Chapter 2 - Derivatives - 2.8 Related Rates - 2.8 Exercises - Page 187: 32

$80cm^{3}/min$

We are given the Boyle's Law equation $PV=C$ We are looking for the rate of change of the volume at the instant where The volume $V=600cm^{3}$ The pressure $P=150kPa$ The pressure $P$ is increasing at a rate of $20kPa/min$, which means we know that $P'=20kPa/min$ So we are looking for $V'$ evaluated at the instant where the conditions provided are true, which means those are the values we will later substitute into the differentiated equation. We will look at the equation provided, as it relates $P$ and $V$. $PV=C$ Working for $V$ we get $V=\frac{C}{P}$. We differentiate this equation so that we find $V'$. $V'=\frac{(C'\times(P))-(C\times(P')}{P^{2}}$ Because $C$ is a constant, we know that $C'=0$, so the equation turns into $V'=\frac{-(C\times(P')}{P^{2}}$ In order to be able to evaluate this differentiated equation, we need to solve for the value of $C$, so we plug it into our original Boyle's Law equation $PV=C$ $150kPa\times(600cm^{3})=C$ $90000=C$ So now we plug in all our given values to evaluate $V'$ $V'=\frac{-(90000\times(20))}{150^{2}}$ $V'=\frac{-(1800000)}{22500}$ $V'=-80cm^{3}/min$ ANSWER. When the volume is $600cm^{3}$, the pressure is $150kPa$, and the pressure is increasing at a rate of $20kPa/min$, the volume is decreasing by $80cm^{3}/min$. Notice that the answer was negative, but when we write the actual answer to the question, the implication that it is *decreasing* is already stated so the value written is positive.