Find T, N, B, Find T, N, B, K, and 7 as functions of t ifr(t) = (sin t)i + (V2 cos t )j + (sin t)k.

Name: Find T, N, B, Find T, N, B, K, and 7 as functions of t ifr(t) = (sin t
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Description: Find T, N, B, Find T, N, B, K, and 7 as functions of t ifr(t) = (sin t)i + (V2 cos t )j + (sin t)k.

Consider the given vector, rt=sinti+2costj+sintk The formula for Tt is given as, Tt=r'tr't Now,…

Answered: Find T, N, B, K, and 7 as functions of…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Find T, N, B,

Expert Solution

Step 1

Consider the given vector,

$r (t) = (\sin t) i + (\sqrt{2} \cos t) j + (\sin t) k$

The formula for $T (t)$ is given as,

$T (t) = \frac{r' (t)}{||r' (t)||}$

Now,

$r' (t) = (\cos t) i + (- \sqrt{2} \sin t) j + (\cos t) k$

and,

$\begin{array}{rcl} ||r' (t)|| & = & \sqrt{{(\cos t)}^{2} + {(- \sqrt{2} \sin t)}^{2} + {(\cos t)}^{2}} \\ = & \sqrt{2 \cos^{2} (t) + 2 \sin^{2} (t)} \\ = & \sqrt{2 [\cos^{2} (t) + \sin^{2} (t)]} \\ = & \sqrt{2} \end{array}$

so, we get,

$\begin{array}{rcl} T (t) & = & \frac{(\cos t) i + (- \sqrt{2} \sin t) j + (\cos t) k}{\sqrt{2}} \\ = & \frac{\cos (t)}{\sqrt{2}} i - \sin (t) j + \frac{\cos (t)}{\sqrt{2}} k \end{array}$

Hence, the value of unit tangent vector $T$ is $\begin{array}{rcl} \frac{\cos (t)}{\sqrt{2}} i - \sin (t) j + \frac{\cos (t)}{\sqrt{2}} k \end{array}$ .

Step 2

Now, formula for unit normal vector is given as,

$N (t) = \frac{T' (t)}{||T' (t)||}$

Since, $T (t) = \begin{array}{rcl} \frac{\cos (t)}{\sqrt{2}} \end{array} \begin{array}{rcl} i \end{array} \begin{array}{rcl} - \end{array} \begin{array}{rcl} \sin \end{array} \begin{array}{rcl} (t) \end{array} \begin{array}{rcl} j \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} \frac{\cos (t)}{\sqrt{2}} \end{array} \begin{array}{rcl} k \end{array}$ , so,

$T' (t) = - \begin{array}{rcl} \frac{\sin (t)}{\sqrt{2}} \end{array} \begin{array}{rcl} i \end{array} \begin{array}{rcl} - \end{array} \begin{array}{rcl} \cos \end{array} \begin{array}{rcl} (t) \end{array} \begin{array}{rcl} j \end{array} \begin{array}{rcl} - \end{array} \begin{array}{rcl} \frac{\sin (t)}{\sqrt{2}} \end{array} \begin{array}{rcl} k \end{array}$

and,

$\begin{array}{rcl} ||T' (t)|| & = & \sqrt{{(- \frac{\sin (t)}{\sqrt{2}})}^{2} + {(- \cos (t))}^{2} + {(- \frac{\sin (t)}{\sqrt{2}})}^{2}} \\ = & \sqrt{\sin^{2} (t) + \cos^{2} (t)} \\ = & 1 \end{array}$

So, we get,

$\begin{array}{rcl} N (t) & = & \frac{- \frac{\sin (t)}{\sqrt{2}} i - \cos (t) j - \frac{\sin (t)}{\sqrt{2}} k}{1} \\ = & - \frac{\sin (t)}{\sqrt{2}} i - \cos (t) j - \frac{\sin (t)}{\sqrt{2}} k \end{array}$

Hence, the value of unit normal vector $N$ is $\begin{array}{rcl} - \frac{\sin (t)}{\sqrt{2}} i - \cos (t) j - \frac{\sin (t)}{\sqrt{2}} k \end{array}$ .

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