The volume of a sphere, $V$ cm³, of radius $r$ is given by the formula $V = \frac{4}{3} \pi r^3$. The surface area of a sphere $A$ cm² of radius $r$ cm is given by the formula $A=4\pi r^2$. Find $\frac{dV}{dA}$ in terms of $r$.
Here's my workings to the question:
$$V= \frac{\frac{4}{3}\pi r^{3}}{4\pi r^2}A = \frac{1}{3}rA$$
So, $$\frac{dV}{dA} = \frac{1}{3}r =\frac{r}{3}.$$
I am not sure about this answer, so it would help to know if anyone got the same answer. Thank you!
Your approach is fine, but remember that in order to take the derivative $\frac{dV}{dA}$ directly you need to completely eliminate $r$ from your equation first. So, since you have $A = 4\pi r^{2}$, you can solve for $r$ to get $r = \frac{A^{1/2}}{2\sqrt{\pi}}$. Then, substituting this into $V = \frac{1}{3}rA$ you get $$V = \frac{1}{3}\frac{A^{3/2}}{2\sqrt{\pi}}.$$ Now you can take the derivative directly, to get $$\frac{dV}{dA} = \frac{A^{1/2}}{4\sqrt{\pi}}.$$ Since the question wanted your answer in terms of $r$, we substitute back: $$\frac{dV}{dA} = \frac{A^{1/2}}{4\sqrt{\pi}} = \frac{1}{2}\frac{A^{1/2}}{2\sqrt{\pi}} = \frac{1}{2}r = \frac{r}{2}.$$
That said, as was pointed out in another post, you could also compute this by finding $\frac{dV}{dr}$ and $\frac{dA}{dr}$ separately, and then using the chain rule: $$\frac{dV}{dA} = \frac{dV}{dr}\cdot \frac{dr}{dA}.$$ However, in either case you should arrive at the same answer.
Given $V= \frac{4\pi}3 r^3$ and $A=4\pi r^2$, we have
$$V= \frac{4\pi}3\left(\frac{A}{4\pi}\right)^{3/2}= \frac{A^{3/2}}{3\sqrt{4\pi}} $$
Then,
$$\frac{dV}{dA}= \frac12\frac{\sqrt A}{\sqrt{4\pi}}= \frac r2 $$
The physicist way. $$\begin{align*} \frac{dV}{dA}&=\lim_{\Delta r\to0}\frac{\Delta V}{\Delta A}\\ &=\lim_{\Delta r\to0}\left(\frac{\Delta V}{\Delta r}\right)/\left(\frac{\Delta A}{\Delta r}\right)\\ &=\lim_{\Delta r\to0}\left(\frac43\pi\frac{(r+\Delta r)^3-r^3}{\Delta r}\right)/\left(4\pi\frac{(r+\Delta r)^2-r^2}{\Delta r}\right)\\ &=\frac{r^2}{2r}\\ &=\frac r2 \end{align*}$$
