Volume of decreasing diameter

Question

A sphere's volume decreases at a rate of $1$ cm$^3$/min. Find the instantaneous rate of which diameter decreases with respect to time when diameter is 10 units.

First, I know radius is $d/2$, so $v=\cfrac{\pi d^3}{6}$. I also know that $\cfrac{dv}{dt}=-1$. Finding the derivative of $v$ I get $$\cfrac{\pi d^2}{2}$$

And at this point is where I'm confused. The expressions here are all in terms of $d$, but I'm taking the derivative with respect to $t$ (time) even though $t$ is nowhere to be found here, so it's throwing me off.

I would prefer your solution to be written with Leibniz's notation.

Answer 1

Both $v$ and $d$ are functions of $t$. So, you may write

$$v(t) =\frac\pi6 d^3(t)$$

and its derivatives with respect to $t$ $$\frac{dv(t)}{dt} = \frac{\pi}2 d^2(t)\frac{dd (t)}{dt}$$

Then, plug in the givens $\frac{dv(t)}{dt} = -1cm^3/min$ and $d(t) =10cm$ to calculate the rate of change for the diameter

$$\frac{dd (t)}{dt}= \frac{2}{\pi d^2(t)}\frac{dv(t)}{dt} =-\frac1{50\pi}cm /min$$

Answer 2

You have that

$$v(t) = \frac\pi 6 {d(t)}^3 \implies d(t) = {\left(\frac 6\pi v(t)\right)}^{1/3}.$$

You also know that $v'(t) = -1$. Therefore:

$$d'(t) = \frac13{\left(\frac 6\pi v(t)\right)}^{-2/3}\cdot \frac 6\pi v'(t) = \frac{-2}\pi{\left(\frac 6\pi v(t)\right)}^{-2/3}$$

But of course, $\frac 6\pi v(t) = d(t)^3$ and thus

$$d'(t) = \frac{-2}{\pi d(t)^2}$$

It follows that when $d(t) = 10$, we have

$$d'(t) = \frac{-1}{50\pi}.$$