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.

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

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