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Calc rate problem 
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#1
Mar1209, 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 


#2
Mar1209, 07:10 PM

Mentor
P: 21,216

You can't start by putting numbers inP 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. 


#3
Mar1209, 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? 


#4
Mar1209, 08:08 PM

Mentor
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, PV^{1.4} = 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. 


#5
Mar1209, 08:31 PM

P: 9

Is this correct ???
dP/dt = [P 1.4*V^.4 dV/dt]/[V^1.4] 


#6
Mar1209, 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. 


#7
Mar1309, 01:01 AM

P: 9

Thank you i got the answer .



#8
Mar1309, 01:29 AM

P: 9

The answer is 32.47422681



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