Rate Change of Radius of Reducing Sphere

Question

Suppose the volume of a sphere is given by $V$ and is declining by $Q$ per second where $Q$ is a constant volume significantly smaller than the sphere.

What function would give us instantaneous velocity at a given radius($v(r)$) where $v$ is velocity?

Work so far:

Initial radius is: $r_0=\sqrt[3]{(V/(4\pi/3))}$

radius after 1 second is: $r_1=\sqrt[3]{(r^34\pi/3 - Q)/(4\pi/3)}$

And then difference in radius is: $r_\Delta =\sqrt[3]{(V/(4\pi/3))}-\sqrt[3]{(r^34\pi/3 - Q)/(4\pi/3)}$

Where do I go from here?

Answer 1

This has to be viewed as a calculus problem: Assume Q is expressed as a positive number.

$$ V=\frac{4\pi r^3}{3}\\ \frac{dV}{dt}=-Q=\frac{4\pi}{3}\frac{d(r^3)}{dt}\\ \frac{d(r^3)}{dt}=-\frac{3Q}{4\pi}\\ r^3=r_0^3-\frac{3Q}{4\pi}t\\ r=\sqrt[3]{r_0^3-\frac{3Q}{4\pi}t}\\ v=\frac{dr}{dt} $$

You should be able to take it from here.

Answer 2

After much thought, I've come to a very straightforward solution to this:

Given that

$V=\frac{4\pi r^3}{3}$

and

$A=4\pi r^2$

Therefore

$Q=dV/dt = (4\pi r^2)(dr/dt)$

Solve for $dr/dt$

$dr/dt=\frac {dV/dt}{4\pi r^2}$

Since $v = dr/dt$ and $Q=dV/dt$ then

$v=\frac{Q}{4\pi r^2}$