Lilliana Livingston

2022-07-18

Finding a rate of sphere area increase given its volume increase
Volume of sphere, $V=\frac{4}{3}\pi {r}^{3}$
Surface area of sphere $S=4\pi {r}^{2}$.
If we know, $\frac{dV}{dt}=R$.
Let us consider both volume and area as composite functions, thus $\frac{dV}{dt}=\frac{dV}{dr}×\frac{dr}{dt}=4\pi {r}^{2}×\frac{dr}{dt}=R$
whence $\frac{dr}{dt}=\frac{R}{4\pi {r}^{2}}$ since $\frac{dS}{dt}=6\pi r\frac{dr}{dt}$, let the value of the $\frac{dr}{dt}$ into the second equation, to get the answer. Is this approach logically correct?

Steppkelk

Expert

Step 1
Everything is fine until you did
whence $\frac{dr}{dt}=\frac{R}{4\pi {r}^{2}}$ since $\frac{dS}{dt}=6\pi r\frac{dr}{dt}$.
$\frac{dS}{dt}=8\pi r\frac{dr}{dt}$
Step 2
Therefore $\frac{dS}{dt}=8\pi r\frac{R}{4\pi {r}^{2}}$
and finally $\frac{dS}{dt}=\frac{2R}{r}$

Do you have a similar question?