Smith Engineering
Queen’s University
APSC 172 Workbook
Winter 2026
© Peter Taylor 2026
peter.taylor@queensu.ca
Table of Contents
1 Multivariable Differential Calculus
1
1.1
Lines and Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.2
Planes in 3D Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
1.3
Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14
1.4
Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
1.5
Tangent Planes and Linear Approximations . . . . . . . . . . . . . . . . . . . . . . .
27
1.6
Multivariable Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35
1.7
Gradient and Directional Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38
1.8
Optimization – Max and Min . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
2 Infinite Series and Power Series
54
2.1
General Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55
2.2
Alternating Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
59
2.3
Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61
2.4
Taylor and Maclaurin Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
67
2.5
Approximating Functions by Polynomials . . . . . . . . . . . . . . . . . . . . . . . . .
74
3 Multiple Integration
77
3.1
Double Integrals: Basic Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
78
3.2
Double Integrals over General Regions . . . . . . . . . . . . . . . . . . . . . . . . . .
84
3.3
Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
88
3.4
Center of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
95
3.5
Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
98
4 Tutorial Problems
109
4.1
Tutorial 1 - Parameterized Lines and Curves . . . . . . . . . . . . . . . . . . . . . . . 110
4.2
Tutorial 2 - Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . . . . 112
4.3
Tutorial 3 - Partial Derivatives and Tangent Planes . . . . . . . . . . . . . . . . . . . 114
i
TABLE OF CONTENTS
ii
4.4
Tutorial 4 - Linear Approximation and Multivariable Chain Rule . . . . . . . . . . . . 117
4.5
Tutorial 5 - Gradient and Directional Derivatives . . . . . . . . . . . . . . . . . . . . 120
4.6
Tutorial 6 - Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
4.7
Tutorial 7 - Infinite Series, Alternating Series and Power Series . . . . . . . . . . . . . 125
4.8
Tutorial 8 - Taylor and Maclaurin Series and Taylor’s Inequality . . . . . . . . . . . . 127
4.9
Tutorial 9 - Double Integrals and Iterated Integrals . . . . . . . . . . . . . . . . . . . 129
4.10 Tutorial 10 - Integration with Polar Coordinates, Center of Mass . . . . . . . . . . . . 131
4.11 Tutorial 11 - Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
Chapter 1
Multivariable Differential Calculus
Contents
1.1
Lines and Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.2
Planes in 3D Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
1.3
Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . . . . .
14
1.4
Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
1.5
Tangent Planes and Linear Approximations . . . . . . . . . . . . . . . .
27
1.6
Multivariable Chain Rule
. . . . . . . . . . . . . . . . . . . . . . . . . . .
35
1.7
Gradient and Directional Derivative . . . . . . . . . . . . . . . . . . . . .
38
1.8
Optimization – Max and Min . . . . . . . . . . . . . . . . . . . . . . . . .
45
1
1.1. LINES AND CURVES
1.1
2
Lines and Curves
Equation of a line. An equation of the form
v = v0 + tu
is a vector parametric equation of a line. It passes through the point v0 and has direction vector u. In
coordinates, we can write the equation:
   
 
x
x0
a
x = x0 + ta
 y  =  y0  + t  b 
or y = y0 + tb
z
z0
c
z = z0 + tc
Example 1.1.1. Find a parametric equation for the line that passes through the point (2, 1, 2) in the
direction of the vector [–1, 1, 3]. Make a sketch of the line and illustrate the parameterization. Interpret t
as time and calculate the velocity and speed of the moving point.
1.1. LINES AND CURVES
3
Example 1.1.2. (a) A line is drawn from the origin to the point (1, 2, 3). What is the slope of the line?
Note: In a coordinate system, “slope” has no meaning unless we specify the vertical axis. In 3 dimensions
we will always take this to be the z-axis.
(b) What is the slope of the line in Example 1.1.1.?
1.1. LINES AND CURVES
4
Example 1.1.3. Parametric equations.
(a) Show that the circle centered at the origin with radius 3 can be parameterized by the equations
x = 3 cos(t),
y = 3 sin(t)
(b) Using this parameterization, calculate the resulting speed in two different ways:
(i) using only geometry and trig.
(ii) using calculus.
1.1. LINES AND CURVES
5
Example 1.1.4. Parametric equations.
(a) Sketch the curve in R3 with parametric equation:
x = 2 cos(t), y = 2 sin(t), z = 3t
(b) Interpreting t as time, find the velocity and the speed of the moving point at any time t.
(c) Take the z-axis as vertical and calculate the slope m of the curve.
Here we use a special purpose slope formula:
m=
rise
vertical velocity
ż
=
=p
run
horizontal speed
ẋ2 + ẏ 2
where x, y, and z are functions of t and their derivatives with respect to t are denoted by ẋ, ẏ and ż.
x = x(t), y = y(t), z = z(t)
ẋ =
dx
,
dt
ẏ =
dy
,
dt
ż =
dz
dt
1.1. LINES AND CURVES
6
Example 1.1.5. (Walking on the hill) I am walking on the hill z = 9 − x2 − y 2 . The projection of my path
on the x − y plane follows the ellipse x2 + 4y 2 = 4.
(a) Find parametric equations for my path (assuming a counterclockwise motion).
(b) Using the above parameterization to find an expression of
the slope of my path at t.
(c) At what point is my path the steepest (going up or going
down) and what is the slope of the climb at that point?
1.1. LINES AND CURVES
7
Example 1.1.5. [continued] At what point is my path the steepest (going up or going down) and what is
the slope of the climb at that point?
1.2. PLANES IN 3D SPACE
1.2
8
Planes in 3D Space
The dot product of two vectors u = [a, b, c] and v = [x, y, z] is defined as:
u • v = [a, b, c] • [x, y, z] = ax + by + cz.
I have given this definition in R3 but it can be formulated in any Rn . Thus
[1, 2, 3] • [4, –1, 1] = 4 − 2 + 3 = 5
[1, 2, 3, 4] • [4, 3, 2, 1] = 4 + 6 + 6 + 4 = 20
[1, 2] • [4, –2] = 4 − 4 = 0
Note that the dot product is a scalar, a real number.
The dot product is called a “product” so you might suspect that it distributes over addition (as any selfrespecting “product” ought to do). And indeed, it does: w • (u + v) = w • u + w • v.
It is also useful to observe that it is commutative: u • v = v • u.
The length of a vector u is defined as
p
∥u∥ = ∥[a, b, c]∥ = a2 + b2 + c2 .
p
√
Thus, the vector ∥[1, −2, 3]∥ has length: 12 + (−2)2 + 32 = 14.
Note that the dot product of a vector with itself is the square of its length:
u • u = [a, b, c] • [a, b, c] = a2 + b2 + c2 = ∥u∥2 .
The distance between two points is the length of their vector difference. For example, the distance between
the points [1, 2, 3] and [4, −2, 1] is:
dist([1, 2, 3], [4, −2, 1]) = ∥[1 − 4, 2 + 2, 3 − 1]∥
= ∥[−3, 4, 2]∥
p
= (−3)2 + 42 + 22
√
= 29.
The geometry of the dot product:
u • v = ∥u∥ · ∥v∥ cos(θ)
where θ is the angle between the two vectors.
1.2. PLANES IN 3D SPACE
9
We’ll prove this in a moment, but a significant (and immediate) consequence is:
The Orthogonality Theorem
Two vectors are orthogonal (perpendicular) if and only if their dot product is zero:
u⊥v ⇔ u • v = 0
Proof of the Geometry Result: The trick is to express the length of the vector difference u-v in two different
ways. The first is algebraic and writes it as the dot product of the vector with itself, and then expands:
∥u − v∥2 = (u − v) • (u − v) = u • u − u • v − v • u + v • v = ∥u∥2 − 2u • v + ∥v∥2 .
The second is geometric and uses the cosine law for the triangle formed by the vectors u, v and u − v:
c2 = a2 + b2 − 2ab cos(θ)
∥u − v∥2 = ∥u∥2 + ∥v∥2 − 2∥u∥ ∥v∥ cos(θ).
Now compare the two right sides to get
2∥u∥ ∥v∥ cos(θ) = 2u • v
and the result follows.
■
Example 1.2.1. Find a non-zero vector that is orthogonal to both u = [1, 1, 1] and v = [1, −1, 2].
Solution. The vector w = [a, b, c] must satisfy the equations:
w•u=a+b+c=0
w • v = a − b + 2c = 0
We have two equations in three unknowns so there will be a “degree of freedom” in the solution. We expect
that as on geometric grounds the solution will be determined only up to a constant. To eliminate one
variable, add the equations:
2a + 3c = 0.
Using the degree of freedom, set a = 3. Then c = −2. And then from either of the original equations,
b = −1. The answer is [3, −1, −2] or any of its multiples.
■
1.2. PLANES IN 3D SPACE
10
An equation of the form
n • (P − P0 ) = 0
[a, b, c] • ([x, y, z] − [x0 , y0 , z0 ]) = 0
describes a plane which passes through the point P0 = [x0 , y0 , z0 ] and is orthogonal to the normal
vector n = [a, b, c].
This equation can be written in two other useful forms:
• The point-normal form a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0
• The standard form ax + by + cz = d where d = ax0 + by0 + cz0 .
Example 1.2.2. The equation z = 3 is a plane in the three dimensional space. Find a normal vector and
sketch the plane.
1.2. PLANES IN 3D SPACE
11
Example 1.2.3. A plane passes through the point (2, 3, 2) with normal vector in the direction of the vector
[1, 2, 4].
(a) Find the equation of the plane.
(b) Find the intersection points of the plane with the three coordinate axes. There are three intercepts.
Sketch a graph of the plane.
1.2. PLANES IN 3D SPACE
12
Example 1.2.4. (Reflection in a plane)
Given the point A = (2, 4, –6) and the plane x − 2y + z = 6, find the reflection B of A in the plane. That
is, if the plane is a mirror, B is the image of A.
1.2. PLANES IN 3D SPACE
13
The distance between two parallel planes is defined as the shortest distance between them. To determine
the distance, we can choose a point A on one plane and find the distance to a point B on the other. This
−−→
distance represents the distance between the two planes if, and only if, the vector AB is perpendicular to
−−→
both planes; equivalently, AB must be parallel to the normal vector of the planes.
