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 $dV/dA$ 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!
On
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,
$$\frac{dV}{dA}= \frac12\frac{\sqrt A}{\sqrt{4\pi}}= \frac r2 $$
On
The physicist way. $$\begin{align*} \frac{dV}{dA}&=\lim_{\Delta r\to0}\frac{\Delta V}{\Delta A}\\ &=\lim_{\Delta r\to0}\left(\frac{\Delta V}{\Delta r}\right)/\left(\frac{\Delta A}{\Delta r}\right)\\ &=\lim_{\Delta r\to0}\left(\frac43\pi\frac{(r+\Delta r)^3-r^3}{\Delta r}\right)/\left(4\pi\frac{(r+\Delta r)^2-r^2}{\Delta r}\right)\\ &=\frac{r^2}{2r}\\ &=\frac r2 \end{align*}$$
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.