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Math Help - Finals. Quick polar questions

  1. #1
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    Finals. Quick polar questions

    Alright. Thank you very much for your help in advance. I went thru all these problems myself and had questions on these particular ones. They shouldn't be too bad.

    1) Convert X^2-y^2=1 to Polar...
    .... so it will = (r^2)(cos^2theta) - (r^2)(sin^2theta) =1
    r= sqrt(1/(cos^2theta-sin^2theta))
    where from here?

    2) Convert r=3sintheta to cartesian.
    would x^2 + y^2 = 3y be correct?

    3) What is the area of r^2=4cos2theta ??

    4) What is the area of the region that lies inside the first curve and outside the second curve
    curve one = r=3costheta. curve 2 = r= 1+costheta

    5) What is the area of the region that lies inside both curves
    1.st = r=sin2theta second = r=sintheta.

    6) Find the solutions of the equations (now into complex number stuff) ..
    2x^2 -2x +1 =0

    7) Converting from complex numbers to polar form...
    does -3+3i = (Sqrt(18))cis(tan(-1))

    8) lastly, can someone provide a problem which wants me to convert from polar to complex? like... how rcostheta = a and risintheta = b



    THANK YOU! If you answer these questions it will help my studying a LOT!
    I just went 4 hrs studying other math stuff.
    THANK YOU again!
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  2. #2
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    4) What is the area of the region that lies inside the first curve and outside the second curve
    curve one = r=3costheta. curve 2 = r= 1+costheta
    You can find the region in the first quadrant and multiply by 2 because of symmetry.

    3cos(t)=1+cos(t)

    Solving this tells us that they intersect at Pi/3.

    \int_{0}^{\frac{\pi}{3}}\left[(3cos(t))^{2}-(1+cos(t))^{2}\right]dt
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  3. #3
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    Galactus, you sure about the starting and the ending? If i remember correctly one of those lines takes 2pi to make a full cycle and the other one 1pi ?
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  4. #4
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    Quote Originally Posted by 3deltat View Post
    Galactus, you sure about the starting and the ending? If i remember correctly one of those lines takes 2pi to make a full cycle and the other one 1pi ?
    Galactus is correct. Draw the two curves (circle and cardioid) and see it for yourself.
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  5. #5
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    Quote Originally Posted by 3deltat View Post
    Alright. Thank you very much for your help in advance. I went thru all these problems myself and had questions on these particular ones. They shouldn't be too bad.

    1) Convert X^2-y^2=1 to Polar...
    .... so it will = (r^2)(cos^2theta) - (r^2)(sin^2theta) =1
    r= sqrt(1/(cos^2theta-sin^2theta))
    where from here?

    Mr F says: Use a double angle formula ....

    2) Convert r=3sintheta to cartesian.
    would x^2 + y^2 = 3y be correct?

    Mr F says: Yes. Myself, I'd then write it in the form x^2 + (y - k)^2 = r^2 ....

    3) What is the area of r^2=4cos2theta ??

    Mr F says: See equations (13) - (15) of Lemniscate -- from Wolfram MathWorld (you should think about why integrate from -pi/4 to pi/4).

    4) What is the area of the region that lies inside the first curve and outside the second curve
    curve one = r=3costheta. curve 2 = r= 1+costheta

    Mr F says: Done by Galactus.

    5) What is the area of the region that lies inside both curves
    1.st = r=sin2theta second = r=sintheta.

    Mr F says: Draw a diagram. Think about how Galactus did the previous one.

    6) Find the solutions of the equations (now into complex number stuff) ..
    2x^2 -2x +1 =0

    Mr F says: Use the quadratic formula.

    7) Converting from complex numbers to polar form...
    does -3+3i = (Sqrt(18))cis(tan(-1))

    Mr F says: Draw an Argand diagram. The argument is clearly 3pi/4.

    8) lastly, can someone provide a problem which wants me to convert from polar to complex? like... how rcostheta = a and risintheta = b

    Mr F says: You mean convert from polar form to Cartesian form. Go to any textbook that covers this material. Or use google.

    [snip]
    ..
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