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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.

25. f(x, y) = 1 2 (x − y)2

To determine

To find: The gradient vector field for equation f(x,y)=12(xy)2 and gradient vector field for equation f(x,y)=12(xy)2 .

Explanation

Given data:

f(x,y)=12(xy)2

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(y)=0x(x)=1

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

y(x)=0y(y)=1

Differentiate the term 12(xy)2 with respect to x .

x(12(xy)2)=12[2(xy)]x(xy)=12[2(xy)](10)=12[2(xy)]=xy

Differentiate the term 12(xy)2 with respect to y .

y(12(xy)2)=12[2(xy)]y(xy)=12[2(xy)](01)=12[2(xy)]=yx

Find the gradient vector field of f(x,y)=12(xy)2 using equation (1).

Modify equation (1) as follows.

f(x,y)=x(12(xy)2)i+y(12(xy)2)j

Substitute (xy) for x(12(xy)2) and (yx) for y(12(xy)2) ,

f(x,y)=(xy)i+(yx)j=(xy),(yx)

Thus, the gradient vector field for 12(xy)2 is (xy)i+(yx)j_ .

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

|f(x,y)|=(xy)2+(yx)2=2|xy|

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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