Explanation Given information:- The given equation is R ( t ) = tan ( π ( t − 1 2 ) ) , a = 1 4 . Calculations:- The given equation is R ( t ) = tan ( π ( t − 1 2 ) ) , a = 1 4 . To find the linearization at the point, firstly will substitute the point a = 1 4 in the given equation and after that compute the equation of tangent line in point-slope form as shown below: R ( t ) = tan ( π ( t − 1 2 ) ) [The given function of the equation] R ( a ) = tan ( π ( a − 1 2 ) ) [Rewrite the equation θ = a ] R ( 1 4 ) = tan ( π ( 1 4 − 1 2 ) ) [Substitute a = 1 4 in the guven equation] R ( 1 4 ) = tan ( π ( 1 − 2 4 ) ) [ ∴ , sin [ 2 π ] = 0 ] R ( a ) = − tan π 4 [ ∴ , tan π 4 = 1 ] R ( a ) = − 1 The point have ( a , R ( a ) ) = ( 1 4 , − 1 ) . For compute the slope, find the R ' ( a ) : R ( a ) = tan ( π ( a − 1 2 ) ) [Find the derivative of the equation] R ' ( a ) = d d a [ tan ( π ( a − 1 2 ) ) ] [Differentiate with respect to a ] R ' ( a ) = sec 2 (

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 4, Problem 12CRE

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The solution of the linearization at the point indicated.

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Chapter 4, Problem 11CRE

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Chapter 4 Solutions

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Ch. 4.1 - Prob. 1PQ Ch. 4.1 - Prob. 2PQ Ch. 4.1 - Prob. 3PQ Ch. 4.1 - Prob. 4PQ Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E

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