Example 1.2.5. How far apart are the following two planes?
z = 3x + 2y
and
z = 3x + 2y + 6
1.3. FUNCTIONS OF SEVERAL VARIABLES
1.3
14
Functions of Several Variables
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, {f (x, y)|(x, y) ∈ D}.
Example 1.3.1. (Temperature in Ontario) On a particular summer day, the daytime high temperature
varied from location to location in Southern Ontario according to the following table.
83
latitude
in ◦
north
46.5
46
45.5
45
44.5
44
43.5
43
42.5
42
25
26
27
27
27.5
82.5
25
26
26.5
27
82
22
24
25
25.5
26.5
27
81.5
21
22
23
25
25.5
26
26.5
27
longitude in ◦ west
81
80.5 80 79.5
79
21
21
21
21
23
24
22
23
23
24
24.5
22
22
22
23
24
23
23
22
22
23
25
24
24
23
23
25.5
25
25
25
25
26
26
26 25.5
26.5
78.5
78
24.5
25
24.5
24
24
24.5
25
25
24
24
77.5
24
24.5
24.5
24.5
24
24
77
24
24
24
24
76.5
24
1.3. FUNCTIONS OF SEVERAL VARIABLES
15
Graph of a function f (x, y)
Recall that the graph of a single variable function y = g(x) can be visualized by means of its graph which
has the form
{(x, y) ∈ R2 | y = g(x)}
The graph of a function of two variables z = f (x, y) is a surface in three dimensional space which is defined
as the set of points
{(x, y, z) ∈ R3 | z = f (x, y)}
Example 1.3.2. Sketch a graph of the following function in 3D.
z = 1 − y2
1.3. FUNCTIONS OF SEVERAL VARIABLES
16
Example 1.3.3. Find the equation for the following surface.
An “elliptical cylinder” with x and y intercepts ±1 and ±2.
To get a more accurate picture of the function f (x, y), we can study it by cutting it with planes. We do
this by selecting sample values of either x, y or z, and visualizing the resulting one variable function, i.e.,
the intersection curve.
Example 1.3.4. Sketch graph of z = 1 + x2 + y 2 in 3D.
1.3. FUNCTIONS OF SEVERAL VARIABLES
17
Level Curves
The level curves of a function z = f (x, y) of two variables are the curves sketched on xy-plane whose
equations have the form f (x, y) = k, where k is a constant (in the range of f (x, y)).
A level curve (contour curve) f (x, y) = k is the set of all points in the domain of f at which f takes on a
given value k, In other words, it shows where the graph of f has height k. One common example of level
curves occurs in topographic maps of mountainous regions.
In principle, the level curves (contour diagram) and the graph of f can each be reconstructed from the other.
Here is a picture illustrating this:
As shown above, the level curve f (x, y) = k is obtained by intersecting the graph of f with the horizontal
plane, z = k, and then dropping (or raising) the resulting curve to the xy–plane. The graph of f is obtained
by raising (or dropping) the contour f (x, y) = k to the level z = k.
1.3. FUNCTIONS OF SEVERAL VARIABLES
18
Example 1.3.5. A collection of level curves (equal z-intervals with ∆z = 2) of the function z = f (x, y) are
drawn below, with the level curves z = 12 and z = 0 in bold. What is the function?
1.3. FUNCTIONS OF SEVERAL VARIABLES
19
Example 1.3.6. The graph of the elliptic paraboloid
z = 4x2 + y 2
is drawn below and a collection of level curves is provided. The inner curve is z = 1 and the outer curve is
z = 16. Find the parameters and study the parabolas obtained by the vertical cuts.
1.3. FUNCTIONS OF SEVERAL VARIABLES
20
Example 1.3.7. The graph on the left belongs to the surface z = x2 − y 2 , and the level curves at the right
are spaced at intervals of ∆z = 1. From the contour map, it is clear that the surface contains two straight
lines: y = x and y = −x.
(a) Find a line passing through the point (0, 2, −4) that lives entirely on the surface.
(b) Draw its projection on the x − y plane.
1.4. PARTIAL DERIVATIVES
1.4
21
Partial Derivatives
Example 1.4.1. The graph of the function
z = 16 − x2 + xy − 4y 2
is drawn at the right. Suppose you are standing on the hill at the point (x, y, z) = (3, 1, 6) and start walking
in the positive x-direction.
What is the slope of your path? Same question for the positive y-direction.
1.4. PARTIAL DERIVATIVES
22
If z = f (x, y) is a function of x and y, and we hold y fixed at y0 , then z = f (x, y0 ) becomes a function
of the single variable x. The derivative of this function at the point x = x0 is called the partial
derivative of f with respect to x and is written
fx (x0 , y0 )
∂z
(x0 , y0 ).
∂x
or
Geometrically, it is the slope of the graph in the x-direction.
Similarly, if we hold x fixed at x0 , then z = f (x0 , y) becomes a function of the single variable y and
the derivative of this function at the point y = y0 is called the partial derivative of f with respect to
y and is written
∂z
fy (x0 , y0 ) or
(x0 , y0 ).
∂y
Geometrically, it is the slope of the graph in the y-direction.
Example 1.4.2. Let
2
.
x2 + y
1 + xy
.
f (x, y) = x3 − xy 3
Calculate fx (2, 1) and fy (2, 1).
Example 1.4.3. Let
f (x, y) = ln
Calculate fx and fy .
1.4. PARTIAL DERIVATIVES
23
Example 1.4.4. (Implicit Differentiation)
Find ∂z/∂x and ∂z/∂y at the point (1, 2, 1) if z is defined implicitly as a function of x and y by the equation
xz 3 + yz − xy + y 2 = 5.
Example 1.4.5. (Second Derivatives)
Find the second partial derivatives of z = f (x, y) = x3 + 2x2 y 3 − 2y 2 .
For example, fxy sometimes written as
∂2z
∂x∂y
means differentiating first with respect to x and then with respect to y.
Note that in the Example above fxy = fyx . This is not a coincidence but is always the case (if the derivatives
exist and are continuous). More generally if you are calculating any mixed higher order partial derivative,
it doesn’t matter in what order you take the variables. This is known as Clairaut’s Theorem.
1.4. PARTIAL DERIVATIVES
Example 1.4.6. Below are diagrammed a set of contour lines for the function
f (x, y) =
18y
x2 + y 2 + 1
on the domain y ≥ 0. The zero level curve is the x-axis
and as we move towards the top of the hill at (0, 1) the
level curves are spaced at intervals of ∆z = 0.5.
(a) Use the diagram to estimate the partial derivatives fx (2, 2) and fy (2, 2).
24
1.4. PARTIAL DERIVATIVES
25
Example 1.4.6 [continued]
(b) Working with the contour map, determine the sign of the partial derivatives fxx and fyy at the point
(2, 2).
1.4. PARTIAL DERIVATIVES
26
Functions of more than two variables
A function of three variables, f , is a rule that assigns to each ordered triple (x, y, z) in a domain D ∈ R3 a
unique real number denoted by f (x, y, z). For instance, to specify a location in a room, you must specify
three coordinates (x, y, z). So, the temperature is a function of three variables: T = T (x, y, z).
What is another variable that we could potentially include in this function, that would make temperature a
function of four variables?
Visualizing functions in three variables
To picture f (x, y, z) by means of a graph is not very useful. The graph would have to have the following
definition:
{(x, y, z, w) ∈ R4 | w = f (x, y, z)}
That is, it would be a set in 4–dimensional space, which is not something we can directly visualize.
However, we can make use of the analogue of the contour diagram. Suppose we pick a constant k and collect
all the points (x, y, z) for which f (x, y, z) = k. In general, this gives a surface called a level surface. By
sketching a number of level surfaces, we can get a sense of the function f .
Example 1.4.7. The temperature in degree C, at a point (x, y, z) is given by T (x, y, z) = x2 + y 2 + z 2 .
Find and sketch several level surfaces.
Partial derivatives of functions of more than two variables
In general, for a function of several variables f (x1 , . . . , xn ), the partial derivative
the derivative obtained by holding all other variables fixed.
Example 1.4.8. If f (x, y, z) = xz 2 + y cos(x2 z), find
∂f
.
∂z
∂f
(x1 , . . . , xn ) is
∂xi
1.5. TANGENT PLANES AND LINEAR APPROXIMATIONS
1.5
27
Tangent Planes and Linear Approximations
Recall that the equation
a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0
describes a plane passing through the point (x0 , y0 , z0 ) with normal vector [a, b, c]. If c = 0, the plane is
vertical (i.e., parallel to the z-axis) but otherwise we can divide by c and rearrange to write the plane as
z = z0 + A(x − x0 ) + B(y − y0 )
where A = −a/c and B = −b/c. This defines z as a linear function of x and y. Note that
∂z
=A
∂x
and
∂z
= B.
∂y
Definition 1.5.1. Given the graph of a function f (x, y) (which is a surface in 3-space), and a point on the
surface, the tangent plane to the surface at that point is the unique plane passing through that point that
has the same slope as the surface in both the x- and the y-directions. Those two slopes are of course the
two partial derivatives of the function at the point.
The equation of the tangent plane to the surface z = f (x, y) at the point (x0 , y0 , z0 ) is
z = z0 + fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 )
Note that this plane has exactly the properties of the definition above: its slopes in the x- and y-direction
are
fx (x0 , y0 ) and fy (x0 , y0 ).
Example 1.5.1. Find the tangent plane to the elliptic paraboloid
3z = x2 + 2y 2
at the point (2, 1, 2).
1.5. TANGENT PLANES AND LINEAR APPROXIMATIONS
Example 1.5.2. Consider the following sphere
x2 + y 2 + z 2 = 9.
(a) Use implicit differentiation to find the tangent plane to the sphere at the point (2, 1, 2).
(b) Use the geometry of the sphere to check your answer.
28
1.5. TANGENT PLANES AND LINEAR APPROXIMATIONS
Example 1.5.3. Find all tangent planes to the elliptic paraboloid
z = 4 − x2 − 2y 2
that contain the line
    

x
0
t
y  = 0 +  t  .
z
7
−2t
29
1.5. TANGENT PLANES AND LINEAR APPROXIMATIONS
Example 1.5.3. [continued]
30
1.5. TANGENT PLANES AND LINEAR APPROXIMATIONS
31
Linear Approximation and Differential
Since the tangent plane lies close to the surface near the point at
which they meet, providing the function is differentiable at (a, b),
we can approximate the function value f (x, y) by the z-value of the
tangent plane.
