Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 32

How fast is the volume increasing at this instant?
Given: Adiabatic Equation: $PV^{1.4} = C$
at a certain instance, $V = 400$ cm$^3$, $P = 80$ kPa, $\displaystyle \frac{dP}{dt} = -10 \frac{\text{kPa}}{\text{min}}$
Required: The rate of increase of the volume at the same instance

$PV^{1.4} = C$
Getting the derivative with respect to time we get,
$\displaystyle P \left( 1.4 V^{1.4-1}\left( \frac{dV}{dt} \right)\right) + V^{1.4} \left( \frac{dP}{dt} \right) = 0$
Solve for $\displaystyle \frac{dV}{dt}$, and simplify the equation

$
\begin{equation}
\begin{aligned}
\frac{dV}{dt} &= \frac{-V^{1.4} \left(\frac{dP}{dt}\right)}{1.4 PV^{0.4}}\\
\\
\frac{dV}{dt} &= \frac{-\left(400 \text{cm}^3\right)^{1.4} \left( -10 \frac{\cancel{\text{kPa}} }{dt}\right)}{1.4\left(80 \cancel{\text{kPa}}\right)\left( 400 \text{cm}^3 \right)^{0.4} }
\end{aligned}
\end{equation}\\
\boxed{\displaystyle \frac{dV}{dt} = \frac{250}{7} \frac{\text{cm}^3}{\text{min}}}
$