Answer
(a) The average rate of change of $P$ is $~~-0.016~lb/in^5$
(b) $V'(P) = \frac{-800}{P^2}$
The instantaneous rate of change of $V$ with respect to $P$ is inversely proportional to the square of $P$.
Work Step by Step
(a) Let $V_1 = 200~in^3$
$P_1 = \frac{800~lb\cdot in}{200~in^3} = 4~lb/in^2$
Let $V_2 = 250~in^3$
$P_2 = \frac{800~lb\cdot in}{250~in^3} = 3.2~lb/in^2$
We can find the average rate of change of $P$ with respect to $V$:
$\frac{\Delta P}{\Delta V} = \frac{3.2~ln/in^2-4~lb/in^2}{250~in^3-200~in^3} = -0.016~lb/in^5$
(b) $V(P) = \frac{800}{P}$
We can find $V'(P)$:
$V'(P) = \lim\limits_{h \to 0}\frac{\frac{800}{P+h}-\frac{800}{P}}{h}$
$V'(P) = \lim\limits_{h \to 0}\frac{800P-800(P+h)}{hP(P+h)}$
$V'(P) = \lim\limits_{h \to 0}\frac{-800h}{hP(P+h)}$
$V'(P) = \lim\limits_{h \to 0}\frac{-800}{P(P+h)}$
$V'(P) = \frac{-800}{P(P+0)}$
$V'(P) = \frac{-800}{P^2}$
The instantaneous rate of change of $V$ with respect to $P$ is inversely proportional to the square of $P$.