The linear function whose graph is the tangent plane, namely
L(x, y) = f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b)
is called the linearization of the function f at the point (a, b)
and the approximation
f (x, y) ≈ L(x, y) = f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b)
is called the linear approximation or tangent plane approximation of f at (a, b).
Note that this form of the equation can be interpreted as expressing the relationship between changes
∆x = x − a and ∆y = y − b in x and y and the resulting change ∆z = f (x, y) − f (a, b):
f (x, y) − f (a, b) ≈ fx (a, b)(x − a) + fy (a, b)(y − b)
∆z ≈ fx (a, b)∆x + fy (a, b)∆y
The right side of the this approximation is a linear function of ∆x and ∆y. We call this linear function the
differential.
The differential df (or dz), at a point (a, b) is the linear function of dx and dy given by the formula
df = fx (a, b)dx + fy (a, b)dy
The differential at a general point is often written df = fx dx + fy dy.
1.5. TANGENT PLANES AND LINEAR APPROXIMATIONS
32
Example 1.5.4. Let z = f (x, y) = x2 + 3xy − y 2 .
(a) Find the linearization of this function the point (1, 3).
(b) Suppose that x changes from 1 to 1.05 and y changes from 3 to 2.96. Calculate the real change ∆z and
use the linearization to find the differential change dz. Compare the real change with the differential.
1.5. TANGENT PLANES AND LINEAR APPROXIMATIONS
33
Linearization for functions of more than two variables
Linear approximation, and differential can be defined in a similar manner for functions of more than two
variables. If a function w = f (x, y, z) is differentiable at (a, b, c), then the linearization is
L(x, y, z) = f (a, b, c) + fx (a, b, c)(x − a) + fy (a, b, c)(y − b) + fz (a, b, c)(z − v)
and the approximation
f (x, y, z) ≈ L(x, y, z) = f (a, b, c) + fx (a, b, c)(x − a) + fy (a, b, c)(y − b) + fz (a, b, c)(z − v)
is called the linear approximation of f at (a, b, c). The differential dw is defined in terms of the differentials
dx, dy, and dz by
dw = fx dx + fy dy + fz dz
q
2
Example 1.5.5. Consider the function T (x, y, z) = x4 + y 2 + z 2 .
(a) Find the linearization of this function at the point (2, 1, 1).
(b) Suppose that x changes from 2 to 2.2, y changes from 1 to 0.9, and z changes from 1 to 1.3, Calculate
the real change ∆T and use the linearization to find the differential change dT . Compare the real
change with the differential.
1.5. TANGENT PLANES AND LINEAR APPROXIMATIONS
34
Error Bounds
If f (x, y) is differentiable at (a, b), local linearity gives us
∆f ≈ fx (a, b) ∆x + fy (a, b) ∆y
To avoid sign confusion, errors are often given in the form “(measured value) ± (error)”. This means
that, if we measured x, |∆x| < (max x error). In these cases we can estimate the maximum
(positive) error in the function output using
|error bound| ≈ |fx (a, b) · (max ∆x)| + |fy (a, b) · (max ∆y)|
The same result applies to 3- and higher-variable functions, with an additional term for each variable.
Example 1.5.6. A projectile fired from the ground at an angle θ and an initial speed v travels a horizontal
distance D before hitting the ground. If v is measured as 10 m/s with a maximum error of 3% and θ is
measured as 1 radian with a maximum error of 2%, what is the maximum error in D? Do this in two ways,
first using the “real” D (as a function of v and θ) and secondly using the differential.
1.6. MULTIVARIABLE CHAIN RULE
1.6
Multivariable Chain Rule
Example 1.6.1. Consider a hillside defined by the function
z = 25 − x2 + xy − 4y 2
where z is in meters, and x and y are in kilometers. Suppose
we are standing on the surface at the point (3, 1, 15) and
begin walking up the hill with
x(t) = 3 − 2t
and
y(t) = 1 − t
where t is in hours.
(a) Give the units of
dx dz
∂z
,
, and
.
dt dt
∂x
(b) How quickly are we moving up or downhill one hour into the hike (at t = 1)?
35
1.6. MULTIVARIABLE CHAIN RULE
36
In this example,
z = f (x, y)
x = g(t)
y = h(t)
=⇒
z = f (g(t), h(t))
dz
To find the rate of change of z with respect to t,
, relates to the derivatives of f, g, and h, we start with
dt
our understanding of linear approximations:
∆z ≈ fx (x, y) ∆x + fy (x, y) ∆y
∆x ≈ g ′ (t) ∆t , ∆y ≈ h′ (t) ∆t
x = g(t) and y = h(t)
Combining these, we get
∆z ≈ fx (g(t), h(t)) · g ′ (t) · ∆t + fy (g(t), h(t)) · h′ (t) · ∆t
Dividing by ∆t and taking limits to get derivatives, we obtain:
∂z dx ∂z dy
dz
=
+
dt
∂x dt
∂y dt
The Chain Rule for z = f (x, y), x = g(t), y = h(t)
Suppose that z = f (x, y) is a differentiable function of x and y where x = g(t) and y = h(t) are both
differentiable functions of t. Then z is a differentiable function of t and
dz
∂z dx ∂z dy
=
+
dt
∂x dt
∂y dt
Functions of several variables can be composed in many different ways. There is a separate chain rule for
every form of composition. By obtaining the chain rule in a few standard cases, it will be straightforward
to see how to get others.
Example 1.6.2. Write out the Chain Rule for the following cases
(a) z = f (x, y),
x = x(s, t), y = y(s, t)
(b) w = f (x, y, z),
x = x(t), y = y(t), z = z(t).
1.6. MULTIVARIABLE CHAIN RULE
37
Example 1.6.3. A function of three variables. The temperature at the point (x, y, z) is
T (x, y, z) = xeyz−x .
(a) A bug is at the point (2, 1, 2) and is flying through the air. Find an expression for the rate at which the
temperature of the air around it changes in terms of the rates of change of its coordinates, ẋ, ẏ, and ż.
(b) Suppose the path of the bug is described by the equation
  

x
t+2
y  = t2 + 1
z
2−t
where t is time. Using the temperature function above, at what rate does it experience the temperature
changing at t = 0?
(c) Suppose the bug in part (a) flies through the point (2, 1, 2) at speed 1 in the direction of the vector
[1, 2, −1]. What is the rate at which the temperature of the air around it changes?
1.7. GRADIENT AND DIRECTIONAL DERIVATIVE
1.7
38
Gradient and Directional Derivative
The Directional Derivative
We return to the paraboloid of Example 1.6.1.
z = f (x, y) = 25 − x2 + xy − 4y 2 .
Suppose you are standing on the surface at the point (x0 , y0 , z0 ) and begin walking up the hill in the direction
of the (horizontal) vector [a, b]. Use the multivariable chain rule to find a general expression for the slope of
your path at that point.
Definition 1.7.1. The directional derivative of f (x, y) at the point (x0 , y0 ) in the direction of the
vector u[a, b] is defined as
fx (x0 , y0 )a + fy (x0 , y0 )b
√
Du f (x0 , y0 ) =
a2 + b 2
This gives us the slope of the surface z = f (x, y) at the point (x0 , y0 ) in the direction of the vector
u = [a, b].
Example 1.7.1. Find the slope of the paraboloid above at the point (3, 1, 15) in the direction [−2, −1].
1.7. GRADIENT AND DIRECTIONAL DERIVATIVE
39
The Gradient
The gradient vector of a differentiable function f = f (x, y) at the point (x0 , y0 ) is defined by
∇f (x0 , y0 ) = [fx (x0 , y0 ), fy (x0 , y0 )]
This concept gives us a natural way of writing the directional derivative at the point (x0 , y0 ) in the direction
of the vector u[a, b] as a simple dot product.
Du f (x0 , y0 ) =
fx (x0 , y0 )a + fy (x0 , y0 )b
∇f • u
√
=
.
2
2
∥u∥
a +b
Example 1.7.2. How does the gradient interact with the graph
of the function f ? What’s the ”geometry” of the configuration?
∇f
Consider the directional derivative of f in any direction u:
Du f =
∇f • u
∥∇f ∥ ∥u∥ cos(φ)
=
= ∥∇f ∥ cos(φ).
∥u∥
∥u∥
This equation shows that Du f depends only on the length of the
vector ∇f and the angle φ between ∇f and the direction vector
u. We can deduce a couple of important things from that.
rφ
X
XXX
z
First, Du f will be zero when φ = π/2. Now note that Du f will
also be zero in the direction of any level curve of f . We deduce
that ∇f is orthogonal to the level curve of f passing through
(x0 , y0 ).
Secondly, Du f will be a maximum when φ = 0, that is when u
points in the same direction as ∇f . Turning this around: points
in the direction of maximum rate of increase of f (x, y).
Further, this maximum rate of increase is:
Du f = ∥∇f ∥ cos(0) = ∥∇f ∥
which is the length of the gradient vector.
To summarize:
(1) ∇f is orthogonal to the level curve
(2) The maximum value that Du f attains is the magnitude ∥∇f ∥ of the gradient.
(3) This maximum is attained in the direction of the gradient ∇f .
u
1.7. GRADIENT AND DIRECTIONAL DERIVATIVE
Example 1.7.3. Below are diagrammed a set of contour lines for the function
f (x, y) =
18y
x2 + y 2 + 1
on the domain y ≥ 0. The zero level curve is the x-axis
and as we move towards the top of the hill at (0, 1) the
level curve are spaced at intervals of ∆z = 0.5.
(a) Starting at (3, 0), sketch my path if I always ascend the hill most steeply.
(b) Starting at (3, 0), sketch a path that ascends the hill with a slope of 1.
40
1.7. GRADIENT AND DIRECTIONAL DERIVATIVE
41
Example 1.7.3. [continued]
(c) Use the spacing of the contour lines to draw the gradient vector of f at the point (1, 2). Check your
work using the given formula for f .
1.7. GRADIENT AND DIRECTIONAL DERIVATIVE
Example 1.7.4. (Properties of the gradient)
(a) In what direction in the xy-coordinate plane does z = f (x, y) = x2 y 2 − xy 3
have its maximum rate of change at the point (2, 1)? What is this maximum rate?
(b) Find the equation of the tangent plane to the surface z = f (x, y) at the point (2, 1).
