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Math Help - parametric equations - tangents and length of the arc

  1. #1
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    parametric equations - tangents and length of the arc

    Hi

    x = integral sign (S) a= 1 b=t (cos u)/u du

    y = S a=1; b=t (sin u)/u du

    question is find the length of arc of the curve from the origin to the nearest point where there is a vertical tangent line

    So I am thinking I can find dx/du and where it equals 0 is where the vertical tangent is

    however when I try and figure out dx/du it gets ugly

    dx/du = (1/u)(-sin u) + (cos u)(-1/sqrt u)
    = - sin u/u - cos u/sqrt u
    = - sin u - cos^2 u

    And I don't have a clue what u is going to equal so that dx/du = 0 so I have a vertical tangent. And then the question becomes do I do that # to 1 or to 0 for the length of the arc?

    Thanks for any guidance you can provide.

    Calculus beginner
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  2. #2
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    Quote Originally Posted by calcbeg View Post
    Hi

    x = integral sign (S) a= 1 b=t (cos u)/u du

    y = S a=1; b=t (sin u)/u du

    question is find the length of arc of the curve from the origin to the nearest point where there is a vertical tangent line

    So I am thinking I can find dx/du and where it equals 0 is where the vertical tangent is

    however when I try and figure out dx/du it gets ugly

    dx/du = (1/u)(-sin u) + (cos u)(-1/sqrt u)
    = - sin u/u - cos u/sqrt u
    = - sin u - cos^2 u

    And I don't have a clue what u is going to equal so that dx/du = 0 so I have a vertical tangent. And then the question becomes do I do that # to 1 or to 0 for the length of the arc?

    Thanks for any guidance you can provide.

    Calculus beginner
    x = \int_1^t \frac{\cos{u}}{u} \, du

    using the Fundamental Theorem of Calculus ...

    \frac{dx}{dt} = \frac{\cos{t}}{t}


    y = \int_1^t \frac{\sin{u}}{u} \, du

    \frac{dy}{dt} = \frac{\sin{t}}{t}


    vertical tangent line will be where \frac{dx}{dt} = 0 ... t = \frac{\pi}{2}


    arclength ...

    S = \int_0^{\frac{\pi}{2}} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt
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