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Math Help - Parametric Equations Horzontal Tangents

  1. #1
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    Parametric Equations Horzontal Tangents

    I have a question that is confusing me.

    Suppose the curve is traced by the parametric equations:
    x = 3sin(t)
    y = 25-6cos^2(t)-12sin(t)
    At what point (x,y) on this curve is the tangent line horizontal??

    I know that I need to find (dy/dt)/(dx/dt)
    I think that the derivative of y is messing me up to start with.
    Then I know I need to find the points for t and sub them into the x and y equations given but I am not getting the correct answers.
    Can anyone help?
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  2. #2
    Moo
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    Hello,

    A tangent line is horizontal when the derivative is 0 (the derivate number is the slope of the tangent).

    So you have to solve for t as you said.

    For the derivative of y, let's do it slightly :

    y = 25-6\cos^2(t)-12\sin(t)

    We know that :

    - (u+v)'=u'+v'

    - the derivative of a constant is 0.

    - \forall \lambda \in \mathbb{R}, \ (\lambda u(x))'=\lambda u'(x)

    - (u^n(x))'=n \ u'(x) \ u^{n-1}(x) or here, what's going to interest you is : (u^2(x))'=2u'(x)u(x)

    - the derivate of cos is -sin and the derivate of sin is cos


    I think you got all you need here =)
    If you still have difficulties show where there is a problem
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