reparametrisation - please check my answer?

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..)

Reply 1

ghostwalker

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.

Reply 2

ellemay96

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.