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Math Help - Need hints to solve antiderivative

  1. #1
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    Need hints to solve antiderivative

    OK I have a slightly new problem that seems much more difficult.

    X'(t) = \mu_{k}gt + \frac{A}{4m} \int X'(t)^2dt


    constants:
    \mu_{k} = kinetic friction coefficient
    g = gravity
    A = cross sectional area
    m = mass

    I am trying to solve X(t) and then solve the specific function as an initial value problem.

    I just need help solving X'(t) for now. I should be OK after that.

    Thanks


    This is for a hobby physics project, not for homework.
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  2. #2
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    Quote Originally Posted by guitardude View Post
    OK I have a slightly new problem that seems much more difficult.

    X'(t) = \mu_{k}gt + \frac{A}{4m} \int X'(t)^2dt


    constants:
    \mu_{k} = kinetic friction coefficient
    g = gravity
    A = cross sectional area
    m = mass

    I am trying to solve X(t) and then solve the specific function as an initial value problem.

    I just need help solving X'(t) for now. I should be OK after that.

    Thanks


    This is for a hobby physics project, not for homework.
    Differentiate both sides with respect to t. Then make the substitution U = X'. Solve the resulting DE for U. Hence solve for X.
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  3. #3
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    I am still confused. If I differentiate both sides I will get X''(t) right? which is what I started with. This equation is my work thus far on taking the antiderivative of X''(t).


    If I substitute u=X' then I get \int u^2 The antiderivative of this part isn't as simple as \frac{1}{3}u^3 is it? Even so I would end up with X(t) = constant*t^2 - constant* X'(t)^3 + C and I don't know X'(t).





    I forgot two negative signs, not that it matters much here but it should be:

    X'(t) = -\mu_{k}gt - \frac{A}{4m} \int X'(t)^2dt


    I previously solved the initial value problem for constant acceleration as a differential equation:

    X''(t) = B
    so X'(t) = Bt + C
    so X(t) = Bt^2 + Ct + D
    which is our familiar physics equation S = S_{i} + V_{i}t  + \frac{1}{2} a(t^2)

    You can the solve the specific function and find B,C,D given 3 X(t) samples.

    Now, instead of assuming a constant acceleration, I am trying to solve this with  B = -\mu_{k}g - \frac{1}{4m}Av^2 and that is how I got X''(t) = -\mu_{k}g - \frac{1}{4m}Av^2 =   -\mu_{k}g - \frac{1}{4m}AX'(t)^2

    It is not obvious to me yet how many X(t) samples I will need to solve this as an initial value problem?
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    Differentiate both sides with respect to t. Then make the substitution U = X'. Solve the resulting DE for U. Hence solve for X.
    Differentiate both sides wrt t: X'' = \mu_k g + \frac{A}{4m} \, (X')^2.

    Substitute U = X': U' = \mu_k g + \frac{A}{4m} \, U^2 \Rightarrow \frac{dU}{dt} - \frac{A}{4m} \, U^2 = \mu_k g.

    Solve this DE for U: I'd suggest using the integrating method.

    Then solve \frac{dX}{dt} = U for X.
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  5. #5
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    I worked through it with my differential equations prof and got to:

    X'(t) = \frac{-\sqrt{\mu_{k}g}}{\sqrt{\frac{A}{4m}}} * \tan((\sqrt{\mu_{k}g})(\sqrt{\frac{A}{4m}})((t-C_{1})))

    I verified the previous equation against simulations as well as with mathematica. Next I used matlab and then simplified to get back to :

    X(t) = \frac{-4m}{A} * \ln(\sec^2((\sqrt{\mu_{k}g})(\sqrt{\frac{A}{4m}})(  t-C_{1}))) + C_{2}

    With the two unknowns, C_{1}, C_{2}

    I was expecting it to have a single \sec not sec^2 ?

    This will be messy to solve as an initial value problem but possible.


    Is there any way to make this function continuous? Tan() and Sec() make these hard to use and I don't see why it is necessary.
    Last edited by guitardude; September 4th 2008 at 10:22 PM.
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  6. #6
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    Quote Originally Posted by guitardude View Post
    I worked through it with my differential equations prof and got to:

    X'(t) = \frac{-\sqrt{\mu_{k}g}}{\sqrt{\frac{A}{4m}}} * \tan((\sqrt{\mu_{k}g})(\sqrt{\frac{A}{4m}})((t-C_{1})))

    I verified the previous equation against simulations as well as with mathematica. Next I used matlab and then simplified to get back to :

    X(t) = \frac{-4m}{A} * \ln(\sec^2((\sqrt{\mu_{k}g})(\sqrt{\frac{A}{4m}})(  t-C_{1}))) + C_{2}

    With the two unknowns, C_{1}, C_{2}

    I was expecting it to have a single \sec not sec^2 ?

    This will be messy to solve as an initial value problem but possible.


    Is there any way to make this function continuous? Tan() and Sec() make these hard to use and I don't see why it is necessary.
    I can't quite understand how you've written the solution for X(t) but it looks like you have a log of a product. So use the usual log rule to break this up into a sum of logs.

    Then it looks like one of those logs has the form \log \sec^2 \theta. Using the usual log rule you get \log \sec^2 \theta = \log \cos^{-2} \theta = -2 \log \cos \theta .....
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    I can't quite understand how you've written the solution for X(t) but it looks like you have a log of a product. So use the usual log rule to break this up into a sum of logs.

    Then it looks like one of those logs has the form \log \sec^2 \theta. Using the usual log rule you get \log \sec^2 \theta = \log \cos^{-2} \theta = -2 \log \cos \theta .....

    Hopefully this helps.... It is of form:
    X(t) = D * \ln( \sec^2(E(t-C_{1})) ) + C_{2}


    For the known constants D,E

    I will have to solve C_{1} which I can do given two pairs of corresponding X,t

    Does your suggestion of splitting \ln(\sec^2(\theta)) up still apply?
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  8. #8
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    Quote Originally Posted by guitardude View Post
    Hopefully this helps.... It is of form:
    X(t) = D * \ln( \sec^2(E(t-C_{1})) ) + C_{2}


    For the known constants D,E

    I will have to solve C_{1} which I can do given two pairs of corresponding X,t

    Does your suggestion of splitting \ln(\sec^2(\theta)) up still apply?
    Yes.
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