The edges of a cube increase at a rate of 4 cm/s. How fast is the volume changing when the length of each edge is 20 cm?

Question

The edges of a cube increase at a rate of 4 cm/s. How fast is the volume changing when the length of each edge is 20 cm?

Write an equation relating the volume of a​ cube, V, and an edge of the​ cube, a.

Answer 1

In this situation, you're looking to utilize implicit differentiation. We'll basically be differentiating the volume formula of a cube with respect to time.

First off, the volume formula for a cube is V = a³, where s is the length of a side (or edge) of the cube.

In order to talk about rates, we'll need to differentiate according to time. While this sounds complex, it's very similar to the derivatives you've likely done in your calc class before.

So, on the right side, if you were to take a derivative, you would get 3a², but since this is differentiated with respect to time, we would also need to add a da/dt next to it.

On the left side, when you take the derivative of a variable like V, the answer is just 1. However, we again are differentiating with respect to t, so it would need to be dV/dt.

That means your equation is:

dV/dt = 3a²(da/dt)

So now, we need to plug in the values that we were given in the problem. We are told that the edge is changing at a rate of 4 cm/s. That will be our da/dt (our change in edge with respect to change in time). Since the edge is 20 cm, then we can plug that in for a. So our solution process looks like this:

dV/dt = 3(20)²(4)

dV/dt = 3(400)(4)

dV/dt = 4800

So, our volume is changing at a rate of 4800 cm³/s.

Answer 2

V = a³ where a is the edge of the cube

da/dt = 4 cm/s - this is the rate at which the edge, a, is changing

Derive: dv/dt = 3a² * da/dt

dv/dt = 3(20)² * 4

= 4800 cm³/s