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Q. 60

Expert-verified
Found in: Page 873

### Calculus

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

# Prove that the tangent vector is always orthogonal to the position vector for the vector-valued function.

Ans:

$\mathbit{r}\mathbf{\left(}\mathbit{t}\mathbf{\right)}\mathbf{·}{\mathbit{r}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbit{t}\mathbf{\right)}\mathbf{=}\left(i\mathrm{sin}t+j\mathrm{sin}t\mathrm{cos}t+k{\mathrm{cos}}^{2}t\right)\mathbf{·}\mathbf{\left(}\mathbit{i}\mathbf{cos}\mathbit{t}\mathbf{+}\mathbit{j}\mathbf{cos}\mathbf{2}\mathbit{t}\mathbf{-}\mathbit{k}\mathbf{sin}\mathbf{2}\mathbit{t}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbit{r}\mathbf{\left(}\mathbit{t}\mathbf{\right)}\mathbf{·}{\mathbit{r}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbit{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$

See the step by step solution

## Step 2. Proving that the tangent vector is always orthogonal to the position vector for the vector-valued function:

Differentiate with respect to t

${r}^{\text{'}}\left(t\right)=i\mathrm{cos}t+j\mathrm{cos}2t-k\mathrm{sin}2t$

Find $r\left(t\right)·{r}^{\text{'}}\left(t\right)$

$r\left(t\right)·{r}^{\text{'}}\left(t\right)=\left(i\mathrm{sin}t+j\mathrm{sin}t\mathrm{cos}t+k{\mathrm{cos}}^{2}t\right)·\left(i\mathrm{cos}t+j\mathrm{cos}2t-k\mathrm{sin}2t\right)\phantom{\rule{0ex}{0ex}}r\left(t\right)·{r}^{\text{'}}\left(t\right)=0$

Thus, the tangent vector is always orthogonal to the position vector for the vector-valued function.