Explanation Given : r → ( t ) = 〈 t − sin t , 1 − cos t 〉 Given point: t = π Formula used: T ∧ ( t ) = r → ( t ) ‖ r → ( t ) ‖ N ∧ ( t ) = T ∧ ' ( t ) ‖ T ∧ ' ( t ) ‖ K ( t ) = ‖ r → ' ( t ) × r → ' ' ( t ) ‖ ‖ r → ' ( t ) ‖ 3 O Q → ( Center of circle ) = r → ( t 0 ) + 1 k ( t 0 ) . N ∧ ( t 0 ) . Equation of circle in parametric form: C → ( t ) = O Q → + 1 k ( t 0 ) 〈 cos t , sin t 〉 Calculation: r → ( t ) = 〈 t − sin t , 1 − cos t 〉 r → ' ( t ) = d d t 〈 t − sin t , 1 − cos t 〉 ( Distribute the differentiation ) = 〈 d d t ( t − sin t ) , d d t ( 1 − cos t ) 〉 = ( d d t ( t − sin t ) = d t d t − d ( sin t ) d t = 1 − cos t . d d t ( 1 − cos t ) = d ( 1 ) d t − d d t ( cos t ) = sin t ) = 〈 1 − cos t , sin t 〉 T ∧ ( t ) = r → ' ( t ) ‖ r → ' ( t ) ‖ = 〈 1 − cos t , sin t 〉 ( 1 − cos t ) 2 + ( sin t ) 2 ( ‖ 〈 a , b 〉 ‖ = a 2 + b 2 ) = 〈 1 − cos t , sin t 〉 1 + cos 2 t − 2 cos t + sin 2 t ( ( 1 − cos t ) 2 = 1 + cos 2 t − 2 cos t ) = 〈 1 − cos t , sin t 〉 2 − 2 cos t ( cos 2 t + sin 2 t = 1 ) T ∧ ( t ) = 〈 1 − cos t , sin t 〉 2 − 2 cos t = 〈 ( 1 − cos t ) 2 ( 1 − cos t ) , sin t 2 ( 2 sin 2 t 2 ) 〉 ( Simplification ) = 〈 1 − cos t 2 , sin t 2 | sin t 2 | 〉 ( 1 − cos t = 2 sin 2 t 2 , & t = π So, sin t 2 is postitive ) = 〈 1 2 ( 1 − cos t ) , sin t 2 sin t 2 〉 ( sin t = 2 sin t 2 , cos t 2 ) = 〈 1 2 ( 1 − cos t ) , 2 sin t 2 . cos t 2 2 sin t 2 〉 = 〈 1 2 ( 1 − cos t ) , cos t 2 〉 T ∧ ' ( t ) = d d t 〈 1 2 ( 1 − cos t ) , cos t 2 〉 = 〈 1 2 d d t ( 1 − cos t ) , d d t ( cos t 2 ) 〉 = 〈 1 2 × 1 2 1 − cos t × ( − x − sin t ) , − sin t 2 × 1 2 〉 = 〈 sin t 2 2 1 − cos t , − 1 2 sin t 2 〉 T ∧ ' ( t ) = 〈 sin t 2 2 ( 1 − cos t ) , − 1 2 sin t 2 〉 ( Put t = π , sin π = 0 , sin π 2 = + 1 ) ...

4th Edition

ISBN: 9781319055844

Author: FRANZOSA

Publisher: VST

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Question

Chapter 14.4, Problem 67E

To determine

The parametrization of the osculating circle at the point indicated

$\vec{r} (t) = 〈 t - \sin t, 1 - \cos t 〉, t = π$

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Chapter 14.4, Problem 66E

Chapter 14.4, Problem 68E

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Chapter 14 Solutions

EBK CALCULUS

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 parametrization of the osculating circle at the point indicated r → ( t ) = 〈 t − sin t , 1 − cos t 〉 , t = π

Solution Summary: The author explains the parametrization equation of the osculating circle at the given point t = 2.

Concept explainers

The parametrization of the osculating circle at the point indicated

$\vec{r} (t) = 〈 t - \sin t, 1 - \cos t 〉, t = π$

Want to see the full answer?

Chapter 14 Solutions

The parametrization of the osculating circle at the point indicated r → ( t ) = 〈 t − sin t , 1 − cos t 〉 , t = π

Solution Summary: The author explains the parametrization equation of the osculating circle at the given point t = 2.

Concept explainers

The parametrization of the osculating circle at the point indicated r→(t)=〈t−sint,1−cost〉,t=π

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Chapter 14 Solutions

The parametrization of the osculating circle at the point indicated

$\vec{r} (t) = 〈 t - \sin t, 1 - \cos t 〉, t = π$