How would one justify the answer to this question on derivatives? I couldn't understand the solution to the following problem:
Suppose a balloon of volume V and radius r is being inflated, so that V and r are both functions of the time t. If dV/dt is constant, what can be said (without calculation)
about the behavior of dr/dt as r increases?
The answers page says that the answer is: 
dr/dt decreases as r increases.
I don't really understand why that answer is true.
I thought that r should has an increasing rate just like the volume, since the volume is dependent on the radius. 
If anyone is interested, the question is from the book Calculus and Analytic Geometry by George Simmons, page 68, section 2.4, problem no. 17
The answer is in page 859.
Thanks in advance.
 A: Well, we have:
$$V=\frac{4}{3}\pi r^3$$
Therefore:
$$\frac{dV}{dr}=4\pi r^2$$
And:
$$\frac{dV}{dt}=k$$
Where $k$ is some constant.
Now, use the fact that:
$$\frac{dr}{dt}=\frac{dV}{dt}/ \frac{dV}{dr}$$
From this, you can clearly see that as $r$ increases, $\frac{dr}{dt}$ decreases.
This is obvious enough to deduce, even 'without calculation' (As $r^2$ is increasing, in consequence its reciprocal must be decreasing).
A: I'm not sure if this counts as "without calculation", but:  $V = \frac{4\pi}{3} r^3$.  This means that $dV/dt = 4 \pi r^2 (dr/dt)$.  Since $dV/dt$ is constant, and $r^2$ is increasing, $dr/dt$ must be decreasing to compensate.
A: Note that $V=\frac{4\pi}{3}r^3(t)$.  Hence,
$$\frac{dV(t)}{dt}=4\pi r^2(t)\frac{dr(t)}{dt} \tag 1$$
We are given that $\frac{dV(t)}{dt}$ is constant and positive.  Let's call this constant $C>0$.  Then, we can write
$$\frac{dr(t)}{dt}=\frac{C}{4\pi r^2(t)} \tag 2$$
Differentiating $(2)$ reveals
$$\frac{d^2r(t)}{dt^2}=-\frac{C}{2\pi r^3(t)} \frac{dr(t)}{dt}<0$$
And we are done!
