Math 21a Midterm I Review

Midterm I Review
Math 21a

March 5, 2008

.

.
Image: Flickr user Mr. Theklan
. . . . . .

Announcements

◮ Midterm covers up to Section 11.4 in text
◮ Any topics not covered in class or on HW will not be on test
(i.e., curvature)
◮ Complete list of learning objectives on the Midterm I Review
sheet
◮ Odd problems, chapter reviews, old exams, reviews

. . . . . .

Outline

Vectors and the Geometry of Derivatives and Integrals of
Space Vector Functions
Three-Dimensional Arc Length (not Curvature)
Coordinate Systems Motion in Space: Velocity and
Vectors Acceleration
The Dot Product Parametric Surfaces
The Cross Product Partial Derivatives
Equations of Lines and Planes Functions of Several Variables
Functions and surfaces Utility Functions and
Cylindrical and Spherical indifference curves
Coordinates Limits and Continuity
Vector Functions Partial Derivatives
Vector Functions and Space Tangent Planes and Linear
Curves Approximations

. . . . . .

Three-Dimensional Coordinate Systems
Section 9.1 Learning Objectives

◮ To understand and be able to apply rectangular coordinate
systems in R3 and the right hand rule.
◮ To understand and be able to ﬁnd the distance d between two
points (x1 , y1 , z1 ) and (x2 , y2 , z2 ) in R3 and to be able to use the
distance formula
√
d = (x1 − x2 )2 + (y1 − y2 )2 + (z1 − z2 )2 .

◮ To understand and be able to apply the equation of a sphere of
radius r and center (x0 , y0 , z0 ),

(x − x0 )2 + (y − y0 )2 + (z − z0 )2 = r2 .

. . . . . .

Axes in three-dimensional space
z
. .
z

y
.

.
y
. x
.
x
.
x
.
y
.
.
y

x
.
z
.

.
z
. . . . . .

The right-hand rule

z
.

x
.

.

.
y

.

. . . . . .

Distance in space

Example
Find the distance between the points P1 (3, 2, 1) and P2 (4, 4, 4).

. . . . . .

Distance in space

Example
Find the distance between the points P1 (3, 2, 1) and P2 (4, 4, 4).
z
.
Solution
..2
P

.
d .
3

. 2
. .
y
1
.. √
P
.1 . 5

x
.
√ √
d= 5 + 32 = 1+4+9

. . . . . .

Distance in space—General
Theorem
The distance between (x1 , y1 , z1 ) and (x2 , y2 , z2 ) is
√
(x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2

In space, the locus of all points which are a ﬁxed distance from a
ﬁxed point is a sphere.

. . . . . .

Munging an equation to see its surface
Example
Find the surface is represented by the equation

x2 + y2 + z2 − 4x + 8y − 10z + 36 = 0

. . . . . .

Munging an equation to see its surface
Example
Find the surface is represented by the equation

x2 + y2 + z2 − 4x + 8y − 10z + 36 = 0

Solution
We can complete the square:

0 = x2 − 4x + 4 + y2 + 8y + 16 + z2 − 10z + 25 + 36 − 4 − 16 − 25
= (x − 2)2 + (y + 4)2 + (z − 5)2 − 9

So

(x − 2)2 + (y + 4) + (z − 5)2 = 9
=⇒ |(x, y, z)(2, −4, 5)| = 3

This is a sphere of radius 3, centered at (2, −4, 5).
. . . . . .

Vectors
Learning objectives for Section 9.2

◮ To understand and be able to apply the definition of a scalar and
a vector.
◮ To be able to represent vectors geometrically as directed line
segments or algebraically as ordered pairs or triples of numbers.
◮ To understand and be able to apply the axioms or vector
addition and scalar multiplication.
◮ To be understand and be able to determine the length of a
vector.
◮ To understand the definitions of unit vectors, standard basis
vectors, and position vectors and to be able to apply these
definitions.

. . . . . .

What is a vector?
Deﬁnition
◮ A vector is something that has magnitude and direction
◮ We denote vectors by boldface (v) or little arrows (⃗ One is
v).
good for print, one for script
◮ Given two points A and B in ﬂatland or spaceland, the vector
which starts at A and ends at B is called the displacement
−→
vector AB.
◮ Two vectors are equal if they have the same magnitude and
direction (they need not overlap)

B
. D
.
.
v .
u

..
A C
.
. . . . . .

Vector or scalar?

Deﬁnition
A scalar is another name for a real number.

Example
Which of these are vectors or scalars?
(i) Cost of a theater ticket
(ii) The current in a river
(iii) The initial ﬂight path from Boston to New York
(iv) The population of the world

. . . . . .

Vector or scalar?

Deﬁnition
A scalar is another name for a real number.

Example
Which of these are vectors or scalars?
(i) Cost of a theater ticket scalar
(ii) The current in a river vector
(iii) The initial ﬂight path from Boston to New York vector
(iv) The population of the world scalar

. . . . . .

Vector addition

Deﬁnition
If u and v are vectors positioned so the initial point of v is the
terminal point of u, the sum u + v is the vector whose initial point is
the initial point of u and whose terminal point is the terminal point
of v.

.
u
.
v .
v
. +v
u . +v
u
.
v
.
u .
u
. .
The triangle law The parallelogram law

. . . . . .

Opposite and difference

Deﬁnition
Given vectors u and v,
◮ the opposite of v is the vector −v that has the same length as
v but points in the opposite direction
◮ the difference u − v is the sum u + (−v)

.
v

.
u
. −
. v
. −v
u

. . . . . .

Scaling vectors

Deﬁnition
If c is a nonzero scalar and v is a vector, the scalar multiple cv is
the vector whose
◮ length is |c| times the length of v
◮ direction is the same as v if c > 0 and opposite v if c < 0
If c = 0, cv = 0.

2
.v
.
v
−2
. 1v .

. . . . . .

Components deﬁned
Deﬁnition
◮ Given a vector a, it’s often useful to move the tail to O and
measure the coordinates of the head. These are called the
components of a, and we write them like this:

a = ⟨a1 , a2 , a3 ⟩

or just two components if the vector is the plane. Note the
angle brackets!
◮ Given a point P in the plane or space, the position vector of
−
→
P is the vector OP.

