Printed Page 1000
Concepts and Vocabulary
True or False If ∇ f=F for some function f that has continuous first-order partial derivatives, then F is a potential function for f.
False
Multiple Choice If ∫C1(Pdx+Qdy)=∫C2(Pdx+Qdy) for any two piecewise-smooth curves C1 and C2 with the same end points lying entirely in a region R, then ∫C(Pdx+Qdy) is said to be [(a) one-to-one, (b) singular, (c) independent of P and Q, (d) independent of the path.]
(d)
True or False The line integral ∫CF⋅dr, where C is a piecewise smooth curve and F is a conservative vector field , is independent of the path.
True
Multiple Choice Suppose F=F(x,y) is a conservative vector field and ∇ f=F. The function f is called a(n) [(a) gradient function, (b) independent function, (c) potential function, (d) gravity function] for F.
(c)
A piecewise-smooth curve C is ____________ if its initial point (x0,y0) and its terminal point (x1,y1) are the same.
Closed
If F is a conservative vector field on some open region R, and C is a closed, piecewise-smooth curve that lies entirely in R, then ∫CF⋅dr= ____________.
0
True or False An open region R is connected if there are two points (x1,y1) and (x2,y2) in R that can be joined by a piecewise-smooth curve C that lies entirely in R.
False
True or False The converse of the Fundamental Theorem of Line Integrals is true if the open region R is connected.
True
True or False Let r=r(t), a≤t≤b, trace out a piecewise-smooth curve C. Then C is closed if there is a least one point where the curve intersects itself.
False
True or False A parabola is a simple curve.
True
True or False A region R is called simply connected if it is closed.
False
Let F=F(x,y)=P(x,y)i+Q(x,y)j be a vector field, where the functions P and Q are continuous on some simply connected,
open region R. Suppose ∂P∂y and ∂Q∂x are also continuous on R.
Then F(x,y)=P(x,y)i+Q(x,y)j is a conservative vector field on R if and only if ____________.
∂P∂y=∂Q∂x
1001
Skill Building
Find the line integral ∫C(excosydx−exsinydy) for each curve C:
Find the line integral ∫C(eycosxdx+eysinxdy) for each curve C:
In Problems 15–20, F=F(x,y)=P(x,y)i+Q(x,y)j is a conservative vector field since F is the gradient of f(x,y). Use this fact to find ∫C[P(x,y)dx+Q(x,y)dy], where C is any piecewise smooth curve joining the two given points.
∫C[(2x−y)dx+(2y−x)dy] from (3,−1) to (5,4);f(x,y)=x2−xy+y2
8
∫C[(y2+2x)dx+2xydy] from (2,2) to (−3,8);(f(x,y)=xy2+x2
∫C(3x2ydx+x3dy) from (0,−1) to (3,4);f(x,y)=x3y
108
∫C(2xex2+y2dx+2yex2+y2dy) from (0,0) to (1,1);f(x,y)=ex2+y2
∫C(xx2+y2dx+yx2+y2dy) from (3,4) to (5,12); f(x,y)=ln√x2+y2
ln13−ln5
∫C(x√x2+y2dx+y√x2+y2dy) from (0,−4) to (3,4); f(x,y)=√x2+y2
In Problems 21–28, F=F(x,y)=P(x,y)i+Q(x,y)j is a conservative vector field. Use this fact to find ∫C[P(x,y)dx+Q(x,y)dy], where C is any piecewise smooth curve joining the two given points.
∫C(x3dx+y3dy) from (1,1) to (2,4)
1352
∫c(2x2dx−4ydy) from (−1,2) to (2,3)
∫C[(x2+y2)dx+(2xy+siny)dy] from (0,0) to (2,0)
83
∫C[(x−cosy)dx+xsinydy] from (1,π2) to (2,π)
∫C(y2exdx+2yexdy) from (0,0) to (ln2,2)
8
∫C[xeydx+12(x2−4)eydy] from (0,0) to (2,1)
∫C(yexdx+exdy) on a closed curve C
0
∫C(ycosxdx+sinxdy) on a closed curve C
In Problems 29–32, F=F(x,y)=P(x,y)i+Q(x,y)j is a conservative vector field. Find a potential function f for F.
