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Homework Help: Curve Length Question

  1. Jan 21, 2008 #1
    Help in Curve Length Question :D

    1. The problem statement, all variables and given/known data
    Find the length of the curve where x=cos(t) and y=t+sin(t) where 0[tex]\leq[/tex]t[tex]\leq[/tex][tex]\pi[/tex]

    2. Relevant equations



    3. The attempt at a solution

    where I am stuck is how to do substitution to find the length of the curve.

    I found the antiderivative:
    (-1/3sint)(2+2cost)^(3/2) but when I substitute the limits pi and zero I get a length of zero :confused:
    any pointers to get me in the right direction? thanks
    Last edited: Jan 21, 2008
  2. jcsd
  3. Jan 21, 2008 #2


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    I do not not think you integrated correctly. Show how you got that please.
  4. Jan 21, 2008 #3

    so, I substituted my two derivatives into the length equation, multiplied out my factors, and added sin(t)^2 with cos(t)^2 to get 1.
    Thus, sqrt of 1+1+2cos(t)

    is there a certain part of the integration that you see an error in?
    Last edited: Jan 21, 2008
  5. Jan 21, 2008 #4


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    [tex]\int \sqrt{2+2cos(t)} dt[/tex]

    [tex]=\sqrt{2} \int \sqrt{1+cost} dt[/tex]

    [tex]=\sqrt{2} \int \sqrt{\frac{1-cos^2t}{1-cost} dt[/tex]

    [tex]=\sqrt{2} \int \sqrt{\frac{sin^2t}{1-cost} dt[/tex]

    [tex]=\sqrt{2} \int \frac{sint}{\sqrt{1-cost}} dt[/tex]

    Did you get that?

    Let u=1-cost => du=sint dt

    [tex]=\sqrt{2} \int u^{\frac{-1}{2}} du[/tex]

    [tex]= 2\sqrt{2}\sqrt{1-cost}[/tex]
  6. Jan 21, 2008 #5
    Oh, I see! It's matter of taking a further step and factoring out the [tex]\sqrt{2}[/tex] from my initial integral.
    One question, how did you go from 1+cost to (1-cost^2)/(1-cost)?

    Thanks rockfreak. I'll try the question using that.
    Last edited: Jan 21, 2008
  7. Jan 21, 2008 #6
    (a - b)*(a + b) = (a^2 - b^2). Always a good one to be able to quote off the top of your head.
  8. Jan 21, 2008 #7


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    yeah it's basically that.

    Multiply the numerator and denominator by 1-cost
  9. Jan 21, 2008 #8


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    I might mention that this is one of those parametric curves where it is important to investigate its behavior before evaluating definite integrals to find arclengths. The integral from 0 to pi for this problem will give a non-zero value for the integral function rock.freak667 shows. However, you would be in trouble if you were asked to go from 0 to 2(pi) and naively used that function.

    I believe this curve is a cycloid; you would have to use a symmetry argument to take the arclength you find in going from 0 to pi and then doubling that result. This is a frequent potential pitfall in evaluating arclengths and enclosed areas for periodic curves (polar curves such as the rosettes often present such difficulties).
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