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Math Help - Polarization online hw help

  1. #1
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    Polarization online hw help

    Use the given equation to answer the following questions. (Enter your answers from smallest to largest.)
    r2 = sin(2θ)
    0 ≤ θ < π

    (a) Find the points on the given curve where the tangent line is horizontal.


    (b) Find the points on the given curve where the tangent line is vertical.
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  2. #2
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    Hello, fogel1497!

    We are expected to know (or be able to derive) this formula:

    . . \frac{dy}{dx} \;=\;\frac{r\cos\theta + r'\sin\theta}{-r\sin\theta + r'\cos\theta} .[1]



    Given: . r^2 \:=\:\sin2\theta\quad 0 \leq \theta < 2\pi
    Differentiate implicitly: . 2r\!\cdot\!r' \:=\:2\cos2\theta \quad\Rightarrow\quad r' \:=\:\frac{\cos2\theta}{r} .[2]

    Substitute [2] into [1]: . \frac{dy}{dx} \;=\;\frac{r\cos\theta + \left(\frac{\cos2\theta}{r}\right)\sin\theta}{-r\sin\theta + \left(\frac{\cos2\theta}{r}\right)\cos\theta}

    Multiply top and bottom by r\!:\;\;\frac{dy}{dx} \;=\;\frac{r^2\cos\theta + \sin\theta\cos2\theta}{-r^2\sin\theta + \cos\theta\cos2\theta}

    Since r^2 = \sin2\theta we have: . \frac{dy}{dx} \;=\;\frac{\sin2\theta\cos\theta + \sin\theta\cos2\theta}{-\sin2\theta\sin\theta + \cos\theta\cos2\theta} . = \;\frac{\sin(2\theta + \theta)}{\cos(2\theta + \theta)}

    . . and we have: . \frac{dy}{dx} \;=\;\frac{\sin3\theta}{\cos3\theta}




    (a) Find the points on the curve where the tangent is horizontal.
    The tangent line is horizontal when \frac{dy}{dx} \,=\,0

    So we have: . \sin3\theta \:=\:0\quad\Rightarrow\quad 3\theta \:=\:0,\:\pi,\:2\pi,\:3\pi,\:4\pi,\:5\pi

    Horizontal tangents when: . x\;=\;0,\:\frac{\pi}{3},\:\frac{2\pi}{3},\:\pi,\:\  frac{4\pi}{3},\:\frac{5\pi}{3}




    (b) Find the points on the curve where the tangent is vertical.
    The tangents line is vertical when \frac{dy}{dx} is undefined.

    So we have: . \cos3\theta \:=\:0 \quad\Rightarrow\quad 3\theta \;=\;\frac{\pi}{2},\:\frac{3\pi}{2},\:\frac{5\pi}{  2},\:\frac{7\pi}{2},\:\frac{9\pi}{2},\:\frac{11\pi  }{2}

    Vertical tangents when: . x \;=\;\frac{\pi}{6},\:\frac{\pi}{2},\:\frac{5\pi}{6  },\:\frac{7\pi}{6},\:\frac{3\pi}{2},\:\frac{11\pi}  {6}


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  3. #3
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    Thank you, now that you explained it i was able to complete the rest of my homework. i missed this day in class, this stuff is actually a lot easier then the trig sub and other stuff we were doing before. thanks a lot! +rep
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