Question

The surface area of a cube is increasing at the constant rate of $0.5 \text{ cm}^2/\text{s}$. Then the rate at which the volume of the cube is increasing (in $\text{cm}^3/\text{s}$), when its surface area has reached $12 \text{ cm}^2$, is

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2\sqrt{2}}$
• C. $\frac{1}{3\sqrt{2}}$
• D. $\frac{1}{4\sqrt{2}}$
• E. $\frac{1}{6\sqrt{2}}$

Hint:

Always express the side $a$ in terms of the given instantaneous measurement ($S=12$) before substituting into the derivative equations. Notice that \( \frac{dV}{dt} = \frac{a}{4} \frac{dS}{dt} \) for any cube, which can save time in such problems.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem involves the application of derivatives as a rate of change.
We need to relate the rate of change of the surface area of a cube to the rate of change of its volume.
Key Formula or Approach:
For a cube of side $a$:
Surface Area $S = 6a^2$
Volume $V = a^3$
Differentiate both with respect to time $t$:
\( \frac{dS}{dt} = 12a \frac{da}{dt} \)
\( \frac{dV}{dt} = 3a^2 \frac{da}{dt} \)

Step 2: Detailed Explanation:
Given \( \frac{dS}{dt} = 0.5 \text{ cm}^2/\text{s} \).
When surface area $S = 12$:
\[ 6a^2 = 12 \implies a^2 = 2 \implies a = \sqrt{2} \text{ cm} \]
Substitute $a = \sqrt{2}$ and \( \frac{dS}{dt} = 0.5 \) into the surface area rate equation:
\[ 0.5 = 12(\sqrt{2}) \frac{da}{dt} \]
\[ \frac{da}{dt} = \frac{0.5}{12\sqrt{2}} = \frac{1}{24\sqrt{2}} \text{ cm/s} \]
Now, calculate the rate of change of volume \( \frac{dV}{dt} \):
\[ \frac{dV}{dt} = 3a^2 \frac{da}{dt} \]
Substitute $a^2 = 2$ and \( \frac{da}{dt} = \frac{1}{24\sqrt{2}} \):
\[ \frac{dV}{dt} = 3(2) \left( \frac{1}{24\sqrt{2}} \right) = \frac{6}{24\sqrt{2}} = \frac{1}{4\sqrt{2}} \text{ cm}^3/\text{s} \]

Step 3: Final Answer:
The rate of increase of the volume is $\frac{1}{4\sqrt{2}} \text{ cm}^3/\text{s}$.