Fact −
→
Given points A(x1 , y1 , z1 ) and B(x2 , y2 , z2 ) in space, the vector AB has
components
−
→
AB = ⟨x2 − x1 , y2 − y1 , z2 − z1 ⟩
. . . . . .

Vector algebra in components

Theorem
If a = ⟨a1 , a2 , a3 ⟩ and b = ⟨b1 , b2 , b3 ⟩, and c is a scalar, then
◮ a + b = ⟨a1 + b1 , a2 + b2 , a3 + b3 ⟩
◮ a − b = ⟨a1 − b1 , a2 − b2 , a3 − b3 ⟩
◮ ca = ⟨ca1 , ca2 , ca3 ⟩

. . . . . .

Properties

Theorem
Given vectors a, b, and c and scalars c and d, we have

1. a+b=b+a 6. (c + d)a = ca + da
2. a + (b + c ) = ( a + b ) + c
7. (cd)a = c(da)
3. a+0=a
4. a + (−a) = 0 8. 1a = a
5. c(a + b) = ca + cb

These can be veriﬁed geometrically.

. . . . . .

Length

Deﬁnition
Given a vector v, its length is the distance between its initial and
terminal points.

Fact
The length of a vector is the square root of the sum of the squares of its
components: √
|⟨a1 , a2 , a3 ⟩| = a2 + a2 + a2
1 2 3

. . . . . .

Useful vectors

Deﬁnition
◮ A unit vector is a vector whose length is one
◮ We deﬁne the standard basis vectors i = ⟨1, 0, 0⟩,
j = ⟨0, 1, 0⟩, k = ⟨0, 0, 1⟩. In script, they’re often written as ˆ, ȷ,
ı ˆ
ˆ
k.

Fact
Any vector a can be written as a linear combination of the standard basis
vectors
⟨a1 , a2 , a3 ⟩ = a1 i + a2 j + a3 k.

. . . . . .

The Dot Product I

◮ To understand and be able to use the deﬁnitions of the dot
product to measure the length of a vector and the angle θ
between two vectors. Given two vectors a = ⟨a1 , a2 , a3 ⟩ and
b = ⟨b1 , b2 , b3 ⟩,

a · b = |a||b| cos θ = a1 b1 + a2 b2 + a3 b3 .

◮ To understand and be able to apply properties of the dot
product.

. . . . . .

The Dot Product II

◮ To understand that two vectors are orthogonal if their dot
product is zero.
◮ To understand and be able to determine the projection of a
vector a onto a vector b,

a·b
proja b = a
|a|2

◮ To be able to use vectors and the dot product in applications.

. . . . . .

Deﬁnition
If a and b are any two vectors in the plane or in space, the dot
product (or scalar product) between them is the quantity

a · b = |a| |b| cos θ,

where θ is the angle between them.
Another way to say this is that a · b is |b| times the length of the
projection of a onto b.

.
a

. · b is |b| times this length.
a
.
b

In components, if a = ⟨a1 , a2 , a3 ⟩ and b = ⟨b1 , b2 , b3 ⟩, then
a · b = a1 b1 + a2 b2 + a3 b3. . . . . .

Algebraic properties of the dot product

Fact
If a, b and c are vectors are c is a scalar, then

1. a · a = |a|2 4. (ca) · b = c(a · b) = a · (cb)
2. a · b = b · a 5. 0 · a = 0
3. a · (b + c) = a · b + a · c

. . . . . .

Geometric properties of the dot product

Fact
The projection of b onto a is given by

a·b
proja b = a
|a|2

Fact
The angle between two nonzero vectors a and b is given by

a·b
cos θ = ,
|a| |b|

where θ is taken to be between 0 and π .

. . . . . .

More geometric properties of the dot product

Fact
The angle between two nonzero vectors a and b is acute if a · b > 0.
obtuse if a · b < 0, right if a · b = 0. The vectors are parallel if
a · b = ± |a| |b|.
◮ b is a positive multiple of a if a · b = |a| |b|
◮ b is a negative multiple of a if a · b = − |a| |b|

. . . . . .

Other uses of the dot product

◮ similarity
◮ Work
W=F·d

. . . . . .

The Cross Product

◮ To understand and be able to use the deﬁnitions of the cross
product to ﬁnd a vector that is orthogoal two vectors. Given
two vectors a = ⟨a1 , a2 , a3 ⟩ and b = ⟨b1 , b2 , b3 ⟩,

a × b = (|a||b| sin θ)n = ⟨a2 b3 − a3 b2 , a3 b1 − a1 b3 , a1 b2 − a2 b1 ⟩,

where n is a unit vector perpendicular to both a and b.
◮ To understand and be able to apply properties of the dot
product, especially a × b = −b × a.
◮ To understand that two vectors are parallel if and only if their
cross product is zero.
◮ To be able to use vectors and the cross product in applications.

. . . . . .

Deﬁnition of the cross product
Deﬁnition
Given vectors a and b in space, the cross product of a and b is
the vector
a × b = |a| |b| (sin θ) n,
where n is a vector perpendicular to a and b such that (a, b, n) is a
right-handed set of three vectors.
In components, if

a = ⟨a1 , a2 , a3 ⟩= a1 i + a2 j + a3 k
b = ⟨b1 , b2 , b3 ⟩= b1 i + b2 j + b3 k

Then

a × b = (a2 b3 − b2 a3 )i + (a3 b1 − b3 a1 )j + (a1 b2 − b1 a2 )k
= ⟨a 2 b 3 − b 2 a 3 , a 3 b 1 − b 3 a 1 , a 1 b 2 − b 1 a 2 ⟩

. . . . . .

Determinant formula

This is only to help you remember, in case you’ve seen determinants
of 3 × 3 matrices:

i j k
a a a a a a
a1 a2 a3 = i 2 3 − j 1 3 + k 1 2
b2 b3 b1 b 3 b1 b2
b1 b2 b3
= (a2 b3 − b2 a3 )i − (b3 a1 − a3 b1 )j + (a1 b2 − b1 a2 )k
=a×b

. . . . . .

