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Arc length of a curve

  1. Sep 27, 2014 #1
    1. The problem statement, all variables and given/known data
    A curve has the equation y2 = x3. Find the length of the arc joining (1, - 1) to (1, 1).

    2. Relevant equations

    3. The attempt at a solution
    I took the integral of the distance and tried to evaluate from -1 to 1.
    L = [intergral (-1 to 1) sqrt (1+(dy/dx x^3/23/2)2 dx]
    Evaluated I got [integral (-1 to 1) sqrt (1+9x/4) dx]
    1/27 [4+9x]3/2 ] exaluated from -1 to 1

    I know the answer is supposed to be (26sqrt(13) - 16)/27 and mine is not coming out to that. Is there a mistake somewhere?
     
  2. jcsd
  3. Sep 27, 2014 #2

    mfb

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    The formatting in your first fomula looks broken.
    This is undefined for some x and not symmetric in x, so the error has to be before that. It would help to see more steps of your work.
     
  4. Sep 27, 2014 #3

    HallsofIvy

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    I presume you mean that the equation of the curve is [itex]y^2= x^3[/itex] (if you don't want to use "Latex", at least write it as y^2= x^3 to indicate the powers). You should know that the arclength is given by the integral [itex]\int \sqrt{1+ (dy/dx)^2} dx[/itex]. Since [itex]y^2= x^3[/itex], [itex]2y (dy/dx)= 3x^2[/itex] so that [itex]dy/dx= (3/2)(x^2/y)= (3/2)(x^2/(x^{3/2})= (3/2)x^{1/2}[/itex]. So [itex]1+ (dy/dx)= 1+ (9/4)x[/itex]

    mfb's complaint that "this I undefined for some x and not symmetric in x" does NOT apply because x is never negative. What is important (and may be what mfb was saying) is that x does NOT go from -1 to 1- it is y that goes from -1 to 1. You can do either of two things:
    1- take the x- integral from 0 to 1 and double. (You can do that because the graph goes from x= 0 to 1 and is symmetric for y> 0 and y< 0.)

    2- use [itex]\int \sqrt{1+ (dx/dy)}dy[/itex] for the arclength instead. [itex]2y= 3x^2 (dx/dy)[/itex] so [itex]dx/dy= (2/3)(y/x^2)= (2/3)(y/y^{4/3})= (2/3)y^{-1/3}[/itex]. (Since the integral from -1 to 1 contains y= 0, that is an "improper" integral, but converges.)
     
  5. Sep 27, 2014 #4

    mfb

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    Ah right, y changes its sign. Well, you still have to pay attention to it in some way.
     
  6. Sep 28, 2014 #5

    haruspex

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    [itex]\int \sqrt{1+ (dx/dy)^2}dy[/itex] ?
     
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