Question:
Consider a rolling clay cylinder. It keeps being cylindrical and its volume remains the same. However, the length increases proportionally to the length. Show that the radius decreases proportionally to the radius.
Attempted solution:
So at time $t_{0}$ it has a certain length h and a certain radius r. Once it starts rolling, the rate at which the length increases at any given time is proportional to the length at that time. Now our job is to show that the rate at which the radius decreases is proportional to the radius at any given time.
It seems like an odd question and I am by far much worse for applied questions like this than other types of questions.
From the question, it is obvious that:
$$\frac{dh}{dt} = ch$$
where h is the length and c is a constant. The volume of a cylinder is also pretty obvious:
$$V = r^2 \pi h$$
Implicit differentiation of this second equation (using the product rule and treating r(t) and h(t) as two functions) gives:
$$0 = r^2 \frac{dh}{dt} + 2r \frac{dr}{dt} h$$
Putting in the first equation and solving for the rate of change in r gives:
$$\frac{dr}{dt} = \frac{-r^2 \frac{dh}{dt}}{2rh} = \frac{-r^2c}{2r} = c \frac{-r^2}{2r}$$
Now I have shown that the rate at which the radius changes is proportional to some negative expression involving $r$ and several other things. But this isn't quite what is being asked. The question is to show how the decrease is proportional to $r$. Here I am, perhaps naively, expecting the answer to come out as $cr$.
What crucial part am I missing?
You're essentially done. $-cr^2/2r = (-c/2)r$. So $dr/dt$ is proportional to $r$ with proportionality constant $-c/2$.