Algebraic Properties of the Cross Product

If a, b, and c are vectors and c is a scalar, then
1. a × b = −b × a
2. (ca) × b = c(a × b) = a × (cb)
3. a × (b + c) = a × b + a × c
4. (a + b) × c = a × c + b × c

. . . . . .

Geometric Properties of the cross product

◮ a × b = 0 exactly when a and b are parallel.
◮ The magnitude of the cross product a × b is the area of the
parallelogram with sides a and b.

.
b |
. b| sin θ

θ
.
. .
a

. . . . . .

Applications of the cross product

◮ Area of a parallelogram
◮ Torque τ = r × F
◮ Volume of a parallelipiped (scalar triple product)

V = |a · (b × c)|

. . . . . .

The scalar triple product in action
Example
Find the volume of the parallelipiped spanned by a = ⟨2, 0, −3⟩,
b = ⟨1, 1, 1⟩, and c = ⟨0, 4, −1⟩

. . . . . .

The scalar triple product in action
Example
Find the volume of the parallelipiped spanned by a = ⟨2, 0, −3⟩,
b = ⟨1, 1, 1⟩, and c = ⟨0, 4, −1⟩
Solution

The scalar triple product can be
calculated by
.
b 2 0 −3
.
c 1 1 1 = −22
. 0 4 −1

It’s negative because the triple
.
a (a, b, c) is left-handed. So the
volume is 22.

. . . . . .

Equations of Lines and Planes I

◮ To understand and be able to represent a line ℓ in R3 using
◮ the vector equation
r = r0 + tv
◮ the parametric equations

x = x0 + at y = y0 + bt z = z0 + ct

◮ the symmetric equations
x − x0 y − y0 z − z0
= =
a b c

. . . . . .

Equations of Lines and Planes II

◮ To be able to represent line segment in R3 from r0 to r1 ,

r(t) = (1 − t)r0 + tr1 ,

where 0 ≤ t ≤ 1.
◮ To understand and be able to represent a plane in R3 given a
normal vector n = ⟨a, b, c⟩ and a point P0 = (x0 , y0 , z0 ) in the
plane,
n · ⟨x − x 0 , y − y 0 , z − z 0 ⟩ = 0
or
a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0.

. . . . . .

Equations of Lines and Planes III

◮ To understand and be able to calculate the distance D from a
point P1 = (x1 , y1 , z1 ) to the plane ax + by + cz + d = 0,

|ax1 + by1 + cz1 + d|
D= √ .
a 2 + b2 + c2

. . . . . .

Applying the deﬁnition
Example
Find the vector, parametric, and symmetric equations for the line
that passes through (1, 2, 3) and (2, 3, 4).

. . . . . .

Applying the deﬁnition
Example
Find the vector, parametric, and symmetric equations for the line
that passes through (1, 2, 3) and (2, 3, 4).
Solution
◮ Use the initial vector ⟨1, 2, 3⟩ and direction vector
⟨2, 3, 4⟩ − ⟨1, 2, 3⟩ = ⟨1, 1, 1⟩. Hence

r(t) = ⟨1, 2, 3⟩ + t ⟨1, 1, 1⟩

◮ The parametric equations are

x=1+t y=2+t z=3+t

◮ The symmetric equations are

x−1=y−2=z−3
. . . . . .

Planes

The set of all points whose displacement vector from a ﬁxed point is
perpendicular to a ﬁxed vector is a plane.

z
.

.
n

.0
r
. y
.

.
x

. . . . . .

Equations for planes
The plane passing through the point with position vector
r0 = ⟨x0 , y0 , z0 ⟩ perpendicular to ⟨a, b, c⟩ has equations:
◮ The vector equation

n · (r − r0 ) = 0 ⇐⇒ n · r = n · r0

◮ Rewriting the dot product in component terms gives the scalar
equation

a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0

The vector n is called a normal vector to the plane.
◮ Rearranging this gives the linear equation

ax + by + cz + d = 0,

where d = −ax0 − by0 − cz0 .
. . . . . .

Example
Find an equation of the plane that passes through the points
P(1, 2, 3), Q(3, 5, 7), and R(4, 3, 1).

. . . . . .

Example
Find an equation of the plane that passes through the points
P(1, 2, 3), Q(3, 5, 7), and R(4, 3, 1).

Solution−
→ −
→ − →
Let r0 = OP = ⟨1, 2, 3⟩. To get n, take PQ × PR:

i j k
−
→ − →
PQ × PR = 2 3 4 = ⟨−10, 16, −7⟩
3 1 −2

So the scalar equation is

−10(x − 1) + 16(y − 2) − 7(z − 3) = 0.

. . . . . .

Distance from point to line

Deﬁnition
The distance between a point and a line is the smallest distance
from that point to all points on the line. You can ﬁnd it by projection.

Q
.
.

b·v
.b − v
v·v
.
b

b·v
. rojv b =
p v
.
v v·v
θ
.
.
P
.0

. . . . . .

Distance from a point to a plane
Deﬁnition
The distance between a point and a plane is the smallest distance
from that point to all points on the line.
.
.
Q

.
b |n · b|
.
|n|
.
n
.
P
.0

To ﬁnd the distance from the a point to a plane, project the
displacement vector from any point on the plane to the given point
onto the normal vector.
. . . . . .

Distance from a point to a plane II

We have
|n · b|
D=
|n|
If Q = (x1 , y1 , z1 ), and the plane is given by ax + by + cz + d = 0, then
n = ⟨a, b, c⟩, and

n · b = ⟨a, b, c⟩ · ⟨x1 − x0 , y1 − y0 , z1 − z0 ⟩
= ax1 + by1 + cz1 − ax0 − by0 − cz0
= ax1 + by1 + cz1 + d

So the distance between the plane ax + by + cz + d = 0 and the point
(x1 , y1 , z1 ) is
|ax1 + by1 + cz1 + d|
D= √
a 2 + b 2 + c2

. . . . . .

Functions and surfaces

◮ To understand and be able to represent a function of two
variables, z = f(x, y), graphically. Section 9.6)
◮ To understand and be able to use the trace (cross-section) of a
function. Section 9.6)
◮ To understand and be able to determine the domain of a
function z = f(x, y). Section 9.6)
◮ To understand and be able to represent a selection of quadric
surfaces graphically.