(c) The diagram at the right displays the contour
lines of f (x, y) for z = 1, 2, 3, 4, 5. Draw the gradient
vector of f based at the point (2, 1).
(d) Now draw the contour lines of the tangent plane
passing through the point (2, 1) and through the
point
(2, 1) + ∇f (2, 1)
(that is, the lines should pass through the tail and
the head of the gradient vector you drew in part (c)).
Label both lines with their z-value. Do this in two
different ways, working with the graph, and working
with the equations.
42
1.7. GRADIENT AND DIRECTIONAL DERIVATIVE
43
The Gradient in Three Dimensions
The directional derivative, the gradient and the idea of level curve extend immediately to functions of three
variables.
If f = f (x, y, z) is differentiable, then the directional derivative of f at (x0 , y0 , z0 ) in the direction of
u = [a, b, c] is
fx (x0 , y0 , z0 )a + fy (x0 , y0 , z0 )b + fz (x0 , y0 , z0 )c
∇f • u
√
Du f =
=
2
2
2
∥u∥
a +b +c
The gradient of f = f (x, y, z) at the point (x0 , y0 , z0 ) is defined by
∇f (x0 , y0 .z0 ) = [fx (x0 , y0 , z0 ), fy (x0 , y0 , z0 ), fz (x0 , y0 , z0 )]
The gradient of f (x, y, z) has the following properties:
(1) ∇f is orthogonal to the level surface.
(2) The maximum value that Du f attains is the magnitude ∥∇f ∥ of the gradient.
(3) This maximum is attained in the direction of the gradient ∇f .
Example 1.7.5. Let the temperature (◦ C) at a point (x, y, z) in space be
T (x, y, z) = 4 + x2 + 2y 3 + 3z 3 .
In which direction from the point (3, 2, 1) does the temperature increase most quickly, and what is this
maximum rate of increase?
1.7. GRADIENT AND DIRECTIONAL DERIVATIVE
44
Example 1.7.6. These diagrams depict a surface in R3 . We have two ways of thinking about this surface:
1) as the graph z = f (x, y) of a function f of two variables;
2) as a level surface T (x, y, z) = k of a function T of three variables. In this case take T (x, y, z) =
z − f (x, y).
Using the diagrams, depict the gradient vector of each function and find an equation for the plane tangent
to the surface in terms of each function. Recall that the gradient of a function f at a point is the vector of
partial derivatives at that point and it points in the direction of the maximum rate of increase of f .
1.8. OPTIMIZATION – MAX AND MIN
1.8
45
Optimization – Max and Min
Review of local max/min of single variable functions
We start with a fast review of the one-variable situation.
A critical point of a function f (x) is a point x = a at which the derivative is zero:
f ′ (a) = 0.
At such point, the tangent line to the graph y = f (x) is horizontal.
A standard result is that an interior local maximum (hilltop) or minimum (valley bottom) of f (x) will
always be a critical point. This is geometrically quite intuitive, but it is rewarding to review how a rigorous
argument might go. Of course, it will depend on the definition of the derivative.
The interesting question goes the other way. Suppose you have a critical point. How do we tell if it’s a local
max? Or a local min? And what other behaviors are possible?
The second derivative test. If x = a is a critical point of f (x) then:
• if f ′′ (a) > 0, then a is a local min;
• if f ′′ (a) < 0, then a is a local max;
• if f ′′ (a) = 0, then the test fails.
Example 1.8.1. Given examples of a critical point a for which f ′′ (a) = 0 and which is
• a local min
• a local max
• neither
1.8. OPTIMIZATION – MAX AND MIN
46
Functions of two variables
The definitions for local maximum, global maximum, local minimum, and global minimum are very similar
to those used for functions of a single variable.
Local Extrema
f has a local maximum/minimum at (a, b) if
• (a, b) is not on the boundary of the domain, and
• f (a, b) ≥ f (x, y) for all points (x, y) near (a, b). (≤ f (x, y) for a local minimum)
Example 1.8.2. A contour diagram is given below. Identify any local min and max points.
At hills and valleys of a graph of f (x, y), the two partial derivatives have to be zero.
A critical point of a function f (x, y) is a point (a, b) at which both partial derivatives vanish:
fx (a, b) = 0
and
fy (a, b) = 0.
At a critical point, the tangent plane to the graph z = f (x, y) is horizontal.
1.8. OPTIMIZATION – MAX AND MIN
47
Max and Min and Saddle
By analogy to the one-variable case (and with a similar argument), a local maximum (hilltop) or minimum
(valley bottom) of f (x, y) will always be a critical point. But how does the converse go now? Suppose we
have a critical point. How do we tell if it’s a local max or min? And what other behaviors are possible?
Example 1.8.3. Let z = f (x, y) = x2 + y 2 − 3xy.
(a) Verify that the origin is a critical point.
(b) Use the second partial derivatives fxx , and fyy to determine the concavities along the line y = 0 and
along the line x = 0. Interpret the behavior of the function around the origin (critical point) using the
contour diagram of f (x, y) below.
1.8. OPTIMIZATION – MAX AND MIN
The two-variable second derivative test
Suppose that (a, b) is a critical point of the function f (x, y), that is, both partial derivatives are zero.
Define the quantity:
D = D(a, b) = fxx (a, b)fyy (a, b) − [fxy (a, b)]2
Then there are three cases:
1. D > 0. In this case note that fxx and fyy cannot be zero and must have the same sign at (a, b),
otherwise D would certainly be negative. If they are positive, then the surface is concave up
and we have a local minimum. If they are negative, then the surface is concave down and we
have a local maximum.
2. D < 0. In this case we have a saddle point, that is, we have a local minimum in some directions
and a local max in others.
3. D = 0. In this case the test fails, and we need to look at the function in more detail.
48
1.8. OPTIMIZATION – MAX AND MIN
49
Example 1.8.4. Find and classify the critical points of
f (x, y) = x4 + y 4 − 4xy + 1.
x
y
fxx
fyy
fxy
D
1.8. OPTIMIZATION – MAX AND MIN
50
Example 1.8.5. Find and classify the critical points of
f (x, y) = x3 + y 3 − 15(x + y) + 3xy 2 .
x
y
fxx
fyy
fxy
D
1.8. OPTIMIZATION – MAX AND MIN
51
Global Extrema
f has a global maximum/minimum at (a, b) if
• f (a, b) ≥ f (x, y) for all points (x, y) in the domain of f .
(f (a, b) ≤ f (x, y) for a global minimum)
Example 1.8.6. A contour diagram is given below. Identify any local min and max and global min and
max points in this closed region D = {(x, y)| − 0.5 ≤ x ≤ 1, 0.5 ≤ y ≤ 2}.
Note that if f has an extreme value at (x0 , y0 ) in a closed region D, then (x0 , y0 ) is either a critical point of
f or a boundary point of D.
Procedure: Finding global Max/Min values on a closed, bounded region D.
1. Find the values of f at all critical points of f in D.
2. Find the extreme values of f on the boundary of D.
3. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of
these values is the absolute minimum value.
1.8. OPTIMIZATION – MAX AND MIN
Example 1.8.7. Find the global maximum and minimum values of the function
f (x, y) = 2x3 + 9y 2 − 24x + 40
in the disk x2 + y 2 ≤ 16.
52
1.8. OPTIMIZATION – MAX AND MIN
Example 1.8.7. [continued]
53
Chapter 2
Infinite Series and Power Series
Contents
2.1
General Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55
2.2
Alternating Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
59
2.3
Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61
2.4
Taylor and Maclaurin Series . . . . . . . . . . . . . . . . . . . . . . . . . .
67
2.5
Approximating Functions by Polynomials . . . . . . . . . . . . . . . . . .
74
54
2.1. GENERAL CONCEPTS
2.1
55
General Concepts
An infinite series is an infinite sum of form:
∞
X
an = a1 + a2 + a3 + · · ·
n=1
Define its nth partial sum sn as the sum of the first n terms:
sn =
n
X
ak = a1 + a2 + a3 + · · · + an
k=1
If the sequence {sn } converges to a finite limit s as n approaches infinity, we say that the series converges,
and that s is the sum of the series and write:
s=
∞
X
an = lim sn
n=1
n→∞
If the series fails to converge, we say that it diverges.
Example 2.1.1. In each case does the series converge or diverge?
(a)
∞
X
n = 1 + 2 + 3 + 4 + ···
n=1
(b)
∞
X
(−1)n−1 = 1 − 1 + 1 − 1 + 1 − · · ·
n=1
2.1. GENERAL CONCEPTS
56
Example 2.1.2. Consider the series:
∞
X
1
n=1
2n
=
1 1 1
1
1
+ + +
+ ··· + n + ···
2 4 8 16
2
(a) Find the n-th partial sum of the series. Determine if the series converges or diverges. If it converges,
find its sum.
(b) Provide an intuitive argument that the series converges and find its sum.
A series in which each term is obtained from the previous term by multiplication by a constant r is
called geometric. If we denote the first term by a, the series has the form:
∞
X
arn−1 = a + ar + ar2 + ar3 + · · ·
n=1
the series converges if and only if |r| < 1, and in this case its sum is
s=
a
.
1−r
As an example, find the sum of the geometric series: 5 −
10 20 40
+
−
+ ···
3
9
27
2.1. GENERAL CONCEPTS
57
Note. Most of our work here will be with series that have a strong geometric-series character, that
is they have the form:
∞
X
an rn = a0 + a1 r + a2 r2 + a3 r3 + · · ·
n=0
for given coefficients an , and this will enable us to decide whether they converge and to find their
sum.
An interesting example is the arithmetic-geometric series:
∞
X
nrn = r + 2r2 + 3r3 + 4r4 + · · ·
n=1
The series converges for |r| < 1. Can you find the sum?
Example 2.1.3. Find the sum of the arithmetic-geometric series:
∞
X
nrn = r + 2r2 + 3r3 + 4r4 + · · ·
n=1
The arithmetic-geometric series:
r + 2r2 + 3r3 + 4r4 + · · · + nrn + · · · =
r
.
(1 − r)2
2.1. GENERAL CONCEPTS
Example 2.1.4. Harmonic series:
58
∞
X
1
n=1
n
=1+
1 1 1
+ + + ···
2 3 4
Does the harmonic series converges?
Example 2.1.5. What’s the story for the following series?
