Explanation Given: The function y = x and the point x = 4 . Calculation: Find the first and the second derivative of the function y = x . The first and the second derivative of the function y = x are y ′ = 1 2 x and y ″ = − 1 4 x 3 / 2 respectively. So, the curvature of the function can be found as shown below. k ( x ) = | y ″ | ( 1 + ( y ′ ) 2 ) 3 / 2 = | − 1 4 x 3 / 2 | ( 1 + ( 1 2 x ) 2 ) 3 / 2 So, at the point x = 4 , the curvature of the function is: k ( 4 ) = | − 1 32 | ( 1 + ( 1 2 4 ) 2 ) 3 / 2 = 1 / 32 ( 17 16 ) 3 / 2 = 1 / 32 17 17 / 64 = 2 17 17 Therefore, radius of curvature of the function at the point x = 4 is r ( 4 ) = 1 k ( 4 ) = 17 17 2 . The parametric equation of the function is r ( t ) = 〈 t , t 〉 . The tangent vector is given as r ′ ( t ) = 〈 1 , 1 2 t 〉 . The vector perpendicular to r ′ ( t ) will be of the form ± 〈 1 2 t , − 1 〉 . So, the unit normal vector is N ( t ) = ± 〈 1 2 t , − 1 〉 ( 1 2 t ) 2 + 1 The normal vector is pointing towards negativey -axis, so it must be N ( t ) = 〈 1 2 t , − 1 〉 1 4 t + 1

4th Edition

ISBN: 9781319279813

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 13, Problem 37CRE

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The equation of the osculating circle of the curve y=x at the point x=4 or state that the osculating circle does not exist.

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Chapter 13, Problem 36CRE

Chapter 13, Problem 38CRE

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CALCULUS:EARLY TRANS.-SAPLINGPLUS

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 at the point x = 4 or state that the osculating circle does not exist.

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

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The equation of the osculating circle of the curve y=x at the point x=4 or state that the osculating circle does not exist.

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Chapter 13 Solutions