# 009A Sample Final 3, Problem 8

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure  ${\displaystyle P}$  and volume  ${\displaystyle V}$  satisfy the equation  ${\displaystyle PV=C}$  where  ${\displaystyle C}$  is a constant. Suppose that at a certain instant, the volume is  ${\displaystyle 600{\text{ cm}}^{3},}$  the pressure is  ${\displaystyle 150{\text{ kPa}},}$  and the pressure is increasing at a rate of  ${\displaystyle 20{\text{ kPa/min}}.}$  At what rate is the volume decreasing at this instant?

Foundations:
Product Rule
${\displaystyle {\frac {d}{dx}}(f(x)g(x))=f(x)g'(x)+f'(x)g(x)}$

Solution:

Step 1:
First, we take the derivative of the equation  ${\displaystyle PV=C.}$
Using the product rule, we get
${\displaystyle P'V+PV'=C'.}$
Since  ${\displaystyle C}$  is a constant,
${\displaystyle C'=0.}$
Therefore, we have
${\displaystyle P'V+PV'=0.}$
Step 2:
Solving for  ${\displaystyle V',}$  we get
${\displaystyle V'={\frac {-P'V}{P}}.}$
Using the information provided in the problem, we have
${\displaystyle V=600{\text{ cm}}^{3},~P=150{\text{ kPa}},~P'=20{\text{ kPa/min}}.}$
Hence, we get

${\displaystyle {\begin{array}{rcl}\displaystyle {V'}&=&\displaystyle {{\frac {-(20)(600)}{150}}{\text{ cm}}^{3}{\text{/min}}}\\&&\\&=&\displaystyle {-80{\text{ cm}}^{3}{\text{/min}}.}\end{array}}}$

Therefore, the volume is decreasing at a rate of  ${\displaystyle 80{\text{ cm}}^{3}{\text{/min}}}$  at this instant.

The volume is decreasing at a rate of  ${\displaystyle 80{\text{ cm}}^{3}{\text{/min}}}$  at this instant.