Evaluate s(t)=∫t−[infinity]||r′(u)||du for the bernoulli spiral r(t)=⟨etcos(8t),etsin(8t)⟩. It is convenient to take −[infinity] as the lower limit since s(−[infinity])=0. Then use s to obtain an arc length parametrization of r(t).

Following are the solution to the given equation:Given:Please find the given question.To find:[tex]\to s(t)= \int^{t}_{-\infty} || r'\ (u) \ || du \\[/tex]Solution:Using the Bernoulli spiral:[tex]\to r(t) = \\\\\to r'(t)=[/tex]Now [tex]\to || r'(t)|| = e^t \sqrt{(\cos 4t- 4 \sin 4t )^2 + (\sin 4t + 4 \cos 4t )^2}[/tex]      [tex]= e^t \sqrt{(\cos^2 4t – 8 \sin 4t\cos 4t +16 \sin^2 4t +\sin^2 4t + 8\sin 4t\cos 4t + 16 \cos^2 4t }\\\\= e^t \sqrt{(\cos^2 t+\sin^2 t)+ 16(\sin^2 4t +\cos^2 4t)}\\\\=\sqrt{17} \cdot e^t\\\\[/tex][tex]\to s(t)=\int^t_{-\infty} || r^t (t)|| dt\\\\[/tex]           [tex]= \sqrt{17}[e^t]^{t}_{-\infty}\\\\ = \sqrt{17} e^t \ \ \ \ \ \ [since e^{-\infty} = 0][/tex]To obtain an arc-length parameterization of [tex]r(t)[/tex] using [tex]s(t)[/tex] obtained above.[tex]\to s(t)= \sqrt{17}e^t\\\\ \to t = ln \frac{s}{\sqrt{17}}[/tex][tex]\therefore \\\\\to r_1(s)= r(\phi (s))= <\frac{s}{\sqrt{17}} \cos (4 ln \frac{s}{\sqrt{17}}), \frac{s}{\sqrt{17}} \sin (4 ln \frac{s}{\sqrt{17}})>[/tex]                             [tex]=\frac{s}{\sqrt{17}}< \cos (4 ln \frac{s}{\sqrt{17}}),\sin ( 4 ln \frac{s}{\sqrt{17}})>[/tex]Learn more:brainly.com/question/11089689