∞
X
1
n=1
n2
=1+
1 1
1
+ +
+ ···
4 9 16
2.2. ALTERNATING SERIES
2.2
59
Alternating Series
A series is alternating if the terms alternate in sign.
For example, the geometric series:
1−
8
2 4
+ −
+ ···
3 9 27
is alternating and converges to 3/5.
Example 2.2.1. An important example is the alternating harmonic series:
∞
X
(−1)n−1
n=1
1
1 1 1 1 1
= 1 − + − + − + ···
n
2 3 4 5 6
Does it converge or diverge?
Convergence Test for Alternating Series
Consider the alternating series
∞
X
(−1)n+1 bn = b1 − b2 + b3 − b4 + · · ·
n=1
If
(i) the bn decrease with n
(ii) and approach zero, that is,
lim bn = 0,
n→∞
then the series converges.
(bn > 0)
2.2. ALTERNATING SERIES
60
If a series converges, then its partial sums Sn converge to its sum s. Thus, by taking sn for a large enough
n, we should get a good estimate of s. The trouble is, we need to know how big to take n (i.e. how many
terms to take) to obtain a desired measure of closeness.
Error Estimation for Alternating Series
Suppose that an alternating series
∞
X
(−1)n+1 bn , bn > 0
satisfies (i) and (ii) of the convergence
n=1
test. Then the “error” in approximating its sum s with the nth partial sum sn is less than the absolute
value of the first omitted term:
|s − sn | ≤ bn+1 .
Example 2.2.2. How many terms of the alternating harmonic series:
∞
X
(−1)n−1
n=1
1
1 1 1 1 1
= 1 − + − + − + ···
n
2 3 4 5 6
do we need to take to get within 1/100th of the sum?
Example 2.2.3. Find the sum of the series
∞
X
(−1)n−1
n=1
n!
correct to three decimal places (i.e., with error less than 0.0005).
2.3. POWER SERIES
2.3
61
Power Series
A power series is an infinite series of the form:
∞
X
cn xn = c0 + c1 x + c2 x2 + c3 x3 + · · ·
n=0
or
∞
X
cn (x − a)n = c0 + c1 (x − a) + c2 (x − a)2 + c3 (x − a)3 + · · ·
n=0
whose terms are ascending powers of x or more generally of (x–a). When such a series converges, its sum
will be a function of x. The question we will be asking here is the reverse – given a function f (x), can we
find a power series that converges to it in some x-interval?
For example, the function
1
1−x
is recognized as the sum of a geometric series and that gives us a power series expansion
f (x) =
∞
X
1
= 1 + x + x2 + x3 + · · · =
xn
1−x
n=0
that converges for |x| < 1. The interval of convergence of a power series is the set of x for which the series
converges.
The interval of convergence is always centered at x = a – except that we may or may not have
convergence at the endpoints. The radius of the interval is called the radius of convergence of the
series.
Example 2.3.1. Express
1
1 + x2
as the sum of a power series in x and find the interval of convergence.
2.3. POWER SERIES
62
Example 2.3.2. Express
x3
x+2
as the sum of a power series in x and find the interval of convergence.
Differentiation and integration.
It turns out that under standard assumptions of continuity, differentiability and integrability, we can
get new power series representations out of old by differentiating and integrating. Also it can be
shown that, except for the end points (which we will treat as a special case) this process leaves the
interval of convergence unchanged.
Example 2.3.3. Obtain a power series for
1
(1 − x)2
by recognizing it as a derivative.
2.3. POWER SERIES
Example 2.3.4. Find a power series representation for ln(1 + x).
63
2.3. POWER SERIES
Example 2.3.5. Find a power series representation for arctan(x).
64
2.3. POWER SERIES
Example 2.3.6. Use a power series representation to approximate
Z 0.5
1
dx
1 + x7
0
correct to within 10−7 .
65
2.3. POWER SERIES
66
General summary of section 3-finding power series
The ”Go To” example is always the geometric series
1 + r + r2 + r3 + · · · =
1
1−r
1. Use substitution:
f (x) =
2. Use derivative:
f (x) =
(converges for |r| < 1)
1
x+2
1
(1 − x)2
3. Use integral:
f (x) = ln(1 + x)
4. Substitution or derivative:
f (x) = ln(1 + x2 )
2.4. TAYLOR AND MACLAURIN SERIES
2.4
67
Taylor and Maclaurin Series
We have been using a power series
∞
X
cn xn = c0 + c1 x + c2 x2 + c3 x3 + · · ·
n=0
to give us a systematic approximation of a given function f (x) by a sequence of polynomials. We have
pointed out that the interval of convergence of such a series is centered at x = 0. But sometimes we are
interested in studying a function in a neighborhood of another point x = a. In that case we use a power
series centered at x = a:
f (x) ∼
∞
X
cn (x − a)n = c0 + c1 (x − a) + c2 (x − a)2 + c3 (x − a)3 + · · ·
n=0
and its interval of convergence is centered at x = a.
The question we ask here is:
Given the function f and the point x = a how do we find the coefficients cn ?
There are different ways we might answer this question depending on where and in what way we want the
approximation to be good. The approach we use is to make the function and the series have the same values
of the derivatives of all orders at x = a.
Example 2.4.1. Write:
f (x) = c0 + c1 (x − a) + c2 (x − a)2 + c3 (x − a)3 + · · ·
and repeatedly differentiate both sides, each time evaluating at x = a, to obtain formulas for the cn .
2.4. TAYLOR AND MACLAURIN SERIES
68
Taylor Series
If a power series centered at x = a converges to a function f (x) in some interval (a–R, a + R) for
R > 0:
n
X
f (x) = lim
ck (x − a)k for a − R < x < a + R
n→∞
k=0
then
ck =
f (k) (a)
k!
and the polynomial
Tn (x) =
n
X
f (k) (a)
k=0
k!
(x − a)k = f (a) +
f ′ (a)
f ′′ (a)
f (n) (a)
(x − a) +
(x − a) + · · · +
(x − a)n
1
2!
n!
is called the nth degree Taylor polynomial for f at the point x = a. The infinite series is simply called
the Taylor series for f centered at x = a.
Note: the special case a = 0 is sufficiently important it is given a name of its own and is called the
Maclaurin series for f .
Example 2.4.2. Consider the function f (x) = ex .
(a) Find the Maclaurin series for f (x).
n
(b) Set x = 1 to obtain a famous series expansion for e.
f (n) (x)
x=0
cn
2.4. TAYLOR AND MACLAURIN SERIES
Summary: The nth Taylor polynomial
Tn (x) = c0 + c1 x + c2 x2 + c3 x3 + · · · + cn xn
Tn (x) = f (0) +
f ′ (0)
f ′′ (0) 2 f ′′′ (0) 3
f (n) (0) n
x+
x +
x + ··· +
x
1!
2!
3!
n!
Describe the graphical significance of Tn (x) for n = 0, 1 and 2.
69
2.4. TAYLOR AND MACLAURIN SERIES
70
Example 2.4.3. Find the Maclaurin series for f (x) = sin(x). Use technology to draw graphs of successive
Maclaurin polynomials and observe the progress of the convergence.
Example 2.4.4. Find the Maclaurin series for f (x) = x cos(x).
Note: Use the simple idea of working with the series for sin which we already have. Of course, this indirect
approach leaves open the question of how we know that what we got is in fact the Maclaurin series. There
is a theorem which tells us that such indirect methods always give us the genuine article.
2.4. TAYLOR AND MACLAURIN SERIES
71
Example 2.4.5. Find an upper bound on the error in the approximation
sin(x) ≈ T5 (x) = x −
x3 x5
+
3!
5!
at x = 2 and x = 3. Check your results by comparing the values of sin and T5 at x = 2 and x = 3 and
interpret these with the graphs drawn below.
2.4. TAYLOR AND MACLAURIN SERIES
72
Example 2.4.6. By calculating successive derivatives, find the Maclaurin series for f (x) = (1 + x)k where
k is any real number.
n
f (n) (x)
x=0
0
(1 + x)k
1
1
k(1 + x)k−1
k
2
k(k − 1)(1 + x)k−2
k(k − 1)
3
k(k − 1)(k − 2)(1 + x)k−3
k(k − 1)(k − 2)
This gives us the series:
k
k(k − 1) 2 k(k − 1)(k − 2) 3
(1 + x)k = 1 + x +
x +
x + ···
1
2!
3!
Note that if k is an integer this will terminate, and we get the familiar binominal theorem. Otherwise, it
goes on forever and we get an infinite series which converges for |x| < 1.
Example 2.4.7. Find the Maclaurin series for
f (x) = √
1
.
4−x
Give an explicit formula for the nth term. The first thing we need to do is turn the 4 into a 1, and we do
that by taking it out of the square root.
This gives us
− 12 − 32 x 2
− 12 − 32 − 25 x 3
1
x −1/2 1
1
x
1−
=
1+ −
−
+
−
+
−
+
2
4
2
2
4
2!
4
3!
4
!
− 21 − 23 − 25 − 72 x 4
−
+ ···
4!
4
1
1 x 1 · 3 x 2 1 · 3 · 5 x 3 1 · 3 · 5 · 7 x 4
=
1+
+ 2
+ 3
+
+ ···
2
2 4
2 2! 4
2 3!
4
24 4!
4
1
1
1·3
1·3·5 3 1·3·5·7 4
=
1 + x + 2 x2 + 3
x +
x + ···
2
8
8 2!
8 3!
84 4!
The nth term is
1
2
1 · 3 · 5 · ... · (2n − 1)
8n n!
xn .
2.4. TAYLOR AND MACLAURIN SERIES
73
Example 2.4.8. Use a Maclaurin series to approximate the value of
Z 1
2
e−x dx
0
with an accuracy of 0.001.
2.5. APPROXIMATING FUNCTIONS BY POLYNOMIALS
2.5
74
Approximating Functions by Polynomials
We can use a power series to get a polynomial approximation to a function on a given interval. The question
is how many terms of the power series we need to get a desired accuracy. We already have one way of
answering this question – if the series is alternating and satisfies the assumptions of the alternating series
test, then the error is less than the size of the first omitted term. This is a very simple test and, in the
examples below, we will use it whenever we can. But if the power series is not alternating, this result won’t
help us. Here’s a theorem that will:
Taylor’s Inequality
For any value of x, the error in the Taylor polynomial approximation:
f (x) ≈ Tn (x) = f (a) +
f ′′ (a)
f (n) (a)
f ′ (a)
(x − a) +
(x − a)2 + · · · +
(x − a)n
1
2!
n!
is no greater than
En =
M
dn+1
(n + 1)!
where d = |x − a| and M is an upper bound of |f (n+1) (x)| on the interval [a − d, a + d].
