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Buncha polar graph/integral stuff

  1. Nov 16, 2009 #1
    1. The problem statement, all variables and given/known data
    1.

    put in polar form

    x2+y2-3x+4y=0

    my work:
    x2+y2=3x-4y
    r2=3rcos[tex]\theta[/tex]-4rsin[tex]\theta[/tex]
    r=3cos[tex]\theta[/tex]-4sin[tex]\theta[/tex]

    2.
    put in cartesian form

    r2=tan[tex]\theta[/tex]
    r2=y/x
    r=sqrt(y/x)

    3.
    find slope at [tex]\theta[/tex]=[tex]\pi[/tex]/2
    then find points where tangent line is horizontal
    3/(2+2cos[tex]\theta[/tex])

    using the long formula which I don't feel like typing out, I get -1 as the slope when I plug in [tex]\pi[/tex]/2 for theta.

    for the part with the horizontal tangent line, I set the numerator of the derivative to 0, but couldnt get an answer. the graph of the original function seems to be like the graph of +-sqrt(x) reflected about the y axis, like the graph of a backwards c. because the graph continuously increased/decreased as i moved left, there were no places where the slope was 0

    4.
    find the area of one loop of the graph
    2sin(5[tex]\theta[/tex])

    first off, i have one question. what does one loop mean? does it mean simply the graph i get when I graph the graph from 0 to 2 pi? or is one loop just one petal of the graph? 2sin5[tex]\theta[/tex] gives me a graph of a rose curve with 5 petals, so to find the area of one loop would I find the area of 5 petals or just one?

    now that that's out of the way, I'm having a little trouble finding the limits of integration. no idea where to start on this one. do I set the original equation to 0 or something?

    5.
    find the area in common b/w both graphs.
    -6cos[tex]\theta[/tex]
    2-2cos[tex]\theta[/tex]

    the 2 graphs where a circle and a limacon.
    kinda hard to really explain the graph but here's my integral

    i worked with only the part above the x axis so everything is mnultipled by 2

    2[[tex]\int[/tex][[tex]\pi[/tex]/2 to [tex]\pi[/tex]] (1/2)(-6cos[tex]\theta[/tex])2 d[tex]\theta[/tex] - [tex]\int[/tex][2[tex]\pi[/tex]/3 to [tex]\pi[/tex]] (1/2)(2-2cos[tex]\theta[/tex])2 d[tex]\theta[/tex] - [tex]\int[/tex][0 to 2[tex]\pi[/tex]/3] (1/2)(-6cos[tex]\theta[/tex])2 d[tex]\theta[/tex]


    dunno if that will come out correctly. basically, if i did it right, there should be 3 integrals there. the first integral gives you the area of the cirlce. the second integral subtracts...fuuu....

    alright, actually, while i am just typing this i realized I possible mistake. I won't erase what i just typed just in case its really correct or w/e. here's my new integral

    2([tex]\int[/tex][2[tex]\pi[/tex]/3 to [tex]\pi[/tex]] (1/2)(2-2cos[tex]\theta[/tex])2 d[tex]\theta[/tex]+[tex]\int[/tex][0 to 2[tex]\pi[/tex]/3](1/2)(-6cos[tex]\theta[/tex])2) d[tex]\theta[/tex])

    here, I find the area of most of the common area using the area of the limacon , and add the area of a small sliver that is missed by using the integral of the circle. i probably didnt explain this well, but w/e.

    anyways, for the first integral i attempted (the one i decided had a mistake), i got -5[tex]\pi[/tex]. should my answer even be negative? for the second integral i get 14 pi....does that seem right?

    I know its a lot of problems, but any help would be appreciated. thanks.


    2. Relevant equations
    I did my work in part 1 b/c it was just easier that way.


    3. The attempt at a solution
    work in part 1...

    EDIT: integrals came out poorly. how do I do the limits of integration in latex again? also, the 2s in fron tof the integrals should be distributed to all the integrals, as in its 2 times all the integrals, which should have been bracketed
     
  2. jcsd
  3. Nov 16, 2009 #2

    Mark44

    Staff: Mentor

    How about you give us one problem at a time? People are more likely to respond if there's not a whole laundry list of stuff to look at.

