Question

The surface area of a spherical balloon is increasing at the rate $2\ \text{cm}^2/\text{sec}$. Then rate of increase in the volume of the balloon is, when the radius of the balloon is $6\ \text{cm}$.

• A. $4\ \text{cm}^3/\text{sec}$
• B. $16\ \text{cm}^3/\text{sec}$
• C. $36\ \text{cm}^3/\text{sec}$
• D. $6\ \text{cm}^3/\text{sec}$

Hint:

Notice the beautiful geometric relationship here: the derivative of the volume of a sphere with respect to its radius is exactly its surface area ($\frac{dV}{dr} = S$). Using the chain rule, $\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}$, which makes relating the rates very straightforward.

Answer 1

The Correct Option is D

Step 1: Write the formulas for a sphere.
Surface area $S = 4\pi r^2$ and volume $V = \frac{4}{3}\pi r^3$.

Step 2: Note what is given.
The surface area grows at $\frac{dS}{dt} = 2\ \text{cm}^2/\text{sec}$. We want $\frac{dV}{dt}$ when $r = 6$ cm.

Step 3: Differentiate the area with time.
\[ \frac{dS}{dt} = 8\pi r \frac{dr}{dt} \] So $2 = 8\pi r \frac{dr}{dt}$, which gives $\frac{dr}{dt} = \frac{1}{4\pi r}$.

Step 4: Differentiate the volume with time.
\[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \]
Step 5: Put the value of $\frac{dr}{dt}$ in.
\[ \frac{dV}{dt} = 4\pi r^2 \times \frac{1}{4\pi r} = r \]
Step 6: Use $r = 6$.
\[ \frac{dV}{dt} = 6\ \text{cm}^3/\text{sec} \] A neat fact: for a sphere $\frac{dV}{dr} = S$, which is why these rates link so cleanly. \[ \boxed{6\ \text{cm}^3/\text{sec (Option 4)}} \]