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Thread: 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 $\displaystyle 6x^2+6y^2=36y$ is the same as $\displaystyle x^2+(y-3)^2=9~?$

    Now I did not check your algebra.
    Last edited by Plato; Dec 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

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

    dropping the t for convenience

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

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

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

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

    As another trickier way of doing this

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

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

    $\displaystyle (x-3)^2+y^2=9$

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

    It's area is $\displaystyle \pi(3)^2=9\pi$
    Last edited by romsek; Dec 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: .$\displaystyle \begin{Bmatrix}x &=& 6\cos^2\!t \\ y &=&6\cos t\sin t \end{Bmatrix}$

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

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

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

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

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

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

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

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

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


    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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