Problem 13 - College Algebra and Calculus: An Applied Approach

Question

All edges of a cube are expanding at a rate of 3 centimeters per second. How fast is the volume changing when each edge is (a) 1 centimeter and (b) 10 centimeters?

Step-by-Step Solution

Answer

(a) When each edge is 1 cm, the volume is increasing at a rate of 9 cubic centimeters per second. (b) When each edge is 10 cm, the volume is increasing at a rate of 900 cubic centimeters per second.

Step 1: Define the relationship

The volume \(V\) of a cube with edge length \(s\) is given by the formula \(V = s^3\). Thus, the relationship between volume and the edge length of the cube is cubic.

Step 2: Differentiate the relationship with respect to time

Differentiating both sides of the equation \(V = s^3\) with respect to time \(t\), we get \(\frac{dV}{dt} = 3s^2\frac{ds}{dt}\). Here, \(\frac{dV}{dt}\) is the rate of change of volume, \(\frac{ds}{dt} = 3\, cm/s\) is the given rate of change of the edge length, and \(s\) is the edge length of the cube.

Step 3: Substitute and solve for (a) 1 cm edge length

Substitute \(s = 1\, cm\) and \(\frac{ds}{dt} = 3\, cm/s\) into the equation to get the rate of change of volume when the edge length is 1 cm: \(\frac{dV}{dt} = 3(1)^2(3) = 9\, cm^3/s\). Hence, when each edge is 1 cm, the volume is increasing at a rate of 9 cubic centimeters per second.

Step 4: Substitute and solve for (b) 10 cm edge length

Substitute \(s = 10\, cm\) and \(\frac{ds}{dt} = 3\, cm/s\) into the equation to get the rate of change of volume when the edge length is 10 cm: \(\frac{dV}{dt} = 3(10)^2(3) = 900\, cm^3/s\). Hence, when each edge is 10 cm, the volume is increasing at a rate of 900 cubic centimeters per second.