Explanation Given: Curve = r ( t ) = 〈 e t cos 4 t , e t sin 4 t 〉 s ( t ) = ∫ − ∞ t | | r ( u ) | | d u Formula used: Curvature = k = | | r ' ( t ) × r ' ' ( t ) | | | | r ' ( t ) | | 3 Radius of curvature = r = 1 k Proof: r ' ( t ) = 〈 e t cos 4 t − 4 e t sin 4 t , e t sin 4 t + 4 e t cos 4 t 〉 = 〈 e t ( cos 4 t − 4 sin 4 t ) , e t ( sin 4 t + 4 cos 4 t ) 〉 Then, ‖ r ' ( t ) ‖ = [ e t ( cos 4 t − 4 sin 4 t ) ] 2 + [ e t ( sin 4 t + 4 cos 4 t ) ] 2 = e 2 t ( cos 4 t − 4 sin 4 t ) 2 + e 2 t ( sin 4 t + 4 cos 4 t ) 2 = e 2 t [ ( cos 4 t − 4 sin 4 t ) 2 + ( sin 4 t + 4 cos 4 t ) 2 ] = e 2 t [ cos 2 4 t + 16 sin 2 4 t − 8 cos 4 t sin 4 t + sin 2 4 t + 16 cos 2 4 t + 8 cos 4 t sin 4 t ] = e 2 t [ cos 2 4 t + 16 sin 2 4 t + sin 2 4 t + 16 cos 2 4 t ] = e 2 t [ 17 cos 2 4 t + 17 sin 2 4 t ] = e 2 t [ 17 { cos 2 4 t + sin 2 4 t } ] = e 2 t [ 17 { 1 } ] = 17 e 2 t Therefore, | | r ' ( t ) | | = 17 e t . And, r ' ' ( t ) = 〈 e t cos 4 t − 4 e t sin 4 t − 4 e t sin 4 t − 16 e t cos 4 t , e t sin 4 t + 4 e t cos 4 t + 4 e t cos 4 t − 16 e t sin 4 t 〉 = 〈 − 8 e t sin 4 t − 15 e t cos 4 t , − 15 e t sin 4 t + 8 e t cos 4 t 〉 = 〈 e t ( − 8 sin 4 t − 15 cos 4 t ) , e t ( − 15 sin 4 t + 8 cos 4 t ) 〉 Now, find r ' ( t ) × r ' ' ( t )

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ISBN: 9781319055905

Author: FRANZOSA

Publisher: VST

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Chapter 13.4, Problem 33E

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Show that the radius of curvature of curve 〈etcos4t,etsin4t〉 is proportional to s(t).

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Chapter 13.4, Problem 32E

Chapter 13.4, Problem 34E

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Ch. 13.1 - Prob. 1PQ Ch. 13.1 - Prob. 2PQ Ch. 13.1 - Prob. 3PQ Ch. 13.1 - Prob. 4PQ Ch. 13.1 - Prob. 5PQ Ch. 13.1 - Prob. 6PQ Ch. 13.1 - Prob. 1E Ch. 13.1 - Prob. 2E Ch. 13.1 - Prob. 3E Ch. 13.1 - Prob. 4E

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Show that the radius of curvature of curve 〈 e t cos 4 t , e t sin 4 t 〉 is proportional to s ( t ) .

Solution Summary: The author demonstrates that the radius of curvature of curve langle 'etmathrmcos4t,' s(t)='displaystyle',

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Show that the radius of curvature of curve 〈etcos4t,etsin4t〉 is proportional to s(t).

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Chapter 13 Solutions