    Also, here's a definite integral in LaTeX:

    [tex]\int_{\theta = \pi/2}^{\pi} (1/2)(2 - cos(\theta))^2 d\theta[/tex]

    Click it to see my code.
     
  4. Nov 16, 2009 #3
    Well, for the first two I think I'm right, I just need someone to verify. I guess I can start with the third one. The only real help I need with the third one is the part with the horizontal tangent line.
     
  5. Nov 16, 2009 #4

    Mark44

    Staff: Mentor

    #1 and #2 look fine.
    For #3, I don't think there are any points where the tangent line is horizontal. I converted the equation r = 3/(2 + 2 cos(theta)) to rectangular coordinates, and got y^2 = -3x + 9/4. I differentiated implicitly to get dy/dx. The graph is a parabola that opens to the left, with its vertex at (3/4, 0). At (0, 3/2) -- which corresponds to theta = pi/2, I got dy/dx = -1, which agrees with what you got.
     
  6. Nov 17, 2009 #5
    alright thanks.

    what about number 4 and 5?
    for number 4 what i need help with the most are understanding what "one loop" is and how to find the limits of integration.

    i might've worded 5 weirdly, but any help would be appreciated. is my integral at least right?
     
  7. Nov 17, 2009 #6

    Mark44

    Staff: Mentor

    For 4, you have the equation r = 2sin(5theta). r = 0 when theta = 0 and r = 0 again when theta = pi/5, and there are no other values of theta in [0, pi/5] for which r = 0. That's a loop. You get one loop on this interval.

    For 5, you need only two integrals, not three. Your idea of calculating half of the area and doubling it is a good one. I believe your integrals should look like this:
    [tex](1/2)\int_{\theta = ?}^? (-6cos(\theta))^2 d\theta~+~(1/2)\int_{\theta = ?}^{?}(-2cos(\theta) + 2)^2 d\theta[/tex]

    EDIT: Revised integrands in integrals above.
    You need to fill in the missing values of theta in the limits of integration and also double the value in the sum above.

    It might be easier to work with r = 6cos(theta) and r= 2cos(theta)-2, so that both graphs are in the first and fourth quadrants. The graphs are just the reflections across the y-axis, the the areas involved are the same. Be careful with the limits of integration. It looks like both graphs start out at the origin, but the circle doesn't.
     
    Last edited: Nov 17, 2009
  8. Nov 17, 2009 #7
    oh so a loop is just one "petal"? that makes sense now.

    as for the integrals you put, why is it just the original function being integrated? shouldnt it be like int of (1/2) r^2 or something?

    the integral i got was

    2(LaTeX Code: \\int [2LaTeX Code: \\pi /3 to LaTeX Code: \\pi ] (1/2)(2-2cosLaTeX Code: \\theta )2 dLaTeX Code: \\theta +LaTeX Code: \\int [0 to 2LaTeX Code: \\pi /3](1/2)(-6cosLaTeX Code: \\theta )2) dLaTeX Code: \\theta )

    which is pretty much the same as yours, except I multiply the functions by 1/2 after squaring them.
     
  9. Nov 17, 2009 #8

    Mark44

    Staff: Mentor

    On 5, you're right. I had forgotten that for polar equatons, dA = 1/2 r^2 d(theta).
     
  10. Nov 18, 2009 #9
    one last thing.
    i looked over number 2 and got confused.

    express in x and y

    r^2=tan(theta)
    i know that tan(theta)= y/x

    so r^2=y/x
    r=sqrt(y/x)

    how do i eliminate the r?

    EDIT:
    also, doesn't r^2 =x^2 +y^2?

    even so, how do i get rid of the r there?

    EDIT2:

    nvm.

    x^2+y^2=tan(theta)
    x^2+y^2=(rsin(theta))/(rcos(theta))
    x&2+y^2=y/x
     
    Last edited: Nov 18, 2009
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