# Equation for a Sphere with Increasing Volume

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!

• Read your problem more carefully: " increases its volume at a rate" Commented Dec 26, 2022 at 19:10

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