. . . . . .

function, domain, range

Deﬁnition
A function f of two variables is a rule that assigns to each
ordered pair of real numbers (x, y) in a set D a unique real number
denoted by f(x, y). The set D is the domain of f and its range is
the set of values that f takes on. That is

range f = { f(x, y) | (x, y) ∈ D }

. . . . . .

Example

Example
√
Find the domain and range of f(x, y) = xy.

. . . . . .

Example

Example
√
Find the domain and range of f(x, y) = xy.

Solution
◮ Working from the outside in, we see that xy must be nonnegative,
which means x ≥ 0 and y ≥ 0 or x ≤ 0 and y ≤ 0. Thus the domain
is the union of the coordinate axes, and the ﬁrst and third quadrants.
◮ The range of f is the set of all “outputs” of f. Clearly the range of f is
restricted to the set of nonnegative numbers. To make sure that we
can get all nonnegative numbers x, notice x = f(x, x).

. . . . . .

Solution continued

Here is a sketch of the domain:

.

. . . . . .

Traces and surfaces

Example
Describe and sketch the surfaces.

(i) z = 4x2 + y2 (vi) y2 − 9x2 − 4z2 = 36
(ii) z = 4 − x 2 − y2
(vii) x2 + 4y2 + 2z2 = 4
(iii) z 2 = 4 (x 2 + y 2 )
(iv) x = 2y2 − z2 (viii) y2 + z2 = 1
(v) x2 + y2 − 9z2 = 9

. . . . . .

Sketch of z = 4x2 + y2
The trace in the xz-plane (y = 0) is the parabola z = 4x2 . The trace
in the yz-plane is the parabola z = y2 . The trace in the xy-plane is the
curve 4x2 + y2 = 0, which is just a point (0, 0). But the trace in the
plane z = k (k > 0) is the ellipse 4x2 + y2 = k. So we get ellipses
crossing the parabolas, an elliptic paraboloid.
2

1

0

1

2
2.0

1.5

1.0

0.5

0.0
1.0
0.5
0.0
0.5
1.0

. . . . . .

Sketch of z2 = 4(x2 + y2 )
The trace in the xz-plane is the equation z2 = 4x2 , or z = ±2x. The
trace in the yz-plane is the equation z = ±2y. Traces in the plane
z = k are circles k2 = 4(x2 + y2 ). So we have a cone.
1.0
0.5

0.0

0.5

1.0
1.0

0.5

0.0

0.5

1.0
1.0
0.5
0.0

0.5

1.0

. . . . . .

Sketch of x = 2y2 − z2
The trace in the xy-plane is a parabola x = 2y2 . The trace in the
xz-plane is a parabola opening the other direction: x = −z2 . Traces
x = k give hyperbolas. So we have a hyperbolic paraboloid.
1.0
0.5

0.0

0.5

1.0
1.0

0.5

0.0

0.5

1.0
1.0
0.5
0.0

0.5

1.0

. . . . . .

Sketch of x2 + y2 − 9z2 = 9

The x2 + y2 piece tells us that this is a surface of revolution, where
the z-axis is the axis of revolution. If we ﬁnd the xz-trace, we get
x2 − 9z2 = 9, a hyperbola. We get a hyperboloid of one sheet.

2

1

0 4

1 2
2
4 0

2

0 2

2
4
4

. . . . . .

Sketch of y2 − 9x2 − 4z2 = 36
Traces in the xy and yz planes are hyperbolas. Traces in the plane
y = k are ellipses 9x2 + 4z2 = k2 − 36. We have a hyperboloid of
two sheets.
10

5

0

5

10
4

2

0

2

4

2
0
2

. . . . . .

Sketch of x2 + 4y2 + 2z2 = 4

The traces in the planes x = 0, y = 0, and z = 0 are all ellipses. We
get an ellipsoid.
1.0 2
0.5 1
0.0 0
0.5 1
1.0
2

2

1

0

1

2

. . . . . .

Traces and surfaces

Example
Describe and sketch the surfaces.

(i) z = 4x2 + y2 elliptic (v) x2 + y2 − 9z2 = 9
paraboloid hyperboloid of one sheet
(ii) z = 4 − x2 − y2 elliptic (vi) y2 − 9x2 − 4z2 = 36
paraboloid hyperboloid of two sheets
(iii) z2 = 4(x2 + y2 ) cone (vii) x2 + 4y2 + 2z2 = 4 ellipsoid
(iv) x = 2y2 − z2 hyperbolic (viii) y2 + z2 = 1 cylinder
paraboloid

. . . . . .

Cylindrical and Spherical Coordinates I

◮ To understand and be able to apply the polar coordinate system in
R2 , and to understand the relationship to rectangular
coordinates:

x = r cos θ y = r sin θ
2 2 2
r =x +y tan θ = y/x,

where r ≥ 0 and 0 ≤ θ ≤ 2π .

. . . . . .

Cylindrical and Spherical Coordinates II

◮ To understand and be able to apply the cylindrical coordinate
system in R3 , and to understand the relationship to rectangular
coordinates:

x = r cos θ y = r sin θ z=z
r 2 = x2 + y 2 tan θ = y/x z = z,

where r ≥ 0 and 0 ≤ θ ≤ 2π .

. . . . . .

Cylindrical and Spherical Coordinates III

◮ To understand and be able to apply the spherical coordinate
system in R3 , and to understand the relationship to rectangular
coordinates:

x = ρ sin φ cos θ y = ρ sin φ sin θ z = ρ cos φ
2 2 2 2
ρ =x +y +z ,

where r ≥ 0, 0 ≤ θ ≤ 2π , and 0 ≤ φ ≤ π .

. . . . . .

Why different coordinate systems?

◮ The dimension of space comes from nature
◮ The measurement of space comes from us
◮ Different coordinate systems are different ways to measure
space

. . . . . .

Vectors and the Geometry of Space

Conversion from polar to
cartesian (rectangular)

x = r cos θ
y = r sin θ

Conversion from cartesian to
.
r .
y
θ polar:
. .
.
x √
r = x2 + y 2

x y y
cos θ = sin θ = tan θ =
r r x

. . . . . .

