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Calc rate problem

by Okie
Tags: calc, rate
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Mar12-09, 07:02 PM
P: 9
hen 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 630 cubic centimeters and the pressure is 97 kPa and is decreasing at a rate of 7 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?

(Pa stands for Pascal -- it is equivalent to one Newton/(meter squared); kPa is a kiloPascal or 1000 Pascals. )

I think i start by doing (97)(630)^1.4 = 805097.5471 what is my next step? How do i solve this problem
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Mar12-09, 07:10 PM
P: 21,216
You can't start by putting numbers in--P and V are functions of time. Find a relationship between the rates of change (i.e., time derivatives) of P and V, and then solve for dV/dt.

Then you can put your numbers in.
Mar12-09, 07:32 PM
P: 9
I don't get it. So i differentiate pv^1.4 ?

so the answer would be 1.4pv' + v^1.4 ... would that be correct?

Mar12-09, 08:08 PM
P: 21,216
Calc rate problem

No. Both p and v are functions of t, so you need to differentiate with respect to t. I would recommend using d/dt (Leibniz) notation rather than ' (Newton) notation.

Start with your equation, PV1.4 = C, and differentiate both sides with respect to t. You should get another equation with P, dP/dt, V, and dV/dt. Solve that equation for dV/dt. Then you can find the value of dV/dt at the particular time in question.
Mar12-09, 08:31 PM
P: 9
Is this correct ???

dP/dt = -[P 1.4*V^.4 dV/dt]/[V^1.4]
Mar12-09, 11:58 PM
P: 81
That can be simplified but, yes, I think it is correct. I have trouble reading the formula like that so...

[tex]\frac{dP}{dt} = \frac{-1.4P\frac{dV}{dt}}{V}[/tex]

Now that you have an equation that relates the rates, the problem is figuring out what these values are from the problem. You solved for dP/dt and I'm not sure why, but it stills works. The problem asks for the rate at which volume is changing, and it gives you a volume, pressure, and a rate at which pressure is changing. Your new equation has 4 variables, Pressure, Volume, the rate at which Pressure changes, and the rate at which Volume changes. Plug in your known variables and solve the problem.
Mar13-09, 01:01 AM
P: 9
Thank you i got the answer .
Mar13-09, 01:29 AM
P: 9
The answer is 32.47422681

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