At the time when the volume and the radius of the sphere are increasing at the same numerical rate, what is the radius?

Question

I can't figure out how to properly solve this. I attempted but got confused.

The radius of a sphere is increasing at a constant rate of 0.04 centimeter per second. At the time when the volume and the radius of the sphere are increasing at the same numerical rate, what is the radius?

Answer 1

Well, you have

$$ 4\pi r^2 (0.04\text{cm/sec})=(0.04\text{cm}^3/\text{sec}) $$

so

$$ r^2=\frac{1}{4\pi}\text{cm}^2 $$

Therefore

$$r=\frac{1}{2\sqrt{\pi}}\text{cm}$$

Answer 2

So you have that $$ V = \frac{4}{3}\pi r^3 \Rightarrow \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}. $$ You want to solve for when $$ \frac{dV}{dt} = \frac{dr}{dt} \Rightarrow \frac{dV}{dt} / \frac{dr}{dt} = 1, $$ so just solve $4\pi r^2 = 1$. The constant rate is irrelevent.