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Math Help - Parametric / polar equations

  1. #1
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    Parametric / polar equations

    I have a few things im stuck on.. <.< I might be able to answere one through the example of others..

    1) Find the polar equation for the curve x^2 + y^2 = 4

    2)Find the equation in x and y for the curve with parametric equations x=(cost) -2 and y= (sin t) +3 for 0 is less or = to t and t is less or = to pie

    3) Sketch the graph of the plar equation r=2-cos feta

    4) Find an equation in x and y that has the same graph as r=2cos feta - 4sin feta
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by 3deltat View Post
    I have a few things im stuck on.. <.< I might be able to answere one through the example of others..

    1) Find the polar equation for the curve x^2 + y^2 = 4

    2)Find the equation in x and y for the curve with parametric equations x=(cost) -2 and y= (sin t) +3 for 0 is less or = to t and t is less or = to pie

    3) Sketch the graph of the plar equation r=2-cos feta

    4) Find an equation in x and y that has the same graph as r=2cos feta - 4sin feta
    x = r*cos(theta)
    y = r*sin(theta)

    r = sqrt(x^2 + y^2)
    theta = tan^(-1)(y/x)
    Two comments about the last equation. It's "theta" not "feta" and tan^(-1) is the inverse tangent function.

    1) r^2 = x^2 + y^2 so,
    r^2 = 4.


    2) 0 <= t <= pi <== It's "pi", not "pie"
    x=(cost) -2
    y= (sin t) +3

    Solve the x equation for cos(t):
    x + 2 = cos(t)

    Set up a right triangle. cos(t) is "adjacent" over hypotenuse so we want the side of the triangle adjecent to angle t to be x + 2 and the hypotenuse 1. Thus the "opposite" side will be sqrt{1 - (x + 2)^2}. So:
    sin(t) = sqrt{1 - (x + 2)^2}/1 = sqrt{1 - (x + 2)^2}

    So
    y= (sin t) + 3 = sqrt{1 - (x + 2)^2} + 3


    3) I can't do this one here. I'll leave it up to you.

    4) r=2cos(theta) - 4sin(theta)
    sqrt{x^2 + y^2} = 2*x/r - 4*y/r <== Square both sides
    x^2 + y^2 = (2*x/r - 4*y/r)^2

    x^2 + y^2 = 4x^2/r^2 + 16y^2/r^2 - 16xy/r^2

    (x^2 + y^2)r^2 = 4x^2 + 16y^2 - 16xy

    But r^2 = x^2 + y^2

    (x^2 + y^2)^2 = 4(x^2 + 4y^2 - 4xy)

    This will do. You can always solve this for y if you wish.

    -Dan
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  3. #3
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    Thanks for the help! One question though...
    For problem 4, when it is " sqrt{x^2 + y^2} = 2*x/r - 4*y/r " Why is the x and y divided by r ?

    Also, for number three; to start out at least, would this be a correct way?
    r = 2-costheta
    r^2 = 2r - rcostheta
    X^2 +Y^2 = 2sqr(X^2 + Y^2) -x
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  4. #4
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    Hello, 3deltat!

    You're expected to know the polar-rectangular conversion formulas:
    . . x = rcosθ, . y = rsinθ, . r .= .x + y


    By the way, you seem to have food on your mind.
    "Pie" is a dessert; you're thinking of "pi": π
    And "feta" is a Greek cheese; you mean "theta": θ


    1) Find the polar equation for the curve: .x + y .= .4
    This one is easy . . . replace x + y with r.

    And we have: .r .= .4 . . r .= .2



    2)Find the equation in x and y for the curve with parametric equations:
    . . x .= .cosθ - 2 .and .y .= .sinθ + 3, .for 0 < t < π

    This one is not easy (unless you know how) . . .

    We have: .x + 2 .= .cosθ
    . . . . . . . . y - 3 .= .sinθ

    Square: .(x + 2) .= .cosθ
    . . . . . . .(y - 3) .= .sinθ

    Add: .(x + 2) + (y - 3) .= .sinθ + cosθ

    Since sinθ + cosθ = 1, we have: .(x + 2) + (y - 3) .= .1

    This is a circle with center at (-2, 3) and radius 1.



    4) Find an equation in x and y that has the same graph as:
    . . . r .= .2cosθ - 4sinθ

    Multiply both sides by r: .r .= .2(rcosθ) - 4(rsinθ)

    Substitute: .x + y .= .2x - 4y


    Basically, we're finished ... but we can show-off a bit.

    We have: .x - 2x + y + 4y .= .0

    . . Complete the square: .x - 2x + 1 + y + 4y + 4 .= .0 + 1 + 4

    And we have: .(x - 1) + (y + 2) .= .5

    This is a circle with center (1, -2) and radius √5.

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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post

    By the way, you seem to have food on your mind.
    "Pie" is a dessert; you're thinking of "pi": π
    And "feta" is a Greek cheese; you mean "theta": θ
    You da man, Soroban! I know he meant to write theta, but the fact that feta is a Greek cheese, is that common knowledge?
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  6. #6
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    mm... food...
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by 3deltat View Post
    mm... food...
    which reminds me, i'm hungry. where's that feta!
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  8. #8
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    Hello again, 3deltat!

    I'll give #3 a try . . .


    3) Sketch the graph of the polar equation: .r .= .2 - cosθ
    You're expected to be familiar with the limacon . . . and the cardioid.

    Plot the four compass-directions to get the orientation.

    θ = 0: . r .= .2 - 1 .= .1

    θ = π/2: . r .= .2 - 0 .= .2

    θ = π: . r .= .2 - (-1) .= .3

    θ = 3π/2: . r .= .2 - 0 .= .2
    Code:
                        |
                  * * * |
              *         |   *
            *           |     *
           *            |     *
                        |    *
          *             |   *
      - - * - - - - - - + - * - - - - - -
          *             |   *
                        |    *
           *            |     *
            *           |     *
              *         |   *
                  * * * |
                        |
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  9. #9
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    Oh I see! For number three I was forgetting i could just graph it as it was. I was trying to get it into x^2 + y^2 form heh. Tyvm
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by 3deltat View Post
    Thanks for the help! One question though...
    For problem 4, when it is " sqrt{x^2 + y^2} = 2*x/r - 4*y/r " Why is the x and y divided by r ?
    x = r*cos(theta)

    Thus
    cos(theta) = x/r

    A similar equation holds for sin(theta).

    -Dan
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