Home / AP Calculus AB : 5.11 Solving Optimization Problems- Exam Style questions with Answer- FRQ

AP Calculus AB : 5.11 Solving Optimization Problems- Exam Style questions with Answer- FRQ

Question

Let f(x) = e2x . Let R be the region in the first quadrant bounded by the graph of f, the coordinate axes, and the vertical line x = k, where k > 0. The region R is shown in the figure above.
(a) Write, but do not evaluate, an expression involving an integral that gives the perimeter of R in terms of k.
(b) The region R is rotated about the x-axis to form a solid. Find the volume, V, of the solid in terms of k.
(c) The volume V, found in part (b), changes as k changes. If \(\frac{dk}{dt}=\frac{1}{3}, determine \frac{dV}{t}when k=\frac{1}{2}.\)

Answer/Explanation

Ans:

(a)

\(\frac{dy}{dx}=2e^{2x}\)

\(p = 1 + k + e^{2x}+ \int_{0}^{k}\sqrt{1+(2e^{2x})^{2}}dx\)

(b)

u = 4x

\(\frac{du}{dx}=4\)

\(v = \int_{0}^{k}\pi r^{2}dx=\int_{0}^{k}\pi (e^{2x})^{2}dx=\pi \int_{0}^{k}e^{4x}dx\)

\(= \pi \int_{0}^{k}\frac{1}{4}e^{4}\cdot du=\frac{\pi }{4}\left [ e^{4x} \right ]_{0}^{k}=\frac{\pi }{4}\left ( e^{4k} -e^{4(0)}\right )\)

\(=\frac{\pi }{4}e^{4k}-1\left ( \frac{\pi }{4} \right )\)

\(v=\frac{\pi }{4}e^{4k}- \frac{\pi }{4}\)

(c)

\(\frac{dV}{dt}=\frac{dV}{dk}\cdot \frac{dk}{dt}=\frac{d}{dk}\left ( \frac{\pi }{4}e^{4k}-\frac{\pi }{4} \right )\cdot \left ( \frac{1}{3} \right )=\frac{\pi }{4}(4)e^{4k}\cdot \left ( \frac{1}{3} \right )\)

\(\frac{dV}{dt}=\frac{\pi }{3}e^{4k}|_{k=\frac{1}{2}}=\frac{\pi }{3}\cdot e^{2}\)

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