Explanation Given: r ( t ) = < t , cos t > Formula used: Let r ( t ) and T ( t ) be a differentiable vector and unit tangent vector respectively. Then the principal normal vector is N ( t ) = T ' ( t ) Here, T ( t ) = r ' ( t ) | r ' ( t ) | Calculation: r ( t ) = < t , cos t > Differentiate w . r . t t, you get r ' ( t ) = d d t < t , cos t > = < 1 , − sin t > Now, | r ' ( t ) | = ( 1 ) 2 + ( − sin t ) 2 = 1 + sin 2 t Therefore, the unit tangent vector is calculated as T ( t ) = r → ( t ) | r → ( t ) | = < 1 , − sin t > 1 + sin 2 t T ( t ) = 〈 1 1 + sin 2 t , − sin t 1 + sin 2 t 〉 Differentiate T → ( t ) with respect to t , you get T ' ( t ) = d d t 〈 1 1 + sin 2 t , − sin t 1 + sin 2 t 〉 = 〈 − sin t . cos t ( 1 + sin 2 t ) 3 2 , − ( 1 + sin 2 t ) cos t − 2 sin 2 t cos t ( 1 + sin 2 t ) 3 2 〉 = 〈 − sin t . cos t ( 1 + sin 2 t ) 3 2 , − cos t − sin 2 t cos t + 2 sin 2 t cos t ( 1 + sin 2 t ) 3 2 〉 = 〈 − sin t . cos t ( 1 + sin 2 t ) 3 2 , − cos t + sin 2 t cos t ( 1 + sin 2 t ) 3 2 〉 = 〈 − sin t . cos t ( 1 + sin 2 t ) 3 2 , − ( 1 − sin 2 t ) cos t ( 1 + sin 2 t ) 3 2 〉 = 〈 − sin t . cos t ( 1 + sin 2 t ) 3 2 , − cos 2 t cos t ( 1 + sin 2 t ) 3 2 〉 = 〈 − sin t . cos t ( 1 + sin 2 t ) 3 2 , − cos 3 t ( 1 + sin 2 t ) 3 2 〉 Now, Plug t = π 4 in to the normal vector equation T ' ( π 4 ) = 〈 − sin ( π 4 ) cos ( π 4 ) ( 1 + sin 2 π 4 ) 3 2 , − cos 3 π 4 ( 1 + sin 2 π 4 ) 3 2 〉 = 〈 − 1 2 × 1 2 ( 1 + 1 2 ) 3 2 , − ( 1 2 ...

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 14.4, Problem 39E

To determine

To find:

The normal vectors $r (t) = < t, \cos t > at t = \frac{π}{4} and t = \frac{3 π}{4}$

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Chapter 14.4, Problem 38E

Chapter 14.4, Problem 40E

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The normal vectors r ( t ) = < t , cos t > at t = π 4 and t = 3 π 4

Solution Summary: The author explains how to find the normal vectors r (t) and T, which are a differentiable vector and unit tangent vector.

Concept explainers

To find:

The normal vectors $r (t) = < t, \cos t > at t = \frac{π}{4} and t = \frac{3 π}{4}$

Want to see the full answer?

Chapter 14 Solutions

The normal vectors r ( t ) = &lt; t , cos t &gt; at t = π 4 and t = 3 π 4

Solution Summary: The author explains how to find the normal vectors r (t) and T, which are a differentiable vector and unit tangent vector.

Concept explainers

To find: The normal vectors r(t)=<t,cost>at t=π4 and t=3π4

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Chapter 14 Solutions

The normal vectors r ( t ) = < t , cos t > at t = π 4 and t = 3 π 4

To find:

The normal vectors $r (t) = < t, \cos t > at t = \frac{π}{4} and t = \frac{3 π}{4}$