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Polar & parametric equations

  1. Nov 19, 2005 #1
    If someone could check my work and make sure I'm doing these problems right, I would really appreciate it.
    1.Eliminate the parameter and obtain the standard form of the rectangular equation.
    Circle: [tex] x= h + r cos \theta , y= k + r sin \theta [/tex]
    [tex] (x-h/r)^2 + (y-k/r)^2 = 1[/tex]
    2.Find the arc length of the given curve on the indicated interval.
    [tex] x=t^2 +1, y=4t^3 + 3 [/tex]
    [tex] 0 \leq t \leq -1 [/tex]
    [tex] S= \int \sqrt (dx/dt)^2 + (dy/dt)^2 [/tex]
    dx/dt= 2t, dy/dt= 12t^2
    so i intregrated from -1 to 0, [tex] \int \sqrt 4t^2 +144t^2 dt [/tex]
    using a u subsitution, I got [tex] 1/432(4+144t^2)^(3/2) [/tex]
    Plugging in -1 and 0 gave me -4.15, which can't be right since it's talking about arclength..
    3.Convert the rectangular equation to polar.
    [tex] x^2 + y^2 - 2ax = 0 [/tex]
    [tex] r^2 = 2ax [/tex]
    [tex] r^2 / r cos \theta = 2a (r cos \theta)/ r cos \theta [/tex]
    Solving for r gave me [tex] r= 2a cos \theta [/tex]
    If these are wrong, any help would be appreciated. Thanks!
  2. jcsd
  3. Nov 20, 2005 #2
    can someone please help?
  4. Nov 20, 2005 #3


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    Homework Helper

    1. Your equation should be

    [tex]\frac {(x-h)^2)}{r^2} + \frac {(y-k)^2}{r^2} = 1[/tex]

    2. Check your inequality. I am guessing it should read [itex] 0 \leq x \leq 1[/itex].
  5. Nov 20, 2005 #4
    Question 2 looks like it gives you an elliptical integral so you will have to numerically approximate it. 3. r=2*a*cos(theta) looks good.
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