If the surface area of a sphere of radius $$r$$ is increasing uniformly at the rate 8 cm$$^2$$/s, then the rate of change of its volume is:

The problem states that the surface area of a sphere with radius $$ r $$ is increasing uniformly at a rate of 8 cm²/s. We need to find how the rate of change of its volume relates to the radius.

First, recall the formulas for the surface area $$ S $$ and volume $$ V $$ of a sphere:

Surface area: $$ S = 4\pi r^2 $$

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

We are given that the surface area increases at a constant rate, so $$ \frac{dS}{dt} = 8 $$ cm²/s.

To find the rate of change of volume $$ \frac{dV}{dt} $$, we need to relate it to the rate of change of radius $$ \frac{dr}{dt} $$, since both $$ S $$ and $$ V $$ depend on $$ r $$.

Start by differentiating the surface area formula with respect to time $$ t $$:

$$ \frac{dS}{dt} = \frac{d}{dt}(4\pi r^2) $$

Apply the chain rule:

$$ \frac{dS}{dt} = 4\pi \cdot 2r \cdot \frac{dr}{dt} = 8\pi r \frac{dr}{dt} $$

Substitute the given rate $$ \frac{dS}{dt} = 8 $$:

$$ 8\pi r \frac{dr}{dt} = 8 $$

Solve for $$ \frac{dr}{dt} $$:

Divide both sides by 8:

$$ \pi r \frac{dr}{dt} = 1 $$

Then:

$$ \frac{dr}{dt} = \frac{1}{\pi r} $$

Now, differentiate the volume formula with respect to time:

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

Apply the chain rule:

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

Substitute the expression for $$ \frac{dr}{dt} $$ we found earlier:

$$ \frac{dV}{dt} = 4\pi r^2 \cdot \frac{1}{\pi r} $$

Simplify:

$$ \frac{dV}{dt} = 4\pi r^2 \cdot \frac{1}{\pi r} = 4 \cdot r^{2-1} \cdot \frac{\pi}{\pi} = 4 \cdot r \cdot 1 = 4r $$

So, $$ \frac{dV}{dt} = 4r $$. This means the rate of change of volume is proportional to $$ r $$, since $$ 4r $$ is directly proportional to $$ r $$ (the constant 4 doesn't affect the proportionality).

Now, examining the options:

A. constant - This would mean independent of $$ r $$, but $$ 4r $$ depends on $$ r $$, so not constant.

B. proportional to $$ \sqrt{r} $$ - This would be like $$ r^{1/2} $$, but we have $$ r^1 $$, so no.

C. proportional to $$ r^2 $$ - This would be $$ r^2 $$, but we have $$ r $$, so no.

D. proportional to $$ r $$ - Yes, $$ 4r $$ is proportional to $$ r $$ (with constant of proportionality 4).

Hence, the correct answer is Option D.