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Thread: parametric dy/dx

  1. #1
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    parametric dy/dx

    Hi
    Can somebody tell me if i am correct?

    find dy/dx in terms of t for the curve with the parametric representation:

    x=(1-t)(1+3t) y=(1+t)(1-3t)

    My solution;

    x=1 + 2t - 3t^2 y= 1 - 2t - 3t^2

    dx/dt= 2-6t dy/dt= -2-6t


    dy/dx=(dy/dt)/ (dx/dt)

    dy/dx=(-2-6t)/(2-6t)

    Can somebody tell me if this is correct-and can somebody recomend a good website with more parametric dy/dx examples-my maths book is not very good.
    thanks
    John
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: parametric dy/dx

    I think it's correct!
    Sorry I don't know any site with more exercices.
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  3. #3
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    Re: parametric dy/dx

    thanks Siron-yet again?

    The problem with my maths book is he examples are really basic and the self tests at the end of the chapters are really difficult. Not all the answers are available.

    thanks for your help
    John
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: parametric dy/dx

    You're welcome! You can create exercices by yourself if you want and make them more difficult .
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  5. #5
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    Re: parametric dy/dx

    You could, of course, cancel. dy/dx= (-1-3t)/(1-3t) might be considered a better answer.
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