1. ## Calculus HW Problem

Oh man... I'm lost.

Boyle's Law states that if the temperature of a gas remains constant, then the pressure, P, and the volume, V, satisfy the equation: PV = c, where c is a constant. If the volume is decreasing at a rate of 12 cubic inches per second, how fast is the pressure increasing when the pressure is 75 pounds per square inch and the volume is 18 cubic inches?

2. This is a related rates problem:

$\displaystyle PV=c$

where

$\displaystyle \frac{dV}{dt}= -12$

$\displaystyle \frac{dP}{dt}=$ what we're trying to find. c is a constant, so you don't have to worry about it.

Take the derivative:
$\displaystyle \frac{dP}{dt}V + \frac{dV}{dt}P = 0$

And now you can substitute (V is given as 18 cubic inches, and P as 75 pounds/square inch):
$\displaystyle \frac{dP}{dt}V = -\frac{dV}{dt}P$

$\displaystyle \frac{dP}{dt}(18) = -(-12)(75)$

And then solve for $\displaystyle \frac{dP}{dt}$.

3. What you are looking for is $\displaystyle \frac {dP}{dV}$.

$\displaystyle PV = c$, so $\displaystyle P = \frac {c}{V}$. This can be rewritten as $\displaystyle P = cV^{-1}$.

So, the derivative of P, or $\displaystyle \frac {dP}{dV}$, is equal to $\displaystyle -cV^{-2}\frac {dV}{dt}$, where $\displaystyle \frac {dV}{dt}$ is the derivative of V with respect to time.

Now, plug 75 and 18 into your original equation to find $\displaystyle c$. Then plug the value for $\displaystyle c$, the value for $\displaystyle \frac {dV}{dt}$, which is -12, and 18 for $\displaystyle V$.

Does this help?

4. So, if I am following you guys... the answer is 50?

(thanks for the quick replies btw, you guys are awesome and so helpful!)

5. Yes, the answer is 50.

6. Thanks so much!

One thing that I am having a hard time with is just understanding the derivative notation.

For example... One of my HW questions gave an example of a tank being filled up with water and made V into the volume of the tank and H into the height of the water level. Then it asked me to explain what dV/dH and dH/dV meant in an english sentence.

I'm not sure really... I can take the derivatives if you give me the problem, but explaining goes right over my head.

7. $\displaystyle \frac {dV}{dH}$ reads as, "The rate of change of V with respect to H".

8. Well, if you look at something like $\displaystyle \frac{dP}{dt}$, we say that that's the rate of change of pressure, or how pressure changes with time. How pressure changes as time changes. You could, I think, apply that reasoning to the two problems you have there. $\displaystyle \frac{dV}{dH}$ could be explained as how volume changes with height. Or, if you want to be mathematical about it, "The rate of change of volume with respect to height." That is, as height changes, how does volume change with it, and vice versa for $\displaystyle \frac{dH}{dV}$