Question

If the rate of increase of surface area of a spherical balloon is \(5\,\text{cm}^2/\text{sec}\) and rate of increase of volume of a spherical balloon is \(10\,\text{cm}^3/\text{sec}\), then the radius of the balloon at that time is...

• A. \(3\) cm
• B. \(5\) cm
• C. \(6\) cm
• D. \(4\) cm

Hint:

Divide dV/dt by dS/dt so that dr/dt cancels, leaving r.

Answer 1

The Correct Option is D

Step 1: Link V and S
Since $V = \frac{r}{3}S$ for a sphere, we can use this relation to link the rates.

Step 2: Differentiate
$\frac{dV}{dt} = \frac{r}{3}\frac{dS}{dt} + \frac{S}{3}\frac{dr}{dt}$. But $\frac{dr}{dt} = \frac{1}{8\pi r}\frac{dS}{dt}$, and $S = 4\pi r^2$, so $\frac{S}{3}\frac{dr}{dt} = \frac{4\pi r^2}{3}\cdot\frac{1}{8\pi r}\frac{dS}{dt} = \frac{r}{6}\frac{dS}{dt}$.

Step 3: Combine
$\frac{dV}{dt} = \left(\frac{r}{3} + \frac{r}{6}\right)\frac{dS}{dt} = \frac{r}{2}\frac{dS}{dt}$.

Step 4: Solve
$10 = \frac{r}{2}\cdot5$, so $r = 4$ cm. Options 3, 5 and 6 give $\frac{dV}{dt}$ of 7.5, 12.5 and 15 for this $\frac{dS}{dt}$.

Final Answer:
The radius is 4 cm. This is option (D). \[ \boxed{\text{(D) }4\ \text{cm}} \]