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: Understanding the Question:
We are given the rate of change of the surface area of a sphere and asked to find the rate of change of its volume at a specific radius.

Step 2: Key Formula or Approach:
The surface area ($S$) and volume ($V$) of a sphere with radius $r$ are:
$$S = 4\pi r^2$$ $$V = \frac{4}{3}\pi r^3$$ We need to differentiate both with respect to time ($t$) and relate $\frac{dV}{dt}$ to $\frac{dS}{dt}$.

Step 3: Detailed Explanation:
Given: $\frac{dS}{dt} = 2\ \text{cm}^2/\text{sec}$.
Differentiating the surface area equation with respect to $t$:
$$\frac{dS}{dt} = 4\pi (2r) \frac{dr}{dt} = 8\pi r \frac{dr}{dt}$$ Substitute the given rate:
$$2 = 8\pi r \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{2}{8\pi r} = \frac{1}{4\pi r}$$ Now, differentiate the volume equation with respect to $t$:
$$\frac{dV}{dt} = \frac{4}{3}\pi (3r^2) \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}$$ Substitute the expression for $\frac{dr}{dt}$ into this equation:
$$\frac{dV}{dt} = 4\pi r^2 \left(\frac{1}{4\pi r}\right)$$ $$\frac{dV}{dt} = r$$ We are asked to find the rate when the radius $r = 6\ \text{cm}$:
$$\frac{dV}{dt} = 6\ \text{cm}^3/\text{sec}$$

Step 4: Final Answer:
The rate of increase in the volume is $6\ \text{cm}^3/\text{sec}$, matching option (D).