Question

the edges of a cube increases at a rate of 3 cm/s. how fast is the volume changing when the lenght of each edge is 60 cm?

Answer 1

\[V = s^3\]
Correct?

When doing related rates, you want to take the derivative with respect to time, t, of the function.

Answer 2

Sorry, I have to go but, taking the derivative with respect to t....
\[\frac{dV}{dt} = 3s^2 \frac{ds}{dt}\]
Where 's' is the edge length and 'ds/dt' is the rate of change of the edge length.
dV/dt will be the rate of change of the volume
You're welcome.