To determine Find the gradient field of the function f ( x , y , z ) = ( x 2 + y 2 + z 2 ) − 1 2 . Explanation Given information: The function is f ( x , y , z ) = ( x 2 + y 2 + z 2 ) − 1 2 . Definition: The gradient field of a differentiable function f ( x , y , z ) to be the field of gradient vectors ∇ f = ∂ f ∂ x i + ∂ f ∂ y j + ∂ f ∂ z k . Calculation: Write the function. f ( x , y , z ) = ( x 2 + y 2 + z 2 ) − 1 2 (1) The function f ( x , y , z ) = ( x 2 + y 2 + z 2 ) − 1 2 is differentiable function. Differentiate the Equation (1) with respect to x . ∂ f ∂ x = ( − 1 2 ) ( x 2 + y 2 + z 2 ) − 3 2 ( 2 x ) = − x ( x 2 + y 2 + z 2 ) − 3 2 (2) Differentiate the Equation (1) with respect to y . ∂ f ∂ y = ( − 1 2 ) ( x 2 + y 2 + z 2 ) − 3 2 ( 2 y ) = − y ( x 2 + y 2 + z 2 ) − 3 2 (3) Differentiate the Equation (1) with respect to z

14th Edition

ISBN: 9780134438986

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

Publisher: PEARSON

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Chapter 16.2, Problem 1E

To determine

Find the gradient field of the function f(x,y,z)=(x2+y2+z2)−12.

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Chapter 16.1, Problem 42E

Chapter 16.2, Problem 2E

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Chapter 16 Solutions

Thomas' Calculus (14th Edition)

Ch. 16.1 - Match the vector equations in Exercises 1–8 with...Ch. 16.1 - Match the vector equations in Exercises 1–8 with...Ch. 16.1 - Match the vector equations in Exercises 1–8 with...Ch. 16.1 - Match the vector equations in Exercises 1–8 with...Ch. 16.1 - Match the vector equations in Exercises 1–8 with...Ch. 16.1 - Prob. 6E Ch. 16.1 - Prob. 7E Ch. 16.1 - Match the vector equations in Exercises 1–8 with...Ch. 16.1 - Evaluate ∫C (x + y) ds, where C is the...Ch. 16.1 - Evaluate ∫C (x − y + z − 2) ds, where C is the...

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