Explanation Given info: Equation of curve y = sin x , point x = π 4 . Formula used: O Q → = r → ( t 0 ) + 1 k ( t 0 ) N ∧ ( t 0 ) where O Q → is the center of circle T ∧ ( t ) = r → ( t ) ‖ r → ( t ) ‖ N ∧ ( t ) = T ∧ ' ( t ) ‖ T ∧ ' ( t ) ‖ K ( t ) = ‖ r → ' ( t ) × r → " ( t ) ‖ ‖ r → ' ( t ) ‖ 3 C → ( t ) = O Q → + 1 K ( t 0 ) 〈 cos t , sin t 〉 Calculation: Equation of space curve is y = sin x Parametric form of this can be written as r → ( x ) = 〈 x , f ( x ) 〉 = 〈 x , sin x 〉 r → ( x ) = d d x ( r → ( x ) ) = d d x 〈 x , sin x 〉 ( d x d x = 1 , d sin x d x = cos x ) r → ( x ) = 〈 1 , cos x 〉 T ∧ ( x ) = r → ' ( x ) ‖ r → ' ( x ) ‖ = 〈 1 , cos x 〉 ‖ 〈 1 , cos x 〉 ‖ = 〈 1 , cos x 〉 1 + cos 2 x N ∧ ( x ) = T ∧ ' ( x ) ‖ T ∧ ' ( x ) ‖ T ∧ ( x ) = 〈 1 1 + cos 2 x cos x 1 + cos 2 x 〉 T ∧ ' ( x ) = d d x 〈 1 1 + cos 2 x cos x 1 + cos 2 x 〉 = 〈 d d t ( 1 1 + cos 2 x ) , d d x ( cos x 1 + cos 2 x ) 〉 Apply chain rule of differentiation and division rule. = 〈 − 1 2 × ( 2 cos x × − sin x ) ( 1 + cos 2 x ) 3 / 2 , 1 + cos 2 x . ( − sin x ) − cos x . ( 2 cos x ) 2 1 + cos 2 x × ( − sin x ) ( 1 + cos 2 x ) 3 / 2 〉 (Simplification) = 〈 cos x . sin x ( 1 + cos 2 x ) 3 / 2 , 2 ( 1 + cos 2 x ) ( − sin x ) + 2 cos 2 x sin x ( 1 + cos 2 x ) 3 / 2 〉 = 〈 cos x . sin x ( 1 + cos 2 x ) 3 / 2 , − 2 ( 1 + cos 2 x ) 3 / 2 〉 T → ' = 〈 cos x . sin x ( 1 + cos 2 x ) 3 / 2 , − 2 ( 1 + cos 2 x ) 3 / 2 〉 N ∧ ( t ) = T ∧ ' ( x ) ‖ T ∧ ' ( x ) ‖ = 〈 cos x . sin x ( 1 + cos 2 x ) 3 / 2 , − 2 ( 1 + cos 2 x ) 3 / 2 〉 ‖ 〈 cos x . sin x ( 1 + cos 2 x ) 3 / 2 , − 2 ( 1 + cos 2 x ) 3 / 2 〉 ‖ = 〈 cos x ...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 14.4, Problem 61E

To determine

The parameterization of osculating circle at the point indicated.

$y = \sin x, x = \frac{π}{2}$

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Chapter 14.4, Problem 60E

Chapter 14.4, Problem 62E

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Chapter 14 Solutions

CALCULUS (CLOTH)

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 6PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E

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The parameterization of osculating circle at the point indicated. y = sin x , x = π 2

Solution Summary: The author explains how to determine the parameterization of osculating circle at the point indicated.

The parameterization of osculating circle at the point indicated.

$y = \sin x, x = \frac{π}{2}$

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Chapter 14 Solutions

The parameterization of osculating circle at the point indicated. y = sin x , x = π 2

Solution Summary: The author explains how to determine the parameterization of osculating circle at the point indicated.

The parameterization of osculating circle at the point indicated. y=sinx,x=π2

Want to see the full answer?

Chapter 14 Solutions

The parameterization of osculating circle at the point indicated.

$y = \sin x, x = \frac{π}{2}$