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The volume of a right circular cylinder of radius $r$ and height $h$ is given by the formula $V=\pi r^2h$. Find the rate of variation of volume with the radius when $r=5.5 \text{in.}$ and $h=20 \text{in}$. If $r=h$, find the dimensions of the cylinder so that a change of $1$ inch in radius causes a change of $400$ cubic inches in the volume. The rate of variation of $V$ with regard to $r$ is $$\frac{dV}{dr}=2\pi rh$$ If $r= 5.5\text{in}$. and $h= 20\text{in}$. this becomes $691.2$. A change of radius of $1\text{in}$. will cause a change of volume of $691.2$ cubic inches. Also, if $h=r$, and $h$ remains constant, $$\frac{dV}{dr}=2\pi r^2=400$$ and $$r=h=\sqrt{\frac{400}{2\pi}}=7.98\text{in}.$$ If, however, $h=r$ and varies with $r$, then $$\frac{dV}{dr}=3\pi r^2=400$$ and $$r=h=\sqrt{\frac{400}{3\pi}}=6.51\text{in}.$$ I'm having trouble understanding the last part, when $h=r$ and varies with $r$. Since both $r$ and $h$ are varying, do I use the product rule? I follow the differentiation of $V$ with respect to $r$ when only $r$ was varying, now that they're both varying I'm stumped. Thanks in advance for any assistance and apologies if this turns out horrendous it's my first question.

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  • $\begingroup$ In the first case you took the partial derivative with respect to r and then set h= r. The partial derivative with respect to r assumes that h is a constant. That is not possible with r= h. You have to replace r with h first, then differentiate. $\endgroup$ – user247327 Jan 19 '17 at 1:18
  • $\begingroup$ If one of the answers is posted, please mark it as the answer with the check mark right beneath the vote ticker in the answers list. $\endgroup$ – Drew Christensen Jan 19 '17 at 3:49
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So, what you're looking at is a problem of "when is $h$ dependent on $r$?" To solve this, we can simply rewrite it as $h(r)$ and differentiate when necessary.

Given $V(r) = \pi r^2h(r)$, we can differentiate with the product rule.

$$\begin{align*} \frac{dV}{dr} &=\pi \frac{d}{dr}\left(r^2\right)h(r)+\pi r^2\cdot \frac{d}{dr}\left(h(r)\right) \\ \frac{dV}{dr}&= 2\pi rh(r) + \pi r^2h'(r) \end{align*}$$

Now, if $h(r)$ is a constant, then $h(r) = h$ and $h'(r)=0$. Simply plugging in will yield your resultant formula, $dV/dr = 2\pi rh$.

However, if $h(r)$ is not a constant, then simply plug in for $h(r)$ and $h'(r)$.

Hope this helps!

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  • $\begingroup$ Thank you big time @Drew Christensen, great help $\endgroup$ – Isosceles Jan 19 '17 at 1:24
  • $\begingroup$ If this is the answer to your question, then please mark it as the answer using the check button right beneath the vote ticker. $\endgroup$ – Drew Christensen Jan 19 '17 at 3:41
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It looks like you have the first two in hand. But don't convert to decimals unless you have been told to do so.

3) $h$ varies with $r.$

You can apply the product rule, but it is not really necessary.

$v = \pi r^2 h\\ \frac {dv}{dr} = (2\pi r)h + (2\pi r^2) \frac {dh}{dr}$

Now you can substitute $r=h$ and $\frac {dh}{dr} = 1$ and $r = h$

However, if $r = h$ and $r$ varies with $h,$ you can merely say $V = \pi r^3$

And differentiating you will get the same result.

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