Cylindrical Coordinates
Just add the vertical dimension

Conversion from cylindrical to
cartesian (rectangular):
z
.
x = r cos θ y = r sin θ
(
. r, θ, z) z=z

z z
. . cylindrical:
. r
. .
y √
θ
. y
.
. r = x2 + y 2
x
. .
x y y
x
. cos θ = sin θ = tan θ =
r r x
z=z

. . . . . .

Spherical Coordinates
like the earth, but not exactly

Conversion from spherical to
cartesian (rectangular):

z
. x = ρ sin φ cos θ
. y = ρ sin φ sin θ
(
. r, θ, φ) z = ρ cos φ
.
φ ρ
. .
z
. spherical:
. .
y
√ √
x
. θ
.
r = x2 + y2 ρ = x2 + y2 + z2
y
. x y y
x
. cos θ = sin θ = tan θ =
r r x
z
cos φ =
ρ
. . . . . .

Vector Functions and Space Curves

◮ To understand and be able to apply the concept of a
vector-valued function,

r(t) = ⟨f(t), g(t), h(t)⟩ = f(t)i + g(t)j + h(t)k.

◮ To be able to apply the concepts of limits and continuity to
vector-valued functions.
◮ To understand that a curve in R3 can be represented
parametrically

x = f(t) y = g(t) z = h (t)

or by a vector-valued function

r(t) = ⟨f(t), g(t), h(t)⟩

. . . . . .

Example
Given the plane curve described by the vector equation

r(t) = sin(t)i + 2 cos(t)j

(a) Sketch the plane curve.

. . . . . .

Solution

r(t) = r(t) = sin(t)i + 2 cos(t)j

y
.

t r (t)
0 2j
π/2 i . (π/4)
r
π −2j . x
.
3π/2 −i
2π 2j

. . . . . .

Derivatives and Integrals of Vector Functions

◮ To understand and be able to calculate the derivative of a
vector-valued function.
◮ To understand and be able to apply the basic rules of
differentiation for vector-valued functions.
◮ To understand and be able to ﬁnd the deﬁnite integral of a
vector-valued function.

. . . . . .

Derivatives of vector-valued functions

Definition
Let r be a vector function.
◮ The limit of r at a point a is defined componentwise:
⟨ ⟩
lim r(t) = lim f(t), lim g(t), lim h(t)
t→a t→a t→a t→a

◮ The derivative of r is defined in much the same way as it is
for real-valued functions:
dr r(t + h) − r(t)
= r′ (t) = lim
dt h→0 h

. . . . . .

Rules for differentiation

Theorem
Let u and v be differentiable vector functions, c a scalar, and f a
real-valued function. Then:
d
1. [u(t) + v(t)] = u′ (t) + v′ (t)
dt
d
2. [cu(t)] = cu′ (t)
dt
d
3. [f(t)u(t)] = f′ (t)u(t) + f(t)u′ (t)
dt
d
4. [u(t) · v(t)] = u′ (t) · v(t) + u(t) · v′ (t)
dt
d
5. [u(t) × v(t)] = u′ (t) × v(t) + u(t) × v′ (t)
dt
d
6. [u(f(t))] = f′ (t)u′ (f(t))
dt

. . . . . .

Example


(b) Find r′ (t)

. . . . . .

Solution

r′ (t) = cos(t)i − 2 sin(t)j

y
.

t r (t)
0 2j
π/2 i . (π/4)
r
π −2j . x
.
3π/2 −i
2π 2j

. . . . . .

Example


(b) Find r′ (t)
(c) Sketch the position vector r(π/4) and the tangent vector
r′ (π/4).

. . . . . .

Solution

r′ (t) = cos(t)i − 2 sin(t)j

y
.

t r (t)
0 2j
π/2 i . (π/4)
r . ′ (π/4)
r
π −2j . .
x
3π/2 −i
2π 2j

. . . . . .

Integrals of vector-valued functions
Definition
Let r be a vector function defined on [a, b]. For each whole number
n, divide the interval [a, b] into n pieces of equal width ∆t. Choose a
point t∗ on each subinterval and form the Riemann sum
i

∑
n
Sn = r(t∗ ) ∆t
i
i=1

Then define
∫ b ∑
n
r(t) dt = lim Sn = lim r(t∗ ) ∆t
i
a n→∞ n→∞
i=1
[ ]
∑ n ∑
n ∑
n
= lim f(t∗ ) ∆ti
i + g(t∗ ) ∆tj
i + h(t∗ ) ∆tk
i
n→∞
i=1 i=1 i=1
(∫ b ) (∫ b ) (∫ b )
= f(t) dt i + g(t) dt j + h(t) dt k
a a a
. . . . . .

FTC for vector functions

Theorem (Second Fundamental Theorem of Calculus)
If r(t) = R′ (t), then
∫ b
r(t) dt = R(b) − R(a)
a

Example ∫ π
Given r(t) = ⟨t, cos 2t, sin 2t⟩, ﬁnd r(t) dt.
0

Answer
⟨∫ π ∫ π ∫ π ⟩ ⟨ ⟩
π2
t dt, cos 2t dt, sin 2t dt = , 0, 0
0 0 0 2

. . . . . .

Arc Length (not Curvature)

◮ To understand and be able to apply the arc length of a curve,
r(t) = ⟨f(t), g(t), h(t)⟩, over an interval a ≤ t ≤ b,
∫ b ∫ b√
′
L= |r (t)| dt = [f′ (t)]2 + [g′ (t)]2 + [h′ (t)]2 dt.
a a

◮ To understand and be able to apply the arc length function of a
curve, r(t) = ⟨f(t), g(t), h(t)⟩, over an interval a ≤ t ≤ b,
∫ t ∫ t√
′
s (t) = |r (u)| du = [f′ (u)]2 + [g′ (u)]2 + [h′ (u)]2 du.
a a

. . . . . .

Length of a curve

Break up the curve into pieces, and approximate the arc length with
the sum of the lengths of the pieces:
y
.

. .
x

. . . . . .

Length of a curve

Break up the curve into pieces, and approximate the arc length with
the sum of the lengths of the pieces:
y
.

∑√
n
. ≈
L (∆xi )2 + (∆yi )2
i=1

. x
.

. . . . . .

