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Math Help - Find The Critical Points

  1. #1
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    Find The Critical Points

    Using derivatives

    for the function where mgu is a constant and if u = 0.2


    \frac{mgu}{sin\theta+ucos\theta}
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    Quote Originally Posted by sk8erboyla2004 View Post
    Using derivatives

    for the function where mgu is a constant and if u = 0.2


    \frac{mgu}{sin\theta+ucos\theta}
    critical points occur where the derivative is zero or undefined.

    so first find your derivative, then you can see where it is undefined. that will be a critical point. then set it equal to zero and solve for \theta, to find the other critical points (what is your dependent variable?)

    to start you off, write as mgu( \sin \theta + u \cos \theta )^{-1}.

    now use the chain rule (remember how to deal with constants)
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    would I consider

    ucos\theta

    A derivative I would need to use the product rule for?

    Ok So I get

    \frac{mgu(cos\theta-usin\theta)}{(sin\theta+ucos\theta)^2}

    So I solve the denom. for 0?

    So if that is correct then

    sin\theta+ucos\theta=0
    ucos\theta=-sin\theta
    -u=\frac{sin\theta}{cos\theta}

    Then it stats u = 0.2 so

    -0.2=tan\theta
    arctan(-0.2)=-.197

    which is the critical pt.?
    Last edited by sk8erboyla2004; April 29th 2009 at 10:03 PM.
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    Quote Originally Posted by sk8erboyla2004 View Post
    would I consider

    ucos\theta

    A derivative I would need to use the product rule for?

    Ok So I get

    \frac{mgu(cos\theta-usin\theta)}{(sin\theta+ucos\theta)^2}
    .
    .
    .
    it should be \frac{mgu( u \sin \theta - \cos\theta)}{(\sin\theta+u \cos\theta)^2}

    then, set BOTH the numerator and denominator equal to zero to find all critical points.
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    Quote Originally Posted by Jhevon View Post
    it should be \frac{mgu( u \sin \theta - \cos\theta)}{(\sin\theta+u \cos\theta)^2}

    then, set BOTH the numerator and denominator equal to zero to find all critical points.

    Shouldnt this \frac{mgu( u \sin \theta - \cos\theta)}{(\sin\theta+u \cos\theta)^2}

    be

    -\frac{mgu( u \sin \theta - \cos\theta)}{(\sin\theta+u \cos\theta)^2}

    Just wanted to make sure?
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    Quote Originally Posted by sk8erboyla2004 View Post
    Shouldnt this \frac{mgu( u \sin \theta - \cos\theta)}{(\sin\theta+u \cos\theta)^2}

    be

    -\frac{mgu( u \sin \theta - \cos\theta)}{(\sin\theta+u \cos\theta)^2}

    Just wanted to make sure?
    \frac d{d \theta} mgu (\sin \theta + u \cos \theta)^{-1} = -mgu (\sin \theta + u \cos \theta)^{-2}(\cos \theta - u \sin \theta) = mgu (\sin \theta + u \cos \theta)^{-2}(- \cos \theta + u \sin \theta) = \frac {mgu(u \sin \theta - \cos \theta)}{(\sin \theta + u \cos \theta)^2}
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  7. #7
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    so you really just distrbuted the negative in the numer. at rewrote it
    Ok well for my critical points I get

    x = 1.373
    x = -.197
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    Quote Originally Posted by sk8erboyla2004 View Post
    so you really just distrbuted the negative in the numer. at rewrote it
    yes

    Ok well for my critical points I get

    x = 1.373
    x = -.197
    yes

    just finding two solutions is enough?
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    Well It said find all critcal points and I assume 2 would be good enough?? sorry for being confusing lol
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    Quote Originally Posted by sk8erboyla2004 View Post
    Well It said find all critcal points and I assume 2 would be good enough?? sorry for being confusing lol
    the thing is, you are dealing with trig functions. there are infinitely many solutions. are you sure they want only two?
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