   Chapter 16.3, Problem 24E

Chapter
Section
Textbook Problem

Find the work done by the force field F in moving an object from P to Q.24. F(x, y) = (2x + y) i + x j; P(1, 1), Q(4, 3)

To determine

To find: The work done by force field F(x,y)=(2x+y)i+xj .

Explanation

Given data:

Force field is F(x,y)=(2x+y)i+xj and points are P(1,1) and Q(4,3) .

Formula used:

Consider a vector field as F(x,y)=P(x,y)i+Q(x,y)j . The condition for vector field F being a conservative field is,

Py=Qx (1)

Here,

Py is continuous first-order partial derivative of P, and

Qx is continuous first-order partial derivative of Q,

Write the expression for work done (W) by force field F .

W=Cfdr (2)

Here,

f is potential function of F such that f=F .

Consider vector function r(t) , atb with a smooth curve C. Consider f is a differentiable function two or three variables of gradient function f and is continuous on curve C. Then,

Cfdr=f(r(b))f(r(a)) (3)

Compare the force field F(x,y)=(2x+y)i+xj with F(x,y)=P(x,y)i+Q(x,y)j .

P=2x+y (4)

Q=x (5)

Apply partial differentiation with respect to y on both sides of equation (4).

Py=y(2x+y)=y(2x)+y(y)=0+1 {t(k)=0,t(t)=1}=1

Apply partial differentiation with respect to x on both sides of equation (5).

Qx=x(x)=1 {t(t)=1}

Substitute 1 for Py and 1 for Qx in equation (1),

1=1

Therefore, force field F(x,y)=(2x+y)i+xj is conservative vector field and there exist a potential function f such that f=F .

Consider f=fx(x,y)i+fy(x,y)j .

Write the relation between the potential function f and vector field F .

f=F

Substitute fx(x,y)i+fy(x,y)j for f ,

F=fx(x,y)i+fy(x,y)j

Compare the equation F=fx(x,y)i+fy(x,y)j with F(x,y)=(2x+y)i+xj .

f=(2x+y)i+xj

Compare the equation f=fx(x,y)i+fy(x,y)j with f=(2x+y)i+xj

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