# Derivatives! Help!

• Mar 17th 2008, 04:39 PM
tennisgirl
Derivatives! Help!
When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV^{1.4}=C where C is a constant. Suppose that at a certain instant the volume is 500 cubic centimeters and the pressure is 93 kPa and is decreasing at a rate of 15 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?

I know that dp/dt=-15
I know that dv/dt=dv/dp dp/dt
I don't know how to use the equation of C in this problem.

Any help is appreciated. Thanks.
• Mar 17th 2008, 06:22 PM
Aryth
Well, we have the following information:

$\displaystyle PV^{1.4} = C$

$\displaystyle \frac{dp}{dt} = -15 \frac{kPa}{min}$

$\displaystyle \frac{dv}{dt} = \frac{dv}{dp}\frac{dp}{dt}$

Well, looks like we need to find $\displaystyle \frac{dv}{dp}$

Let's take a look at it, if we implicitly differentiate the first equation, we are left with a $\displaystyle \frac{dv}{dp}$ are we not?

$\displaystyle \frac{dv}{dp}[PV^{1.4} = C] = V + 1.4V^{0.4}P\frac{dv}{dp} = 0$

$\displaystyle 1.4V^{0.4}P\frac{dv}{dp} = -V$

$\displaystyle \frac{dv}{dp} = -\frac{V^{0.6}}{1.4P}$

Looks like it's something we can work with, now we plug it back in to the equation given for $\displaystyle \frac{dv}{dt}$

$\displaystyle \frac{dv}{dt} = \frac{15V^{0.6}}{1.4P}$

Now we plug in the values:

$\displaystyle V = 500 cm^3$

$\displaystyle P = 93 kPa$

$\displaystyle \frac{dv}{dt} = \frac{15(500)^{0.6}}{1.4(93)}$

$\displaystyle \frac{dv}{dt} = 4.78 \frac{cm^3}{min}$

Notice that it's positive, as the question states, it is increasing.

Just to see how the units work here's the equation without the numbers:

$\displaystyle \frac{dv}{dt} = \frac{dv}{dp}\frac{dp}{dt} = \frac{cm^3}{kPa}\frac{kPa}{min} = \frac{cm^3}{min}$