The Problem:

A sphere with radius $r$ increases in volume at a rate proportional to the surface area of the sphere at that time, with proportionality constant $k$. Write a function for the sphere’s volume at any time $t$.

My Attempt:

We can write an ODE to solve for $r(t)$, and then plug that into $V=\frac{4}{3}\pi r^3$. Since the rate of change of the radius is equal to the surface area $4\pi r^2$ times $k$, we get:$$\frac{dr}{dt}=4\pi kr^2$$

I separated the variables and got:$$r(t)=-\frac{1}{4\pi kt+C}$$Let alone the volume function, this radius function makes no sense. Help please!

  • 4
    $\begingroup$ Read your problem more carefully: " increases its volume at a rate" $\endgroup$ Dec 26, 2022 at 19:10

2 Answers 2


You've gotten the main equation wrong: it should be $\dfrac{d\color{red}V}{dt}=4\pi kr^2$, not $\dfrac{d\color{red}r}{dt}=4\pi kr^2$. By the chain rule, we get $\dfrac{dV}{dt}=\dfrac{dV}{dr}\dfrac{dr}{dt}$. But from the known relation $V=\dfrac43\pi r^3$, we have $\dfrac{dV}{dr}=4\pi r^2$. Substituting this we get $\dfrac{dr}{dt}=k$, from where $r=kt+C$, and $C=r(0)$.


$$\frac d{dt}\left(\frac43\pi r^3\right)=k\,4\pi r^2\iff\frac{dr}{dt}=k\iff r(t)=kt+r(0).$$


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