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Math Help - dy/dx for paramatic

  1. #1
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    Post dy/dx for paramatic

    Hey All,

    find dy/dx

    where y(t)=sin(t+7), x(t)=tln|t|

    I'm having trouble with the |t| get abit confusing. hope someone can help. Thank you
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by dadon View Post
    Hey All,

    find dy/dx

    where y(t)=sin(t+7), x(t)=tln|t|

    I'm having trouble with the |t| get abit confusing. hope someone can help. Thank you
    d/dt(|t|) = {1 for t >= 0 and -1 for t < 0}, so you need to split your real line into intervals.

    dx/dt = ln|t| + t*(1/t)*1

    -Dan

    (Corrected version! Sorry about the mistake.)
    Last edited by topsquark; April 30th 2007 at 12:24 PM. Reason: Made an oopsie!
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  3. #3
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    re:

    are you applying the chain rule?

    sorry im still abit confused
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    Re:

    RE:

    No for

    x(t)=tln|t| you would use the product rule....

    f'(x)*g(x)+f(x)*g'(x)
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    RE:

    No for

    x(t)=tln|t| you would use the product rule....

    f'(x)*g(x)+f(x)*g'(x)
    i think he was talking about using the chain rule to find dy/dx. if so, then the answer is yes. you would use dy/dx = dy/dt * dt/dx
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  6. #6
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    I disagree with Danís derivative of (t)[ln(|t|)], maybe he misread it.
    Yes, we must use the chain rule.
    Attached Thumbnails Attached Thumbnails dy/dx for paramatic-dydy.gif  
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    Quote Originally Posted by topsquark View Post
    d/dt(|t|) = {1 for t >= 0 and -1 for t < 0}, so you need to split your real line into intervals.
    Or you can use the fact that,

    (|x|) ' = sgn (x) for x!=0
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  8. #8
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    Quote Originally Posted by ThePerfectHacker View Post
    Or you can use the fact that,
    (|x|) ' = sgn (x) for x!=0
    Yes, that is true enough.
    But the point is: [ln(|x|)]'=(1/x), x!=0.
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  9. #9
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    Post

    thank you thats great guys
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  10. #10
    Senior Member ecMathGeek's Avatar
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    Are you sure? If they go un-interrupted, I bet the people on this forum could go on for another 12 pages on how to take the derivative of |x|.
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  11. #11
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    Quote Originally Posted by ecMathGeek View Post
    Are you sure? If they go un-interrupted, I bet the people on this forum could go on for another 12 pages on how to take the derivative of |x|.
    lol

    yeah i was thinking this question got quite a good response.
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  12. #12
    TD!
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    And if you don't see right away why the derivative of |x| would be sgn(x) (for x nonzero), rewrite |x| = sqrt(x≤). Now take the derivative:

    1/(2.sqrt(x≤) * 2x = x/sqrt(x≤) = x/|x| = sgn(x), by definition.
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