All edges of a cube are expanding at a rate of 8 mm per hour. At what rate is the volume of the cube changing when each edge is 16 mm?

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Problem:**

All edges of a cube are expanding at a rate of 8 mm per hour. At what rate is the volume of the cube changing when each edge is 16 mm?

**Solution:**

To solve this problem, we will use related rates of change in calculus. Let \( x \) be the length of each edge of the cube, and \( V \) be the volume of the cube. The volume of a cube is given by the formula:
\[ V = x^3 \]

To find the rate at which the volume is changing, we need to differentiate \( V \) with respect to time \( t \):
\[ \frac{dV}{dt} = 3x^2 \frac{dx}{dt} \]

Given:
- The rate at which the edge length is increasing: \( \frac{dx}{dt} = 8 \) mm per hour.
- The edge length at the specific time: \( x = 16 \) mm.

Substitute these values into the differentiated equation:
\[ \frac{dV}{dt} = 3(16)^2 (8) \]

Calculate:
\[ \frac{dV}{dt} = 3 \times 256 \times 8 \]
\[ \frac{dV}{dt} = 6144 \]

**Conclusion:**

The volume of the cube is increasing at a rate of 6144 cubic millimeters per hour when each edge is 16 mm.
Transcribed Image Text:**Problem:** All edges of a cube are expanding at a rate of 8 mm per hour. At what rate is the volume of the cube changing when each edge is 16 mm? **Solution:** To solve this problem, we will use related rates of change in calculus. Let \( x \) be the length of each edge of the cube, and \( V \) be the volume of the cube. The volume of a cube is given by the formula: \[ V = x^3 \] To find the rate at which the volume is changing, we need to differentiate \( V \) with respect to time \( t \): \[ \frac{dV}{dt} = 3x^2 \frac{dx}{dt} \] Given: - The rate at which the edge length is increasing: \( \frac{dx}{dt} = 8 \) mm per hour. - The edge length at the specific time: \( x = 16 \) mm. Substitute these values into the differentiated equation: \[ \frac{dV}{dt} = 3(16)^2 (8) \] Calculate: \[ \frac{dV}{dt} = 3 \times 256 \times 8 \] \[ \frac{dV}{dt} = 6144 \] **Conclusion:** The volume of the cube is increasing at a rate of 6144 cubic millimeters per hour when each edge is 16 mm.
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