Answer
$\frac{\dfrac{dP}{dt}}{\dfrac{dV}{dt}}=-\dfrac{3}{2}\dfrac{P}{V}$ is proved.
Yes, we could have predicted that $\frac{\dfrac{dP}{dt}}{\dfrac{dV}{dt}}$ is negative.
Work Step by Step
Differentiate $PV^{\frac{3}{2}} = C$ with respect to time $t$.
We get, $\dfrac{d}{dt}\left(PV^{\frac{3}{2}}\right)=0$
Now use the product rule.
We get, $V^{\frac{3}{2}}\dfrac{dP}{dt}+P\dfrac{d(V^{\frac{3}{2}})}{dt}=0$
Now use the chain rule.
We get, $V^{\frac{3}{2}}\dfrac{dP}{dt}+P\times\dfrac{3}{2}V^{\frac{1}{2}}\dfrac{dV}{dt}=0$
Now take common $V^{\frac{1}{2}}$.
We get, $V^{\frac{1}{2}}\left(V\dfrac{dP}{dt}+P\times\dfrac{3}{2}\dfrac{dV}{dt}\right)=0$
$V\dfrac{dP}{dt}+P\times\dfrac{3}{2}\dfrac{dV}{dt}=0$
Now divide by $\dfrac{dV}{dt}$ in both sides.
$V\frac{\dfrac{dP}{dt}}{\dfrac{dV}{dt}}+\dfrac{3}{2}P=0$
Now solve for $\frac{\dfrac{dP}{dt}}{\dfrac{dV}{dt}}$.
We get, $\frac{\dfrac{dP}{dt}}{\dfrac{dV}{dt}}=-\dfrac{3}{2}\dfrac{P}{V}$
Hence proved.
If we simplify $\frac{\dfrac{dP}{dt}}{\dfrac{dV}{dt}}$.
We get, $\dfrac{dP}{dV}$
Since, $PV^{\frac{3}{2}}$ is constant $P$ and $V$ are inversely related.
So, If $V$ increases $P$ decreases.
Thus, the slope of the graph between $P$ and $V$ will always be negative.
That is, $\dfrac{dP}{dV}$ will be negative.
Since, $\frac{\dfrac{dP}{dt}}{\dfrac{dV}{dt}}=\dfrac{dP}{dV}$
$\frac{\dfrac{dP}{dt}}{\dfrac{dV}{dt}}$ will also is negative.
So, we could have predicted that $\frac{\dfrac{dP}{dt}}{\dfrac{dV}{dt}}$ is negative.