F(x,y)=(5x−2y+1)i+(4−2x)j
f(x,y)=52x2−2xy+x+4y+K
F(x,y)=(4xy)i+(2x2+y)j
F(x,y)=(lny+2x)i+(xy)j
f(x,y)=xlny+x2+K
F(x,y)=(yex)i+exj
In Problems 33–36, determine whether each vector field F is a conservative vector field.
F(x,y)=x2i+y2j
f is a conservative vector field.
F(x,y)=xyi+xyj
F(x,y)=xeyi+12x2eyj
f is a conservative vector field.
F(x,y)=(x2+y2)i+(2xy−siny)j
In Problems 37–42:
∫C(xdx+ydy), where C is a piecewise-smooth curve joining the points (1,3) and (2,5)
∫C[2xydx+(x2+1)dy], where C is a piecewise-smooth curve joining the points (1,−4) and (−2,3)
∫C[(x2+3y)dx+3xdy], where C is a piecewise-smooth curve joining the points (1,2) and (−3,5)
∫C[(2x+y+1)dx+(x+3y+2)dy], where C is a piecewise-smooth curve joining the points (0,0) and (1,2)
∫C[(4x3+20xy3−3y4)dx+(30x2y2−12xy3+5y4)dy] , where C is a piecewise-smooth curve joining the points (0,0) and (1,1)
∫C[(2yx−1)dx+(lnx2)dy], where C is a piecewise-smooth curve in the first quadrant joining the points (1,1) and (5,5)
In Problems 43–54:
∫C(3x2y2dx+2x3ydy)
∫C[(2x+y)dx+(x−2y)dy]
∫C[(x+3y)dx+3xdy]
∫C[(2x+y)dx+(2y+x)dy]
1002
∫C[(2xy−y2)dx+(x2−2xy)dy]
∫C[y2dx+(2yx−ey)dy]
∫C[(x2−x+y2)dx−(yey−2xy)dy]
∫C[(3x2y+xy2+ex)dx+(x3+x2y+siny)dy]
∫C[(ycosx−2siny)dx−(2xcosy−sinx)dy]
∫C[(exsiny+2ysinx)dx+(excosy−2cosx)dy]
∫C[(2x+ycosx)dx+sinxdy]
∫C[(cosy−cosx)dx+(ey−xsiny)dy]
Applications and Extensions
Show that ∫C(−yx2+y2dx+xx2+y2dy) is independent of the path in the rectangle R whose vertices are (1a,−a), (a,−a), (1a,−1a), and (a,−1a), a>1. Find a potential function f for F.
P,Q,∂P∂y, and ∂Q∂x are continuous everywhere in the rectangle, and ∂P∂y=∂Q∂x=y2−x2(x2+y2)2; f(x,y)=tan−1(yx)+K.
Given the constant vector c, show that ∇(c⋅r)=c, where r=xi+yj+zk. Use the result to prove that ∫Cc⋅dr=0, where C is any closed piecewise smooth curve in space.
Suppose that F(x,y) is a force field directed toward the origin with magnitude inversely proportional to the square of the distance from the origin. (Such “inverse square law” forces are common in nature. See Examples 4 and 5 in Section 15.1, pp. 975-976.)
Suppose f and g are differentiable functions of one variable. Show that ∫C[f(x)dx+g(y)dy]=0, where C is any circle in the xy-plane.
Challenge Problem
Let f and g have continuous partial derivatives in a plane region R, and let C be a piecewise-smooth curve in R going from A to B. Show that ∫Cf∇g⋅dr=f(B)g(B)−f(A)g(A)−∫Cg∇ f⋅dr
See Student Solutions Manual.