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Homework Help: Parametric question

  1. Apr 3, 2005 #1
    Find the area of the region enclosed by the parametric equation
    [tex]x=t^3-2t[/tex]
    [tex]y=9t^2[/tex]

    [tex]dx/dt = 3t^2-2[/tex]
    9t^2 - 1 = 0
    [tex]t=\pm \sqrt {1/9}[/tex]

    [tex]\int_{-1/3}^{1/3} (9t^2)*(3t^2-2) dt[/tex]

    = -2/5

    anyone know where i went wrong?
     
  2. jcsd
  3. Apr 3, 2005 #2

    quasar987

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    I don't understand what you did in line 3, 4, 5...

    What I recommand doing is to find a explicit relation between x and y. This can be done in the following manner.

    [tex]y = 9t^2 \Rightarrow t = \pm \sqrt{y/9}[/tex]

    In (1):

    [tex]x(y) = \pm (y/9)^{3/2} \mp \frac{2}{3}\sqrt{y}[/tex]

    Now you got two curves. See where they intersect. See at which values of t does this corespond. Integrate acordingly (you did that right, except with the wrong bounds).
     
  4. Apr 4, 2005 #3
    [tex]x(y) = \pm (y/9)^{3/2} \mp \frac{2}{3}\sqrt{y}[/tex]

    what am i suppose to do after this? solve for y? can you help me out a bit more? i'm just so confuse on how to find the bounds.
     
  5. Apr 4, 2005 #4

    quasar987

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    See where they intersect. Set them equal:

    [tex](y/9)^{3/2} - \frac{2}{3}\sqrt{y} = - (y/9)^{3/2} + \frac{2}{3}\sqrt{y} [/tex]

    and solve for y.

    You'll find 2 y. Using [itex]y=9t^2[/itex], find to which values of t these two y corespond. These are your bounds.
     
  6. Apr 4, 2005 #5
    setting those equal to each other and solving for y, i get [tex]\pm\sqrt{973}[/tex]

    those are the two y's right?

    "Using y=9t^2, find to which values of t these two y corespond. These are your bounds"

    are you saying to plug in [tex]\pm\sqrt{973}[/tex] into the y equation?
     
  7. Apr 4, 2005 #6

    quasar987

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    I don't get the same thing as you do. I get

    [tex](y/9)^{3/2} - \frac{2}{3}\sqrt{y} = - (y/9)^{3/2} + \frac{2}{3}\sqrt{y} [/tex]

    Right there, a solution is obviously y = 0. Let's find another.

    [tex]2(y/9)^{3/2} = \frac{4}{3}\sqrt{y} [/tex]

    [tex]\frac{4}{729}y^3 = \frac{16}{9}y [/tex]

    [tex]\frac{4}{729}y^2 = \frac{16}{9} [/tex]

    [tex]y = \pm\sqrt{ \frac{16}{9}\frac{729}{4}} = \pm \sqrt{324} = \pm 18 [/tex]

    This is a bit weird because [itex]y=-18[/itex] is not a selution because [itex]\sqrt{-18}[/itex] and [itex]\sqrt{(-18)^3}[/itex] are undefined. So y = +18 is our other solution. There are no other.

    By "Using y=9t^2, find to which values of t these two y corespond. These are your bounds" I mean...

    for which t will we have y=0 and for which t will be have y=18?

    Well,

    [tex]y= 0 = 9t^2 \Leftrightarrow t=0[/tex]

    and

    [tex]y=18 = 9t^2 \Leftrightarrow t = \pm \sqrt{2}[/tex]

    We only need one. So let's discard the minus one.

    Now integrate dxdy from 0 to sqrt{2} like you did before.
     
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