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Math Help - practice work help

  1. #1
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    practice work help

    It's a word problem.

    When air expands without gaining or losing heat, it's pressure P and volume V are related by the equation PV^1.4=C (c is constant). Instant volume is 400 cm^3 and pressure is 80kPa and is decreasing at a rate of 10kPa/min. At what rate is the volume increasing at this instant?

    Should I solve for V and the take the derivative? What is the answer worked out?

    Thankies, first time doing these types of problems.
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  2. #2
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    Quote Originally Posted by mathman66 View Post
    It's a word problem.

    When air expands without gaining or losing heat, it's pressure P and volume V are related by the equation PV^1.4=C (c is constant). Instant volume is 400 cm^3 and pressure is 80kPa and is decreasing at a rate of 10kPa/min. At what rate is the volume increasing at this instant?

    Should I solve for V and the take the derivative? What is the answer worked out?

    Thankies, first time doing these types of problems.
    PV^{1.4} = C
    implies

    dP/dt * V^{1.4} + P * 1.4*V^{0.4}*dV/dt = 0

    So:
    dV/dt = - [dP/dt * V^{1.4}]/[1.4*P*V^{0.4}*dV/dt] = -V/1.4 * [dP/dt]/P

    dV/dt = -(400 cm^3)/1.4 * [-10 kPa/min]/(80 kPa) = 35.714285714 cm^3/min

    So the volume is increasing at a rate of about 35.7 cm^3/min.

    -Dan

    PS Yes, you CAN solve for V first, though I think in this case it's a bit harder:
    PV^{1.4} = C

    V^{1.4} = C/P

    V = (C/P)^{1/1.4} = (C/P)^{5/7}

    dV/dt = (5/7)(C/P)^{-2/7}*(-C/P^2)*dP/dt

    Now C = PV^{1.4} = PV^{7/5}:

    dV/dt = -(5/7)([PV^{7/5}]/P)^{-2/7}*([PV^{7/5}]/P^2)*dP/dt

    dV/dt = -(5/7)V*(1/P)*dP/dt

    which is the same as I had above.

    -Dan
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