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Math Help - Parametric equations and the lengths of curves.

  1. #1
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    Parametric equations and the lengths of curves.

    1. Find parametric equations for the semicircle: x^2 + y^2 = a^2 using as parameter the arc length s measured counterclockwise from the point (a, 0) to the point (x, y).

    We were not taught this, and I don't think the book really helps. I was wondering if somebody could clearly explain this concept and problem to me.

    2. Find the length of the curve: x = cos t, y = t + sin t, [0, [pi]].

    Here's where I get stuck. I know that the length of a curve is the integral evaluated from a to b (or, in this case, from zero to pi) of the quadrature of dx/dt and dy/dt. However, upon doing that, I get the integral of the square root of (-sin t)^2 + (1 + cos t)^2. I cannot use u substitution (although that does not rule out it's possibility of use) and if I make (-sin t)^2 into (sin t)^2 and expand (1 + cos t)^2, I get the integral of sqrt(2 + 2cos t), which then, by pulling out the square root of 2, makes the integral of the square root of 1 + cos t. I...don't know what to do then.

    I would appreciate any help.
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  2. #2
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    Quote Originally Posted by niyati View Post
    1. Find parametric equations for the semicircle: x^2 + y^2 = a^2 using as parameter the arc length s measured counterclockwise from the point (a, 0) to the point (x, y).

    We were not taught this, and I don't think the book really helps. I was wondering if somebody could clearly explain this concept and problem to me.

    ...
    Hello,

    I've attached a drawing.

    A point on the semicircle, the centre of the semicircle and the diameter form an angle \theta which can be calculated in radians by

    \theta = \frac sa

    Therefore you get:

    \left| \begin{array}{l}x=a\cdot \cos\left(\frac sa\right) \\ y = a \cdot \sin\left(\frac sa\right) \end{array} \right.
    Attached Thumbnails Attached Thumbnails Parametric equations and the lengths of curves.-halbkreis_parametr.gif  
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  3. #3
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    Second problem

    On the second part, you said you obtained the equation L=\sqrt{2}\int_{0}^{\pi}\sqrt{1+\cos{x}}\,dx
    To solve that, remember the trig identity \cos{2\theta}=2\cos^2\theta-1. Thus 1+\cos{2\theta}=2\cos^2\theta, or using \theta=\frac{x}{2}, we obtain 1+\cos{x}=2\cos^2\frac{x}{2}, so
    L=\sqrt{2}\int_{0}^{\pi}\sqrt{1+\cos{x}}\,dx
    =\sqrt{2}\int_{0}^{\pi}\sqrt{2\cos^2\frac{x}{2}}\,  dx
    =2\int_{0}^{\pi}\left|\cos\frac{x}{2}\right|\,dx
    Note that for your region of integration, \cos\frac{x}{2}\ge0, so you can remove the absolute value:
    L=2\int_{0}^{\pi}\cos\frac{x}{2}\,dx
    You should be able to do that integral.

    --Kevin C.
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