Explanation Given: Approximate value of, 5.03 2 + 11.99 2 . Approach: We use the differential approximation. f ( x + d x , y + d y ) ≈ f ( x , y ) + d z f ( x + d x , y + d y ) ≈ f ( x , y ) + f x ( x , y ) ⋅ d x + f y ( x , y ) ⋅ d y f ( x + d x , y + d y ) ≈ f ( x , y ) + ∂ f ∂ x ⋅ d x + ∂ f ∂ x ⋅ d y ... ( 1 ) We first choose a number x which is close to 5.03 and y which is close to 11.99. Let f ( x , y ) = x 2 + y 2 , x = 5 , y = 12 , d x = .03 , d y = ( − .01 ) f ( x , y ) = x 2 + y 2 ... ( 2 ) Calculation: The absolute difference between differential and calculator approximation is calculated as, Partial differentiation of the equation (2) with respect to x . ∂ f ∂ x = ∂ ∂ x ( x 2 + y 2 ) ∂ f ∂ x = 1 2 x 2 + y 2 ⋅ 2 x ∂ f ∂ x = x x 2 + y 2 ... ( 3 ) Partial differentiation of the equation (2) with respect to y

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Author: Lial

Publisher: PEARSON EDUCATION (COLLEGE)

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Question

Chapter 9.4, Problem 2YT

To determine

To find:

Differential approximation value of 5.032+11.992.

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Chapter 9.4, Problem 1YT

Chapter 9.4, Problem 3YT

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Chapter 9 Solutions

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