Example 2.5.1. Use Taylor’s inequality to decide how many terms of the series
e=1+
1
1
1
1
1
+ + + + + ···
1! 2! 3! 4! 5!
we need to take to approximate e to within 10−6 .
2.5. APPROXIMATING FUNCTIONS BY POLYNOMIALS
75
Example 2.5.2. Approximate the function f (x) = x1/3 by
a Taylor polynomial of degree 2 at x = 8. How accurate is
this approximation on the interval [7, 9]?
Note: When x < 8, you’ll need to use Taylor’s Inequality.
Solution.
n
f (n) (x)
f (n) (x) at x = 8
cn
0
x1/3
2
2
1
(1/3)x−2/3
(1/3)8−2/3
Div f ′ (8) by 1!
2
(−2/9)x−5/3
(−2/9)8−5/3
Div f ′′ (8) by 2!
3
(10/27)x−8/3
(10/27)8−8/3
Div f (3) (8) by 3!
We use a table to calculate the coefficients. I have not simplified the expressions as I want to be able to
focus on the form of the terms of the series. First note that in the first column (the derivatives of f ) the
signs alternate from n = 1 on. Here are the Taylor polynomials of degree 2 and 3:
T2 (x) = 2 +
(2/9)8−5/3
(1/3)8−2/3
(x − 8) −
(x − 8)2
1!
2!
(1/3)8−2/3
(2/9)8−5/3
(10/27)8−8/3
(x − 8) −
(x − 8)2 +
(x − 8)3
1!
2!
3!
Is the Taylor series of this function alternating? The answer is yes as long as all the terms (x − 8)n are
positive. Otherwise it is not. Thus, we have an alternating series for x > 8 but not for x < 8. Thus, our
error analysis for x > 8 can use the alternating series estimate, but for x < 8 we will use the Taylor estimate.
T3 (x) = 2 +
Case 1 (x > 8). The series is alternating. Here our error estimate will apply to the interval [8, 9].
2.5. APPROXIMATING FUNCTIONS BY POLYNOMIALS
76
Case 2 (x < 8). The series is not alternating.
n
f (n) (x)
x=8
cn
0
x1/3
2
2
1
(1/3)x−2/3
(1/3)8−2/3
Div by 1!
2
(−2/9)x−5/3
(−2/9)8−5/3
Div by 2!
3
(10/27)x−8/3
(10/27)8−8/3
Div by 3!
Here our error estimate will apply to the interval [7, 8]. As a reminder, here is the degree-3 Taylor polynomial:
T3 (x) = 2 +
(2/9)8−5/3
(10/27)8−8/3
(1/3)8−2/3
(x − 8) −
(x − 8)2 +
(x − 8)3
1!
2!
3!
x=7
x=9
x1/3
1.91293
2.08008
T2 (x)
1.91319
2.07986
Difference
0.00026
0.00022
Error Est
0.00034
0.00024
Chapter 3
Multiple Integration
Contents
3.1
Double Integrals: Basic Concepts . . . . . . . . . . . . . . . . . . . . . . .
78
3.2
Double Integrals over General Regions . . . . . . . . . . . . . . . . . . .
84
3.3
Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
88
3.4
Center of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
95
3.5
Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
98
77
3.1. DOUBLE INTEGRALS: BASIC CONCEPTS
3.1
78
Double Integrals: Basic Concepts
If f (x, y) ≥ 0, the volume V of the solid lying below the surface z = f (x, y) and above the region D
in the x − y plane is called the double integral of f over D and is written
ZZ
V =
f (x, y) dA.
D
Note: If f (x, y) is allowed to be negative, the integral gives us a “signed” volume where volumes below the
xy-coordinate plane are counted negative.
Example 3.1.1. The volume over the disk D = {(x, y) | x2 + y 2 ≤ 1}
lying below the plane
y−x
+3
z=
2
and above the xy-plane is written
ZZ y−x
V =
+ 3 dA
2
D
3.1. DOUBLE INTEGRALS: BASIC CONCEPTS
The volume V of the solid lying below the surface
z = f (x, y)
above the surface
z = g(x, y)
and “over” the base region D in the xy-plane is written
ZZ
V =
f (x, y) − g(x, y) dA
D
Example 3.1.2. The volume of the solid above the parabolic
bowl z = x2 + y 2 and below the plane z = 4 is written
ZZ
V =
4 − (x2 + y 2 ) dA
D
where D is the disk:
D = {(x, y) | x2 + y 2 ≤ 4}
79
3.1. DOUBLE INTEGRALS: BASIC CONCEPTS
80
Here is a wonderful general principle for calculating such volumes. You’ve already seen special case of this
in the method of calculating volumes of revolution. For example, you already know how to calculate the
volume in Example 3.1.2 above by slicing the solid horizontally into disks and integrating the area of the
slices.
The Sliced Potato Principle
Suppose we have a general solid in 3-space (think of it as a
potato). Suppose that if we slice it perpendicular to some
suitable axis, say x, we have a way of calculating the crosssection area A(x) at x. Then the volume of the solid is the
integral of these areas A(x) over the interval containing
the solid.
Z
b
V =
A(x) dx
a
Note that this is a natural generalization of the standard formula
for calculating the area under a graph with an integral or (perhaps
more generally) of calculating the area between two curves.
Z b
f (x) − g(x) dx
A=
a
Example 3.1.3. Use the sliced potato principle to calculate the
volume of the solid in Example 3.1.2:
ZZ
V =
4 − (x2 + y 2 ) dA
D
where D is the disk of radius 2:
D = {(x, y) | x2 + y 2 ≤ 4}
3.1. DOUBLE INTEGRALS: BASIC CONCEPTS
Example 3.1.4. Use the sliced potato principle to evaluate the double integral
ZZ
(3x2 + 4xy) dA
D
where D is the rectangle {(x, y) | 0 ≤ x ≤ 2, 0 ≤ y ≤ 4}. Do this in two ways:
(a) Slice perpendicular to the y-axis;
(b) Slice perpendicular to the x-axis.
81
3.1. DOUBLE INTEGRALS: BASIC CONCEPTS
82
Example 3.1.4. [continued]
Observe that this approach has calculated the volume essentially as an iterated integral. The two calculations
are:
(a)
R 4 hR 2
(b)
R 0 hR 4
2
2
i
2 + 4xy)dx dy where the inner integral treats y as constant
(3x
0
i
2 + 4xy)dy dx where the inner integral treats x as constant.
(3x
0
3.1. DOUBLE INTEGRALS: BASIC CONCEPTS
Example 3.1.5. Use an iterated integral to evaluate the double integral
ZZ
y sin(xy) dA
I=
D
where D is the rectangle {(x, y) | 1 ≤ x ≤ 2, 0 ≤ y ≤ π/2}
(a) Write the integral that you get by slicing perpendicular to the x-axis.
(b) Do the calculation by slicing perpendicular to the y-axis. Note that order sometimes matters.
83
3.2. DOUBLE INTEGRALS OVER GENERAL REGIONS
3.2
84
Double Integrals over General Regions
Example 3.2.1. Find the volume of the solid that lies under the plane z = x + 2y and above the region D
in the xy-plane bounded by the line y = 2x and the parabola y = x2 .
The big idea. The sliced potato principle allows us to convert a double integral:
ZZ
f (x, y) dA
D
into an iterated integral
Z b
Z h(x)
a
g(x)
!
f (x, y)dy dx
where in the inner y−integral, x is held constant.
3.2. DOUBLE INTEGRALS OVER GENERAL REGIONS
Example 3.2.2. Evaluate the integral
ZZ
x dA
D
where D is the region bounded by the parabolas
y = 4 − x2
and
y = x2 − 2x.
85
3.2. DOUBLE INTEGRALS OVER GENERAL REGIONS
Example 3.2.3. Find the volume of the tetrahedron bounded by the planes
z = 2 − x − 2y, x = 2y, x = 0, and z = 0.
86
3.2. DOUBLE INTEGRALS OVER GENERAL REGIONS
Example 3.2.4. Evaluate the iterated integral
Z 1Z 1
0
2
ey dy dx
x
by interchanging the order of integration (represent it first as a double integral).
87
3.3. POLAR COORDINATES
3.3
Polar Coordinates
Polar Coordinates.
For the point (x, y), we take r to be the distance of the point from
the origin, and θ be the angle the line from the origin makes with the
positive x-axis. Then
x = r cos(θ)
y = r sin(θ)
• r is called the modulus of (x, y).
• θ is called the argument of (x, y).
Example 3.3.1. Let (x, y) have modulus r = 4 and argument θ = 7π/6. Find x and y.
Let’s go the other way. How do we find r and θ in terms of x and y?
Example 3.3.2.
(a) Let x = 4 and y = 3. Find r and θ.
(b) Let x = −4 and y = 3. Find r and θ.
88
3.3. POLAR COORDINATES
89
Functions in Polar Coordinates - Polar Curves
The graph of a polar equation r = f (θ), or more generally F (r, θ) = 0, consists of all points P that have at
least one polar representation (r, θ) whose coordinates satisfy the equation.
Example. What curves are represented by the following polar equations?
(a) r = 2
(b) θ = π4
Example. Sketch the following polar curve by first finding a Cartesian equation for the curve.
r = 2 cos θ
π
π
Example. Sketch the region D = {(r, θ) | 1 ≤ r ≤ 3, − ≤ θ ≤ }. Such a region is often refered to be a
4
4
polar rectangle.
3.3. POLAR COORDINATES
90
Integrating in Polar Coordinates
When calculating a double integral
ZZ
f (x, y) dA
D
it will be sometimes much better to use polar coordinates to work with the base region D and in this case,
the question arises as to what the differential area dA
should be. When working with Cartesian coordinates,
dA was the area of a differential rectangle with sides
dx and dy and thus, when we moved to an iterated integral, dA turned into the area dxdy of this rectangle.
