Find the normal vectors to r(t) = \left \langle 4t,cos t \right \rangle at\, t=\frac{\pi}{4}...


Find the normal vectors to r(t) = {eq}\left \langle 4t,cos t \right \rangle at\, t=\frac{\pi}{4} and t= \frac{3\pi}{4}).\\T(\frac{\pi}{4})=\\T(\frac{3\pi}{4})= {/eq}

Finding Normal Vector:

The normal vector is obtained by differentiating the unit tangent vector. The unit tangent vector is containing the derivative of the vector function and the norm of derivative.

Finally, the normal vector is evaluated at the given parameter values.

Answer and Explanation:

Given vector function is {eq}\displaystyle r(t) = \left \langle 4t, \cos t \right \rangle {/eq} at {eq}\displaystyle t=\frac{\pi}{4} {/eq} and {eq}\displaystyle t= \frac{3\pi}{4} {/eq}.

Finding the unit tangent vector:

{eq}\begin{align*} \displaystyle \frac{dr}{dt} &=\frac{d}{dt} \left \langle 4t, \cos t \right \rangle \\ \displaystyle {r}'(t) &=\left \langle 4, -\sin \left(t\right) \right \rangle \\ \displaystyle \left \| {r}'(t) \right \| &=\sqrt{(4)^{2}+(-\sin \left(t\right))^{2}} \\ \displaystyle \left \| {r}'(t) \right \| &=\sqrt{16+\sin ^2\left(t\right)} \\ \displaystyle T(t) &=\frac{{r}'(t)}{\left \| {r}'(T) \right \|} \\ \displaystyle T(t) &=\left \langle \frac{4}{\sqrt{16+\sin ^2\left(t\right)}}, \frac{-\sin \left(t\right)}{\sqrt{16+\sin ^2\left(t\right)}} \right \rangle \end{align*} {/eq}

Finding the normal vectors:

{eq}\begin{align*} \displaystyle \frac{dT}{dt} &=\frac{d}{dt}\left \langle \frac{4}{\sqrt{16+\sin ^2\left(t\right)}}, \frac{-\sin \left(t\right)}{\sqrt{16+\sin ^2\left(t\right)}} \right \rangle \\ \displaystyle {T}'(t) &=\left \langle -\frac{2\sin \left(2t\right)}{\left(16+\sin ^2\left(t\right)\right)^{\frac{3}{2}}}, -\frac{2\cos \left(t\right)\left(16+\sin ^2\left(t\right)\right)-\sin \left(2t\right)\sin \left(t\right)}{2\left(16+\sin ^2\left(t\right)\right)^{\frac{3}{2}}} \right \rangle \\ \displaystyle {T}'\left( \frac{\pi}{4} \right) &=\left \langle -\frac{2\sin \left(2\left( \frac{\pi}{4} \right)\right)}{\left(16+\sin ^2\left(\frac{\pi}{4}\right)\right)^{\frac{3}{2}}}, -\frac{2\cos \left(\frac{\pi}{4}\right)\left(16+\sin ^2\left(\frac{\pi}{4}\right)\right)-\sin \left(2\left( \frac{\pi}{4} \right)\right)\sin \left(\frac{\pi}{4}\right)}{2\left(16+\sin ^2\left(\frac{\pi}{4}\right)\right)^{\frac{3}{2}}} \right \rangle \\ \displaystyle {T}'\left( \frac{\pi}{4} \right) &=\left \langle -\frac{4\sqrt{2}}{33\sqrt{33}}, -\frac{32}{33\sqrt{33}} \right \rangle \\ \\ \displaystyle {T}'\left( \frac{3\pi}{4} \right) &=\left \langle -\frac{2\sin \left(2\left( \frac{3\pi}{4} \right)\right)}{\left(16+\sin ^2\left(\frac{3\pi}{4}\right)\right)^{\frac{3}{2}}}, -\frac{2\cos \left(\frac{3\pi}{4}\right)\left(16+\sin ^2\left(\frac{3\pi}{4}\right)\right)-\sin \left(2\left( \frac{3\pi}{4} \right)\right)\sin \left(\frac{3\pi}{4}\right)}{2\left(16+\sin ^2\left(\frac{3\pi}{4}\right)\right)^{\frac{3}{2}}} \right \rangle \\ \displaystyle {T}'\left( \frac{3\pi}{4} \right) &=\left \langle \frac{4\sqrt{2}}{33\sqrt{33}}, \frac{32}{33\sqrt{33}} \right \rangle \end{align*} {/eq}

The normal vector at {eq}\displaystyle t=\frac{\pi}{4} {/eq} is {eq}\ \displaystyle \mathbf{\color{blue}{ {T}'\left( \frac{\pi}{4} \right)=\left \langle -\frac{4\sqrt{2}}{33\sqrt{33}}, -\frac{32}{33\sqrt{33}} \right \rangle }} {/eq}.

The normal vector at {eq}\displaystyle t= \frac{3\pi}{4} {/eq} is {eq}\ \displaystyle \mathbf{\color{blue}{ {T}'\left( \frac{3\pi}{4} \right)=\left \langle \frac{4\sqrt{2}}{33\sqrt{33}}, \frac{32}{33\sqrt{33}} \right \rangle }} {/eq}.

Learn more about this topic:

Vectors: Definition, Types & Examples

from Common Entrance Test (CET): Study Guide & Syllabus

Chapter 57 / Lesson 3

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