Explanation Given: The function y = sin x and the point x = π 2 . Calculation: Find the first and the second derivative of the function y = sin x . The first and the second derivative of the function y = sin x are y ′ = cos x and y ″ = − sin x respectively. So, the curvature of the function can be found as shown below. k ( x ) = | y ″ | ( 1 + ( y ′ ) 2 ) 3 / 2 = | − sin x | ( 1 + ( 2 cos x ) 2 ) 3 / 2 = | sin x | ( 1 + 4 cos 2 x ) 3 / 2 So, at the point x = π 2 , the curvature of the function is: k ( π 2 ) = | sin ( π 2 ) | ( 1 + 4 cos 2 ( π 2 ) ) 3 / 2 = 1 ( 1 + 4 ⋅ 0 ) 3 / 2 = 1 Therefore, radius of curvature of the function at the point x = π 2 is r ( π 2 ) = 1 . The parametric equation of the function is r ( t ) = 〈 t , sin t 〉 . The tangent vector is given as r ′ ( t ) = 〈 1 , cos t 〉 . The vector perpendicular to r ′ ( t ) will be of the form ± 〈 cos t , − 1 〉 . So, the unit normal vector is N ( t ) = ± 〈 cos t , − 1 〉 cos 2 t + 1

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Author: Rogawski

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Chapter 13.4, Problem 61E

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The equation of the osculating circle of the curve y=sinx at the point x=π2 or state that the osculating circle does not exist.

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Chapter 13.4, Problem 60E

Chapter 13.4, Problem 62E

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Find the indefinite integral. Check Answer: 7x 4 + 1x dx

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Chapter 13 Solutions

CALCULUS W/ACHIEVE PLATFORM

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 = sin x at the point x = π 2 or state that the osculating circle does not exist.

Solution Summary: The author calculates the first and second derivative of the function y=mathrmsinx.

The equation of the osculating circle of the curve y=sinx at the point x=π2 or state that the osculating circle does not exist.

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Chapter 13 Solutions