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reparametrisation - please check my answer? watch

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    The question asked to reparametrise the curve
    r(t)=(2/(1+t^2)-1)i + 2t/(1+t^2)j , t>=0
    with respect to the arc length measured from the point (1,0).

    I first figured out what value of t I would get from (1,0) by satisfying each entry to the corresponding value, where I found t=0 being the initial value.

    Then I differentiated r(t), giving r'(t)=(-(4t/(1+t^2)^2)i + (-(2t^2-2)/(1+t^2)^2)j
    To get s(t), I squared the coefficients of i and j, took the square root and then integrated, hence integrating (2/1+t^2) between t and 0.
    Thus s=arctan(t) and hence t=tan(s/2)
    Substituting this into r(t) and simplifying gave me
    r(s)=(2cot^2(s/2))i + (sin(s))j

    Can someone please tell me if I have done this correctly?
    (I didn't type out all lines of working out and I apologise for the lack of latex..)
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    (Original post by ellemay96)
    Substituting this into r(t) and simplifying gave me
    r(s)=(2cot^2(s/2))i + (sin(s))j

    Can someone please tell me if I have done this correctly?
    You made a slip in substituting in at the end, for the i-direction.

    PS: You can always plot these to see what they give. Wolfram or whatever software you have.
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    (Original post by ghostwalker)
    You made a slip in substituting in at the end, for the i-direction.

    PS: You can always plot these to see what they give. Wolfram or whatever software you have.
    Yes, I made a really silly mistake. I now have cos(s) for the i direction. Thank you for checking it.
 
 
 
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