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I'm solving the following related rates problem (formatted as an image).

Part (a) is easily found to be $A = A(x) = 2xe^{-x^2/2}$.

Part (b), I thought, would be easy, too, but my answer is differing from the textbook. Let me share my findings.

$\frac{dA}{dt} = 2 \frac{dx}{dt}e^{-x^2/2} + 2x e^{-x^2/2}(-x)\frac{dx}{dt}$. Cleaning this up a bit, we have $\frac{dA}{dt} = 2e^{-x^2/2}(1-x^2)\frac{dx}{dt}.$ Then, evaluating at $x = 4$ and $\frac{dx}{dt} = 4$, we have $\frac{dA}{dt} = -120e^{-8} \approx -0.04026$ cm$^2$/sec.

However, the textbook claims the answer to be $-3.25$ cm$^2$/sec. Am I doing something wrong in my solution?

Thanks in advance.

Related rates problem.

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    $\begingroup$ Looks fine to me. Do you have an instructor you could ask? $\endgroup$ – Andrew Li Mar 14 '18 at 4:02
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Hope this isn't too late to be useful to someone!

Almost everything you did looks good. Remember that $\frac{dx}{dt}=4$ centimeters per minute, not per second, so the answer should be $-2e^{-8}\:\mathrm{cm^2/sec}\approx-6.709\times10^{-4}\:\mathrm{cm^2/sec}$, which, unfortunately, is even further away from the book's answer.

You can be pretty confident that the book is wrong, though, because when $x=4$, the height of the rectangle is already so small that the rectangle is nearly impossible to see, so there's no way its area could be decreasing at a rate anywhere near even $-1\:\mathrm{cm^2/sec}$.

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