Solve the problem. Round your answer, if appropriate.
-Boyle's law states that if the temperature of a gas remains constant, then $\mathrm { PV } = \mathrm { c }$, where $\mathrm { P } =$ pressure, $V =$ volume, and $c$ is a constant. Given a quantity of gas at constant temperature, if $V$ is decreasing at a rate of 9 in. $3 / \mathrm { sec }$, at what rate is $\mathrm { P }$ increasing when $\mathrm { P } = 90 \mathrm { lb } / \mathrm { in } ^ { 2 }$ and $\mathrm { V } = 70 \mathrm { in } .3$ ? (Do not round your answer.)
A) $700 \mathrm { lb } /$ in. $^ { 2 }$ per sec
B) $\frac { 81 } { 7 } \mathrm { lb } /$ in. $^ { 2 }$ per sec
C) $\frac { 81 } { 49 } \mathrm { lb } / \mathrm { in } ^ { 2 }$ per sec
D) $7 \mathrm { lb } /$ in. $^ { 2 }$ per sec

Question

Quizplus · Accepted Answer

Boyle's law states that $\mathrm{PV}=c$ where $c$ is a constant. Differentiating this equation with respect to time, we get $\mathrm{P}\frac{dV}{dt}+V\frac{dP}{dt}=0$.
Given $V$ is decreasing at a rate of $9$ in$^3$/sec. Hence, $\frac{dV}{dt}=-9$. Also, when $\mathrm{P}=90\mathrm{lb/in^2}$ and $\mathrm{V}=70.3$, we have $\mathrm{PV}=90	imes70.3=6330.9$.
Substituting the given values, we get $\mathrm{90}	imes{(-9)}+70.3\frac{dP}{dt}=0$, which simplifies to $\frac{dP}{dt}=\frac{81}{7}=11.57\mathrm{lb/in^2/sec}$. Therefore, the answer is B.

Solve the Problem PV=c\mathrm { PV } = \mathrm { c }PV=c

Solve the Problem $\mathrm { PV } = \mathrm { c }$