# 009A Sample Final 3, Problem 8

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

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

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

${\begin{array}{rcl}\displaystyle {V'}&=&\displaystyle {{\frac {-(20)(600)}{150}}{\text{ cm}}^{3}{\text{/min}}}\\&&\\&=&\displaystyle {-80{\text{ cm}}^{3}{\text{/min}}.}\end{array}}$ Therefore, the volume is decreasing at a rate of  $80{\text{ cm}}^{3}{\text{/min}}$ at this instant.

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