Explanation Given information: The equation of the circular helix is r = a cos t i + a sin t j + c t k . Formula used: The unit binomial vector is given as the cross product of unit tangent vector and unit normal vector, B s = T s × N s . Recall that the cross product of two vectors v = c 1 i + d 1 j + e 1 k and u = c 2 i + d 2 j + e 2 k is given by v × u = i j k c 1 d 1 e 1 c 2 d 2 e 2 . Calculation: It is known that the given circular helix can be represented using the arc length parameter by r s = a cos a w i + a sin a w j + c s w k . where s is arc length parameter and s has reference point a , 0 , 0 . Differentiate both sides in with the respect to s it follows: r ′ s = − a w sin a w i + a w cos a w j + c w k Since T s = r ′ s the result implies T s = − a w sin a w i + a w cos a w j + c w k . Now, find N s by using the fact that N s = r ″ t r ″ t . Differentiate both sides in r ′ s = − a w sin a w i + a w cos a w j + c w k with the respect to s . It follows r ″ s = − a w 2 cos a w i − a w 2 sin a w j + 0 k and the magnitude of r ″ s is given by, r ″ s = a 2 w 4 cos 2 a w + a 2 w 4 sin 2 a w + 0 2 = a 2 w 4 cos 2 a w + sin 2 a w = a 2 w 4 = a w 2 Then, the result obtained is, N s = w 2 a − a w 2 cos a w i − a w 2 sin a w j + 0 k which implies N s = − cos a w i − sin a w j + 0 k

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ISBN: 9781119778189

Author: Anton, Howard, Bivens, Irl C., Davis, Stephen

Publisher: WILEY

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Chapter 12, Problem 5MC

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To find: For the circular helix r=acosti+asintj+ctk with a>0 , the torsion and the binormal vector are τ=cw2 and B=cwsinswi−cwcosswj+awk where w=a2+c2 and s has reference point a,0,0 .

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Chapter 12, Problem 4MC

Chapter 12, Problem 6MC

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Chapter 12 Solutions

Calculus: Early Transcendentals

Ch. 12.1 - Prob. 1QCE Ch. 12.1 - Describe the graph of rt=1+2t,1+3t.Ch. 12.1 - Prob. 3QCE Ch. 12.1 - Prob. 4QCE Ch. 12.1 - Prob. 1ES Ch. 12.1 - Find the domain of r(t) and the value of rt0. 2....Ch. 12.1 - Find the domain of rt and the value of rt0....Ch. 12.1 - Find the domain of rt and the value of rt0....Ch. 12.1 - Prob. 5ES Ch. 12.1 - Prob. 6ES

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For the circular helix r = a cos t i + a sin t j + c t k with a > 0 , the torsion and the binormal vector are τ = c w 2 and B = c w sin s w i − c w cos s w j + a w k where w = a 2 + c 2 and s has reference point a , 0 , 0 .

Solution Summary: The author explains that the torsion and the binormal vector are tau =cw2.

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To find: For the circular helix r=acosti+asintj+ctk with a>0 , the torsion and the binormal vector are τ=cw2 and B=cwsinswi−cwcosswj+awk where w=a2+c2 and s has reference point a,0,0 .

Want to see the full answer?

Chapter 12 Solutions

For the circular helix r = a cos t i + a sin t j + c t k with a &gt; 0 , the torsion and the binormal vector are τ = c w 2 and B = c w sin s w i − c w cos s w j + a w k where w = a 2 + c 2 and s has reference point a , 0 , 0 .

Solution Summary: The author explains that the torsion and the binormal vector are tau =cw2.

Videos

To find: For the circular helix r=acosti+asintj+ctk with a>0 , the torsion and the binormal vector are τ=cw2 and B=cwsinswi−cwcosswj+awk where w=a2+c2 and s has reference point a,0,0 .

Want to see the full answer?

Chapter 12 Solutions

For the circular helix r = a cos t i + a sin t j + c t k with a > 0 , the torsion and the binormal vector are τ = c w 2 and B = c w sin s w i − c w cos s w j + a w k where w = a 2 + c 2 and s has reference point a , 0 , 0 .