Explanation Given: c ( t ) = ( e − t cos t , e − t sin t ) for 0 ≤ t < ∞ . Formula Used: The length of a curve c ( t ) from t = a to t = b will be given by: L = ∫ t = a t = b ( d x d t ) 2 + ( d y d t ) 2 d t Calculation: It is given that c ( t ) = ( e − t cos t , e − t sin t ) that means we have x ( t ) = e − t cos t , y ( t ) = e − t sin t Now, differentiate both the equations x ( t ) = e − t cos t , y ( t ) = e − t sin t with respect to t and we get: x ( t ) = e − t cos t , y ( t ) = e − t sin t d x d t = − e − t ( cos t + sin t ) , d y d t = e − t ( cos t − sin t ) So, ( d x d t ) 2 + ( d y d t ) 2 = ( − e − t ( cos t + sin t ) ) 2 + ( e − t ( cos t − sin t ) ) 2 = e − 2 t ( cos t + sin t ) 2 + e − 2 t ( cos t − sin t ) 2 = e − t cos 2 t + sin 2 t + 2 cos t sin t + cos 2 t + sin 2 t − 2 cos t sin t = 2 e − t Now, the slope of the tangent line at any point t will be given by: d y d x = d y d t d x d t = e − t ( cos t − sin t ) − e − t ( cos t + sin t ) = ( cos t − sin t ) − ( cos t + sin t ) The tangent line to c ( t 0 ) is vertical, if the denominator of the above expression is 0...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 12, Problem 24CRE

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The first positive value of $t_{0}$ such that the tangent line to $c (t_{0})$ is vertical, and to calculate the speed at $t = t_{0}$ .

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Chapter 12, Problem 23CRE

Chapter 12, Problem 25CRE

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CALCULUS (CLOTH)

Ch. 12.1 - Prob. 1PQ Ch. 12.1 - Prob. 2PQ Ch. 12.1 - Prob. 3PQ Ch. 12.1 - Prob. 4PQ Ch. 12.1 - Prob. 5PQ Ch. 12.1 - Prob. 6PQ Ch. 12.1 - Prob. 7PQ Ch. 12.1 - Prob. 1E Ch. 12.1 - Prob. 2E Ch. 12.1 - Prob. 3E

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The first positive value of t 0 such that the tangent line to c ( t 0 ) is vertical, and to calculate the speed at t = t 0 .

Solution Summary: The author calculates the first positive value of t 0 such that the tangent line to c is vertical, and the speed.

Concept explainers

The first positive value of $t_{0}$ such that the tangent line to $c (t_{0})$ is vertical, and to calculate the speed at $t = t_{0}$ .

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Chapter 12 Solutions

The first positive value of t 0 such that the tangent line to c ( t 0 ) is vertical, and to calculate the speed at t = t 0 .

Solution Summary: The author calculates the first positive value of t 0 such that the tangent line to c is vertical, and the speed.

Concept explainers

The first positive value of t0 such that the tangent line to c(t0) is vertical, and to calculate the speed at t=t0.

Want to see the full answer?

Chapter 12 Solutions

The first positive value of $t_{0}$ such that the tangent line to $c (t_{0})$ is vertical, and to calculate the speed at $t = t_{0}$ .