Explanation Given data: The curve is r ( t ) = sin t i + cos t j + tan t k and point is ( 2 2 , 2 2 , 1 ) . Formula used: Write the expression for tangent vector of a vector function r ( t ) ( T ( t ) ) . T ( t ) = r ′ ( t ) | r ′ ( t ) | (1) Here, r ′ ( t ) is first derivative of function r ( t ) . Write the equation of plane with vector r ′ ( t ) = 〈 r 1 , r 2 , r 3 〉 at point ( x 1 , y 1 , z 1 ) . r 1 ( x − x 1 ) + r 2 ( y − y 1 ) + r 3 ( z − z 1 ) = 0 (2) Consider the two three-dimensional vector functions such as u ( t ) = 〈 u 1 ( t ) , u 2 ( t ) , u 3 ( t ) 〉 and v ( t ) = 〈 v 1 ( t ) , v 2 ( t ) , v 3 ( t ) 〉 . Cross product of vectors: Write the expression for cross product of vectors u ( t ) and v ( t ) ( u ( t ) × v ( t ) ) . u ( t ) × v ( t ) = | i j k u 1 ( t ) u 2 ( t ) u 3 ( t ) v 1 ( t ) v 2 ( t ) v 3 ( t ) | = 〈 [ ( u 2 ( t ) v 3 ( t ) − v 2 ( t ) u 3 ( t ) ) ] , [ − ( u 1 ( t ) v 3 ( t ) − v 1 ( t ) u 3 ( t ) ) ] , [ ( u 1 ( t ) v 2 ( t ) − v 1 ( t ) u 2 ( t ) ) ] 〉 Write the expression for magnitude of vector a ( | a | ) . | a | = a 1 2 + a 2 2 + a 3 2 Here, a 1 , a 2 and a 3 are the x , y , and z -coordinates of vector respectively. The rectifying plane contains vectors T and B , so the normal vector of rectifying plane is N since it is orthogonal to the vectors T and B . Equate the components of r ( t ) with point ( 2 2 , 2 2 , 1 ) . sin t = 2 2 t = sin − 1 ( 2 2 ) t = π 4 cos t = 2 2 t = cos − 1 ( 2 2 ) t = π 4 tan t = 1 t = tan − 1 ( 1 ) t = π 4 Hence, the value of t is π 4 . Find the value of r ′ ( t ) . r ′ ( t ) = d d t ( sin t i + cos t j + tan t k ) = d d t ( sin t ) i + d d t ( cos t ) j + d d t ( tan t ) k = cos t i + ( − sin t ) j + sec 2 t k { ∵ d d x ( sin x ) = cos x , d d x ( cos x ) = sin x , d d x ( tan x ) = sec 2 x } = cos t i − sin t j + sec 2 t k Find the value of | r ′ ( t ) | . | r ′ ( t ) | = ( cos t ) 2 + ( − sin t ) 2 + ( sec 2 t ) 2 = cos 2 t + sin 2 t + sec 4 t = 1 + sec 4 t { ∵ cos 2 x + sin 2 x = 1 } Substitute cos t i − sin t j + sec 2 t k for r ′ ( t ) and 1 + sec 4 t for | r ′ ( t ) | in equation (1), T ( t ) = cos t i − sin t j + sec 2 t k 1 + sec 4 t = cos t 1 + sec 4 t i − sin t 1 + sec 4 t j + sec 2 t 1 + sec 4 t k Apply differentiation with respect to t on both sides of equation

Bundle: Multivariable Calculus, 8th + WebAssign Printed Access Card for Stewart's Calculus, 8th Edition, Single-Term

8th Edition

ISBN: 9781305779198

Author: James Stewart

Publisher: Cengage Learning

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Question

Chapter 13.3, Problem 58E

To determine

To find: The rectifying plane of curve r(t)=sinti+costj+tantk at point (22,22,1) .

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Chapter 13 Solutions

Bundle: Multivariable Calculus, 8th + WebAssign Printed Access Card for Stewart's Calculus, 8th Edition, Single-Term

Ch. 13.1 - Prob. 1E Ch. 13.1 - Prob. 2E Ch. 13.1 - Find the limit. 3. limt0(e3ti+t2sin2tj+cos2tk)Ch. 13.1 - Find the limit. 4. limt1(t2-tt-1i+t+8j+sintlntk)Ch. 13.1 - Find the limit. 5. limt1+t21t2,tan-1t,1e2tt Ch. 13.1 - Prob. 6E Ch. 13.1 - Prob. 7E Ch. 13.1 - Prob. 8E Ch. 13.1 - Prob. 9E Ch. 13.1 - Prob. 10E

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The rectifying plane of curve r ( t ) = sin t i + cos t j + tan t k at point ( 2 2 , 2 2 , 1 ) .

Concept explainers

To find: The rectifying plane of curve r(t)=sinti+costj+tantk at point (22,22,1) .

Chapter 13 Solutions