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Answers without the blur. Sign up and see all textbooks for free! Q. 31

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 in Exercises , find the unit tangent vector and the principal unit normal vector at the specified value of t. $r\left(t\right)=\left(\mathrm{cos}\left(\alpha t\right),\mathrm{sin}\left(\alpha t\right)\right),where\alpha >0,t=\mathrm{\pi }$

The unit tangent vector and principle unit normal vector are

$T\left(\mathrm{\pi }\right)=\left(-\mathrm{sin}\left(\alpha \mathrm{\pi }\right),\mathrm{cos}\left(\alpha \mathrm{\pi }\right)\right)\phantom{\rule{0ex}{0ex}}N\left(\mathrm{\pi }\right)=\left(-\mathrm{cos}\left(\alpha \mathrm{\pi }\right),-\mathrm{sin}\left(\alpha \mathrm{\pi }\right)\right)$

See the step by step solution

## Step 1. Given information

Given $r\left(t\right)=\left(\mathrm{cos}\left(\alpha t\right),\mathrm{sin}\left(\alpha t\right)\right),t=\mathrm{\pi }$

## Step 2..Parta. Find tangent vector Partb. Find principle unit normal vector

$r\left(t\right)=\left(\mathrm{cos}\left(\alpha t\right),\mathrm{sin}\left(\alpha t\right)\right)\phantom{\rule{0ex}{0ex}}r\text{'}\left(t\right)=\left(-\alpha \mathrm{sin}\left(\alpha t\right),\alpha \mathrm{cos}\left(\alpha t\right)\right)\phantom{\rule{0ex}{0ex}}\parallel r\text{'}\left(t\right)\parallel =\parallel \left\{-\alpha \mathrm{sin}\left(\alpha t\right),\alpha \mathrm{cos}\left(\alpha t\right)\right\}\parallel \phantom{\rule{0ex}{0ex}}=\sqrt{{\left(-\alpha \mathrm{sin}\left(\alpha t\right)\right)}^{2}+{\left(\alpha \mathrm{cos}\left(\alpha t\right)\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\alpha }^{2}\left({\mathrm{sin}}^{2}\left(\alpha t\right)+{\mathrm{cos}}^{2}\left(\alpha t\right)\right)}\phantom{\rule{0ex}{0ex}}=\alpha \phantom{\rule{0ex}{0ex}}Part\left(a\right)\phantom{\rule{0ex}{0ex}}T\left(t\right)=\frac{r\text{'}\left(t\right)}{\parallel r\text{'}\left(t\right)\parallel }\phantom{\rule{0ex}{0ex}}=\frac{\left(-\alpha \mathrm{sin}\left(\alpha t\right),\alpha \mathrm{cos}\left(\alpha t\right)\right)}{\alpha }\phantom{\rule{0ex}{0ex}}=\left(-\mathrm{sin}\left(\alpha t\right),\mathrm{cos}\left(\alpha t\right)\right)\phantom{\rule{0ex}{0ex}}Att=\pi ,\phantom{\rule{0ex}{0ex}}T\left(\mathrm{\pi }\right)=\left(-\mathrm{sin}\left(\alpha \mathrm{\pi }\right),\mathrm{cos}\left(\alpha \mathrm{\pi }\right)\right)\phantom{\rule{0ex}{0ex}}T\text{'}\left(t\right)=\left(-\alpha \mathrm{cos}\left(\alpha t\right),-\alpha \mathrm{sin}\left(\alpha t\right)\right)\phantom{\rule{0ex}{0ex}}\parallel T\text{'}\left(t\right)\parallel =\sqrt{{\alpha }^{2}\left({\mathrm{cos}}^{2}\left(\alpha t\right)+{\mathrm{sin}}^{2}\left(\alpha t\right)\right)}\phantom{\rule{0ex}{0ex}}=\alpha \phantom{\rule{0ex}{0ex}}Part\left(b\right)\phantom{\rule{0ex}{0ex}}N\left(t\right)=\frac{T\text{'}\left(t\right)}{\parallel T\text{'}\left(t\right)\parallel }\phantom{\rule{0ex}{0ex}}=\frac{\left(-\alpha \mathrm{cos}\left(\alpha t\right),-\alpha \mathrm{sin}\left(\alpha t\right)\right)}{\alpha }\phantom{\rule{0ex}{0ex}}=\left(-\mathrm{cos}\left(\alpha t\right),-\mathrm{sin}\left(\alpha t\right)\right)\phantom{\rule{0ex}{0ex}}Att=\mathrm{\pi }\phantom{\rule{0ex}{0ex}}\mathrm{N}\left(\mathrm{\pi }\right)=\left(-\mathrm{cos}\left(\mathrm{\alpha \pi }\right),-\mathrm{sin}\left(\mathrm{\alpha \pi }\right)\right)$

## Step 3. The solution

The unit tangent vector and principle unit normal vector is

$T\left(\mathrm{\pi }\right)=\left(-\mathrm{sin}\left(\alpha \mathrm{\pi }\right),\mathrm{cos}\left(\alpha \mathrm{\pi }\right)\right)\phantom{\rule{0ex}{0ex}}N\left(\mathrm{\pi }\right)=\left(-\mathrm{cos}\left(\alpha \mathrm{\pi }\right),-\mathrm{sin}\left(\alpha \mathrm{\pi }\right)\right)$ ### Want to see more solutions like these? 