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Calc 3 (parametric equations) My answers won't match/can't find arithmetic error

  1. Sep 4, 2011 #1
    two questions.
    I know I'm doing the work right, but I can't get my answers to match and they are pretty close. I think it's just some arithmetic errors.
    Help? I've been trying to solve my mistakes forever. I can't find them!

    1. The problem statement, all variables and given/known data
    Use the parametric equation of an ellipse x = acos(t) and y=bsin(t) from 0<t<2PI (those are or equal to signs)


    2. Relevant equations
    A=INTEGRAL from 0 - 2PI ydx


    3. The attempt at a solution

    INTEGRAL from 0-2PI bsin(t)(-asin(t) dt
    = -ab INTEGRAL 0 -2PI sin^2(t) dt
    = -ab EVALUATED 1/2 (t) - 1/4*sin(2t) from 0-2PI
    = -ab ((1/2*(2PI) - 1/4 (sin(2*2PI) - 0)
    = -abPI

    But the answer is abPI. I don't know where the negative sign goes... or if I am supposed to have another one that cancels mine out

    1. The problem statement, all variables and given/known data
    Find the exact length of the curve


    2. Relevant equations
    x = 1+3t^2
    y= 4+2t^3
    0<t<1 (or equal)
    )^2)
    L= INTEGRAL from a - b SQUARE ROOT (dx/dt)^2 +(dy/dt)^2 END SQUARE ROOT dt


    3. The attempt at a solution
    dx/dt = 6t
    dy/dt = 6t^2

    L= INTEGRAL from 0-1 SQARE ROOT (6t)^2 + (6t^2)^2 END SQUARE ROOT dt
    = 6 INTEGRAL from 0-1 t* SR 1+t^2 dt
    using u substitution

    u = 1+t^2
    du = 2t dt
    boundaries change to 1 - 2
    = 3 * integral from 1-2 square root u du
    = 3 * 2*u^3/2 evaluated from 1-2
    = 3 * ( 2(2)^3/2 - 2(1)^3/2 )
    = 3 * (4 square root (2) - 2)
    = 12 square root (2) - 6


    answer is supposed to be
    4 square root (2) - 1
     
  2. jcsd
  3. Sep 4, 2011 #2
    Solved #2.
    I was doing the integral wrong. DUH! I can't believe it took me so long to figure out. I was bringing 2 (reversed of 1/2) down, not 2/3 (reversed of 3/2) and I wrote the answer down wrong. It was supposed to be 4 square root (2) - 2.

    So I solved that one after an 'aha!' moment. Still working on the first.
     
  4. Sep 4, 2011 #3
    For the integral

    [tex]-ab\int_{0}^{2Pi} \sin^2 t dt[/tex]

    use [tex]\sin^2 t = \frac{1 - \cos(2t)}{2}[/tex]
     
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