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Math Help - Parametric Calc

  1. #1
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    Parametric Calc

    Given that y=x^2-x, {dx}/{dt}>0 and that the particle is traveling at a constant 2\sqrt{10}, what is dy/dt?

    So far I've used a formula for speed as \sqrt{(dy/dt)^2+(dx/dt)^2} and set it equal to (40)^(1/2). Any help would be hot.

    Thanks.
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  2. #2
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    Re: Parametric Calc

    perhaps I should learn how to type those math symbols in text.
    Sorry for bad handwriting, and I was lazy to simplify the last expression. I hope its right )Parametric Calc-untitled.png
    Thanks from Xeritas
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  3. #3
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    Re: Parametric Calc

    Quote Originally Posted by pyromania View Post
    perhaps I should learn how to type those math symbols in text.
    Sorry for bad handwriting, and I was lazy to simplify the last expression. I hope its right )Click image for larger version. 

Name:	Untitled.png 
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ID:	23502
    It's right. I forgot to mention that the point of interest is the rectangular point (2,2), and a friend of mine mentioned the answer was 6, which is concurrent with what you said. No worries about your handwriting either. Thanks again!
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  4. #4
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    Re: Parametric Calc

    Try this
    dy/dt =dy/dx times dx/dt=(2x-1)dx/dt
    So (2x-1)^2(dx/dt)^2+(dx/dt)^2=40
    Now put x=2
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  5. #5
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    Re: Parametric Calc

    Quote Originally Posted by biffboy View Post
    Try this
    dy/dt =dy/dx times dx/dt=(2x-1)dx/dt
    So (2x-1)^2(dx/dt)^2+(dx/dt)^2=40
    Now put x=2
    Don't put x= 2 until after you have solved the differential equation!
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  6. #6
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    Re: Parametric Calc

    We are not solving a differential equation we just want dx/dt when x=2. So I think x=2 can be substituted as I said. I get the answer dx/dt=2
    So dy/dt=3.2=6
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