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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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BuyFindarrow_forward

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Find the gradient vector field ∇f of f and sketch it.

26. f(x, y) = 1 2 (x2 – y2)

To determine

To find: The gradient vector field for equation f(x,y)=12(x2y2) and gradient vector field for equation f(x,y)=12(x2y2) .

Explanation

Given data:

f(x,y)=12(x2y2)

Formula used:

Write the expression for gradient vector field of two dimensional vector.

f(x,y)=fxi+fyj (1)

Consider a two-dimensional vector F=x,y .

Write the expression for length of the two dimensional vector.

|F(x,y)|=x2+y2 (2)

Write the required differentiation formulae with respect to x as follows.

x(x2)=2xx(y2)=0

Write the required differentiation formulae with respect to y as follows.

y(x2)=0y(y2)=2y

Differentiate the term 12(x2y2) with respect to x .

x(12(x2y2))=12[x(x2)x(y2)]=12[2x0]=2x2=x

Differentiate the term 12(x2y2) with respect to y .

y(12(x2y2))=12[y(x2)y(y2)]=12[02y]=2y2=y

Find the gradient vector field of f(x,y)=12(x2y2) using equation (1).

Modify equation (1) as follows.

f(x,y)=x(12(x2y2))i+y(12(x2y2))j

Substitute x for x(12(x2y2)) and y for y(12(x2y2)) ,

f(x,y)=xiyj=x,y

Thus, the gradient vector field for 12(x2y2) is xiyj_ .

Find the length of f(x,y) using equation (2).

|f(x,y)|=(x)2+(y)2=x2+y2

Consider a certain interval of x as (2,2) and y as (2,2) to plot f(x,y)

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Chapter 16 Solutions

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