# Related Rates Question (Rectangle Under Normal Curve)

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? 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 $$-1\:\mathrm{cm^2/sec}$$.