Question

The volume of a cube is increasing at the rate of 9 cm\(^3\)/s. If the length of its edge is 10 cm, then its surface area is increasing with which rate?

Hint:

In related rates problems, the key is to find an intermediate rate that connects the given rate and the required rate. Here, \( \frac{dx}{dt} \) was the crucial link between \( \frac{dV}{dt} \) and \( \frac{dS}{dt} \). Always start by writing the geometric formulas, then differentiate with respect to time.

Answer 1

Step 1: Understanding the Concept:
This is a related rates problem. We are given the rate of change of the volume of a cube (\( \frac{dV}{dt} \)) and need to find the rate of change of its surface area (\( \frac{dS}{dt} \)) at a specific instant when the edge length is known. The link between these rates is the rate of change of the edge length (\( \frac{dx}{dt} \)).
Step 2: Key Formula or Approach:
Let \( x \) be the edge length of the cube.
Volume of the cube: \( V = x^3 \)
Surface area of the cube: \( S = 6x^2 \)
We will differentiate both formulas with respect to time (t) to relate the rates.
Step 3: Detailed Explanation or Calculation:
First, let's write down the given information:
- Rate of increase of volume: \( \frac{dV}{dt} = 9 \) cm\(^3\)/s
- Edge length: \( x = 10 \) cm
We need to find \( \frac{dS}{dt} \).
Differentiate the volume formula with respect to time \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt}(x^3) = 3x^2 \frac{dx}{dt} \] We can use this equation to find \( \frac{dx}{dt} \), the rate at which the edge length is increasing. Substitute the known values:
\[ 9 = 3(10)^2 \frac{dx}{dt} \] \[ 9 = 3(100) \frac{dx}{dt} \] \[ 9 = 300 \frac{dx}{dt} \] \[ \frac{dx}{dt} = \frac{9}{300} = \frac{3}{100} \text{ cm/s} \] Now, differentiate the surface area formula with respect to time \( t \):
\[ \frac{dS}{dt} = \frac{d}{dt}(6x^2) = 12x \frac{dx}{dt} \] Substitute the known values of \( x \) and the calculated value of \( \frac{dx}{dt} \):
\[ \frac{dS}{dt} = 12(10) \left(\frac{3}{100}\right) \] \[ \frac{dS}{dt} = 120 \left(\frac{3}{100}\right) = \frac{360}{100} = 3.6 \] Step 4: Final Answer:
The surface area of the cube is increasing at a rate of 3.6 cm\(^2\)/s.