Sum goes to integral
If ⟨x, y⟩ is given by a vector-valued function r(t) = ⟨f(t), g(t), ⟩ with
domain [a, b], we can approximate:
∆xi ≈ f′ (ti )∆ti ∆xi ≈ g′ (ti )∆ti
So
∑√
n ∑ √[
n
]2
L≈ (∆xi )2 + (∆yi )2 ≈ f′ (ti )∆ti + [g′ (ti )∆ti ]2
i=1 i=1
∑ √[
n
]2
= f′ (ti ) + [g′ (ti )]2 ∆ti
i=1
As n → ∞, this converges to
∫ b√
L= [f′ (t)]2 + [g′ (t)]2 dt
a
In 3D, r(t) = ⟨f(t), g(t), h(t)⟩, and
∫ b√
L= [f′ (t)]2 + [g′ (t)]2 + [h′ (t)]2 dt
a
. . . . . .

Example

Example
Find the length of the parabola y = x2 from x = 0 to x = 1.

. . . . . .

Example

Example
Find the length of the parabola y = x2 from x = 0 to x = 1.

Solution ⟨ ⟩
Let r(t) = t, t2 . Then
∫ 1√
√
5 1 √
L= 1 + (2t)2 = + ln 2 + 5
0 2 4

. . . . . .

Motion in Space: Velocity and Acceleration

◮ To understand the ﬁrst derivative as the velocity of a curve and
the second derivative as the acceleration of a curve and be able
to apply these deﬁnitions.

. . . . . .

Velocity and Acceleration

Deﬁnition
Let r(t) be a vector-valued function.
◮ The velocity v(t) is the derivative r′ (t)
◮ The speed is the length of the derivative |r′ (t)|
◮ The acceleration is the second derivative r′′ (t).

. . . . . .

Example
Find the velocity, acceleration, and speed of a particle with position
function
r(t) = ⟨2 sin t, 5t, 2 cos t⟩

. . . . . .

Example
Find the velocity, acceleration, and speed of a particle with position
function
r(t) = ⟨2 sin t, 5t, 2 cos t⟩

Answer
◮ r′ (t) = ⟨2 cos(t), 5, −2 sin(t)⟩
√
◮ r′ (t) = 29
◮ r′′ (t) = ⟨−2 sin(t), 0, −2 cos(t)⟩

. . . . . .

Parametric Surfaces

◮ To understand and be able to apply the concept of a parametric
surface. A parametric surface may be deﬁned by a vector
equation,
r(t) = x(u, v) i + y(u, v) j + z(u, v) k
or by a set of parametric equations

x = x (u , v ) y = y(u, v) z = z(u, v),

where u and v are variables with a domain D contained in R2 .
◮ To understand and be able to represent surfaces of revolutions
parametrically.

. . . . . .

Surfaces with easy parametrizations

◮ graph
◮ plane
◮ sphere
◮ surface of revolution

. . . . . .

Parametrizing a graph

Example
Parametrize the surface described by

z = x 2 + y2 , −2 ≤ x ≤ 2 , −3 ≤ y ≤ 3

. . . . . .

Parametrizing a graph

Example
Parametrize the surface described by

z = x 2 + y2 , −2 ≤ x ≤ 2 , −3 ≤ y ≤ 3

Solution
Let
x = u, y = v, z = u 2 + v2
on the domain −2 ≤ u ≤ 2, −3 ≤ v ≤ 3.

. . . . . .

parametrizing a sphere

Example
Find a parametrization for the unit sphere.

. . . . . .

parametrizing a sphere

Example
Find a parametrization for the unit sphere.

Solution
Use spherical coordinates. Let:

x = cos θ sin φ, y = sin θ sin φ, z = cos φ

The domain of parametrization is 0 ≤ θ ≤ 2π , 0 ≤ φ ≤ π .

. . . . . .

Parametrizing a surface of revolution

Example
Find a parametrization of the surface described by

x2 − y 2 + z 2 = 1 , −3 ≤ y ≤ 3

. . . . . .

Parametrizing a surface of revolution

Example
Find a parametrization of the surface described by

x2 − y 2 + z 2 = 1 , −3 ≤ y ≤ 3

Solution
The surface is the graph of z2 − y2 = 1 revolved around the y-axis. So let
√ √
x= 1+y 2 cos θ, y = u, z = 1 + y2 sin θ

The domain is −3 ≤ u ≤ 3, 0 ≤ θ ≤ 2π .

. . . . . .

Functions of Several Variables

◮ To understand functions of several variables and be able to
represent these functions using level sets.

. . . . . .

A contour plot is a topographic map of a graph
√
Consider the graph z = x2 + y2 Intersect the cone with planes
z = c and what do you get? Circles. A contour plot shows evenly
spaced circles.
3

2

1

0
4
3
-1 2 2
1
0
-2 0
-2

0
-2
-3
-3 -2 -1 0 1 2 3 2

. . . . . .

The paraboloid

Example
Graph z = x2 + y2 .
3

2

1

0

15
10
-1 2
5
0
0
-2 -2

0
-2
-3
-3 -2 -1 0 1 2 3 2

. . . . . .

The hyperbolic paraboloid

Example
Graph z = x2 − y2 .
3

2

1

0

5
-1 0 2
-5

-2 0
-2

0
-2
-3
-3 -2 -1 0 1 2 3 2

. . . . . .

Plotting a Cobb-Douglas function

Example
Plot z = x1/2 y1/2 .

3

2.5

2

1.5 3

2 3
1 1
2
0
0
0.5
1 1

2
0
0 0.5 1 1.5 2 2.5 3 30

. . . . . .

Utility Functions and indifference curves

◮ If u is a utility function, a level curve of u is a curve along which
all points have the same u value.
◮ We also know this as an indifference curve

. . . . . .

Limits and Continuity

◮ To understand and be able to apply the concept of a limit of a
function of several variables.
◮ To understand and be able to apply the deﬁnition of continuity
for a function of several variables.

. . . . . .

Deﬁnition
We write
lim f(x, y) = L
(x,y)→(a,b)

and we say that the limit of f(x, y) as (x, y) approaches (a, b) is
L if we can make the values of f(x, y) as close to L as we like by taking
the point (x, y) to be sufﬁciently close to (a, b).