For the rectangular region at the right:
Z dZ b
ZZ
f (x, y) dA =
f (x, y) dx dy
c
D
a
Suppose, however, that D is a polar rectangle, defined
by the inequalities θ1 ≤ θ ≤ θ2 and r1 ≤ r ≤ r2 . Then,
if f (x, y) could be readily expressed in polar coordinates, we might want to use those coordinates to do
the integration. Thinking about Riemann sums, we’d
decompose the region into differential polar rectangles
with angular width dθ and distance from the origin
increased by dr. But what is the corresponding differential area dA in terms of dθ and dr? Is it the product
dθdr?
Convince yourself that when integrating over a region
in the x − y plane using polar coordinates, the area dA
of a differential polar rectangle must be converted to
rdrdθ.
Calculating a double integral using polar coordinates:
ZZ
Z β
Z r=h(θ)
α
r=g(θ)
f (x, y) dA =
D
!
f (r cos(θ), r sin(θ)) r dr dθ
3.3. POLAR COORDINATES
91
Example 3.3.3. Use polar coordinates to calculate
ZZ p
x2 + y 2 dA
D
where D is the disk {(x, y) | x2 + y 2 ≤ 1}.
3.3. POLAR COORDINATES
92
Example 3.3.4. Let E be the solid bounded by the plane z = 0 and the paraboloid z = 1 − x2 − y 2 . Write
an iterated integral for the volume of E.
3.3. POLAR COORDINATES
93
Example 3.3.5. Write an iterated integral in polar coordinates for the area enclosed by one loop of the
four-leaved rose r = cos(2θ).
Z
[Tech-help:
1
1
cos2 (u) du = u + sin(2u)]
2
4
3.3. POLAR COORDINATES
94
Example 3.3.6. Use polar coordinates to find the volume of the solid bounded by the xy-plane, the cylinder
x2 + y 2 = 2x, and the paraboloid z = x2 + y 2 .
Z
[Tech-help:
cos4 (u) du =
1
3
3
cos3 (u) sin(u) + cos(u) sin(u) + u.]
4
8
8
3.4. CENTER OF MASS
3.4
95
Center of Mass
Imagine we have a seesaw running between x = 2 and
x = 6, and an adult at x = 2 weighing 3 times as much
as a child at x = 6. Where should the fulcrum be so
that the adult will balance the child? The answer is
that it should be 3 times as far from the child as from
the adult. And that places it at x̄ = 3.
If the adult is 3 times as big as the child, the child
should be 3 times as far from the “center”.
Another way to say this is that the products of the
mass and the distance from the center must be same
for adult and child:
m1 (x̄ − x1 ) = m2 (x2 − x̄)
where x̄, the position of the balance point, is called
the center of mass. Solving:
x̄ =
m1 x1 + m2 x2
m1 + m2
This equation displays the center of mass x̄ as a
weighted average of x1 and x2 where the mi are used
as weights:
For our example:
x̄ =
3·2+1·6
= 3.
3+1
More generally, suppose the mass is continuously distributed over the x-axis with density ρ(x). Then the
center of mass formula becomes:
Ra
Z a
xρ(x) dx
1
b
x̄ = R a
=
xρ(x) dx
M b
b ρ(x) dx
where the total mass M is the integral of the density.
Finally suppose the mass is distributed over a planar region R. Then the centroid will have two components
(x̄, ȳ). The first will balance the masses in the x-direction and the second will balance the masses in the
y-direction. The formulas are:
RR
xρ(x, y) dA
ZZ
1
R
RR
x̄ =
=
xρ(x, y) dA
M
ρ(x, y) dA
R
R
RR
R
ȳ = RR
R
yρ(x, y) dA
1
=
M
ρ(x, y) dA
ZZ
yρ(x, y) dA
R
3.4. CENTER OF MASS
96
Example 3.4.1. Let E be a triangular lamina with vertices (0, 0), (1, 0) and (0, 2) and density function is
ρ(x, y) = x.
(a) Find the mass E.
(b) Calculate the coordinates of the center of mass.
3.4. CENTER OF MASS
Example 3.4.2. Find the mass and center of mass of the semicircular lamina
(x, y) | x2 + y 2 ≤ 1, y ≥ 0
if the density equals the distance r from the origin.
97
3.5. TRIPLE INTEGRALS
3.5
98
Triple Integrals
Here we look at problems in which we have a function f (x, y, z) of three variables and we want to integrate
it over a solid region E in R3 . This will be written as the triple integral:
ZZZ
f (x, y, z) dV.
E
For example, if the temperature at a point in the region E is T (x, y, z) then the average temperature over
the region will be:
ZZZ
1
T̄ =
T (x, y, z) dV
V
E
where
ZZZ
dV is the volume of E.
V =
E
In the examples we will work with it will always be the
case that there is an axis (say z) for which (if we think
of it as pointing “up”) the region E will be described
as all points lying below the top surface z = H(x, y)
and above the bottom surface z = L(x, y). Then the
triple integral can be written:
!
ZZZ
ZZ Z
H(x,y)
f (x, y, z) dV =
E
f (x, y, z) dz
D
dA
L(x,y)
Once the dz-integral is calculated, the integral over D can be done in the usual way, either with Cartesian or
polar coordinates. If we use polar coordinates in the x-y place the 3-D coordinate system is called cylindrical
coordinates.
3.5. TRIPLE INTEGRALS
99
Example 3.5.1. Evaluate the triple integral
ZZZ
xyz 2 dV
B
where B is the rectangular box:
B = {(x, y, z) | 0 ≤ x ≤ 1, −1 ≤ y ≤ 2, 0 ≤ z ≤ 3} .
3.5. TRIPLE INTEGRALS
100
Example 3.5.2. Consider the tetrahedron E bounded by the four planes:
x = 0, y = 0, z = 0, and x + y + z = 1.
(a) Evaluate the triple integral
ZZZ
z dV.
E
3.5. TRIPLE INTEGRALS
Example 3.5.2. [continued]
(b) Does your answer allow you to easily find the centroid of E?
101
3.5. TRIPLE INTEGRALS
102
Example 3.5.3. Evaluate the triple integral
ZZZ p
x2 + y 2 dV
E
where E is the region bounded by the plane z = 4 and the paraboloid z = x2 + y 2 .
3.5. TRIPLE INTEGRALS
103
Example 3.5.4. Find the centroid of the region bounded by the parabolic cylinder x = y 2 , and the planes
z = x, z = 0, and x = 1.
3.5. TRIPLE INTEGRALS
Example 3.5.4. [continued]
104
3.5. TRIPLE INTEGRALS
105
Example 3.5.5. A solid E lies inside the cylinder x2 + y 2 = 1, and between the plane z = 4, and the
paraboloid z = 1 − x2 − y 2 . The density at any point is equal to its distance from the axis of the cylinder.
Find the mass of E.
3.5. TRIPLE INTEGRALS
Example 3.5.6. Evaluate
106
Z 2 Z √4−x2
−2
√
− 4−x2
!
Z 2
√
(x2 + y 2 ) dz
x2 +y 2
dy dx.
3.5. TRIPLE INTEGRALS
107
Example 3.5.7. For the integral
Z 1 Z 1 Z 1−y
0
change the order of integration to dy dx dz.
√
dz dy dx,
x
0
3.5. TRIPLE INTEGRALS
Example 3.5.7. [continued]
108
Chapter 4
Tutorial Problems
Contents
4.1
Tutorial 1 - Parameterized Lines and Curves . . . . . . . . . . . . . . . . 110
4.2
Tutorial 2 - Functions of Several Variables . . . . . . . . . . . . . . . . . 112
4.3
Tutorial 3 - Partial Derivatives and Tangent Planes . . . . . . . . . . . . 114
4.4
Tutorial 4 - Linear Approximation and Multivariable Chain Rule . . . 117
4.5
Tutorial 5 - Gradient and Directional Derivatives . . . . . . . . . . . . . 120
4.6
Tutorial 6 - Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
4.7
Tutorial 7 - Infinite Series, Alternating Series and Power Series . . . . 125
4.8
Tutorial 8 - Taylor and Maclaurin Series and Taylor’s Inequality . . . . 127
4.9
Tutorial 9 - Double Integrals and Iterated Integrals . . . . . . . . . . . . 129
4.10 Tutorial 10 - Integration with Polar Coordinates, Center of Mass . . . 131
4.11 Tutorial 11 - Triple Integrals
. . . . . . . . . . . . . . . . . . . . . . . . . 134
109
4.1. TUTORIAL 1 - PARAMETERIZED LINES AND CURVES
4.1
110
Tutorial 1 - Parameterized Lines and Curves
1. A bug moves on the surface z = x2 − 2y 2 such that its projection down on the xy-coordinate plane traces
the path x = t2 , y = t3 where t is time. What is the speed of the bug and the slope of its path on the surface
at t = 1?
2. An elliptical water slide has equation
x = 3 cos(πt), y = 2 sin(πt),
z = 3t + 1.
(a) Find the slope of the slide at t = 0 (at the point (3, 0, 1)).
(b) Find parametric equations for the tangent line to the water slide at the point (3, 0, 1).
4.1. TUTORIAL 1 - PARAMETERIZED LINES AND CURVES
3. Find the equation of the plane that passes through the point (6, 0, −2) and contains the line
x = 4 − 2t, y = 3 + 5t, z = 7 + 4t
111
4.2. TUTORIAL 2 - FUNCTIONS OF SEVERAL VARIABLES
4.2
Tutorial 2 - Functions of Several Variables
1. A function in 3D is given by z = f (x, y) = 1 − y 2 .
(a) Sketch the surface.
(b) Draw a contour map of the function showing several level curves for z = −2, −1, 0, 1.
2. A contour diagram of the function
f (x, y) = 3x2 − x3 + 2y − 2y 2
is given at the right. The contour lines
are drawn at intervals of z = 1.
(a) Select some points on the graph and
determine the level curve values. Label
level curves.
(b) On xz-plane, sketch the intersection curve of the surface with the vertical plane y = k.
112
4.2. TUTORIAL 2 - FUNCTIONS OF SEVERAL VARIABLES
113
3. Consider the curve in R3 formed by the intersection of the cylinder x2 +y 2 = 1 with the plane x+y+z = 1.
This curve can be parameterized by
x = cos(t)
y = sin(t)
z = 1 − cos(t) − sin(t)
Interpreting t as time, find the times at which the moving point has maximum and minimum speed. Identify
these points on the diagram. What are the corresponding speeds?
