Explanation Given: The function y = x 2 − x 3 and the point x = 1 . Calculation: Find the first and the second derivative of the function y = x 2 − x 3 . The first and the second derivative of the function y = x 2 − x 3 are y ′ = 2 x − 3 x 2 and y ″ = 2 − 6 x respectively. So, the curvature of the function can be found as shown below. k ( x ) = | y ″ | ( 1 + ( y ′ ) 2 ) 3 / 2 = | 2 − 6 x | ( 1 + ( 2 x − 3 x 2 ) 2 ) 3 / 2 So, at the point x = 1 , the curvature of the function is: k ( 1 ) = | 2 − 6 | ( 1 + ( 2 − 3 ) 2 ) 3 / 2 = 4 2 3 / 2 = 4 2 2 = 2 Therefore, radius of curvature of the function at the point x = 1 is r ( 1 ) = 1 2 . The parametric equation of the function is r ( t ) = 〈 t , t 2 − t 3 〉 . The tangent vector is given as r ′ ( t ) = 〈 1 , 2 t − 3 t 2 〉 . The vector perpendicular to r ′ ( t ) will be of the form ± 〈 2 t − 3 t 2 , − 1 〉 . So, the unit normal vector is N ( t ) = ± 〈 2 t − 3 t 2 , − 1 〉 ( 2 t − 3 t 2 ) 2 + 1 The normal vector is pointing towards negativey -axis, so it must be N ( t ) = 〈 2 t − 3 t 2 , − 1 〉 ( 2 t − 3 t 2 ) 2 + 1

Loose-leaf Version For Calculus: Early Transcendentals

4th Edition

ISBN: 9781319055912

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Chapter 13, Problem 36CRE

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The equation of the osculating circle of the curve y=x2−x3 at the point x=1 or state that the osculating circle does not exist.

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Chapter 13, Problem 35CRE

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Chapter 13 Solutions

Loose-leaf Version For 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 = x 2 − x 3 at the point x = 1 or state that the osculating circle does not exist.

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The equation of the osculating circle of the curve y=x2−x3 at the point x=1 or state that the osculating circle does not exist.

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Chapter 13 Solutions