Explanation Given: The function y = ln x and the point x = 1 . Calculation: Find the first and the second derivative of the function y = ln x . The first and the second derivative of the function y = ln x are y ′ = 1 x and y ″ = − 1 x 2 respectively. So, the curvature of the function can be found as shown below. k ( x ) = | y ″ | ( 1 + ( y ′ ) 2 ) 3 / 2 = | − 1 x 2 | ( 1 + ( 1 x ) 2 ) 3 / 2 = x ( x 2 + 1 ) 3 / 2 So, at the point x = 1 , the curvature of the function is: k ( 1 ) = 1 ( 1 + 1 ) 3 / 2 = 1 2 3 / 2 Therefore, radius of curvature of the function at the point x = 1 is r ( 1 ) = 2 3 / 2 = 2 2 . The parametric equation of the function is r ( t ) = 〈 t , ln t 〉 . The tangent vector is given as r ′ ( t ) = 〈 1 , 1 t 〉 . The vector perpendicular to r ′ ( t ) will be of the form ± 〈 1 t , − 1 〉 . So, the unit normal vector is N ( t ) = ± 〈 1 t , − 1 〉 ( 1 t ) 2 + 1 = ± 〈 t , − t 〉 t 2 + 1

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ISBN: 9781319050740

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Question

Chapter 13.4, Problem 64E

To determine

The equation of the osculating circle of the curve y=lnx at the point x=1 or state that the osculating circle does not exist.

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Chapter 13.4, Problem 63E

Chapter 13.4, Problem 65E

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Chapter 13 Solutions

Calculus: Early Transcendentals

Ch. 13.1 - Prob. 1PQ Ch. 13.1 - Prob. 2PQ Ch. 13.1 - Prob. 3PQ Ch. 13.1 - Prob. 4PQ Ch. 13.1 - Prob. 5PQ Ch. 13.1 - Prob. 6PQ Ch. 13.1 - Prob. 1E Ch. 13.1 - Prob. 2E Ch. 13.1 - Prob. 3E Ch. 13.1 - Prob. 4E

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The equation of the osculating circle of the curve y = ln x at the point x = 1 or state that the osculating circle does not exist.

Solution Summary: The author states that the osculating circle of the curve y=mathrmlnx is not present.

The equation of the osculating circle of the curve y=lnx at the point x=1 or state that the osculating circle does not exist.

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Chapter 13 Solutions