4.3. TUTORIAL 3 - PARTIAL DERIVATIVES AND TANGENT PLANES
4.3
Tutorial 3 - Partial Derivatives and Tangent Planes
1. Consider the function
f (x, y) = arctan
x+y
xy
(a) Find and simplify both partial derivatives of f .
(b) Find all points at which fx = fy .
2. Calculate and simplify:
p
∂
ln x x2 + y 2
∂x
114
4.3. TUTORIAL 3 - PARTIAL DERIVATIVES AND TANGENT PLANES
115
3. The equation x2 + y 2 + z 2 = 9 describes a sphere of radius 3 centred at the origin. The point P = (2, 2, 1)
is on this sphere. Find a parametric equation for the line that is tangent to the sphere at P and that
intersects the z-axis.
4.3. TUTORIAL 3 - PARTIAL DERIVATIVES AND TANGENT PLANES
116
4. Show that there are two different lines that pass through the point (1, 0, 1) and are entirely contained
in the surface z = x2 − y 2 . Find parametric equations for these lines and draw their projection on the x-y
plane.
4.4. TUTORIAL 4 - LINEAR APPROXIMATION AND MULTIVARIABLE CHAIN RULE
4.4
117
Tutorial 4 - Linear Approximation and Multivariable
Chain Rule
1. Let the function f (x, y) at the point (3, −1) take the value f (3, −1) = 5 and have partial derivatives
fx (3, −1) = 2, fy (3, −1) = 4.
(a) Find the equation of the tangent plane to the surface z = f (x, y) at the point (3, −1).
(b) Find the linearization L(x, y) of f at the point (3, −1).
(c) Suppose that x changes from 3 to 3.1 and y changes from −1 to −0.8. Find the ”first-order approximate
change” in f (x, y). [This is denoted dz and is also called the ”linearized change”.]
4.4. TUTORIAL 4 - LINEAR APPROXIMATION AND MULTIVARIABLE CHAIN RULE
118
2. A bug is crawling on a surface such that at the time t in seconds, its location is given by
x(t) = t2 − 4t + 5,
y(t) = t2 − 3t + 1,
z = 9 − x2 + 3y 2
where x, y, z are measured in centimeters, and t is in minutes. include units in your answers.
(a) How fast is the height z of the bug changing at t = 3?
√
(b) Suppose the temperature at a location (x, y, z) is given by T (x, y, z) = 2x−y 2 + xz measured in degrees
Celsius. How fast is the temperature changing along the bug’s path when t = 3?
4.4. TUTORIAL 4 - LINEAR APPROXIMATION AND MULTIVARIABLE CHAIN RULE
119
3. [Winter 2021 Midterm Q6] The objective of this problem is to find the shortest distance between the
paraboloid z = 1 − 2x2 − y 2 and the plane x + y + z = 4.
(a) sketch a graph.
(b) Find a tangent plane of the paraboloid that is parallel to the plane x + y + z = 4. What is the tangent
point P?
(c) The shortest distance between the paraboloid z = 1 − 2x2 − y 2 and the plane x + y + z = 4 is the distance
between the tangent point P (above) on the paraboloid and the plane x + y + z = 4. Find this shortest
distance.
4.5. TUTORIAL 5 - GRADIENT AND DIRECTIONAL DERIVATIVES
4.5
120
Tutorial 5 - Gradient and Directional Derivatives
1. For the function f (x, y, z) = y 2 ez/x , find the derivative of f at the point (2, 1, 0) in the direction of the
vector u = [1, 2, −2].
2. Let z be defined as a function of x and y implicitly by the equation xz = z y . Suppose that x = y = 2
and z = 4, and x and y are both increasing at rate 1. How fast is z changing?
4.5. TUTORIAL 5 - GRADIENT AND DIRECTIONAL DERIVATIVES
121
3. Let
F (x, y, z) = x2 + y 2 + z 2 .
Note that F (3, 1, 2) = 14. Suppose the equation F (x, y, z) = 14 defines z as a function of x and y in a
neighbourhood of the point (3, 1, 2):
z = f (x, y).
Calculate both the gradient of F and the gradient of f at the point x = 3, y = 1, z = 2, and discuss the
relationship between these two measures of change.
4.5. TUTORIAL 5 - GRADIENT AND DIRECTIONAL DERIVATIVES
122
4. In two different ways, find a parametric equation for the line that is orthogonal to the surface z 3 xy 2 =
xy − z + 3 and passes through the point (1, 2, 1) on the surface.
(a) Method 1. The surface defines z as a function f of x and y. Use implicit differentiation to find the
partial derivatives of f and evaluate them at the point (x, y) = (1, 2). Then use these to write the equation
of the tangent plane, and then take its normal vector.
(b) Method 2 (Use gradients). Treat the equation as defining the level surface of a function T of three
variables.
4.6. TUTORIAL 6 - OPTIMIZATION
4.6
123
Tutorial 6 - Optimization
1. Find and classify the critical points of
f (x, y) = (x2 − 4)(y − 2)(x + y).
x
y
fxx
fyy
fxy
D
4.6. TUTORIAL 6 - OPTIMIZATION
124
2. Below is the contour diagram for the surface
f (x, y) = (x2 − 4)(y − 2)(x + y).
(a) The contour lines are spaced at intervals of z = 1. Use the formula to identify the z value of a number
of the contours.
(b) You are walking on the surface along the line y = 2–x. Parameterize the line as x = t and draw the
graph of your altitude z against t in two different ways. First use the contour diagram to obtain as good an
estimate as you can to locate the max or min points and the inflection points of your journey. Second use
the equation of the surface to check your estimates.
4.7. TUTORIAL 7 - INFINITE SERIES, ALTERNATING SERIES AND POWER SERIES
4.7
125
Tutorial 7 - Infinite Series, Alternating Series and Power
Series
1. (a) Start with the geometric series:
1
= 1 + x + x2 + x3 + x4 + x5 + x6 + ...
1−x
and differentiate it twice to get a series for
1
.
(1 − x)3
(b) If we take x = −1/4 in the series of (a) we get a series converging to (4/5)3 . Write the series and find a
formula for the nth term.
(c) How many terms of that series do we need to take to get within 1/1000th of (4/5)3 ?
4.7. TUTORIAL 7 - INFINITE SERIES, ALTERNATING SERIES AND POWER SERIES
2. By any method find an expression for the sum of the infinite series
∞
X
(−1)n+1
n=1
(x − 1)2n
n
for |x − 1| < 1. Simplify your expression as much as possible.
126
4.8. TUTORIAL 8 - TAYLOR AND MACLAURIN SERIES AND TAYLOR’S INEQUALITY 127
4.8
Tutorial 8 - Taylor and Maclaurin Series and Taylor’s
Inequality
1. Find the Maclaurin series for arctan(x2 ). Write an expression for the general term and explicitly display
the first four non-zero terms.
Tech help:
d
1
arctan(x) =
dx
1 + x2
4.8. TUTORIAL 8 - TAYLOR AND MACLAURIN SERIES AND TAYLOR’S INEQUALITY 128
2. (a) Find the Taylor series for the function f (x) = ln(x) centered at x = 1. Simplify your expression as
much as you can. Formulate the general term and explicitly display the first four non-zero terms.
(b) Use Taylor’s Inequality to get as good as estimate as possible for the maximum error if the Taylor
polynomial of degree 3 is used to approximate ln(x) on the interval [0.5, 1.5].
(c) Note that if x − 1 is positive the Taylor series will be alternating and then we can use the alternating
series test. Calculate the error in this manner and compare with your answer in (b).
4.9. TUTORIAL 9 - DOUBLE INTEGRALS AND ITERATED INTEGRALS
4.9
129
Tutorial 9 - Double Integrals and Iterated Integrals
1. (a) Let D be the region bounded by the line y = x and the parabola y = 3x − x2 . Write down and
evaluate an iterated integral for the volume of the solid lying above D and below the plane z = 4x + 5. Slice
the solid parallel to the y-axis.
(b) Now change the order of integration but do not evaluate the new integral.
4.9. TUTORIAL 9 - DOUBLE INTEGRALS AND ITERATED INTEGRALS
130
2. (a) Find the volume of the solid S lying above the region D bounded by the lines y = x, y = x/2, and
y = 2, and below the plane z = 2x + 3y. Take slices parallel to the x-axis.
(b) Now draw a picture of a typical cross-section belonging to a particular y. Because of its simple shape,
you can calculate the area A(y) of the cross-section using simple geometric reasoning. Check with your work
of part (a).
4.10. TUTORIAL 10 - INTEGRATION WITH POLAR COORDINATES, CENTER OF MASS131
4.10
Tutorial 10 - Integration with Polar Coordinates, Center of Mass
1. Find the mass and center of mass of the region D bounded by y = 1 − x2 and the x-axis if the density at
any point is ρ(x, y) = y.
4.10. TUTORIAL 10 - INTEGRATION WITH POLAR COORDINATES, CENTER OF MASS132
2. At the right is a graph of the cardioid
r = 1– sin(θ)
Use polar coordinates for the following calculations. Perform the r-integration but not the θ-integration. Thus your
answer will be a θ-integral.
(a) Calculate the area of the cardioid.
(b) Let the density of the cardioid be ρ(x, y) = r. That is, its density increases linearly with distance from
the origin. Calculate the mass of the cardioid. (Mass is integral of density).
(c) Calculate ȳ, the y-coordinate of the center of mass.
4.10. TUTORIAL 10 - INTEGRATION WITH POLAR COORDINATES, CENTER OF MASS133
(d) [from 2014 final exam] Find the width and the height of the box which exactly encloses the cardioid.
4.11. TUTORIAL 11 - TRIPLE INTEGRALS
4.11
134
Tutorial 11 - Triple Integrals
1. Let C be a right circular cone of height 2 with base radius 1. Set it in a coordinate system with vertex
at the origin. Find its centroid following two different approaches:
(a) Use a triple integral.
(b) Slice it horizontally treating it as a volume of revolution.
4.11. TUTORIAL 11 - TRIPLE INTEGRALS
135
2. A solid is bounded by the planes x = 0, z = 0 and y = 2x, and the surface z = 4–y 2 in the first octant.
Express its volume as a triple integral in three ways, by integrating:
(a) in the order dz dy dx
(b) in the order dx dz dy
(c) in the order dy dz dx