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Math Help - Finding the function

  1. #1
    Member Jones's Avatar
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    Finding the function

    Hello,

    I need help with this one.

    "During a boat trip, the engine suddenly stops. If the breaking force is proportional to the velocity it is possible to set up a differential equation m*dv/dt=c*v where m is the mass of the boat, v is the velocity, t is the time and c is the braking force"

    a) Determine v as a function of time if t=0 yields v= v_0

    How am i supposed to do this?
    Last edited by Jones; December 15th 2006 at 06:12 AM.
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  2. #2
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    Hello, Jones!

    During a boat trip, the engine suddenly stops.

    If the braking force is proportional to the velocity,
    it is possible to set up a differential equation: m\cdot\frac{dv}{dt}\:=\:kv
    where m is the mass of the boat, v is the velocity,
    t is the time and k is the braking force.

    a) Determine v as a function of time if t=0 yields v = v_0

    We are given: . m\cdot\frac{dv}{dt} \:=\:kv

    Separate variables: . \frac{dv}{v}\:=\:\frac{k}{m}dt

    Integrate: . \ln v \:=\:\frac{k}{m}t + c

    . . and we have: . v \:=\:e^{\frac{k}{m}t + c} \:=\:e^{\frac{k}{m}t}\cdot e^c \:=\:e^{\frac{k}{m}t}\cdot C

    Hence: . v(t)\:=\:Ce^{\frac{k}{m}t}


    We are told that: . v(0) = v_o
    . . So we have: . v_o \:=\:Ce^{\frac{k}{m}(0)}\quad\Rightarrow\quad C = v_o


    Therefore: . \boxed{v(t)\;=\;v_oe^{\frac{k}{m}t}}

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  3. #3
    Member Jones's Avatar
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    Im i supposed to know this stuff?

    Thanks for the help.
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  4. #4
    Member Jones's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, Jones!

    [size=3]
    We are given: . m\cdot\frac{dv}{dt} \:=\:kv

    Separate variables: . \frac{dv}{v}\:=\:\frac{k}{m}dt
    Isn't dt just with respect to the variable you are differentiating?

    Why are we substituting for dt?
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jones View Post
    Im i supposed to know this stuff?

    Thanks for the help.
    Quote Originally Posted by Jones View Post
    Isn't dt just with respect to the variable you are differentiating?

    Why are we substituting for dt?
    All Soroban is doing is the "separation of variables" method of solving a differential equation.

    -Dan
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