155f 100755 --- a/source/previews/PA-changing-aroc.xml +++ b/source/previews/PA-changing-aroc.xml @@ -28,7 +28,15 @@ - + +

+ First compute s(1.5) = 64 - 16(1.5-1)^2 = 60 and + s(2.5) = 64 - 16(2.5-1)^2 = 28. Then + + AV_{[1.5,2.5]} &= \frac{s(2.5)-s(1.5)}{2.5-1.5} = \frac{28-60}{1} = -32. + +

+ @@ -39,7 +47,13 @@ - + +

+ The units are feet per second. The value AV_{[1.5,2.5]} = -32 means that + between t = 1.5 and t = 2.5 seconds, the ball's height decreases + at an average rate of 32 feet per second. +

+ @@ -60,7 +74,16 @@ - + +

+ Without additional context specifying when the ball is launched or lands, the + domain and range of the model are not fully determined by the equation alone. + The formula s(t) = 64 - 16(t-1)^2 is a classical free-fall equation, + and negative values of t or s are not necessarily unphysical. +

+ + + @@ -74,7 +97,19 @@ - + +

+ The slope is m = AV_{[1.5,2.5]} = -32. Using the point (1.5, 60) + to find the vertical intercept: + + 60 &= -32(1.5) + b + b &= 108. + + The equation of the line is y = -32t + 108. +

+ + + @@ -85,7 +120,12 @@ - + +

+ AV_{[1.5,2.5]} is the slope of the secant line connecting the points + (1.5, s(1.5)) and (2.5, s(2.5)) on the graph of s. +

+ @@ -96,7 +136,32 @@ - + +

+ Interval [0.25, 0.75]: + s(0.25) = 55, s(0.75) = 63, so + AV_{[0.25,0.75]} = \frac{63-55}{0.5} = 16 \text{ ft/sec.} + The secant line through (0.25, 55) with slope 16 is y = 16t + 51. +

+

+ Interval [0.5, 1.5]: + s(0.5) = 60 and s(1.5) = 60, so + AV_{[0.5,1.5]} = \frac{60-60}{1} = 0 \text{ ft/sec.} + The secant line is the horizontal line y = 60. +

+

+ Interval [1, 3]: + s(1) = 64, s(3) = 0, so + AV_{[1,3]} = \frac{0-64}{2} = -32 \text{ ft/sec.} + The secant line through (3, 0) with slope -32 is y = -32t + 96. +

+ + + + + + + diff --git a/source/previews/PA-changing-combining.xml b/source/previews/PA-changing-combining.xml index c72e14fb..c96b2d87 100755 --- a/source/previews/PA-changing-combining.xml +++ b/source/previews/PA-changing-combining.xml @@ -68,7 +68,12 @@ - + +

+ From the table, f(3) = 20 and g(3) = 2, so + h(3) = f(3) + g(3) = 20 + 2 = 22. +

+ @@ -78,7 +83,12 @@ - + +

+ From the graph, p(-1) = \frac{1}{5} and q(-1) = \frac{7}{2}, so + r(-1) = p(-1) - q(-1) = \frac{1}{5} - \frac{7}{2} = \frac{2}{10} - \frac{35}{10} = -\frac{33}{10}. +

+ @@ -88,7 +98,18 @@ - + +

+ Yes. Setting r(x) = 0 requires p(x) = q(x). + On the relevant interval, p(x) = -\frac{2}{5}x - \frac{1}{5} and + q(x) = 4x + 12, so: + + -\frac{2}{5}x - \frac{1}{5} &= 4x + 12 + -\frac{22}{5}x &= \frac{61}{5} + x &= -\frac{61}{22}. + +

+ @@ -98,7 +119,12 @@ - + +

+ From the table, f(0) = 5 and g(0) = 9, so + k(0) = f(0) \cdot g(0) = 5 \cdot 9 = 45. +

+ @@ -108,7 +134,12 @@ - + +

+ From the graph, p(1) = -\frac{3}{5} and q(1) = \frac{5}{2}, so + s(1) = \frac{p(1)}{q(1)} = \frac{-3/5}{5/2} = -\frac{3}{5} \cdot \frac{2}{5} = -\frac{6}{25}. +

