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Math Help - polar/cartesian curves

  1. #1
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    polar/cartesian curves

    For each of the described curves, decide if the curve would be more easily given by a polar or a Cartesian equation. Then write an equation for the curve:

    1. A circle with radius 5 and center (2,3).

    2. A circle centered at the origin with radius 4.

    A length of curve problem:

    Find the length of the curve x = 3cos(t) - cos(3t), y = 3sin(t) - sin(3t) for
    0 <= t <= pi.

    Thanks.
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  2. #2
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    Hello, eigenvector11!

    For each of the described curves, decide if the curve would
    be more easily given by a polar or a Cartesian equation.
    Then write an equation for the curve:

    1. A circle with radius 5 and center (2,3).
    Cartesian: . (x-2)^2 + (y-3)^2 \:=\:25


    2. A circle centered at the origin with radius 4.
    Polar: . r \:=\:4


    A length of curve problem:
    Find the length of the curve: . \begin{Bmatrix}x &= & 3\cos(t) - \cos(3t) \\ y & = & 3\sin(t) - \sin(3t)\end{Bmatrix}\quad\text{ for }0 \leq  t \leq \pi
    Formula: . L \;=\;\int^b_a\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt

    We have: . \frac{dx}{dt} \:= \: -3\sin(t) + 3\sin(3t) \:=\:-3[\sin(t) - \sin(3t)]

    . . . . . . . . \frac{dy}{dt} \: = \: 3\cos(t) - 3\cos(3t) \: = \:  3[\cos(t) - \cos(3t)]


    Then: . \left(\frac{dx}{dt}\right)^2\;=\;9\left[\sin^2(t) - 2\sin(t)\sin(3t) + \sin^2(3t)\right]

    . . . . . \left(\frac{dy}{dt}\right)^2 \;=\;9\left[\cos^2(t) - 2\cos(t)\cos(3t) + \cos^2(3t)\right]


    \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 \;= . 9\left\{[\underbrace{\sin^2(t) + \cos^2t}_{\text{This is 1}} - 2\underbrace{[\cos(t)\cos(3t) + \sin(t)\sin(3t)]}_{\text{This is }\cos(2t)} + \underbrace{\sin^2(3t) + \cos^2(3t)}_{\text{This is 1}} ]\right\}

    . . . . . . = \;9\left[2 - 2\cos(2t)\right] \;=\;9\left[2 - 2(1 - 2\sin^2t)\right] \;=\;9\cdot4\sin^2t


    Hence: . \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \;=\;\sqrt{36\sin^2t} \;=\;6\sin(t)


    Therefore: . L \;=\;6\int^{2\pi}_0 \sin(t)\,dt



    Since this definite integral comes out to zero,
    . . I suspect the graph has symmetry about the origin.
    I'll let you work out the details . . .
    .
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