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Math Help - finding minimal point

  1. #1
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    finding minimal point

    I'm having trouble with this question...

    To find the minimal point, is there an easier way other than plugging in all possible values?

    Compute the polar coordinates and the xy-coordinates for the point where y is maximal
    on the graph of the polar coordinate equation r = 1 + \cos(\theta).
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  2. #2
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    Quote Originally Posted by centenial View Post
    I'm having trouble with this question...

    To find the minimal point, is there an easier way other than plugging in all possible values?

    Compute the polar coordinates and the xy-coordinates for the point where y is maximal
    on the graph of the polar coordinate equation r = 1 + \cos(\theta).
    maximum y or minimum?

    if maximum ...

    r = 1 + \cos{t}

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

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

    \frac{dy}{dt} = \sin{t}(-\sin{t}) + (1 + \cos{t})\cos{t}

    simplified ...

    \frac{dy}{dt} = 2\cos^2{t} + \cos{t} - 1<br />

    set \frac{dy}{dt} = 0 and determine the angle t that maximizes y ...

    I get t = \frac{\pi}{3}
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  3. #3
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    Sorry, yes, I meant maximum.

    Could you explain why you set dy/dt = 0.. is that because when dy/dt = 0, then the slope is horizontal, and we must be at a maximum point? I just want to make sure I understand.

    If I were asked to find the minimum, would I first multiply by cos(t) and then try to calculate dx/dt? Would I also set that to 0 to find the minimal point?

    Thanks for your help!
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  4. #4
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    Quote Originally Posted by centenial View Post
    Sorry, yes, I meant maximum.

    Could you explain why you set dy/dt = 0.. is that because when dy/dt = 0, then the slope is horizontal, and we must be at a maximum point? I just want to make sure I understand.

    If I were asked to find the minimum, would I first multiply by cos(t) and then try to calculate dx/dt? Would I also set that to 0 to find the minimal point?

    Thanks for your help!
    \frac{dy}{d\theta} is not the slope of the polar curve ... the idea was to get y as a function of \theta and find the value of \theta that maximizes y , just as you would with any other optimization problem.
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