The volume of a spherical balloon grows at a rate of $100\ cm^3/s$ The volume of a spherical balloon grows at a rate of $100\  cm^3/s\ $,what is the growing rate when the radius measures $50cm$.
I already know how to work this out, But I can't understand the problem 100%.
I was thinking about this problem and some doubts came up.
In the following example, the problem already give a related rate, that is, Volume over Time.
But then I thought, If the volume increases 100cm^3/s, The radius is also increasing, as the volume depends on the radius. But the radius will also grow only if the volume increases that is both measures are related.
As both rates are constants, if I change 50cm for 20 cm wouldn't the dr/dt be the same ?
I worked it out for both measures, and the results are not the same, I'm wondering why.
While the volume is increasing, the radius is increasing as well, so If  the radius is increasing the dr/dt is changing ?
I thought only the radius could variate, not the dr/dt.
I'm very confused.
______edit__________
@RoddyMacPhee As the sphere depends on the $R^3$ and not only on the R, I can say that the variation of the
radius of the sphere will never be proportional to the variation of the volume of the sphere,
because the radius will never grow proportionally to the rate of variation of the volume, because
the volume will increase 100 unidades of $\ cm^3 \ $ per second.
and the radius will always be increasing in a exponential way, that is impossible to determine
a ratio between your variation per second.
So, the rate of increase of the radius when it measures 50cm,
means how much it increased to get to 50cm, and not a default increase per time until it measures 50 cm.
 A: From the formula for the volume of a sphere, we have $V=\dfrac{4}{3} \pi r^{3}$. This shows how the volume depends on the radius - when the volume is constant, the radius is constant. However, it does not show that when the rate of change of the volume is constant, the rate of change of the radius is constant.
If we differentiate with respect to t, we get the relationship between the rate of increase in the volume and the rate of increase of the radius (with respect to time) $\dfrac{dV}{dt}=4 \pi r^{2}\dfrac{dr}{dt}$. This shows that both cannot be constant as the radius is not constant. 
A: By differentiating the volume formula,
$$\begin{align*}
V &= \frac23 (2\pi)r^3\\
\frac {dV}{dt} &= \frac23(2\pi)\cdot 3r^2\frac{dr}{dt}
\end{align*}$$
If $\frac{dV}{dt}$ is constant and non-zero, $\frac{dr}{dt}$ is not constant and depends on the $r$ at that moment.
A: hint
The volume is given by
$$V (t)=\frac {4\pi}{3}R^3 (t) $$
thus
$$\frac {dV}{dt}=4\pi R^2\frac {dR}{dt} $$
with $$\frac {dV}{dt}=100\;\; cm^3/s $$
and $$R=50 \;\;cm $$
hence
$$\frac {dR}{dt}=\frac {1}{100\pi} \;\;cm/s $$
A: You may be overthinking this. The volume of the sphere is
$$V=4\pi r^3/3\\
\&\\
V=\frac{dV}{dt}t
$$
since $dV/dt$ is constant. Therefore,
$$r=\left(\frac{3}{4\pi}\frac{dV}{dt}t\right)^{1/3}\\
\frac{dr}{dt}=\left(\frac{3}{4\pi}\frac{dV}{dt}\right)^{1/3}\frac{1}{3t^{2/3}}
$$
Here we notice that $dr/dt$ is infinite at $t=0$ when the growth begin.
