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Math Help - Separation of Variables

  1. #1
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    Separation of Variables

    i need help with a problem involving separation of variables. here's the entire problem:

    let v(t) be the velocity, in feet per second, of a skydiver at time t seconds, t > or = 0. after her parachute opens, her velocity satisfies the differential equation dv/dt = -2v - 32, with initial condition v(0) = -50

    use separation of variables to find an expression for v in terms of t, where t is measured in seconds

    anyone mind helping out?
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  2. #2
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    Quote Originally Posted by junkedly View Post
    i need help with a problem involving separation of variables. here's the entire problem:

    let v(t) be the velocity, in feet per second, of a skydiver at time t seconds, t > or = 0. after her parachute opens, her velocity satisfies the differential equation dv/dt = -2v - 32, with initial condition v(0) = -50

    use separation of variables to find an expression for v in terms of t, where t is measured in seconds

    anyone mind helping out?
    hi jun

    \frac{dv}{dt} = -2v - 32, then

    \frac{dv}{-2v - 32} = dt

    solve this differential equation
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  3. #3
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    \frac{dv}{dt} = -2v - 32

    \frac{dv}{-2v - 32} = dt

    Now integrate both sides.
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  4. #4
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    thanks for the quick replies

    i had that much so far, but i when i integrate dv/-2v-32 i get an antiderivative that is a nonreal answer

    i end up with -2ln(-2v - 32) = t, but then the initial condition doesn't work because a natural logarithm can't equal zero

    thanks again for the help
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  5. #5
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    \frac{dv}{-2v - 32} = dt

    \frac{dv}{-2(v + 16)} = dt

    \frac{dv}{v + 16} = -2 \, dt

    \ln|v+16| = -2t+C

    finish ...
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  6. #6
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    a huuuuuge thank you skeeter

    i always manage to overlook possible factoring

    you saved me
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  7. #7
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    so adding the initial conditions of v(0) = -50 i get


    ln|-50+16| = -2(0) + C

    C = 3.526

    therefore ln|v+16| = -2t + 3.526

    does this look correct?
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  8. #8
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    Quote Originally Posted by junkedly View Post
    so adding the initial conditions of v(0) = -50 i get


    ln|-50+16| = -2(0) + C

    C = 3.526

    therefore ln|v+16| = -2t + 3.526

    does this look correct?
    \ln|v+16| = -2t+C

    v+16 = e^{-2t+C} = e^{-2t} \cdot e^C ... let e^C = A

    v+16 = Ae^{-2t}

    v = -16+Ae^{-2t}

    v(0) = -50 ...

    finish
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