+ @@ -118,7 +149,12 @@ - + +

+ Yes: s(x) is undefined wherever q(x) = 0. + From the graph, q(-3) = 0, so s(-3) is undefined. +

+ diff --git a/source/previews/PA-changing-composite.xml b/source/previews/PA-changing-composite.xml index be75c29b..0408ec83 100755 --- a/source/previews/PA-changing-composite.xml +++ b/source/previews/PA-changing-composite.xml @@ -27,7 +27,15 @@ - + +

+ + r(t) &= p(q(t)) = p(t^2 - 1) + &= 3(t^2 - 1) - 4 + &= 3t^2 - 7. + +

+ @@ -37,7 +45,14 @@ - + +

+ In the introductory example, the inner function g(t) = 3t-4 is linear + and the outer function f(x) = x^2 - 1 is quadratic. In part (a), the + roles are reversed: the inner function q(t) = t^2-1 is quadratic and the + outer function p(x) = 3x-4 is linear. +

+ @@ -47,7 +62,11 @@ - + +

+ q(s(z)) = \left(\frac{1}{z+4}\right)^2 - 1 = \frac{1}{(z+4)^2} - 1. +

+ @@ -57,7 +76,12 @@ - + +

+ One natural decomposition: f(x) = \sqrt{x} and g(t) = 2t^2 + 5, + so that f(g(t)) = \sqrt{2t^2 + 5} = h(t). +

+ diff --git a/source/previews/PA-changing-functions-crickets.xml b/source/previews/PA-changing-functions-crickets.xml index 7b3953f5..d915a540 100755 --- a/source/previews/PA-changing-functions-crickets.xml +++ b/source/previews/PA-changing-functions-crickets.xml @@ -27,7 +27,9 @@ - + +

We seek T when N = 92: T = 40 + 0.25(92) = 63. So the temperature should be 63^\circ F.

+ @@ -37,7 +39,9 @@ - + +

We seek N when T = 77. Solving 77 = 40 + 0.25N gives N = \frac{77 - 40}{0.25} = 148 chirps per minute.

+ @@ -47,7 +51,9 @@ - + +

The model is known to be accurate only for temperatures from 50^\circ to 85^\circ F, so 35^\circ F lies outside the valid range and we should not expect reasonable results at this temperature.

+ @@ -57,7 +63,9 @@ - + +

At 65 chirps per minute the temperature is T = 40 + 0.25(65) = 56.25^\circ F, and at 75 chirps per minute the temperature is T = 40 + 0.25(75) = 58.75^\circ F. The temperature has risen by 58.75 - 56.25 = 2.5 degrees Fahrenheit.

+ @@ -67,7 +75,9 @@ - + +

At 50^\circ F we get N = \frac{50 - 40}{0.25} = 40 chirps per minute; at 85^\circ F we get N = \frac{85 - 40}{0.25} = 180 chirps per minute. The fewest expected is 40 and the greatest is 180 chirps per minute.

+ diff --git a/source/previews/PA-changing-inverse-F-C.xml b/source/previews/PA-changing-inverse-F-C.xml index ec046ad7..f98707e9 100755 --- a/source/previews/PA-changing-inverse-F-C.xml +++ b/source/previews/PA-changing-inverse-F-C.xml @@ -27,7 +27,13 @@ - + +

+ Subtracting 32 from both sides: F - 32 = \frac{9}{5}C. + Multiplying both sides by \frac{5}{9}: + C = \frac{5}{9}(F-32). +

+ @@ -35,22 +41,36 @@ Note that the equation C = \frac{5}{9}(F-32) expresses C as a function of F. Call this function h so that C = h(F) = \frac{5}{9}(F-32).

- Find the simplest expression that you can for the composite function j(C) = h(g(C)). + Find the simplest expression that you can for the composite function j(C) = h(g(C)).

- + +

+ + j(C) &= h(g(C)) = \frac{5}{9}\!\left(\frac{9}{5}C + 32 - 32\right) + &= \frac{5}{9} \cdot \frac{9}{5}C = C. + +

+

- Find the simplest expression that you can for the composite function k(F) = g(h(F)). + Find the simplest expression that you can for the composite function k(F) = g(h(F)).

