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Thread: Converting from Cartesian to Polar Coordinates

  1. #1
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    Mar 2011
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    Converting from Cartesian to Polar Coordinates

    Hey there. I've had a problem that I've been stumbling over for the past 5 hours or so. Any help would be really, really appreciated.

    Make the following change of variables from rectangular coordinates to polar coordinates:

    x = r*cos(@), y = r*sin(@), r^2 = x^2 + y^2, @ = arctan(y/x)

    Then Show:

    r * dr/dt = x * dx/dt + y * dy/dt and (r^2) * d@/dt = x * dy/dt - y * dx/dt

    Then express the following in terms of polar coordinates, where dr/dt = r^3 and d@/dt = -1

    dx/dt = y + x * (x^2 + y^2)
    dy/dt = -x + y * (x^2 + y^2)


    Here's the work I have so far. I fear a good percentage of it is wrong.


    r^2 = x^2 + y^2
    d/dt(r^2 = x^2 + y^2) = [2r(dr/dt) = 2x(dx/dt) + 2y(dy/dt)] -> r(dr/dt) = x(dx/dt) + y(dy/dt)

    As for (r^2) * d@/dt = x * dy/dt - y * dx/dt, I have no idea.


    dx/dt = y + x * (x^2 + y^2)
    rsin(t) + (r^3)cos(t) = rsin(t) + rcos(t)[(rcos(t))^2 + (rsin(t))^2]
    rsin(t) + (r^3)cos(t) = rsin(t) + (r^3)cos(t)
    0 = 0


    dy/dt = -x + y * (x^2 + y^2)
    -rcos(t) + (r^3)sin(t) = -rcos(t) + (r^3)sin(t)
    0 = 0

    Obviously those last two parts are wrong, as from there I am supposed to graph the phase line diagram for r, then find a solution in terms of r and @ when x(0)=1 and y(0)=0. Any help would be greatly appreciated, as I've been working on this problem since 3:30 today. Seeing as it is now almost 1am, I'm getting some rest. I have another problem I'm stumped on as well, but it's rather difficult, and I already feel like a bother for asking help on this at such a late hour. Thank you to anyone who does help. I'll be awake again in about 7 hours to answer any questions regarding this.

  2. #2
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    Mar 2011
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    Re: Converting from Cartesian to Polar Coordinates

    Well, I've worked on it a bit more, though it doesn't really seem to help.

    r*dr/dt = x*dx/dt + y*dy/dt -> r^4 = r*cos(t) * [r*sin(t) + (r^3)*cos(t)] + r*sin(t)*[-r*cos(t) + (r^3)*sin(t)]
    r^4 = (r^4) * [cos(t)^2 + sin(t)^2]
    r^4 = r^4
    0 = 0

  3. #3
    Elite Member
    Join Date
    Jun 2007
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    12,508

    Re: Converting from Cartesian to Polar Coordinates

    Quote Originally Posted by Rukar
    Hey there. I've had a problem that I've been stumbling over for the past 5 hours or so. Any help would be really, really appreciated.

    Make the following change of variables from rectangular coordinates to polar coordinates:

    x = r*cos(@), y = r*sin(@), r^2 = x^2 + y^2, @ = arctan(y/x)

    Then Show:

    r * dr/dt = x * dx/dt + y * dy/dt and (r^2) * d@/dt = x * dy/dt - y * dx/dt

    Then express the following in terms of polar coordinates, where dr/dt = r^3 and d@/dt = -1

    dx/dt = y + x * (x^2 + y^2)
    dy/dt = -x + y * (x^2 + y^2)
    _____________________________________

    ? = tan[sup:bvegt3uf]-1[/sup:bvegt3uf](y/x)

    d?/dt = 1/[1+(y/x)[sup:bvegt3uf]2[/sup:bvegt3uf]] * [(-y*(dx/dt) + x*(dy/dt)]/x[sup:bvegt3uf]2[/sup:bvegt3uf]

    Now finish the algebra to get

    r[sup:bvegt3uf]2[/sup:bvegt3uf] * d?/dt = -y*(dy/dt) + x*(dx/dt)

    _________________________________________


    Here's the work I have so far. I fear a good percentage of it is wrong.


    r^2 = x^2 + y^2
    d/dt(r^2 = x^2 + y^2) = [2r(dr/dt) = 2x(dx/dt) + 2y(dy/dt)] -> r(dr/dt) = x(dx/dt) + y(dy/dt)

    As for (r^2) * d@/dt = x * dy/dt - y * dx/dt, I have no idea.


    dx/dt = y + x * (x^2 + y^2)
    rsin(t) + (r^3)cos(t) = rsin(t) + rcos(t)[(rcos(t))^2 + (rsin(t))^2]
    rsin(t) + (r^3)cos(t) = rsin(t) + (r^3)cos(t)
    0 = 0


    dy/dt = -x + y * (x^2 + y^2)
    -rcos(t) + (r^3)sin(t) = -rcos(t) + (r^3)sin(t)
    0 = 0

    Obviously those last two parts are wrong, as from there I am supposed to graph the phase line diagram for r, then find a solution in terms of r and @ when x(0)=1 and y(0)=0. Any help would be greatly appreciated, as I've been working on this problem since 3:30 today. Seeing as it is now almost 1am, I'm getting some rest. I have another problem I'm stumped on as well, but it's rather difficult, and I already feel like a bother for asking help on this at such a late hour. Thank you to anyone who does help. I'll be awake again in about 7 hours to answer any questions regarding this.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  4. #4
    Elite Member
    Join Date
    Jun 2007
    Posts
    12,508

    Re: Converting from Cartesian to Polar Coordinates

    Quote Originally Posted by Rukar
    Hey there. I've had a problem that I've been stumbling over for the past 5 hours or so. Any help would be really, really appreciated.

    Make the following change of variables from rectangular coordinates to polar coordinates:

    x = r*cos(@), y = r*sin(@), r^2 = x^2 + y^2, @ = arctan(y/x)

    Then Show:

    r * dr/dt = x * dx/dt + y * dy/dt and

    (r^2) * d@/dt = x * dy/dt - y * dx/dt

    Then express the following in terms of polar coordinates, where dr/dt = r^3 and d@/dt = -1

    dx/dt = y + x * (x^2 + y^2)
    dy/dt = -x + y * (x^2 + y^2)

    r * dr/dt = x * dx/dt + y * dy/dt ..........................(1)

    and

    (r^2) * d@/dt = x * dy/dt - y * dx/dt .......................(2)



    Treat equations (1) & (2) as two equations with two unknowns (unknowns being dy/dt and dx/dt)

    Then solve for dx/dt and dy/dt


    .
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

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