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 is a related rates problem. We relate the rate of change of the surface area to the rate of change of the side length, and then use that to find the rate of change of the volume.
Step 2: Key Formula or Approach:
Surface area of a cube: \(S = 6x^2\).
Volume of a cube: \(V = x^3\).
Differentiate both with respect to time \(t\) using the chain rule.
Step 3: Detailed Explanation:
Given \(\frac{dS}{dt} = 0.5 = \frac{1}{2}\).
We want to find \(\frac{dV}{dt}\) when \(S = 12\).
First, find the side length \(x\) when \(S = 12\):
\[ 6x^2 = 12 \implies x^2 = 2 \implies x = \sqrt{2} \] Differentiate the surface area equation with respect to \(t\):
\[ \frac{dS}{dt} = 12x \frac{dx}{dt} \] Substitute the known values to find \(\frac{dx}{dt}\):
\[ \frac{1}{2} = 12(\sqrt{2}) \frac{dx}{dt} \] \[ \frac{dx}{dt} = \frac{1}{24\sqrt{2}} \] Differentiate the volume equation with respect to \(t\):
\[ \frac{dV}{dt} = 3x^2 \frac{dx}{dt} \] Substitute \(x = \sqrt{2}\) and \(\frac{dx}{dt} = \frac{1}{24\sqrt{2}}\):
\[ \frac{dV}{dt} = 3(\sqrt{2})^2 \left( \frac{1}{24\sqrt{2}} \right) \] \[ \frac{dV}{dt} = 3(2) \left( \frac{1}{24\sqrt{2}} \right) = \frac{6}{24\sqrt{2}} \] Simplify the fraction:
\[ \frac{dV}{dt} = \frac{1}{4\sqrt{2}} \] Step 4: Final Answer:
The rate of increase is \(\frac{1}{4\sqrt{2}}\).