A balloon, which always remains spherical, has a variable radius. The rate at which its volume is increasing with respect to its radius $r$ when $r = 5$ cm is:
Show Hint
Differentiating volume with respect to radius always yields the surface area of the sphere: $\frac{dV}{dr} = 4\pi r^2$. Simply plug in the radius to get the area directly!
Step 1: Understand the question. The balloon is a sphere. We want how fast its volume changes as the radius changes, which is $\frac{dV}{dr}$, when the radius is $5$ cm.
Step 2: Write the volume formula. The volume of a sphere with radius $r$ is a standard result. \[ V = \frac{4}{3}\pi r^3 \]
Step 3: Differentiate with respect to $r$. Bring down the power $3$ and reduce it by one. The constant $\frac{4}{3}\pi$ stays. \[ \frac{dV}{dr} = \frac{4}{3}\pi\,(3r^2) \]
Step 4: Simplify. The $3$ cancels with the $\frac{1}{3}$, leaving a clean form. \[ \frac{dV}{dr} = 4\pi r^2 \]
Step 5: Put in $r = 5$. Square the radius first: $5^2 = 25$. Then multiply. \[ \frac{dV}{dr} = 4\pi(25) = 100\pi \]
Step 6: State the answer. So when $r = 5$ cm the volume grows at the rate $100\pi$ per unit radius.
\[ \boxed{100\pi} \]