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Thread: 2 questions on parametric and implicit differntiations

  1. #1
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    2 questions on parametric and implicit differntiations

    Ok the curve has parametric equations:

    x = cos t y = sin 2t, 0 ≤ t < 2pi

    a) Find an expression for dy/dx in terms of parameter t
    b) Find the values of the parameter t at the points where dy/dx = 0

    An example in my book has confused me and so i was wondering if somebody could show me how to work through these two questions.
    Thanks
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  2. #2
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    Quote Originally Posted by sharp357 View Post
    the curve has parametric equations:

    x = cos t y = sin 2t, 0 ≤ t < 2pi

    a) Find an expression for dy/dx in terms of parameter t
    b) Find the values of the parameter t at the points where dy/dx = 0

    thanks
    a) note that $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

    b) set $\displaystyle \frac{dy}{dx}$ found in (a) equal to 0 and solve for t.
    Last edited by mr fantastic; Feb 11th 2009 at 05:27 PM. Reason: No edit - just flagging it as having been moved from a duplicate post
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    Quote Originally Posted by sharp357 View Post
    Ok the curve has parametric equations:

    x = cos t y = sin 2t, 0 ≤ t < 2pi

    a) Find an expression for dy/dx in terms of parameter t
    First find $\displaystyle y'$ then find $\displaystyle x'$ and to get $\displaystyle \tfrac{dy}{dx}$ form the fraction $\displaystyle \frac{y'}{x'}$.

    b) Find the values of the parameter t at the points where dy/dx = 0
    You want to solve $\displaystyle \frac{dy}{dx} = 0 \implies \frac{y'}{x'} = 0 \implies y' = 0$.
    Can you finish?
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    ok i differentiated so dy/dx = cos 2t / - sin t

    just wondering if this part is right?
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  5. #5
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    $\displaystyle y = \sin(2t)$

    chain rule ...

    $\displaystyle \frac{dy}{dt} = 2\cos(2t)$

    $\displaystyle \frac{dx}{dt} = -\sin{t}$ is correct
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