Question

The volume of a cube is increasing at a rate of 56 in^3/sec. At what rate is the length of each edge of the cube changing when the edges are 4 in. long? (Recall that for a cube,

V = x^3.)

Answer 1

Explanation:

Let's denote the volume of the cube as V and the length of each edge as x. Given that the volume of a cube is V = x^3, we can find the rate at which the length of each edge is changing.

We're given that the rate of change of the volume is dV/dt = 56 in³/sec. We want to find the rate of change of the length of each edge, which is dx/dt, when the length of each edge is 6 inches.

First, we differentiate the volume equation with respect to time t:

V = x^3

dV/dt = d(x^3)/dt

Using the chain rule:

dV/dt = 3x^2 * (dx/dt)

Now, we know that dV/dt = 56 in³/sec and x = 6 in. Plugging these values into the equation, we get:

56 = 3 * (6)^2 * (dx/dt)

Solving for dx/dt:

56 = 108 * (dx/dt)

dx/dt = 56 / 108

dx/dt ≈ 0.5185 in/sec (rounded to four decimal places)

So, the rate at which the length of each edge is changing is approximately 0.5185 inches per second when the edges are 6 inches long.