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Math Help - Differentials and Integrating

  1. #1
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    Differentials and Integrating

    *EDIT: Whoops, wrong forum. This can be moved if needed. Sorry.

    Hello,

    I'm really having a bit of trouble here. I'm doing an online course with limited recorded lectures and a textbook which is helpful only some of the time, so I'm hoping for a little bit of guidance here.

    I have problem in which I've managed to write a differential equation to represent the problem. In the problem, C is defined as calories per day, so it's not an 'arbitrary constant'. I'm supposed to solve the equation, but I don't really know how to go about it.



    \frac {dw}{dt} = \frac {C-17.5w}{3500}
    = \frac {1}{C-17.5w} \, \frac {dw}{dt} = \frac {1}{3500}
    = \int \frac {1}{C-17.5w}\,dw = \int 3500^{-1} dt

    I fall down a little bit here, as I don't know exactly how to deal with C (the constant), whether the dt on the right, is just t??

    = \frac{1}{C} \, \int \frac {1}{-17.5w} dw = \frac {1}{3500t} + M
    = \frac {1}{C} \, -17.5\,ln|w| = \frac {1}{3500t} + M

    From here it just kind of falls to pieces, and I don't know whether I am supposed to get the 1/C to the other side, then divide by 17.5, then get rid of the natural log, or whether I messed up an earlier step.

    I really wish the lectures explained this sort of question a little better...
    Last edited by astuart; October 20th 2012 at 09:47 PM.
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  2. #2
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    Re: Differentials and Integrating

    Quote Originally Posted by astuart View Post
    *EDIT: Whoops, wrong forum. This can be moved if needed. Sorry.

    Hello,

    I'm really having a bit of trouble here. I'm doing an online course with limited recorded lectures and a textbook which is helpful only some of the time, so I'm hoping for a little bit of guidance here.

    I have problem in which I've managed to write a differential equation to represent the problem. In the problem, C is defined as calories per day, so it's not an 'arbitrary constant'. I'm supposed to solve the equation, but I don't really know how to go about it.



    \frac {dw}{dt} = \frac {C-17.5w}{3500}
    = \frac {1}{C-17.5w} \, \frac {dw}{dt} = \frac {1}{3500}
    = \int \frac {1}{C-17.5w}\,dw = \int 3500^{-1} dt

    I fall down a little bit here, as I don't know exactly how to deal with C (the constant), whether the dt on the right, is just t??

    = \frac{1}{C} \, \int \frac {1}{-17.5w} dw = \frac {1}{3500t} + M
    = \frac {1}{C} \, -17.5\,ln|w| = \frac {1}{3500t} + M

    From here it just kind of falls to pieces, and I don't know whether I am supposed to get the 1/C to the other side, then divide by 17.5, then get rid of the natural log, or whether I messed up an earlier step.

    I really wish the lectures explained this sort of question a little better...
    First of all \displaystyle \begin{align*} \frac{1}{3500} \end{align*} is a constant, so its integral will be \displaystyle \begin{align*} \frac{1}{3500}t \end{align*}. As for the LHS, rewrite it as \displaystyle \begin{align*} -\frac{1}{17.5}\int{\frac{-17.5}{C - 17.5w}\,dw} \end{align*} so that you can use the substitution \displaystyle \begin{align*} u = C - 17.5w \implies du = -17.5\,dw \end{align*} and the integral becomes \displaystyle \begin{align*} -\frac{1}{17.5}\int{\frac{1}{u}\,du}  \end{align*}.
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: Differentials and Integrating

    Your separation of the variables is good up to here:

    \int\frac{1}{C-17.5w}\,dw=\int 3500^{-1}\,dt

    although I would write:

    -\int\frac{1}{17.5w-C}\,dw=\frac{1}{3500}\int\,dt

    Now, in your next step, you pulled a \frac{1}{C} out in front of the integral as if it were a factor, which is is not. You did not have:

    \int\frac{1}{-17.5Cw}\,dw=\int 3500^{-1}\,dt

    If you did, then what you did would have been fine. So, let's go back to:

    -\int\frac{1}{17.5w-C}\,dw=\frac{1}{3500}\int\,dt

    Let's multiply the integral on the left by 1=\frac{17.5}{17.5}

    -\frac{17.5}{17.5}\int\frac{1}{17.5w-C}\,dw=\frac{1}{3500}\int\,dt

    -\frac{1}{17.5}\int\frac{17.5}{17.5w-C}\,dw=\frac{1}{3500}\int\,dt

    -\frac{2}{35}\int\frac{17.5\,dw}{17.5w-C}=\frac{1}{3500}\int\,dt

    Now, observe that the integral on the left is of the form \int\frac{du}{u}=\ln|u|+c_1...can you finish?
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    Re: Differentials and Integrating

    Quote Originally Posted by MarkFL2 View Post
    Your separation of the variables is good up to here:

    \int\frac{1}{C-17.5w}\,dw=\int 3500^{-1}\,dt

    although I would write:

    -\int\frac{1}{17.5w-C}\,dw=\frac{1}{3500}\int\,dt

    Now, in your next step, you pulled a \frac{1}{C} out in front of the integral as if it were a factor, which is is not. You did not have:

    \int\frac{1}{-17.5Cw}\,dw=\int 3500^{-1}\,dt

    If you did, then what you did would have been fine. So, let's go back to:

    -\int\frac{1}{17.5w-C}\,dw=\frac{1}{3500}\int\,dt

    Let's multiply the integral on the left by 1=\frac{17.5}{17.5}

    -\frac{17.5}{17.5}\int\frac{1}{17.5w-C}\,dw=\frac{1}{3500}\int\,dt

    -\frac{1}{17.5}\int\frac{17.5}{17.5w-C}\,dw=\frac{1}{3500}\int\,dt

    -\frac{2}{35}\int\frac{17.5\,dw}{17.5w-C}=\frac{1}{3500}\int\,dt

    Now, observe that the integral on the left is of the form \int\frac{du}{u}=\ln|u|+c_1...can you finish?
    I think so...However, I don't understand why \frac {1}{17.5} became \frac{2}{35}. I realize they're equivalent, but is there a reason for doing it?

    -\frac{1}{-17.5}\int\frac{17.5}{17.5w-C}\,dw=\frac{1}{3500}\int\,dt

    = ln |17.5w - C| = \frac {\frac{t}{3500}}{\frac{1}{-17.5}}

    = ln |17.5w - C| = -0.005t

    = 17.5w - C = e^{-0.005t}

    = w = \frac {e^{-0.005t} + c} {17.5}

    Is this right? When I ran it through Wolfram, it gave me a different answer, which I just can't see how it got to that point..
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  5. #5
    MHF Contributor MarkFL's Avatar
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    Re: Differentials and Integrating

    You forgot to include a parameter representing the constant of integration.
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