2013 ap calculus ab free - response questions\ncalculus ab\nsection ii, part a\ntime—30 minutes\nnumber of…

2013 ap calculus ab free - response questions\ncalculus ab\nsection ii, part a\ntime—30 minutes\nnumber of problems—2\na graphing calculator is required for these problems.\n1. on a certain workday, the rate, in tons per hour, at which unprocessed gravel arrives at a gravel processing plant is modeled by $g(t)=90 + 45cos(\frac{t^{2}}{18})$, where $t$ is measured in hours and $0leq tleq8$. at the beginning of the workday ($t = 0$), the plant has 500 tons of unprocessed gravel. during the hours of operation, $0leq tleq8$, the plant processes gravel at a constant rate of 100 tons per hour.\n(a) find $g(5)$. using correct units, interpret your answer in the context of the problem.\n(b) find the total amount of unprocessed gravel that arrives at the plant during the hours of operation on this workday.\n(c) is the amount of unprocessed gravel at the plant increasing or decreasing at time $t = 5$ hours? show the work that leads to your answer.\n(d) what is the maximum amount of unprocessed gravel at the plant during the hours of operation on this workday? justify your answer.

2013 ap calculus ab free - response questions\ncalculus ab\nsection ii, part a\ntime—30 minutes\nnumber of problems—2\na graphing calculator is required for these problems.\n1. on a certain workday, the rate, in tons per hour, at which unprocessed gravel arrives at a gravel processing plant is modeled by $g(t)=90 + 45cos(\frac{t^{2}}{18})$, where $t$ is measured in hours and $0leq tleq8$. at the beginning of the workday ($t = 0$), the plant has 500 tons of unprocessed gravel. during the hours of operation, $0leq tleq8$, the plant processes gravel at a constant rate of 100 tons per hour.\n(a) find $g(5)$. using correct units, interpret your answer in the context of the problem.\n(b) find the total amount of unprocessed gravel that arrives at the plant during the hours of operation on this workday.\n(c) is the amount of unprocessed gravel at the plant increasing or decreasing at time $t = 5$ hours? show the work that leads to your answer.\n(d) what is the maximum amount of unprocessed gravel at the plant during the hours of operation on this workday? justify your answer.

Answer

Explanation:

Step1: Differentiate (G(t))

Using the chain - rule, if (G(t)=90 + 45\cos(\frac{t^{2}}{18})), then (G'(t)=45(-\sin(\frac{t^{2}}{18}))\cdot\frac{2t}{18}=- 5t\sin(\frac{t^{2}}{18})).

Step2: Evaluate (G'(5))

Substitute (t = 5) into (G'(t)): (G'(5)=-5\times5\sin(\frac{5^{2}}{18})=-25\sin(\frac{25}{18})\approx - 25\times0.979=-24.475) tons per hour. The negative sign indicates that the rate at which unprocessed gravel is arriving at the plant is decreasing at (t = 5) hours.

Step3: Find the total amount of unprocessed gravel that arrives

Use the definite integral (\int_{0}^{8}G(t)dt=\int_{0}^{8}(90 + 45\cos(\frac{t^{2}}{18}))dt). We know that (\int_{0}^{8}90dt=90t\big|{0}^{8}=90\times8 - 90\times0 = 720) and (\int{0}^{8}45\cos(\frac{t^{2}}{18})dt). Using a graphing calculator, (\int_{0}^{8}45\cos(\frac{t^{2}}{18})dt\approx45\times1.77 = 79.65). So (\int_{0}^{8}G(t)dt\approx720 + 79.65=799.65) tons.

Step4: Determine if the amount of unprocessed gravel is increasing or decreasing at (t = 5)

We found (G'(5)\approx - 24.475) tons per hour. The plant processes gravel at a rate of 100 tons per hour. The net rate of change of the amount of unprocessed gravel is (G'(5)-100\approx - 24.475-100=-124.475) tons per hour. Since this value is negative, the amount of unprocessed gravel is decreasing at (t = 5) hours.

Step5: Find the maximum amount of unprocessed gravel

Let (A(t)) be the amount of unprocessed gravel at time (t). Then (A(t)=500+\int_{0}^{t}G(s)ds-100t). First, find the critical points of (A(t)) by setting (A'(t)=G(t)-100 = 0), i.e., (90 + 45\cos(\frac{t^{2}}{18})-100=0), so (45\cos(\frac{t^{2}}{18}) = 10), (\cos(\frac{t^{2}}{18})=\frac{10}{45}=\frac{2}{9}). Using a graphing calculator to solve for (t) in the interval (0\leq t\leq8), we get (t\approx2.63) and (t\approx7.16). Evaluate (A(t)) at the critical points and endpoints: (A(0)=500), (A(2.63)=500+\int_{0}^{2.63}(90 + 45\cos(\frac{s^{2}}{18}))ds-100\times2.63), (A(7.16)=500+\int_{0}^{7.16}(90 + 45\cos(\frac{s^{2}}{18}))ds-100\times7.16), (A(8)=500+\int_{0}^{8}(90 + 45\cos(\frac{s^{2}}{18}))ds-100\times8). Using a graphing calculator, we find that the maximum value of (A(t)) occurs at (t\approx2.63) and (A(2.63)\approx500+\int_{0}^{2.63}(90 + 45\cos(\frac{s^{2}}{18}))ds-263\approx564.5) tons.

Answer:

(a) (G'(5)\approx - 24.475) tons per hour. The rate at which unprocessed gravel is arriving at the plant is decreasing at (t = 5) hours. (b) Approximately (799.65) tons. (c) Decreasing, since (G'(5)-100\approx - 124.475) tons per hour (negative value). (d) Approximately (564.5) tons at (t\approx2.63) hours.