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Parametric curve and some irregular integtrals

  1. Jan 18, 2007 #1
    1. The problem statement, all variables and given/known data

    1) Consider the paramerric curve given by x = t^2 + 3t and y = 4 - t^2
    a) Find an equation of the tangent lne to the curve at the point (x,y) = (0,-5)
    b) Determine the equation of every vertical tangent line to this parametric curve.

    2)For each of the following definite integrals, determine its value if it convereges, otherwise, explain why it diverges.

    a) integral (e to infinity) [dx/(x(ln(x))^(3/2))]
    b) integral (-2 to -3) [dx/(sqrt(x^2 - 4))]

    3. The attempt at a solution

    1)

    a) dy/dx = (-2t)/(4-t^2)

    since x = t^2 + 3t and y = 4 - t^2 and (x,y) = (0,-5)

    t = -3

    dy/dx = (-2*-3)/(4 - (-3)^2) = -(6/5)

    y = -6/5(x) -5

    b) 4 - t^2 = 0

    t = -2, 2

    (x,y) = (-2, 0) and (10, 0)

    I can figure out the equation if these points are correct

    2)

    a) integral (e to infinity) [dx/(x(ln(x))^(3/2))]

    I get

    -2 lim (t -> inf) [ 1 / ln(x)^1/2 ] ( e - t)



    b) integral (-3 to -2) [dx/(sqrt(x^2 - 4))]

    I got as far as lim ( t -> -2) [ln | x + sqrt(x^2 - 4) | - ln 2 ] ( -3 to t)

    any support is much appreciated.

    Thanks
     
  2. jcsd
  3. Jan 19, 2007 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    How did you get dy/dx= -2t/(4-t^2)? That is
    [tex]\frac{\frac{dy}{dt}}{y}[/tex]
    Shouldn't x have something to do with dy/dx?

    Those points are NOT correct because you have the denominator of the derivative wrong. What is the equation of ANY vertical line?

    Again, how did you get that? The result of a definite integral (from e to t) with respect to x cannot contain an "x". And even if that were correct, you haven't yet answered the question! Does the integeral exist and if so what is it? Try the substitution u= ln(x)


    Again, that is obviously incorrect- the result of the integral cannot contain "x"! What is the derivative of arcsin(x)?
     
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