Q 39. - Calculus

Question

Suppose the sides of a cube are expanding at a rate of $2$ inches per minute.

How fast is the volume of the cube changing at the moment that the cube has a side length of $8$ inches? 

Step-by-Step Solution

Answer

The rate of change in volume is $384 i{n}^3/min$

Step 1. Given Information

It is given that 

$\frac{ds}{dt}=2 in/min, s=8 in$

Step 2. Setting up the rate of change of the volume equation

The formula for the volume of a cube is

$V=s^3$

where 

V is the volume of a cube 

s is the side length

Find the derivative with respect to t

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

Step 3. Finding the rate of change in Volume

Plug values $\frac{ds}{dt}=2 in/min , s=8 in$ into volume derivative

$$\begin{gathered} \frac{dV}{dt}=3s^2\frac{ds}{dt} \\ \frac{dV}{dt}=3{\left(8\right)}^2\times 2 \\ \frac{dV}{dt}=384 i{n}^3/in \end{gathered}$$