Explanation Given: The function y = x 2 and the point ( x 0 , x 0 2 ) . Proof: The first and the second derivative of the given function are y ′ = 2 x and y ″ = 2 . So, at the point ( x 0 , x 0 2 ) , y ′ 0 = 2 x 0 and y ″ 0 = 2 . Let the equation of the osculating circle be ( x − a ) 2 + ( y − b ) 2 = R 2 , where ( a , b ) is the center of the osculating circle. Differentiate the equation ( x − a ) 2 + ( y − b ) 2 = R 2 twice implicitly. 2 ( x − a ) + 2 ( y − b ) y ′ = 0 , .......(1) and 2 + 2 ( y ′ ) 2 + 2 ( y − b ) y ″ = 0 . .......(2) Solving equation (2) forb gives, 2 + 2 ( y ′ 0 ) 2 + 2 ( y 0 − b ) y ″ 0 = 0 1 + ( y ′ 0 ) 2 + ( y 0 − b ) y ″ 0 = 0 y 0 − b = − 1 + ( y ′ 0 ) 2 y ″ 0 b = y 0 + 1 + ( y ′ 0 ) 2 y ″ 0 Substitute the values y ′ 0 = 2 x 0 and y ″ 0 = 2 , to find the value of b

ACHIEVE FOR CALCULUS 4 TERM >CSI CUSTOM<

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ISBN: 9781319438333

Author: Rogawski

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Chapter 13.4, Problem 57E

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To show that the center of the osculating circle to the curve y=x2 at the point (x0,x02) is given as (−4x03,12+3x02).

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Chapter 13.4, Problem 56E

Chapter 13.4, Problem 58E

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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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To show that the center of the osculating circle to the curve y = x 2 at the point ( x 0 , x 0 2 ) is given as ( − 4 x 0 3 , 1 2 + 3 x 0 2 ) .

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To show that the center of the osculating circle to the curve y=x2 at the point (x0,x02) is given as (−4x03,12+3x02).

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