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Math Help - Length of a curve

  1. #1
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    Length of a curve

    Question is: Find the length of the curve x = t-sint, y = 1-cost for 0<=t<=2pi

    I've started the question using the formula \int_{0}^{2\pi}\sqrt{{(\frac{dx}{dt}})^2 + (\frac{dy}{dt})^2}} dt

    differentiated and subbed back into the formula: \int_{0}^{2\pi}\sqrt{{(1-cost)^2 + (sint)^2}} dt

    expanded simplified using trig identity: \int_{0}^{2\pi}\sqrt{{2-2cost}} dt

    did some u substitution. u = 2-2cost du/dt=2sint du=2sint dt

    \int_{0}^{2\pi}\sqrt{{u}} dt

    this is where im getting stuck. I'm trying to change to dt to a du and i get:

    \int_{0}^{2\pi}\frac{\sqrt{u}}{sint} du

    looking at wolframalpha I should be getting
    integrate sqrt&#40;2-2cost&#41; dt - Wolfram|Alpha

    \int_{0}^{2\pi}- \frac{1}{\sqrt{4-u}} du

    can anyone help me see how they completed this step please?
    Interestingly the final solution to the question is zero(i think), is there any insight that would mean you wouldnt need to complete the integration but come to the solution earlier.. ?

    again, any help would be great(man latex code takes forever to write your first time) thanks
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: Length of a curve

    \int_{0}^{2\pi}\sqrt{2-2cost} dt=\int_{0}^{2\pi}\sqrt{2(1-cost)}=\int_{0}^{2\pi}\sqrt{2*2sin^2(t/2)}=\int_{0}^{2\pi}2sin(t/2)
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  3. #3
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    Re: Length of a curve

    Quote Originally Posted by spaz84 View Post
    Question is: Find the length of the curve x = t-sint, y = 1-cost for 0<=t<=2pi
    I've started the question using the formula \int_{0}^{2\pi}\sqrt{{(\frac{dx}{dt}})^2 + (\frac{dy}{dt})^2}} dt
    differentiated and subbed back into the formula: \int_{0}^{2\pi}\sqrt{{(1-cost)^2 + (sint)^2}} dt
    expanded simplified using trig identity: \int_{0}^{2\pi}\sqrt{{2-2cost}} dt
    I would write it as \int_0^{2\pi } {\sqrt {2 - 2\cos (t)} dt}  = \sqrt 2 \int_0^{2\pi }{\sqrt {1 - \cos (t)} dt}
    Then look this Be sure to click "Show Steps".
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  4. #4
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    Re: Length of a curve

    Quote Originally Posted by alexmahone View Post
    \int_{0}^{2\pi}\sqrt{2-2cost} dt=\int_{0}^{2\pi}\sqrt{2(1-cost)}=\int_{0}^{2\pi}\sqrt{2*2sin^2t}=\int_{0}^{2  \pi}2sint
    correction ...

    2(1 - \cos{t}) = \frac{4(1 - \cos{t})}{2} = 4\sin^2\left(\frac{t}{2}\right)

    \int_0^{2\pi} 2\sin\left(\frac{t}{2}\right) \, dt
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  5. #5
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    Re: Length of a curve

    Quote Originally Posted by alexmahone View Post
    \int_{0}^{2\pi}\sqrt{2-2cost} dt=\int_{0}^{2\pi}\sqrt{2(1-cost)}=\int_{0}^{2\pi}\sqrt{2*2sin^2(t/2)}=\int_{0}^{2\pi}2sin(t/2)
    Outstanding, thank u. It looks like u are using a trig identity I am not familiar with, would u mind clarifying how u went from step 2 to 3
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  6. #6
    MHF Contributor alexmahone's Avatar
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    Re: Length of a curve

    Quote Originally Posted by spaz84 View Post
    Outstanding, thank u. It looks like u are using a trig identity I am not familiar with, would u mind clarifying how u went from step 2 to 3
    cost=1-2sin^2(t/2)

    1-cost=2sin^2(t/2)
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  7. #7
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    Re: Length of a curve

    Quote Originally Posted by spaz84 View Post
    Outstanding, thank u. It looks like u are using a trig identity I am not familiar with, would u mind clarifying how u went from step 2 to 3
    You're expected to recognise that 1 - cos(t) = 2sin^2(t/2) follows from the double angle formula cos(2A) = 1 - 2 sin^2(A). Let A = t/2 ....
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  8. #8
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    Re: Length of a curve

    Quote Originally Posted by mr fantastic View Post
    You're expected to recognise that 1 - cos(t) = 2sin^2(t/2) follows from the double angle formula cos(2A) = 1 - 2 sin^2(A). Let A = t/2 ....
    brilliant, thank you to everyone for their time and willingness to help...
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