# Instantaneous rates of change

I am having problems solving the following question.

The volume, $V$, of a sphere of radius r is given by $V=f(r)=\frac{4}{3}\pi r^3$. Calculate the instantaneous rate of change of the volume, $V$, with the respect to change of the radius, $r$, at $r=36.4$.

I assume the answer to this question would be $f\prime(36.4)$

where $f\prime$ is equal to;

$f\prime(x) = 4\pi x^2 \\ f\prime(36.4) = 4\pi (36.4)^2 \\= 16649.93$

Although this is not the solution. Please advise me where I have went wrong.

• Sorry, my appologies, I left a typo while formatting my question. I meant to put $\frac{4}{3}$ and not $\frac{3}{4}$ – Miroslav Glamuzina Oct 12 '15 at 23:03
• Then you were doing right! – Benicio Oct 12 '15 at 23:04
• That is what I believed, but the solution I have is apparently incorrect? I just wanted to confirm my answer. Thankyou. – Miroslav Glamuzina Oct 12 '15 at 23:05
• Looks good to me. You've been asked to compute for $\frac{dV(r)}{dr}$ for some value of $r$ and that's what you did. – Aldon Oct 12 '15 at 23:08
• Closer to $16649.94$ (kidding). Maybe they wanted you to round. – André Nicolas Oct 12 '15 at 23:11

This is a related rates problem, and it seems your trouble is in the formula you're using. The volume of a sphere is $V = \frac{4}{3}\pi r^3$, not $\frac{3}{4}$.
$V(r)=\frac{4}{3}\pi r^3$. Using the chain rule, we have $\frac{dV}{dr}=4\pi r^2\cdot\frac{dr}{dt}$. We know $r=36.4$. However, the radius depends on time and you did not provide that info. In your problem, they'll give you $r(t)$, and take it's derivative. Then your answer is $\frac{dV}{dr}=4\pi\cdot 36.4^2\cdot\frac{dr}{dt}$