Section 16.1  Vector Fields
Denition. Let D be a set in Rn . A vector eld is a function F that assigns a vector F (x1 , . . . , xn )
in Vn to each point (x1 , . . . , xn ) in D.
F (x, y ) = P (x, y ), Q(x, y )
F (x, y, z ) = P (x, y, z ), Q(x, y, z ), R(
Section 16.4  Greens Theorem
Let C be a simple closed curve traversed once by a position vector r(t) = x(t), y (t) for a t b.
Let D be the region in the xy plane enclosed by C .
Denition. We sometimes refer to C as the boundary of the region D, denoted
Section 16.3  The Fundamental Theorem for Line Integrals
The Fundamental Theorem of Calculus. If F is continuous on [a, b] then
b
F (x) dx = F (b) F (a)
a
Theorem. Let C be a smooth (or piecewise smooth) curve given by the vector function r(t),
a t b. Le
Section 15.4  Double Integrals in Polar Coordinates
Polar Coordinates (r, )
r is the signed distance from the origin
is the angle measured counterclockwise
from the positive xaxis
x = r cos()
y = r sin()
r 2 = x2 + y 2
tan() =
y
x
(r, ) = (r, + )
Exam
Section 16.2  Line Integrals
Arc Length
Let C be a smooth curve traced out by the
position vector r(t) for a t b.
Plane curve:
r(t) = x(t), y (t) , a t b
Space curve:
r(t) = x(t), y (t), z (t) , a t b
Arc Length :
t
r (u)du
s(t) =
a
Two Dimensions:
Thr
Section 14.8  Lagrange Multipliers
Example 1. Find the absolute maximum and minimum values of f (x, y ) = x2 + 2y 2 on D =
cfw_(x, y ) : x2 + y 2 1.
In the interior of the disk: x2 + y 2 < 1
fx = 2x
fy = 4y
2x = 0
4y = 0
x=0
y=0
so there is one critical
Section 12.6  Cylinders and Quadric Surfaces
Denition. The traces (or crosssections) of a surface are the curves of intersection of the surface
with planes parallel to the coordinate planes.
Plane Sections of Surfaces from the Wolfram Demonstrations Pro
Section 15.1  Double Integrals Over Rectangles
Denite Integrals of Functions of One Variable
If f (x) 0 then
a and b.
b
a f (x)dx
represents the area under the curve y = f (x) above the xaxis between
n
b
f (x )x
i
f (x) dx = lim
a
n
i=1
Riemann Sum
Divi
Section 15.3  Double Integrals over General Regions
Denition
Let f (x, y ) be a function of two variables and suppose we want to nd the double integral
f (x, y ) dA
D
over a general bounded region D, not necessarily a rectangle.
Since D is bounded, D is
Section 15.2  Iterated Integrals
Let z = f (x, y ) be a function of two variables dened on a rectangle R = [a, b] [c, d].
The double integral
f (x, y ) dA is the
R
signed volume of the region between the
graph of f and the xy plane.
The volume is counte
Section 14.4  Tangent Planes and Linear Approximations
Let z = f (x, y ) be a function with continuous partial derivatives and S be the graph of f .
P = (x0 , y0 , z0 )
C1 is the trace of f in the plane y = y0 .
C2 is the trace of f in the plane x = x0 .
Section 14.7  Maximum and Minimum Values
Denition. A function f (x, y ) has a local maximum at (a, b) if f (x, y ) f (a, b) when (x, y ) is
near (a, b). The number f (a, b) is called a local maximum value.
If f (x, y ) f (a, b) when (x, y ) is near (a, b
Section 14.6  Directional Derivatives and the Gradient Vector
Partial Derivatives
If f is a function of two variables, its partial derivatives at a point (x0 , y0 ) are
f (x0 + h, y ) f (x0 , y0 )
f (x0 , y0 + h) f (x0 , y0 )
fy (x0 , y0 ) = lim
h0
h0
h
Section 14.2  Limits and Continuity
Recall: Let f (x) be a function of one variable dened on some open interval that contains the
number a, except possibly a itself. Then we say the limit of f (x) as x approaches a is L and write
lim f (x) = L
xa
if for
Section 14.3  Partial Derivatives
Example 1. Consider the function of two variables
f (x, y ) = sin(x y ) + cos(x).
