Explanation Given: Equation of the curve in vector form is r ( t ) = 〈 ln t , t , t 2 2 〉 and t = 1. Formula used: Velocity, v ( t ) = r ' ( t ) Acceleration, a ( t ) = r ' ' ( t ) Tangent vector, T = v ∥ v ∥ Normal vector, N = a N N a N , where a N = ∥ a N N ∥ , a N N = a − a T T and a T = a . T Binomial vector, B = T × N Calculation: We have r ( t ) = 〈 ln t , t , t 2 2 〉 v ( t ) = r ' ( t ) = 〈 1 t , 1 , t 〉 , at t = 1 v ( 1 ) = 〈 1 , 1 , 1 〉 T = v ∥ v ∥ = 〈 1 , 1 , 1 〉 3 = 〈 1 3 , 1 3 , 1 3 〉 a ( t ) = r ' ' ( t ) = 〈 − 1 t 2 , 0 , 1 〉 ,at t = 1 a ( 1 ) = 〈 − 1 , 0 , 1 〉 a T = a . T = 〈 − 1 , 0 , 1 〉 ...

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 14, Problem 39CRE

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Equation of the osculating plane corresponding to t=1 for $r (t) = 〈 \ln t, t, \frac{t^{2}}{2} 〉$

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Chapter 14, Problem 38CRE

Chapter 14, Problem 40CRE

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Chapter 14 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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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Equation of the osculating plane corresponding to t =1 for r ( t ) = 〈 ln t , t , t 2 2 〉

Solution Summary: The author explains the equation of the osculating plane corresponding to t =1.

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Equation of the osculating plane corresponding to t=1 for $r (t) = 〈 \ln t, t, \frac{t^{2}}{2} 〉$

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Chapter 14 Solutions

Equation of the osculating plane corresponding to t =1 for r ( t ) = 〈 ln t , t , t 2 2 〉

Solution Summary: The author explains the equation of the osculating plane corresponding to t =1.

Concept explainers

Equation of the osculating plane corresponding to t=1 for r(t)=〈lnt,t,t22〉

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Chapter 14 Solutions

Equation of the osculating plane corresponding to t=1 for $r (t) = 〈 \ln t, t, \frac{t^{2}}{2} 〉$