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Math Help - Related Rates Word Problem with Derivation

  1. #1
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    Related Rates Word Problem with Derivation

    I keep getting the wrong answer to this related rate problem, so I would greatly appreciate it if someone could go over my work and tell me where I messed up and why. I'm not sure if I made a mistake early on in the problem, or if it's a little mistake somewhere at the end. Thanks a ton:

    When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4 = C, where C is a constant. Suppose that at a certain instant the volume is 750 cm3 and the pressure is 80 kPa and is decreasing at a rate of 10 kPa/min. At what rate is the volume increasing at this instant?

    So I am given V = 750, P = 80, and dP/dt = -10.

    PV^1.4 = C

    1.4 lnPV = lnC

    1.4 (1/PV) ( P(dP/dt) + V(dV/dt)) = 1/C (dC/dt)
    [but since it is a constant, dC/dt goes to 0, right? Or is this wrong?]

    1.4 (1/60000) ( 80(dV/dt) + 750(-10)) = 0

    (2.333e-5) (80(dV/dt) - 7500) = 0

    .0018667(dV/dt) - .175 = 0

    .0018667(dV/dt) = .175

    dV/dt = 93.75
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  2. #2
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    I'm not sure about your method, honestly I didn't really review it (once you started taking the logs I zoned out).

    But here's what I would do. Take the derivative of your equation with respect to time = t;

    \frac{d}{dt} [ PV^{1.4} = C ]

    Using product rule gets you:

    P'V^{1.4} + 1.4PV' = 0

    Solve for V' to get the rate the volume is increasing.

     <br />
V' = \frac{ -P'V^{1.4}}{1.4P}<br />

    The problem gives you P, P' and V, so you are all set to substitute in the values and get the answer. I got 945.949 \frac{cm^3}{min}.

    Patrick
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  3. #3
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    Quote Originally Posted by Maziana View Post
    I keep getting the wrong answer to this related rate problem, so I would greatly appreciate it if someone could go over my work and tell me where I messed up and why. I'm not sure if I made a mistake early on in the problem, or if it's a little mistake somewhere at the end. Thanks a ton:

    When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4 = C, where C is a constant. Suppose that at a certain instant the volume is 750 cm3 and the pressure is 80 kPa and is decreasing at a rate of 10 kPa/min. At what rate is the volume increasing at this instant?

    So I am given V = 750, P = 80, and dP/dt = -10.

    PV^1.4 = C

    1.4 lnPV = lnC
    This is incorrect. It is ln P+ 1.4ln V= ln C. Since the "1.4" powere is only on V, only ln(V) is multiplied by 1.4. But I can see no reason to take logarithms anyway. By the product rule \frac{dP}{dt}V^{1.4}+ 1.4PV^{.4}dV/dt= 0. Now set P= 80, dP/dt= -10, and V= 750 and solve for dV/dt.

    1.4 (1/PV) ( P(dP/dt) + V(dV/dt)) = 1/C (dC/dt)
    [but since it is a constant, dC/dt goes to 0, right? Or is this wrong?]

    1.4 (1/60000) ( 80(dV/dt) + 750(-10)) = 0

    (2.333e-5) (80(dV/dt) - 7500) = 0

    .0018667(dV/dt) - .175 = 0

    .0018667(dV/dt) = .175

    dV/dt = 93.75
    No, this is not correct.
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  4. #4
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    Quote Originally Posted by PatrickFoster View Post
    I'm not sure about your method, honestly I didn't really review it (once you started taking the logs I zoned out).

    But here's what I would do. Take the derivative of your equation with respect to time = t;

    \frac{d}{dt} [ PV^{1.4} = C ]

    Using product rule gets you:

    P'V^{1.4} + 1.4PV' = 0
    No, you left out the " V^{0.4}"! It should be
    P'V^{1.4}+ 1.4PV^{0.4}V'= 0


    Solve for V' to get the rate the volume is increasing.

     <br />
V' = \frac{ -P'V^{1.4}}{1.4P}<br />

    The problem gives you P, P' and V, so you are all set to substitute in the values and get the answer. I got 945.949 \frac{cm^3}{min}.

    Patrick
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  5. #5
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    No, you left out the ""! It should be

    Agreed. Thanks for the correction.

    The new answer is 66.9643.
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