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Math Help - Finding dy/dx for parametric equations

  1. #1
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    Finding dy/dx for parametric equations

    Consider the parametric curve x = [1+cos(t)]cos(t), y = [1+sin(t)]cos(t).

    Find dy/dx for a general t.

    I have done the work and gotten:
    dy/dx = y'(t)/x'(t) = [cos(t) - 1]/[-sin(t) + (1 + 2 cos(t))]

    To me, this seems too long, but I can't see any way of simplifying it. If there is a way of simplifying it, could someone point it out to me. Thanks.
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  2. #2
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    Quote Originally Posted by kiddopop View Post
    Consider the parametric curve x = [1+cos(t)]cos(t), y = [1+sin(t)]cos(t).

    Find dy/dx for a general t.

    I have done the work and gotten:
    dy/dx = y'(t)/x'(t) = [cos(t) - 1]/[-sin(t) + (1 + 2 cos(t))]

    To me, this seems too long, but I can't see any way of simplifying it. If there is a way of simplifying it, could someone point it out to me. Thanks.
    recheck your derivatives ...

    x = (1+\cos{t})\cos{t} = \cos{t} + \cos^2{t}

    x' = -\sin{t} - 2\cos{t}\sin{t} = -\sin{t}(1 + 2\cos{t})<br />

    y = (1 + \sin{t})\cos{t}

    y' = (1 + \sin{t})(-\sin{t}) + \cos^2{t} = -sin{t} - \sin^2{t} + \cos^2{t} = 1 - \sin{t} - 2\sin^2{t} = (1 - 2\sin{t})(1 + \sin{t})

    \frac{dy}{dx} = \frac{y'}{x'} = \frac{(2\sin{t}-1)(1 + \sin{t})}{\sin{t}(1 + 2\cos{t})}

    that's about all you can do with this one.
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