The radius, $r$, of the base of a circular cylinder increases by 2 feet per second while the height, $h$, decreases by 1 foot per second. How fast is the surface area of the cylinder changing when the height of the cylinder is $50$ feet and the radius of the base is 40 feet? (Hint: The surface area, $A$, of such a cylinder is given by $A = 2πr² + 2πrh$.)

I was given this question to solve. I took the information that I know: $\frac{{dr}}{{dt}}$ = 2 feet per second
$\frac{{dh}}{{dt}}$ = −1 foot per second $r_0 = 40$ and $h_0=50$. Differentiating the area formula gives: $$4\pi r r'+2\pi(r'h+h'r)$$ Plugging it all in gives me: $$A'=4\pi(40)(2)+2\pi(2(50)-1(40))$$Solving that gives $440\pi$.
I cannot find an error in my work. However, the answer key gives something else. This is what the answer key says:

This is the correct answer. We are given the following information that is pertinent at the time of interest:
$\frac{{dr}}{{dt}}$ = 2 feet per second $\frac{{dh}}{{dt}}$ = −1 foot per second
We are interested in the value of $\frac{{dA}}{{dt}}$ when $h = 50$ feet and $r = 40$ feet. To find this, differentiate both sides of the surface area formula implicitly with respect to t and substitute in the given information, as follows: $\frac{{dA}}{{dt}} = 2π(2r)\frac{{dr}}{{dt}} + 2π(r\frac{{dh}}{{dt}} + h\frac{{dr}}{{dt}})= 4π(40)(2) + 2π(2(−1) + 50(2)) = 320π + 196π = 516π$
So, the surface area is increasing at a rate of $516π$ square feet per second.

The difference I found was in their $2π(2(−1) + 50(2))$ were they seem to have a 2 instead of a 1. Am I correct in thinking that they made the error? Or is the error mine? Did I do the implicit differentiation correctly?

  • 1
    $\begingroup$ I see the difference as they put for $r=2$ instead of $r=40$. Your implicit differentiation is right $\endgroup$ – imranfat Aug 6 '19 at 1:00

You did the implicit differentiation correctly. The solution appears to have used 2 in lieu of $r = 40$ (for what reason is unbeknownst to me).

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