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Math Help - nature of turning points

  1. #1
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    nature of turning points

    dy/dx = (cos t - sin t)/(sin t - cos t)

    The second derivative is 0.

    I am trying to find the nature of turning points but the second derivative can't be used and a nature table can't be made because dy/dx always equals -1 for any value of t.
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  2. #2
    Super Member Quacky's Avatar
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    Re: nature of turning points

    \frac{dy}{dx}=-1

    When \frac{dy}{dx}=0, you have turning points. What can you deduce?
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    Re: nature of turning points

    The turning points are where 0=cos t - sin t. I already have the points. How do I find their nature?
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    Re: nature of turning points

    Quote Originally Posted by Stuck Man View Post
    The turning points are where 0=cos t - sin t. I already have the points. How do I find their nature?
    You clearly didn't read Quacky's Post.

    Notice that \displaystyle \begin{align*} \frac{\cos{t} - \sin{t}}{\sin{t} - \cos{t}} &= \frac{-\left(\sin{t} - \cos{t}\right)}{\sin{t} - \cos{t}} \\ &= -1 \end{align*}

    So the derivative is \displaystyle \begin{align*} -1 \end{align*} . You only have stationary points when the derivative is \displaystyle \begin{align*} 0 \end{align*} . What does this tell you about the function?
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  5. #5
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    Re: nature of turning points

    The curve is a spiral and it does have tuning points between 0<=t<=2pi. It is x=e^tsint, y=e^tcost.
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    Re: nature of turning points

    The bottom line of dy/dx should have a plus symbol. I have finished this question now, thanks.
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    Re: nature of turning points

    Quote Originally Posted by Stuck Man View Post
    The bottom line of dy/dx should have a plus symbol. I have finished this question now, thanks.
    This is why we always advise everyone to post the ENTIRE question, so that our correct guidance isn't completely useless.
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