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Math Help - Parametric equations?

  1. #1
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    Parametric equations?

    Find the area of the graph enclosed by the parametric equations,

    y = 6 cos(t) sin(t)

    x= 6 cos^2(t)




    Thus far I did some algebra and came up with 6x^2+6y^2= 36y, but am stuck after that.

    Thank you!
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  2. #2
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    Re: Parametric equations?

    Quote Originally Posted by orange View Post
    Find the area of the graph enclosed by the parametric equations,
    y = 6 cos(t) sin(t)
    x= 6 cos^2(t)
    Thus far I did some algebra and came up with 6x^2+6y^2= 36y, !
    Can you show 6x^2+6y^2=36y is the same as x^2+(y-3)^2=9~?

    Now I did not check your algebra.
    Last edited by Plato; December 16th 2013 at 12:26 PM.
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  3. #3
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    Re: Parametric equations?

    Quote Originally Posted by orange View Post
    Find the area of the graph enclosed by the parametric equations,

    y = 6 cos(t) sin(t)

    x= 6 cos^2(t)




    Thus far I did some algebra and came up with 6x^2+6y^2= 36y, but am stuck after that.

    Thank you!
    Did you plot the curve? The answer is pretty obvious if you do.

    To do it formally though

    y^2(t)=36cos^2(t)sin^2(t)=6x(t)(1-\frac{x(t)}{6})

    dropping the t for convenience

    y^2=6x(1-\frac{x}{6})=6x-x^2

    y=\pm\sqrt{6x-x^2}

    so find the area under y(x)=\sqrt{6x-x^2} and double it to include the area under the -\sqrt{6x-x^2}

    This will all be much clearer if you plot these functions.

    As another trickier way of doing this

    y^2(t)=36cos^2(t)sin^2(t)=36cos^2(t)(1-cos^2(t))

    y^2=6x(1-\frac{x}{6})=6x-x^2=-(x^2-6x+9-9)=-(x-3)^2+9

    (x-3)^2+y^2=9

    This is immediately recognized as a circle of radius 3 centered at (3,0).

    It's area is \pi(3)^2=9\pi
    Last edited by romsek; December 16th 2013 at 01:06 PM.
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  4. #4
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    Re: Parametric equations?

    Hello, orange!

    Find the area of the area enclosed by the parametric equations: . \begin{Bmatrix}x &=& 6\cos^2\!t \\ y &=&6\cos t\sin t \end{Bmatrix}

    Square both equations: . \begin{Bmatrix}x^2 &=& 36\cos^4\!t \\ y^2 &=& 36\cos^2\!t\sin^2\!t \end{Bmatrix}

    Add: . . . . . x^2+y^2 \;=\;36\cos^4\!t + 36\cos^2\!t\sin^2\!t

    . . . . . . . . . x^2+y^2\;=\;36\cos^2\!t\underbrace{(\cos^2\!t + \sin^2\!t)}_{\text{This is 1}}

    . . . . . . . . . x^2 + y^2\;=\;36\cos^2\!t

    . . . . . . . . . x^2+y^2 \;=\;6(6\cos^2\!t)

    . . . . . . . . . x^2 + y^2 \;=\;6x

    . . . . . x^2 - 6x + y^2 \;=\;0

    . . x^2 - 6x + 9 + y^2 \:=\:9

    . . . . (x-3)^2 + y^2 \;=\;9


    This is a circle with radius 3.
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  5. #5
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    Re: Parametric equations?

    Thank you for all these wonderful explanations!
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