Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 31

How fast is the volume decreasing at this instant?
Given: Boyle's Law = $PV = C$ where; $\begin{array}{c}
P & = & \text{pressure}\\
V & = & \text{volume}\\
C & = & \text{constant}\\
\end{array}$
at a certain instance, $P = 150$ kPa, $V = 600$ cm$^3$, $\displaystyle \frac{dP}{dt} = 20 \frac{\text{kPa}}{\text{min}}$


Required: $\displaystyle \frac{dV}{dt}$ at the same distance
Solution: when $P = 150$ kPA, $V = 600$ cm$^3$, $C = 150$ kPA (600 cm$^3$)
$C$ = 90,000,000

We get the derivative of the Boyle's Law with respect to time.

$\displaystyle P \frac{dV}{dt} + V \frac{dP}{dt} = 0 $

$
\begin{equation}
\begin{aligned}
\frac{dV}{dt} &= \frac{-V \frac{dP}{dt}}{P} \\
\\
\frac{dV}{dt} &= \frac{-\left(600\text{cm}^3\right)\left(20 \frac{\text{kPa}}{\text{min}}\right)}{150 \text{kPa}}\\
\\
\frac{dV}{dt} &= -80 \frac{\text{cm}^3}{\text{min}} && \text{,negative value because the rate is decreasing}
\end{aligned}
\end{equation}
$


Since the unknown in the problem is the rate of volume decreasing the final answer would be $\boxed{\displaystyle\frac{dV}{dt} = 80 \frac{\text{cm}^3}{\text{min}}}$

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