Question

Please help with this two part question (Calculus)^^

Answer 1

Answer:

1) -0.016 pounds per square inch per cubic inch.

2) $V'(P)=-\frac{800}{P^2}$

Step-by-step explanation:

We have the equation $PV=800$.

Part A)

We want to find the average rate of change of P as V increases from 200 cubic inches to 250 cubic inches.

The average rate of change is simply another way of saying slope. So, we simply need to find the slope between the two points when V=200 and V=250.

Let's first write the equation as a function of V. We can divide both sides by V. This yields:

$P(V)=\frac{800}{V}$

So, let's find P(V) for each of these values:

$P(200)=\frac{800}{200}=4$

And:

$P(250)=\frac{800}{250}=3.2$

So, we will have the two points (200, 4) and (250, 3.2).

Now, we will use the slope formula to find the average rate of change. Therefore:

$m=\frac{3.2-4}{250-200}=\frac{-0.8}{50}=-0.016$

Therefore, our rate of change is -0.016 pounds per square inch per cubic inch.

Part B)

We want to express V as a function of P. So, let's divide both sides by P. This yields:

$V(P)=\frac{800}{P}$

We want to show that the instantaneous change of V with respect to P is inversely proportional to the square of P. So, let's take the derivative of both sides with respect to P:  

$\frac{d}{dP}[V(P)]=\frac{d}{dP}[\frac{800}{P}]$

The left will just be V'(P). On the right, we can first move the constant-multiple outside and rewrite it as:

$V'(P)=800\frac{d}{dP}[P^{-1}]$

Then using the power rule, we get:

$V'(P)=800(-1P^{-2})=-\frac{800}{P^2}$

So, our derivative is:

$V'(P)=-\frac{800}{P^2}$

Therefore, the instantaneous change of V is indeed inversely proportional to the square of P.