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Q. 28

Expert-verified
Found in: Page 880

Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

For each of the vector-valued functions, find the unit tangent vector. $r\left(t\right)=\left(\mathrm{sin}2t,\mathrm{cos}2t,t\right)$

$\frac{1}{\sqrt{5}}⟨2\mathrm{cos}2t,-2\mathrm{sin}2t,1⟩$

See the step by step solution

Step2. Continue

$\begin{array}{l}∥{r}^{\mathrm{\prime }}\left(t\right)∥=\parallel ⟨2\mathrm{cos}2t,-2\mathrm{sin}2t,1⟩\parallel \\ =\sqrt{\left(2\mathrm{cos}2t{\right)}^{2}+\left(-2\mathrm{sin}2t{\right)}^{2}+\left(1{\right)}^{2}}\\ =\sqrt{4{\mathrm{cos}}^{2}2t+4{\mathrm{sin}}^{2}2t+1}\\ =\sqrt{4\left({\mathrm{cos}}^{2}2t+{\mathrm{sin}}^{2}2t\right)+1}\end{array}\phantom{\rule{0ex}{0ex}}=\sqrt{5}\mathrm{sin}ce{\mathrm{sin}}^{2}2t+{\mathrm{cos}}^{2}2t=1\phantom{\rule{0ex}{0ex}}\text{The unit tangent vector to}r\left(t\right)\text{is}\phantom{\rule{0ex}{0ex}}\begin{array}{l}T\left(t\right)=\frac{{r}^{\mathrm{\prime }}\left(t\right)}{∥{r}^{\prime \prime }\left(t\right)∥}\\ =\frac{⟨2\mathrm{cos}2t,-2\mathrm{sin}2t,1⟩}{\sqrt{5}}\\ \text{Thus the unit tangent vector to the given vector - valued function is}\\ \frac{1}{\sqrt{5}}⟨2\mathrm{cos}2t,-2\mathrm{sin}2t,1⟩\end{array}$