- + +

+ + k(F) &= g(h(F)) = \frac{9}{5}\!\left(\frac{5}{9}(F-32)\right) + 32 + &= (F-32) + 32 = F. + +

+ @@ -60,7 +80,15 @@ - + +

+ The function g converts Celsius to Fahrenheit, and h converts + Fahrenheit back to Celsius. Composing them in either order simply undoes the + conversion: applying both functions in sequence returns the original input. + This is why j(C) = C and k(F) = F — the two functions are + inverses of each other. +

+ diff --git a/source/previews/PA-changing-linear-3-ex.xml b/source/previews/PA-changing-linear-3-ex.xml index 4c76a343..1c1d1325 100755 --- a/source/previews/PA-changing-linear-3-ex.xml +++ b/source/previews/PA-changing-linear-3-ex.xml @@ -24,7 +24,14 @@ - + +

+ AV_{[-3,-1]} = \frac{f(-1)-f(-3)}{-1-(-3)} = \frac{10-16}{2} = -3, + AV_{[2,5]} = \frac{f(5)-f(2)}{5-2} = \frac{-8-1}{3} = -3, + AV_{[4,10]} = \frac{f(10)-f(4)}{10-4} = \frac{-23-(-5)}{6} = -3. + All three average rates of change equal -3. +

+ @@ -118,7 +125,15 @@ - + +

+ Reading values from the table: + AV_{[-5,-2]} = \frac{g(-2)-g(-5)}{-2-(-5)} = \frac{-1.25-(-2.75)}{3} = \frac{1.5}{3} = 0.5, + AV_{[-1,1]} = \frac{g(1)-g(-1)}{1-(-1)} = \frac{0.25-(-0.75)}{2} = \frac{1}{2} = 0.5, + AV_{[0,4]} = \frac{g(4)-g(0)}{4-0} = \frac{1.75-(-0.25)}{4} = \frac{2}{4} = 0.5. + All three average rates of change equal 0.5. +

+ @@ -135,7 +150,15 @@ - + +

+ Reading values from the graph: + AV_{[-5,-2]} = \frac{h(-2)-h(-5)}{-2-(-5)} = \frac{3-4}{3} = -\frac{1}{3}, + AV_{[-1,1]} = \frac{h(1)-h(-1)}{1-(-1)} = \frac{2-\frac{8}{3}}{2} = -\frac{1}{3}, + AV_{[0,4]} = \frac{h(4)-h(0)}{4-0} = \frac{1-\frac{7}{3}}{4} = -\frac{1}{3}. + All three average rates of change equal -\frac{1}{3}. +

+ @@ -145,12 +168,19 @@ - + +

+ All three functions have a constant average rate of change: the average rate of + change is the same regardless of which interval is chosen. They differ in their + specific rate of change (-3, 0.5, and -\frac{1}{3} respectively) + and in how they were presented (formula, table, graph). +

+

- For the function y = f(x) = 7 - 3x from (a), find the simplest expression you can for + For the function y = f(x) = 7 - 3x from (a), find the simplest expression you can for AV_{[a,b]} = \frac{f(b)-f(a)}{b-a} @@ -159,7 +189,15 @@ - + +

+ + AV_{[a,b]} &= \frac{f(b)-f(a)}{b-a} = \frac{(7-3b)-(7-3a)}{b-a} + &= \frac{3a-3b}{b-a} = \frac{3(a-b)}{b-a} = -3. + + For any a \ne b, AV_{[a,b]} = -3. +

+ diff --git a/source/previews/PA-changing-quadratic.xml b/source/previews/PA-changing-quadratic.xml index b0e44115..3950fa1c 100755 --- a/source/previews/PA-changing-quadratic.xml +++ b/source/previews/PA-changing-quadratic.xml @@ -25,7 +25,7 @@

- Execute appropriate computations to complete both of the following tables: values of the function h on the left, average rates of change for h on the right. + Execute appropriate computations to complete both of the following tables: values of the function h on the left, average rates of change for h on the right.