Calc Rate Problem: At What Rate is Volume Increasing?

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In summary, when air expands adiabatically, its pressure and volume are related by the equation PV^{1.4}=C, where C is a constant. At a certain instant, the volume is 630 cubic centimeters and the pressure is 97 kPa, decreasing at a rate of 7 kPa/minute. Using the equation dP/dt = (-1.4PV^0.4*dV/dt)/V, we can find the rate at which the volume is increasing to be 32.47422681 cubic centimeters per minute.
  • #1
Okie
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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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  • #2
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.
 
  • #3
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
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.
 
  • #5
Is this correct ?

dP/dt = -[P 1.4*V^.4 dV/dt]/[V^1.4]
 
  • #6
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
Thank you i got the answer . :smile:
 
  • #8
The answer is 32.47422681
 

1. What is a "Calc Rate Problem"?

A "Calc Rate Problem" involves finding the rate at which a certain variable is changing over time. In this specific problem, we are trying to determine the rate at which volume is increasing.

2. How is the rate of change calculated?

The rate of change, also known as the derivative, is calculated by finding the slope of the tangent line to the curve at a specific point. This can be done using the formula: dy/dx = lim (Δx → 0) [f(x + Δx) - f(x)] / Δx. In the context of this problem, we would find the derivative of the volume function with respect to time, which would give us the rate of change of volume over time.

3. What information do we need to solve this problem?

To solve this problem, we would need to know the function that represents the volume as a function of time, as well as the specific time at which we want to find the rate of change. We would also need to have a basic understanding of calculus and the concept of derivatives.

4. What units would the rate of change be expressed in?

The rate of change would be expressed in units of volume per unit of time. For example, if the volume is measured in cubic centimeters and time is measured in seconds, the rate of change would be expressed in cubic centimeters per second.

5. How can this problem be applied in real life?

Calc Rate Problems, including the problem of finding the rate at which volume is increasing, have many real-life applications. For example, they can be used in engineering to calculate the rate of growth of a population, or in economics to determine the rate at which a company's profits are increasing. They can also be used in physics to calculate the velocity of an object or in chemistry to determine the rate of a chemical reaction.

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