. . . . . .

easy limits

◮ lim x=a
(x,y)→(a,b)
◮ lim y=b
(x,y)→(a,b)
◮ lim c=c
(x,y)→(a,b)

. . . . . .

Like regular limits, limits of multivariable functions can be
◮ added
◮ subtracted
◮ multiplied
◮ composed
◮ divided, provided the limit of the denominator is not zero.

. . . . . .

Limit of a Polynomial

Example
Find lim (x5 + 4x3 y − 5xy2 )
(x,y)→(5,−2)

. . . . . .

Limit of a Polynomial

Example
Find lim (x5 + 4x3 y − 5xy2 )
(x,y)→(5,−2)

Solution

lim (x5 + 4x3 y − 5xy2 ) = (5)5 + 4(5)3 (−2) − 5(5)(−2)2
(x,y)→(5,−2)

= 3125 + 4(125)(−2) − 5(5)(4)
= 2025.

. . . . . .

Limit of a Rational Expression

Example
Compute
x2
lim .
(x,y)→(1,2) x2 + y2

. . . . . .

Limit of a Rational Expression

Example
Compute
x2
lim .
(x,y)→(1,2) x2 + y2

Solution

x2 (1)2
lim =
(x,y)→(1,2) x2 + y2 (1)2 + (2)2
1
=
5

. . . . . .

What can go wrong?

The only real problem is a limit where the denominator goes to zero.

◮ If the numerator goes to some number and the denominator
goes to zero then the quotient cannot have a limit.

. . . . . .

What can go wrong?

The only real problem is a limit where the denominator goes to zero.

◮ If the numerator goes to some number and the denominator
goes to zero then the quotient cannot have a limit.
◮ If on the other hand the numerator and denominator both go to
zero we have no clue.

. . . . . .

Showing a limit doesn’t exist

Theorem
Suppose lim f(x, y) = L. Then the limit of f as (x, y) → (a, b) is L
(x,y)→(a,b)
along all paths through (a, b).
There are two contrapositives to this statement:
◮ If there is a path through (a, b) along which the limit does not exist,
the two-dimensional limit does not exist
◮ If there are two paths through (a, b) along which the limits exist but
disagree, the two-dimensional limit does not exist

. . . . . .

Continuity

Deﬁnition
A function f of two variables is called continuous at (a, b) if

lim f(x, y) = f(a, b).
(x,y)→(a,b)

We say f is continuous on D if f is continuous at every point (a, b)
in D.

. . . . . .

Partial Derivatives

◮ To understand and be able to apply the deﬁnition of a partial
derivative.
◮ To be able to compute partial derivatives.
◮ To understand and be able to apply Clairaut’s Theorem. If f is
deﬁned on a disk D that contains the point (a, b) and the
functions fxy and fyx are continuous on D, then

fxy (a, b) = fyx (a, b).

◮ To understand the idea of a partial differential equation, and to
be able to verify solutions to partial differential equations.

. . . . . .

Deﬁnition
Let f(x, y) be a function of two variables. We deﬁne the partial
∂f ∂f
derivatives and at a point (a, b) as
∂x ∂y

∂f f (a + h, b) − f (a, b)
(a, b) = lim
∂x h→0 h
∂f f (a, b + h) − f (a, b)
(a, b) = lim
∂y h→0 h

In other words, we temporarily treat the other variable as constant
and differentiate the resulting one-variable function as in Calculus I.

. . . . . .

Example
Let f(x, y) = x3 − 3xy2 . Find its partial derivatives.

. . . . . .

Example
Let f(x, y) = x3 − 3xy2 . Find its partial derivatives.

Solution
∂f
When ﬁnding , we hold y constant. So
∂x
∂f ∂
= 3x2 − (3y2 ) (x) = 3x2 − 3y2
∂x ∂x
Similarly,
∂f
= 0 − 3x(2y) = −6xy
∂y

. . . . . .

Second derivatives

If f(x, y) is a function of two variables, each of its partial derivatives
are function of two variables, and we can hope that they are
differentiable, too. So we deﬁne the second partial derivatives.
( )
∂2f ∂ ∂f
= = fxx
∂ x2 ∂x ∂x
( )
∂2f ∂ ∂f
= = fxy
∂y ∂x ∂y ∂x
( )
∂2f ∂ ∂f
= = fyx
∂x ∂y ∂x ∂y
( )
∂2f ∂ ∂f
= = fyy
∂ y2 ∂y ∂y

. . . . . .

Don’t worry about the mixed partials

The “mixed partials” bookkeeping may seem scary. However, we are
saved by:
Theorem (Clairaut’s Theorem/Young’s Theorem)
If f is deﬁned near (a, b) and fxy and fyx are continuous at (a, b), then

fxy (a, b) = fyx (a, b).

The upshot is that we needn’t worry about the ordering.

. . . . . .

Example (Continued)
Let f(x, y) = x3 − 3xy2 . Find the second derivatives of f.

. . . . . .

Example (Continued)
Let f(x, y) = x3 − 3xy2 . Find the second derivatives of f.

Solution
We have

fxx = (3x2 − 3y2 )x = 6x
fxy = (3x2 − 3y2 )y = −6y
fyx = (−6xy)x = −6y
fyy = (−6xy)y = −6x

Notice that fyx = fxy , as predicted by Clairaut (everything is a
polynomial here so there are no concerns about continuity). The fact
that fxx = fyy is a coincidence.

. . . . . .

Partial Differential Equations

Deﬁnition
A partial differential equation (PDE) is a differential equation
for a function of more than one variable. So the derivatives involved
are partial derivatives.

. . . . . .


Deﬁnition

Example
Let f(x, y) = x3 − 3xy2 . Show that f satisﬁes the Laplace equation

∂2f ∂2f
+ 2 =0
∂ x2 ∂ y

. . . . . .


Deﬁnition

Example
Let f(x, y) = x3 − 3xy2 . Show that f satisﬁes the Laplace equation

∂2f ∂2f
+ 2 =0
∂ x2 ∂ y

Solution
We already showed that fxx = 6x and fyy = −6x. So fxx + fyy = 0.

. . . . . .

Tangent Planes and Linear Approximations I

◮ To understand the concept of a tangent plane to a surface
z = f(x, y) and to be able to compute the equation of tangent
planes
◮ To understand and be able to ﬁnd a linear approximation to a
function z = f(x, y).

