To determine Show that the center of the oscillating circle for the parabola y = x 2 at the point ( a , a 2 ) is located at ( − 4 a 3 , 3 a 2 + 1 2 ) . Explanation Given information: The center of the oscillating circle for the parabola y = x 2 at the point ( a , a 2 ) is located at ( − 4 a 3 , 3 a 2 + 1 2 ) . Formula used: Write a general expression to calculate the unit tangent vector . T = v | v | (1) Here, v is the constant velocity vector. Write a general expression to calculate the curvature. κ = 1 | v | | d T d t | (2) Calculation: Consider the expression of constant vector ( v ) and magnitude of constant vector for the parabola y = x 2 as v = i + 2 t j and 1 + 4 t 2 respectively using Example 4 in section 12.4 in the textbook. v = i + 2 t j | v | = 1 + 4 t 2 Substitute i + 2 t j for v , and 1 + 4 t 2 for | v | in equation (1) to find T . T = i + 2 t j 1 + 4 t 2 = ( 1 + 4 t 2 ) − 1 2 ( i + 2 t j ) Differentiate above equation with respect to t . d T d t = ( − 1 2 ) ( 1 + 4 t 2 ) − 1 2 − 1 ( 0 + 4 ( 2 t ) ) ( i + 2 t j ) + ( 1 + 4 t 2 ) − 1 2 ( 0 i + 2 j ) = ( 1 + 4 t 2 ) − 3 2 ( − 4 t ) ( i + 2 t j ) + ( 1 + 4 t 2 ) − 1 2 ( 2 j ) = ( 1 + 4 t 2 ) − 1 2 [ ( 1 + 4 t 2 ) − 1 ( − 4 t ) ( i + 2 t j ) + ( 2 j ) ] = ( 1 + 4 t 2 ) − 1 2 [ ( − 4 t ) ( i + 2 t j ) 1 + 4 t 2 + ( 2 j ) ] Simplify the above equation as follows: d T d t = ( 1 + 4 t 2 ) − 1 2 [ ( − 4 t ) ( i + 2 t j ) + ( 2 j ) ( 1 + 4 t 2 ) 1 + 4 t 2 ] = ( 1 + 4 t 2 ) − 1 2 [ ( − 4 t i − 8 t 2 j + 2 j + 8 t 2 j ) 1 + 4 t 2 ] = ( 1 + 4 t 2 ) − 1 2 [ ( 1 + 4 t 2 ) − 1 ( − 4 t i + 2 j ) ] = 2 ( 1 + 4 t 2 ) − 3 2 ( − 2 t i + j ) Take magnitude for above equation as follows: | d T d t | = | 2 ( 1 + 4 t 2 ) − 3 2 ( − 2 t i + j ) | = 2 ( 1 + 4 t 2 ) − 3 2 ( − 2 t ) 2 + 1 2 = 2 ( 1 + 4 t 2 ) − 3 2 4 t 2 + 1 = 2 ( 1 + 4 t 2 ) 3 2 1 + 4 t 2 Substitute 1 + 4 t 2 for | v | , and 2 ( 1 + 4 t 2 ) 3 2 1 + 4 t 2 for | d T d t | in equation (2) to find κ

Thomas' Calculus: Early Transcendentals, Books a la Carte Edition (14th Edition)

14th Edition

ISBN: 9780134439440

Author: Joel R. Hass, Christopher E. Heil, Maurice D. Weir

Publisher: PEARSON

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Chapter 13.4, Problem 29E

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Show that the center of the oscillating circle for the parabola y=x2 at the point (a,a2) is located at (−4a3,3a2+12).

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Chapter 13.4, Problem 28E

Chapter 13.4, Problem 30E

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