Question

Reparametrize the curve \(r(t) = \left( {\frac{2}{{{t^2} + 1}} - 1} \right)i + \frac{{2t}}{{{t^2} + 1}}j\)with respect to arc length measured from the point (1, 0) in the direction of increasing . Express the reparametrization in its simplest form. What can you conclude about the curve?

Short answer

Thus we found that r(t(s)) is a unit circle : \(r(t(s)) = \left\langle {\cos (s),\sin (s)} \right\rangle \)

Step-by-step solution

Step 1: Convert fractions into exponents and find the tangent function

We have given the following curve and we have to reparametize its length from point (1,0).

\(r(t) = \left( {\frac{2}{{{t^2} + 1}} - 1} \right)i + \left( {\frac{{2t}}{{{t^2} + 1}}} \right)j\)

Now let us convert the fractions into exponents so that we can use the product rule.

\(\begin{aligned}{l}\left( {\frac{2}{{{t^2} + 1}} - 1} \right) = 2{({t^2} + 1)^{ - 1}} - 1\\\frac{{2t}}{{{t^2} + 1}} = 2{({t^2} + 1)^{ - 1}}\end{aligned}\)

Next let us find the derivative or tangent function r’(t) of the vector function r(t), that is

\(\begin{aligned}{l}|r'(t)| = 2{({t^2} + 1)^{ - 2}}(2t)i + ( - 2t{({t^2} + 1)^{ - 2}}(2t) + 2{({t^2} + 1)^{ - 1}})j\\ = - 4t{({t^2} + 1)^{ - 2}}i + (( - 4{t^2}){({t^2} + 1)^{ - 2}} + 2{({t^2} + 1)^{ - 1}})j\end{aligned}\)

Step 2: Magnitude

Find the magnitude of vector equation r’(t) that is

\(\begin{aligned}{l}\left| {r'(t)} \right| = \sqrt {16{t^2}{{({t^2} + 1)}^{ - 4}} + 16{t^4}{{({t^2} + 1)}^{ - 4}} + 4{{({t^2} + 1)}^{ - 3}}( - 4{t^2}) + 4{{({t^2} + 1)}^{ - 2}}} \\ = \sqrt {\frac{{16{t^2}}}{{{{({t^2} + 1)}^4}}} + \frac{{16{t^4}}}{{{{({t^2} + 1)}^4}}} - \frac{{16{t^2}}}{{{{({t^2} + 1)}^3}}} + \frac{4}{{{{({t^2} + 1)}^2}}}} \\ = \sqrt {\frac{{16{t^2} + 16{t^4} - 16{t^2}({t^2} + 1) + 4{{({t^2} + 1)}^2}}}{{{{({t^2} + 1)}^4}}}} \\ = \sqrt {\frac{4}{{{{({t^2} + 1)}^2}}}} \\ = \frac{2}{{({t^2} + 1)}}\end{aligned}\)

Find the magnitude of vector equation r’(t) that is

\(\begin{aligned}{l}\left| {r'(t)} \right| = \sqrt {16{t^2}{{({t^2} + 1)}^{ - 4}} + 16{t^4}{{({t^2} + 1)}^{ - 4}} + 4{{({t^2} + 1)}^{ - 3}}( - 4{t^2}) + 4{{({t^2} + 1)}^{ - 2}}} \\ = \sqrt {\frac{{16{t^2}}}{{{{({t^2} + 1)}^4}}} + \frac{{16{t^4}}}{{{{({t^2} + 1)}^4}}} - \frac{{16{t^2}}}{{{{({t^2} + 1)}^3}}} + \frac{4}{{{{({t^2} + 1)}^2}}}} \\ = \sqrt {\frac{{16{t^2} + 16{t^4} - 16{t^2}({t^2} + 1) + 4{{({t^2} + 1)}^2}}}{{{{({t^2} + 1)}^4}}}} \\ = \sqrt {\frac{4}{{{{({t^2} + 1)}^2}}}} \\ = \frac{2}{{({t^2} + 1)}}\end{aligned}\)

Step 3: The arclength

Next we have to find s, then solve for t, that is,

\(\begin{aligned}{l}s = \int\limits_0^t {\frac{2}{{{t^2} + 1}}dt} \\ = 2{\tan ^{ - 1}}(t)\end{aligned}\)

This implies we have

\(t = \tan \left( {\frac{s}{2}} \right)\).

Plug in for t. Using trigonometric identities, simplify

\(\begin{aligned}{l}r(t(s)) = \left\langle {\frac{2}{{{{\sec }^2}\left( {\frac{s}{2}} \right)}} - 1,\frac{{2\tan \left( {\frac{s}{2}} \right)}}{{{{\sec }^2}\left( {\frac{s}{2}} \right)}}} \right\rangle \\ = \left\langle {2{{\cos }^2}\left( {\frac{s}{2}} \right) - 1,2\sin \left( {\frac{s}{2}} \right)\cos \left( {\frac{s}{2}} \right)} \right\rangle \\ = \left\langle {\cos (s),\sin (s)} \right\rangle \end{aligned}\)

Thus we found that r(t(s)) is a unit circle : \(r(t(s)) = \left\langle {\cos (s),\sin (s)} \right\rangle \)