. . . . . .

Tangent Planes and Linear Approximations II

◮ To understand the concept of a tangent plane to a parametrically
deﬁned surface

r(u, v) = x(u, v) i + y(u, v) j + z(u, v) k

and to be able to compute the equation of tangent planes.

. . . . . .

Tangent Planes and Linear Approximations III

◮ To understand and be able to ﬁnd the differential to a function
z = f(x, y),
∂z ∂z
dz = dx + dy
∂x ∂y
To be able to use the differential to estimate maximum error.

. . . . . .

Deﬁnition
If f is a function and P0 = (x0 , y0 , z0 = f(x0 , y0 ) a point on its graph,
then
◮ The linearization of f near (x0 , y0 ) is the function

∂f ∂f
L(x, y) = f(x0 , y0 ) + (x0 , y0 )(x − x0 ) + (x0 , y0 )(y − y0 )
∂x ∂y
◮ The tangent plane to the graph at P0 is the graph of L

The linearization is the best possible linear approximation to a
function at a point.

. . . . . .

Example
Find an equation of the tangent plane to the surface z = y cos(x − y)
at (2, 2, 2).

. . . . . .

Example
Find an equation of the tangent plane to the surface z = y cos(x − y)
at (2, 2, 2).

Solution
We have
∂z ∂z
= −y sin(x − y), = cos(x − y) + y sin(x − y)
∂x ∂y

So
∂z ∂z
= 0, =1
∂x (2,2) ∂y (2,2)

Therefore the equation of the tangent plane is

z = 2 + 0(x − 2) + 1(y − 2) = y

. . . . . .

Example
The number of units of output per day at a factory is
[ ]−1/2
1 −2 9
P(x, y) = 150 x + y−2 ,
10 10

where x denotes capital investment (in units of $1000), and y denotes
the total number of hours (in units of 10) the work force is employed
per day. Suppose that currently, capital investment is $50,000 and the
total number of working hours per day is 500. Estimate the change in
output if capital investment is increased by $5000 and the number of
working hours is decreased by 10 per day.

. . . . . .

Solution
( )[ ] ( )
∂P 1 1 −2 9 −2 −3/2 −2 −3
(x, y) = 150 − x + y x
∂x 2 10 10 10
[ ]
1 −2 9 −2 −3/2 −3
= 15 x + y x
10 10
∂P
(50, 50) = 15
∂x

( )[ ]− 3 / 2 ( )
∂P 1 1 −2 9 9
(x, y) = 150 − x + y−2 (−2)y−3
∂y 2 10 10 10
[ ]
1 −2 9 −2 −3/2 −3
= 15 x + y x
10 10
∂P
(50, 50) = 135
∂y
So the linear approximation is L = 7500 + 15(x − 50) + 135(y − 50)
. . . . . .

Solution, continued

So the linear approximation is

L = 7500 + 15(x − 50) + 135(y − 50)

If ∆x = 5 and ∆y = −1, then

L = 7500 + 15 · 5 + 135 · (−1) = 7440

The actual value is
P(55, 49) ≈ 7427
13
So we are off by ≈ 1.75%
7427

. . . . . .

Tangent planes to parametrized surfaces
◮ If r(u, v) parametrizes a surface S, and P0 = r(u0 , v0 ), then the
vector
r(u0 + h, v0 ) − r(u0 , v0 )
ru (u0 , v0 ) = lim
h→0 h
is tangent to S at P.
◮ If r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k, then

∂x ∂y ∂z
ru (u0 , v0 ) = (u0 , v0 )i + (u0 , v0 )j + (u0 , v0 )k
∂u ∂u ∂u
◮ We have another tangent vector

∂x ∂y ∂z
rv (u0 , v0 ) = (u0 , v0 )i + (u0 , v0 )j + (u0 , v0 )k
∂v ∂v ∂v
◮ Their cross product is normal to the tangent plane.
. . . . . .

Example
Find an equation of the tangent plane to the parametric surface

r(u, v) = u2 i + 2u sin v j + u cos v k

at the point u = 1 and v = 0.

. . . . . .

Example
Find an equation of the tangent plane to the parametric surface

r(u, v) = u2 i + 2u sin v j + u cos v k

at the point u = 1 and v = 0.
Solution
The point we are ﬁnding the tangent plane at is r(1, 0) = ⟨1, 0, 1⟩. The
tangent plane is spanned by the two vectors ru (1, 0) and rv (1, 0):

ru (u, v) = ⟨2u, 2 sin v, cos v⟩ =⇒ ru (1, 0) = ⟨2, 0, 1⟩
rv (u, v) = ⟨0, 2u cos v, −u sin v⟩ =⇒ rv (1, 0) = ⟨0, 2, 0⟩

So a normal vector to the tangent plane is

ru (1, 0) × rv (1, 0) = ⟨2, 0, 1⟩ × ⟨0, 2, 0⟩ = ⟨−2, 0, 4⟩

This means an equation for the tangent plane is

−2(x − 1) + 4(z − 1) = 0. . . . . . .

Differentials

This is really just another way to express linear approximation.
Deﬁne new variables dx = ∆x, dy = ∆y, dz. Then the equation for
the tangent plane through (x0 , y0 , z0 = f(x0 , y0 )) is

∂f ∂f
dz = dx + dy
∂x ∂y

And the concept that this is a good linear approximation is expressed
as
∆z = f(x + dx, y + dy) − f(x, y) ≈ dx
when dx and dy are “small enough”

. . . . . .

Example
If z = 5x2 + y2 and (x, y) changes from (1, 2) to (1.05, 2.1), compare
the values of ∆z and dz.

. . . . . .

Example
If z = 5x2 + y2 and (x, y) changes from (1, 2) to (1.05, 2.1), compare
the values of ∆z and dz.

Solution
We have
dz = 10x dx + 2y dy
When x = 1, y = 2, dx = 0.05, and dy = 0.1, we get dz = 0.9. On the
other hand
z(1.05, 2.1) = 9.9225
So ∆z = 0.9225. The difference is 0.0225, or 2.4%.

. . . . . .

Math 21a Midterm I Review

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Math 21a Midterm I Review