Results 1 to 5 of 5

Thread: Finding the function

  1. #1
    Member Jones's Avatar
    Joined
    Apr 2006
    From
    Norway
    Posts
    170

    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 $\displaystyle m*dv/dt=c*v$ 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=$\displaystyle v_0$

    How am i supposed to do this?
    Last edited by Jones; Dec 15th 2006 at 06:12 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    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: $\displaystyle m\cdot\frac{dv}{dt}\:=\:kv$
    where $\displaystyle m$ is the mass of the boat, $\displaystyle v$ is the velocity,
    $\displaystyle t$ is the time and $\displaystyle k$ is the braking force.

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

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

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

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

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

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


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


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

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Jones's Avatar
    Joined
    Apr 2006
    From
    Norway
    Posts
    170
    Im i supposed to know this stuff?

    Thanks for the help.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member Jones's Avatar
    Joined
    Apr 2006
    From
    Norway
    Posts
    170
    Quote Originally Posted by Soroban View Post
    Hello, Jones!

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

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

    Why are we substituting for $\displaystyle dt$?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,122
    Thanks
    719
    Awards
    1
    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 $\displaystyle dt$ just with respect to the variable you are differentiating?

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

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding a Cumulative Distribution Function and Density Function
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: Nov 17th 2011, 09:41 AM
  2. Replies: 14
    Last Post: Oct 7th 2011, 09:45 PM
  3. Finding probability function of moment generating function
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: Jul 4th 2011, 04:03 PM
  4. finding a function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 30th 2011, 11:24 AM
  5. Help finding this function?
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Dec 10th 2009, 03:03 PM

Search Tags


/mathhelpforum @mathhelpforum