Explanation Given info: θ : Angle of indignation between unit tangent vector T ^ and the x − axis r → ( s ) Is are length parameterization θ = θ ( S ) Be the angle of inclination at r ( s ) Formula used: From the figure unit tangent vector can be written as T ^ = cos θ i ^ + sin θ j ^ = 〈 cos θ , sin θ 〉 θ is function of S So it can be rewrite as T ^ = 〈 cos θ ( s ) , sin θ ( s ) 〉 d T ∧ d s = d d s 〈 cos θ ( s ) , sin θ ( s ) 〉 Differentiation is distributive d T ∧ d s = 〈 d ( cos θ ( s ) ) d s , d ( sin θ ( s ) ) d s 〉 = 〈 d ( cos θ ( s ) ) d ( θ ( s ) ) × d θ ( s ) d s , d ( sin θ ( s ) ) d ( θ ( s ) ) × d ( Q ( s ) ) d s 〉 (Apply chain rule of differentiation)

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 14.4, Problem 71E

To determine

To prove:

Prove that $K (S) = | \frac{d θ}{d S} |$

Where $Q :$ angle indignation between unit tangent vectors $\hat{T} and the x - a x i s$

Assume that $\vec{r} (s)$ is are length parameterization and let $θ = θ (S)$ be the angle of indignation at $r (s)$

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Chapter 14.4, Problem 70E

Chapter 14.4, Problem 72E

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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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To prove: Prove that K ( S ) = | d θ d S | Where Q : angle indignation between unit tangent vectors T ^ and the x − a x i s Assume that r → ( s ) is are length parameterization and let θ = θ ( S ) be the angle of indignation at r ( s )

Solution Summary: The author explains how to prove that K ( S ) = | d s | where Q : angle of indignation between unit tangent vectors T

Videos

To prove:

Prove that $K (S) = | \frac{d θ}{d S} |$

Where $Q :$ angle indignation between unit tangent vectors $\hat{T} and the x - a x i s$

Assume that $\vec{r} (s)$ is are length parameterization and let $θ = θ (S)$ be the angle of indignation at $r (s)$

Want to see the full answer?

Chapter 14 Solutions

To prove: Prove that K ( S ) = | d θ d S | Where Q : angle indignation between unit tangent vectors T ^ and the x − a x i s Assume that r → ( s ) is are length parameterization and let θ = θ ( S ) be the angle of indignation at r ( s )

Solution Summary: The author explains how to prove that K ( S ) = | d s | where Q : angle of indignation between unit tangent vectors T

Videos

To prove: Prove that K(S)=|dθdS| Where Q: angle indignation between unit tangent vectors T^ and the x−axis Assume that r→(s) is are length parameterization and let θ=θ(S) be the angle of indignation at r(s)

Want to see the full answer?

Chapter 14 Solutions

To prove:

Prove that $K (S) = | \frac{d θ}{d S} |$

Where $Q :$ angle indignation between unit tangent vectors $\hat{T} and the x - a x i s$

Assume that $\vec{r} (s)$ is are length parameterization and let $θ = θ (S)$ be the angle of indignation at $r (s)$