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Q. 33

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.$r\left(t\right)=\left(3\mathrm{sin}\left(t\right),5\mathrm{cos}\left(t\right),4\mathrm{sin}\left(t\right)\right),t=\mathrm{\pi }$

The unit tangent vector and the principle unit normal vector are

$T\left(\mathrm{\pi }\right)=\left(\frac{-3}{5},0,\frac{-4}{5}\right)\phantom{\rule{0ex}{0ex}}N\left(\mathrm{\pi }\right)=\left(0,1,0\right)$

See the step by step solution

## Step 1. Given information

Given $r\left(t\right)=\left(3\mathrm{sin}\left(t\right),5\mathrm{cos}\left(t\right),4\mathrm{sin}\left(t\right)\right),t=\mathrm{\pi }$

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

$r\left(t\right)=\left(3\mathrm{sin}\left(t\right),5\mathrm{cos}\left(t\right),4\mathrm{sin}\left(t\right)\right)\phantom{\rule{0ex}{0ex}}r\text{'}\left(t\right)=\left(3\mathrm{cos}\left(t\right),-5\mathrm{sin}\left(t\right),4\mathrm{cos}\left(t\right)\right)\phantom{\rule{0ex}{0ex}}\parallel r\text{'}\left(t\right)\parallel =\sqrt{\left({3}^{2}{\mathrm{cos}}^{2}\left(t\right)+{5}^{2}{\mathrm{sin}}^{2}\left(t\right)+{4}^{2}{\mathrm{cos}}^{2}\left(t\right)\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{25{\mathrm{sin}}^{2}\left(t\right)+25{\mathrm{cos}}^{2}\left(t\right)}\phantom{\rule{0ex}{0ex}}=5\phantom{\rule{0ex}{0ex}}\mathbit{P}\mathbit{a}\mathbit{r}\mathbit{t}\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(3\mathrm{cos}\left(t\right),-5\mathrm{sin}\left(t\right),4\mathrm{cos}\left(t\right)\right)}{5}\phantom{\rule{0ex}{0ex}}T\left(\mathrm{\pi }\right)=\frac{\left(3\mathrm{cos}\left(\mathrm{\pi }\right),-5\mathrm{sin}\left(\mathrm{\pi }\right),4\mathrm{cos}\left(\mathrm{\pi }\right)\right)}{5}\phantom{\rule{0ex}{0ex}}=\left(\frac{-3,0,-4}{5}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{-3}{5},0,\frac{-4}{5}\right)\phantom{\rule{0ex}{0ex}}T\text{'}\left(t\right)=\frac{1}{5}\left(3\mathrm{sin}\left(t\right),-5\mathrm{cos}\left(t\right),4\mathrm{sin}\left(t\right)\right)\phantom{\rule{0ex}{0ex}}\parallel T\text{'}\left(t\right)\parallel =\frac{1}{5}\sqrt{\left({3}^{2}{\mathrm{sin}}^{2}\left(t\right)+\left(-{5}^{2}{\mathrm{cos}}^{2}\left(t\right)\right)+{4}^{2}{\mathrm{sin}}^{2}\left(t\right)\right)}\phantom{\rule{0ex}{0ex}}=\frac{1}{5}\sqrt{25{\mathrm{sin}}^{2}\left(t\right)+25{\mathrm{cos}}^{2}\left(t\right)}\phantom{\rule{0ex}{0ex}}=1\phantom{\rule{0ex}{0ex}}\mathbit{P}\mathbit{a}\mathbit{r}\mathbit{t}\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{\frac{1}{5}\left(3\mathrm{sin}\left(t\right),-5\mathrm{cos}\left(t\right),4\mathrm{sin}\left(t\right)\right)}{1}\phantom{\rule{0ex}{0ex}}N\left(t\right)=\frac{1}{5}\left(3\mathrm{sin}\left(t\right),-5\mathrm{cos}\left(t\right),4\mathrm{sin}\left(t\right)\right)\phantom{\rule{0ex}{0ex}}Att=\mathrm{\pi }\phantom{\rule{0ex}{0ex}}\mathrm{N}\left(\mathrm{\pi }\right)=\frac{1}{5}\left(3\mathrm{sin}\left(\mathrm{\pi }\right),-5\mathrm{cos}\left(\mathrm{\pi }\right),4\mathrm{sin}\left(\mathrm{\pi }\right)\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{5}\left(0,5,0\right)\phantom{\rule{0ex}{0ex}}=\left(0,1,0\right)$

## Step 3. The solution

The unit tangent vector and principle unit normal vector are

$T\left(\mathrm{\pi }\right)=\left(\frac{-3}{5},0,\frac{-4}{5}\right)\phantom{\rule{0ex}{0ex}}N\left(\mathrm{\pi }\right)=\left(0,1,0\right)$