Question

Reparametrize the curve with respect to arc length measured from the point where t=0 in the direction of increasing t.

\(r(t) = {e^{2t}}cos 2t\;i + 2\;j + {e^{2t}}sin\;2t\;k\)

Short answer

\(r(t(s)) = \left( {\frac{s}{{\sqrt 2 }} + 1} \right)\cos \left( {\ln \left( {\frac{s}{{\sqrt 2 }} + 1} \right)} \right)i + 2j + \left( {\frac{s}{{\sqrt 2 }} + 1} \right)\sin \left( {\ln \left( {\frac{s}{{\sqrt 2 }} + 1} \right)} \right)k\)

Step-by-step solution

Step 1: Given data and reparametrization of the curve

Let us reparametrize the curve

\(r(t) = {e^{2t}}\cos 2ti + 2j + {e^{2t}}sin2tk\)

With respect to arc length measured from the point where t=0 in the direction of increasing t. As in example 2 we see that the initial point is (1,0,0) corresponds to t=0. To reparametrize the given curve we will first compute the derivative of r(t) (and its intensity):

\(\begin{aligned}{l}r'(t) = ({e^{2t}}\cos 2t)'i + 0 + ({e^{2t}}\sin 2t)'k\\\left| {r'(t)} \right| = (2{e^{2t}}\cos 2t + {e^{2t}}( - \sin 2t)2)i + (2{e^{2t}}\sin 2t + {e^{2t}}2\cos 2t)k\\ = (2{e^{2t}}\cos 2t - 2{e^{2t}}\sin 2t)i + (2{e^{2t}}\sin 2t + 2{e^{2t}}\cos 2t)k\end{aligned}\)

Step 2: Magnitude (Intensity)

Its magnitude (intensity) is

\(\begin{aligned}{l}\left| {r'(t)} \right| = \sqrt {{{(2{e^{2t}}\cos 2t - 2{e^{2t}}\sin 2t)}^2} + {{(2{e^{2t}}\sin 2t + 2{e^{2t}}\cos 2t)}^2}} \\\left| {r'(t)} \right| = \sqrt {{{(2{e^{2t}})}^2}({{(\cos 2t - \sin 2t)}^2} + {{(\sin 2t + \cos 2t)}^2})} \\\left| {r'(t)} \right| = \sqrt {4{e^{4t}}({{\cos }^2}2t - 2\cos (2t)\sin (2t) + {{\sin }^2}(2t) + {{\sin }^2}(2t) + 2\sin (2t)\cos (2t) + {{\cos }^2}(2t))} \end{aligned}\)

\(\begin{aligned}{l}\left| {r'(t)} \right| = \sqrt {4{e^{4t}}(2{{\cos }^2}(2t) + 2{{\sin }^2}(2t))} \\\left| {r'(t)} \right| = 2\sqrt {2{e^{4t}}({{\cos }^2}2t + {{\sin }^2}2t)} \\\left| {r'(t)} \right| = 2{e^{2t}}\sqrt 2 \end{aligned}\)

Since \(\frac{{ds}}{{dt}} = \left| {r'(t)} \right|\) it follows that:

\(\begin{aligned}{l}\frac{{ds}}{{dt}} = 2{e^{2t}}\sqrt 2 = \int\limits_0^t {2\sqrt 2 {e^{2u}}du} \\ = 2\sqrt 2 \int\limits_0^t {{e^{2u}}du} = \left| {\begin{aligned}{*{20}{c}}{2u = x}\\{du = \frac{1}{2}dx}\\{0 \to 0}\\{t \to 2t}\end{aligned}} \right|\end{aligned}\)

\(\begin{aligned}{l} = 2\sqrt 2 \int\limits_0^{2t} {{e^x}\frac{1}{2}dx} \\ = \sqrt 2 \left. {{e^x}} \right|_0^{2t} = \sqrt 2 ({e^{2t}} - 1)\end{aligned}\)

Form here we have that,

\(\begin{aligned}{l}s = \sqrt 2 ({e^{2t}} - 1)\\\frac{s}{{\sqrt 2 }} + 1 = {e^{2t}}\\\ln \left( {\frac{s}{{\sqrt 2 }} + 1} \right) = \ln {e^{2t}} \Rightarrow t = \frac{1}{2}\ln \left( {\frac{s}{{\sqrt 2 }} + 1} \right)\end{aligned}\)

Step 3: Substituting t in initial equation we obtain

\(\begin{aligned}{l}r(t(s)) = {e^{2\left( {\frac{1}{2}\ln \left( {\frac{s}{{\sqrt 2 }} + 1} \right)} \right)}}\cos \left( {2\left( {\frac{1}{2}\ln \left( {\frac{s}{{\sqrt 2 }} + 1} \right)} \right)} \right)i + 2j + \\{e^{2\left( {\frac{1}{2}\ln \left( {\frac{s}{{\sqrt 2 }} + 1} \right)} \right)}}\sin \left( {2\left( {\frac{1}{2}\ln \left( {\frac{s}{{\sqrt 2 }} + 1} \right)} \right)} \right)k\\r(t(s)) = \left( {\frac{s}{{\sqrt 2 }} + 1} \right)\cos \left( {\ln \left( {\frac{s}{{\sqrt 2 }} + 1} \right)} \right)i + 2j + \left( {\frac{s}{{\sqrt 2 }} + 1} \right)\sin \left( {\ln \left( {\frac{s}{{\sqrt 2 }} + 1} \right)} \right)k\end{aligned}\)