Fixing one of the variables and letting the other vary, we can think of f as a function of one variable.
For example, setting y = 0 we obt
Section 14.5  The Chain Rule
Functions of One Variable
Theorem. If y = f (x) and x = g (t) are dierentiable then y is indirectly a function of t and
dy
dy dx
=
dt
dx dt
Example 1. If y = sin(x) and x = t2 + 1, nd
dy
dt .
Functions of Several Variables
Th
Section 13.4  Motion in Space: Velocity and Acceleration
Suppose that a particle moves along the curve C traced by the vector function r(t).
TEC V13.4: Velocity and Acceleration Vectors: http:/www.stewartcalculus.com/tec/
r(t) is the position vector at t
Section 13.3  Arc Length
Recall: (Section 10.2) If C is a plane curve described by the parametric equations x = f (t),
y = g (t) for a t b with f and g continuous on [a, b], and C traversed exactly once between a
and b, then the length of C is
b
b
dx
dt
Section 13.1  Vector Functions and Space Curves
A function f : X Y is a rule that maps each element of a domain X to an element of the range Y .
Denition. A vectorvalued function (or vector function) is a function whose domain is a set of
real numbers a
Chapter 12 Review
ThreeDimensional Coordinate Systems
Right hand rule: to choose an orientation on R3 = R R R
Curl ngers around the z axis in the direction o a counterclockwise rotation from the positive xaxis to the positive
y axis. The thumb points
Section 13.2  Derivatives and Integrals of Vector Functions
Denition. The derivative r of a vector function r is dened to be
dr
r(t + h) r(t)
= r (t) = lim
.
h0
dt
h
If the derivative r (t) exists, then r(t) is called dierentiable.
Denition. The second d
Chapter 16: Solutions
(1) (a) Find parametric equations for the line segment C from the point (1, 5, 0) to the point
(1, 6, 4).
x(t) = 1 + 2t
y (t) = 5 + t
z (t) = 4t
0t1
(b) Evaluate the line integral
xz 2 dy + y dz
C
where C is the line segment from par
Section 12.5: Equations of Lines and Planes
Equations of Lines
To nd the equation of a line, we need to know:
1. a point on the line
2. the direction of the line
In R2 , the slope m =
rise
run
=
y2 y1
x2 sx1
tells us the direction of the line.
SlopeInter
Section 12.4: The Cross Product
ab
, with row vectors u = a, b and
cd
Denition. The determinant of a 2 2 matrix, M =
v = c, d is
ab
= ad bc.
cd
The determinant of M is also denoted by det(M ) or M .
Geometrically, the absolute value of the determinant o
Chapter 14 Review
Partial Derivatives
First Partial Derivatives
fx =
f
x
and
fy =
f
y
Rule for Finding Partial Derivatives
To nd fx , regard y as constant and dierentiate f (x, y ) with respect to x.
To nd fy , regard x as constant and dierentiate f (x,
Chapter 15: Solutions
(1) Evaluate the iterated integrals.
(a)
4
2
(x +
1
4
y ) dx dy =
0
1
4
=
x=2
12
x +x y
2
dy
x=0
2 + 2 y dy
1
43
= 2y + y 2
3
y =4
=
y =1
46
3
(b)
2x
1
1
(x y ) dy dx =
0
0
x
1
x y y2
2
y =2x
2
2
1
=
0
dx
y =x
1
1
x2 (2 x) (2 x)2 x2
Chapter 16 Exercises
(1) (a) Find parametric equations for the line segment C from the point (1, 5, 0) to the point
(1, 6, 4).
(b) Evaluate the line integral
xz 2 dy + y dz
C
where C is the line segment from part (a).
(2) Determine whether the following v
In Class Exercises  Part 1 Solutions
(1) Let f (x, y ) = x2 + 4y 2
(a) Find the domain and range of f and sketch a contour map (or a family of level curves).
Domain = R R = R2 . Range=cfw_z : z 0.
Level curves are ellipses in the xy plane.
3
2
1
5
4
