# Related Rates volume change

1. Nov 15, 2015

### Michele Nunes

1. The problem statement, all variables and given/known data
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?
2. Relevant equations

3. The attempt at a solution
I used the equation for the volume of a cube: V = s3 but I'm not sure if side and edge can be considered the same thing. Anyways, I implicitly differentiated it with respect to time t and got: dV/dt = 3s2(ds/dt) and since they give ds/dt = 3 cm/sec and values for s I just plugged all that in and for (a) I got 9 cm3/sec and for (b) I got 900 cm3/sec but I'm not sure if I did it correctly though.

2. Nov 15, 2015

### Staff: Mentor

To answer your other question, a side and an edge are the same thing here.

3. Nov 15, 2015

### Michele Nunes

Thank you for double checking my work

4. Nov 15, 2015

### Staff: Mentor

You're welcome!

5. Nov 15, 2015

### Staff: Mentor

As you mentioned is $s = s(t) = 3t + s_0$ where $s_0$ is the starting length of the cube. Isn't the volume's vvelocity then accelerating quadratic in time?
And if so when will be measured? At $3t + s_0 = 1$ and $3t + s_0 = 100$ or is $s_0 = 0$, $s_0 = 100$ resp.?

6. Nov 15, 2015

### Staff: Mentor

No, you are misinterpreting the problem. The cube isn't moving through space. It is expanding. This is a typical problem in calculus textbooks in the section on Related Rates.

7. Nov 15, 2015

### Staff: Mentor

@Mark44 I got that. $s(t)$ has been noted the length of the cube's edges in the OP.

My misunderstanding was that I first thought the given lengths were those of the original cube when expansion started.
In that case it would have been just a formula of time and time of measurement needed to be specified.

But meanwhile I understood it: $s(t) = 3t + s_0 = 1$ or $100$ defines the measurement via the actual edges then.
I have to admit that I sometimes tend to make things more complicated than they are.