If the volume of a sphere is increasing at a constant rate, then the rate at which its radius is increasing is

The correct option is: inversely proportional to the surface area.

To understand why this is correct, let us analyze the relationship between the volume, radius, and surface area of a sphere over time.

First, let $V$ represent the volume, $r$ represent the radius, and $A$ represent the surface area of the sphere at any time $t$.

The formula for the volume of a sphere is given by:

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

We are given that the volume of the sphere is increasing at a constant rate. This means the derivative of the volume with respect to time, $\frac{dV}{dt}$, is a constant. Let this constant rate be $k$:

$\frac{dV}{dt}=k$

Next, we differentiate the volume formula with respect to time $t$ using the chain rule:

$\frac{dV}{dt}=\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)$

Applying the chain rule gives:

$\frac{dV}{dt}=\frac{4}{3}\pi \cdot 3r^2\frac{dr}{dt}$

Simplifying the equation, we get:

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

Recall that the formula for the surface area $A$ of a sphere is:

$A=4\pi r^2$

Substituting $A$ into our differentiated equation yields:

$\frac{dV}{dt}=A\frac{dr}{dt}$

Since $\frac{dV}{dt}=k$, we can substitute this constant in:

$k=A\frac{dr}{dt}$

Now, we solve for the rate at which the radius is increasing, $\frac{dr}{dt}$:

$\frac{dr}{dt}=\frac{k}{A}$

Because $k$ is a constant value, this equation directly establishes that the rate of increase of the radius, $\frac{dr}{dt}$, is inversely proportional to the surface area $A$ of the sphere.