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Math Help - Parametric and Polar Questions

  1. #1
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    Parametric and Polar Questions

    Reviewing for a test tomorrow, and most of the homework is pretty simple, but I just wanted to ask if you guys would check a few that I'm unsure on.

    Turn this equation from Cartesian to polar.

    x + y = 9

    What I did..

    square both sides,
    x^2 + 2xy + y^2 = 81
    x^2 + y^2 is r^2, so
    r^2 + (2(r)cos(theta) times (r)sin(theta)) = 81
    r^2 (1+ 2costheta sintheta) = 81
    r = 9/sq.root (1+2cos(theta)sin(theta))

    Next one..

    find the area inside of the larger loop and outside of the smaller loop of the limacon r = 1/2 + cos(theta)

    I solved this one through, and got pi/2 + (3root3)/4

    but the answer in the back of the book has pi/4 instead of pi/2.. Could someone just show where I may have missed an extra 1/2 that would change my 1/2 into 1/4?

    this was my integral
    2 integral from 0 to 4pi/6 of [1/2 (1/2 + cos(theta))^2 dtheta]
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  2. #2
    MHF Contributor Calculus26's Avatar
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    In the first why not simply rcos(t) +r sin(t) =9

    then r= 9/[cos(t)+sin(t)] ? By the way there are pblms with your algebra --no need to square both sides

    In the second your integral Calculates the area inside the outer loop

    you have to subtract the area of the inner loop

    2integral 1/2( 2pi/3 to pi) f(theta)^2(dtheta)
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  3. #3
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    alright, I fixed up those two, thanks a lot

    I have one more question, a little easier this time

    find all the points of intersection,

    r=cos(3theta), r = sin(3theta)

    cos3theta = sin3theta
    1 = tan(theta)

    so, 3theta = pi/4, 5pi/4
    so theta = pi/12, 5pi/12

    but when I graph it on my calculator (now allowed for the test, just doing it to check), there is one more. What did I miss?
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  4. #4
    MHF Contributor Calculus26's Avatar
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    I assume you are looking for all solutions from 0 to 2pi. you only missed one?

    This compresses the graph and so you have to extend beyond the usual

    2 solutions . for 3theta go out six

    3theta = pi/4, 5pi/4,9pi/4,13pi/4, 17pi/4, 21pi/4, etc

    t = pi/12,5pi/12,3pi/4,13pi/12,17pi/12,7pi/4 the next one would be 25pi/12

    which is outside the 2pi range.
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  5. #5
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    alright, that makes sense.

    I lied though, it looks like I have one more question..

    find the area between a large loop and the enclosed small lopp of the curve

    r = 1 + 2cos(3theta)

    I graphed it, it's three clovers, and inside each clover is a smaller clover. But how do you determine which bounds to use to get the outside loop, and then just the inside loop?
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  6. #6
    MHF Contributor Calculus26's Avatar
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    For half of a big loop find the zero of 1+ 2 cos(3t)

    Find the next zero --- between these 2 zeroes an entire small loop is generated


    subtract the small loop from the big loop
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