Q 41. - 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’s volume is $55$ cubic inches? 

Step-by-Step Solution

Answer

The rate of change in volume is $86.77457 i{n}^3/min$

Step 1. Given Information

It is given that

$\frac{ds}{dt}=2 in/min , V=55 i{n}^3$

Step 2. Finding the edge of the cube

The volume is 

$V=55 i{n}^3$

Apply the volume of the cube formula

$V=s^3$

$s^3=55$

$$\begin{gathered} s=\sqrt[3]{55} \\ s=3.80295 \end{gathered}$$

Step 3. Finding the rate of change in Volume

The volume equation is 

$V=s^3$

Find derivative with respect to t

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

Plug values $s=3.80295,\frac{ds}{dt}=2$ into the derivative

$$\begin{gathered} \frac{dV}{dt}=3{(3.80295)}^2\times 2 \\ \frac{dV}{dt}=86.77457 i{n}^3/min \end{gathered}$$