The volume of a cube is decreasing at a constant rate of 1334 cubic centimeters per second. At the instant when the side length of the cube is 7 centimeters, what is the rate of change of the surface area of the cube? Round your answer to three decimal places (if necessary).

To solve this problem, let’s break it down into steps.

1. **Define the volume and surface area of a cube in terms of its side length:**

   - Let \( s \) be the side length of the cube (in cm).
   - The volume \( V \) of a cube is given by:
     \[
     V = s^3
     \]
   - The surface area \( A \) of a cube is given by:
     \[
     A = 6s^2
     \]

2. **Differentiate volume with respect to time:**

   Since the volume is decreasing, \( \frac{dV}{dt} = -1334 \, \text{cm}^3/\text{s} \).
   
   Now, using the formula \( V = s^3 \), differentiate both sides with respect to time \( t \):
   \[
   \frac{dV}{dt} = 3s^2 \frac{ds}{dt}
   \]
   
   Substitute \( \frac{dV}{dt} = -1334 \) and \( s = 7 \) cm:
   \[
   -1334 = 3 \cdot 7^2 \cdot \frac{ds}{dt}
   \]
   \[
   -1334 = 3 \cdot 49 \cdot \frac{ds}{dt}
   \]
   \[
   -1334 = 147 \cdot \frac{ds}{dt}
   \]
   Solving for \( \frac{ds}{dt} \):
   \[
   \frac{ds}{dt} = \frac{-1334}{147} \approx -9.0748 \, \text{cm/s}
   \]

3. **Differentiate surface area with respect to time:**

   Now, differentiate \( A = 6s^2 \) with respect to \( t \) to find \( \frac{dA}{dt} \):
   \[
   \frac{dA}{dt} = 12s \cdot \frac{ds}{dt}
   \]
   
   Substitute \( s = 7 \) cm and \( \frac{ds}{dt} \approx -9.0748 \):
   \[
   \frac{dA}{dt} = 12 \cdot 7 \cdot (-9.0748)
   \]
   \[
   \frac{dA}{dt} \approx 12 \cdot 7 \cdot -9.0748
   \]
   \[
   \frac{dA}{dt} \approx -762.2864
   \]
   Rounded to three decimal places, the rate of change of the surface area is:
   \[
   \frac{dA}{dt} \approx -762.286 \, \text{cm}^2/\text{s}
   \]

So, the rate of change of the surface area at the instant when the side length is 7 cm is approximately \(-762.286\) square centimeters per second.

Would you like more details or have any questions?

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Here are five related questions to expand understanding:
1. How would the rate of change in surface area vary if the side length of the cube was larger?
2. If the side length of the cube was increasing instead, how would that affect the surface area rate?
3. How is the rate of change of the side length related to the rate of change of the volume?
4. How would this problem differ if the shape was a rectangular prism rather than a cube?
5. Can the rate of surface area change ever exceed the rate of volume change?

**Tip:** Always check the units of rates carefully, especially when dealing with derivatives involving different powers like surface area (square units) and volume (cubic units).

Calculate the rate of change in the surface area of a cube when its volume is decreasing at a rate of 1334 cm³/s and the side length is 7 cm. This solution demonstrates related rates, differentiation, and geometric formulas. Suitable for students in Grades 11-12.

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