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Thread: Derivatives! Help!

  1. #1
    Mar 2008

    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.
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  2. #2
    Super Member Aryth's Avatar
    Feb 2007
    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:

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