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Q. 21

Expert-verified
Found in: Page 880

### Calculus

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

# Let C be the graph of a vector-valued function r. The plane determined by the vectors T(t0) and B(t0) and containing the point r(t0) is called the rectifying plane for C at r(t0). Find the equation of the rectifying plane to the helix determined by $r\left(t\right)=\left(\mathrm{cos}\left(t\right),\mathrm{sin}\left(t\right),t\right)$ when t = π.

$x+1=0$

See the step by step solution

## Step1. Given Information

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## Step2. Rectifying plane

$\text{The plane determined by the vectors}T\left({t}_{0}\right)\text{and}B\left({t}_{0}\right)\text{containing the point}r\left({t}_{0}\right)\text{is called the}\phantom{\rule{0ex}{0ex}}\text{rectifying plane for}\mathrm{C}\text{at}r\left({t}_{0}\right)\text{.}\phantom{\rule{0ex}{0ex}}\text{Thus the equation of the rectifying plane to}r\left(t\right)\text{at}t=\pi \text{is}\phantom{\rule{0ex}{0ex}}\left(T\left(\pi \right)×B\left(\pi \right)\right)\cdot ⟨x-x\left(\pi \right),y-y\left(\pi \right),z-z\left(\pi \right)⟩=0\phantom{\rule{0ex}{0ex}}\text{First computing}T\left(\pi \right)×B\left(\pi \right)\text{:}\phantom{\rule{0ex}{0ex}}$

$\begin{array}{r}=\left|\begin{array}{ccc}i& j& k\\ 0& \frac{-\sqrt{2}}{2}& \frac{\sqrt{2}}{2}\\ 0& \frac{\sqrt{2}}{2}& \frac{\sqrt{2}}{2}\end{array}\right|\\ =\left|\begin{array}{ccc}i& j& k\\ 0& \frac{-\sqrt{2}}{2}& \frac{\sqrt{2}}{2}\\ 0& \frac{\sqrt{2}}{2}& \frac{\sqrt{2}}{2}\end{array}\right|\end{array}$

$\text{Evaluating}\phantom{\rule{0ex}{0ex}}\begin{array}{l}\left(T\left(\pi \right)×B\left(\pi \right)\right)\cdot ⟨x-x\left(\pi \right),y-y\left(\pi \right),z-z\left(\pi \right)⟩=0\\ ⟨-1,0,0⟩\cdot ⟨x+1,y,z-\pi ⟩=0\\ ⇒-1\left(x+1\right)=0\\ ⇒x+1=0\\ \text{Thus the equation of the rectifying plane to}r\left(t\right)\text{at}\mathrm{t}=\pi \text{is}\\ x+1=0\end{array}$