A spherical snowballs radius is decreasing by 4% per second. Find the percentage rate at which its volume is decreasing.

Question

For this question I have to find the rate at which the volume of the sphere decreases, $\frac{dV}{dt}$. I already have $\frac{dr}{dt}$, the rate at which the radius decreases, which is $-\frac{4}{100}$. In order to be able to find $\frac{dV}{dt}$ I need to use the chain rule, seeing as I already know $\frac{dr}{dt}$. Thus we see that

$$\frac{dV}{dt}=?*\frac{dr}{dt}$$

and deduce that we need to find $\frac{dV}{dr}$. Using the equation for the volume of a shpere($\frac{4}{3}\pi r^3$) we can relate the two variables, $V$ and $r$. And so we see that we need to find $$=\frac{d}{dr}(\frac{4}{3}\pi r^3)$$

$$\frac{dV}{dr}=\frac{3*4\pi r^2}{3}=4\pi r^2$$

Thus we have $$\frac{dV}{dt}=-\frac{4}{100}4\pi r^2=-\frac{16\pi r^2}{100}$$

However, we need $\frac{dV}{dt}$ to be in terms of $V$. Any ideas?

Answer 1

You want $\frac{dV}{dt}$ in terms of $V$ $$V=\frac{4}{3}\pi r^3$$ $$r^3=\frac{3V}{4 \pi }$$ $$r=(\frac{3V}{4 \pi })^{1/3}$$

$$\frac{dV}{dt}=4 \pi r^2 \frac{-4r}{100}=-\frac{16 \pi r^3}{100}=-\frac{16 \pi \cdot 3V}{100 \cdot 4 \pi}=-\frac{3V}{25}$$

Answer 2

After one second,

$$\frac{V'-V}{V\cdot1}=\frac{(0.96r)^3-r^3}{r^3\cdot1}=-0.115264.$$

After one millisecond

$$\frac{V'-V}{V\cdot0.001}=\frac{(0.99996r)^3-r^3}{r^3\cdot0.001}=-0.11999520\cdots.$$

After infinitesimal time,

$$\frac{V'-V}{V\cdot \theta}=\frac{((1-0.04\theta)r)^3-r^3}{r^3\theta}=-0.12+0.0048\theta-0.000064\theta^2\to-0.12.$$