Explanation “The binomial vector of a curve is B = T × N , where T is the unit tangent vector and N is the principal unit normal vector.” For example, consider a space curve r ( t ) = ( 3 sin t ) i + ( 3 cos t ) j + 4 t k with T = ( 3 5 cos t ) i − ( 3 5 sin t ) j + 4 5 k and N = ( − sin t ) i + ( − cos t ) j . Then the binomial vector B is obtained by using the formula B = T × N as follows. B = T × N = | i j k 3 5 cos t − 3 5 sin t 4 5 − sin t − cos t 0 | = ( 0 + 4 5 cos t ) i − ( 0 + 4 5 sin t ) j + ( − 3 5 cos 2 t − 3 5 sin 2 t ) k = ( 4 5 cos t ) i − ( 4 5 sin t ) j + ( − 3 5 ( cos 2 t + sin 2 t ) ) k = ( 4 5 cos t ) i − ( 4 5 sin t ) j − 3 5 k Therefore, the binomial vector is B = ( 4 5 cos t ) i − ( 4 5 sin t ) j − 3 5 k . “The torsion of a curve is given by the formula τ = − d B d s ⋅ N = | x ˙ y ˙ z ˙ x ¨ y ¨ z ¨ x ⃛ y ⃛ z ⃛ | | v × a | 2 .” The above formula implies that the torsion of a curve depends on the binomial vector. Consider the same space curve r ( t ) = ( 3 sin t ) i + ( 3 cos t ) j + 4 t k . Obtain the velocity and acceleration as shown below. v = d r d t = ( 3 cos t ) i + ( − 3 sin t ) j + 4 k a = d v d t = ( − 3 sin t ) i + ( − 3 cos t ) j Compute the vector | v × a | as follows

Thomas' Calculus: Early Transcendentals, Single Variable (14th Edition)

14th Edition

ISBN: 9780134439419

Author: Joel R. Hass, Christopher E. Heil, Maurice D. Weir

Publisher: PEARSON

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Chapter 13, Problem 11GYR

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To explain: The binomial vector B and the torsion τ of a curve with examples.

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Chapter 13, Problem 10GYR

Chapter 13, Problem 12GYR

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