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MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Vectors
Lecture 1 · vector arithmetic revision · dot product · cross productText Reference: §4.1 - 4.2
Vectors and Lines, a quick review.
Many quantities in nature are completely specified by one number (called the magnitude of the
quantity) and are usually referred to as scalar quantities. Some examples are temperature, time,
length, and mass.
However, certain quantities require both a magnitude and a direction to specify them. To
say that a boat sailed 10 kilometers (km) does not specify where it went. It is necessary to
give the direction too; perhaps it sailed 10 km northwest. We then describe the position of the
boat by giving its displacement relative to some point, a quantity that involves distance as
well as direction. Quantities that require both a magnitude and a direction to describe them are
called vectors. Other examples include velocity and force. Vector quantities will be denoted by
boldface type: u,v,w, and so on. In handwritten work vectors are denoted by v˜ or by −→v . Thevector that joins the two points A and B is denoted
−−→AB or by AB.
A vector v can be represented geometrically as a directed line segment or arrow. The magnitude
of a vector v will be denoted by ‖v‖ and is sometimes referred to as the length of v because itis represented by the length of the arrow.
Two vectors v and w are equal (written v = w)
if they have the same length and the same di-
rection. Thus, for example, the two vectors in
the diagram are equal even though the initial
and terminal points are different!
vw
There is one vector that has no direction whatsoever-the zero vector 0.
Given a vector v the vector that has the same
length as v but opposite in direction is the neg-
ative of vector v, denoted −v.v
− v
When we multiply a vector by a scalar we
multiply the length of the vector by the rel-
evant amount, without changing its direction
(unless the scalar is negative and then the di-
rection is opposite). Two vectors are parallel if
one is a scalar multiple of the other.
That is, if a = λb then a is parallel to b.
−½vv2v
ENG1091 Mathematics for Engineering page 1
If u and v are two vectors we define their sum u+ v by adding the vectors ‘head to tail’which
is to say we attach the tail of the second vector,v, to the head of the first u, the sum u + v is
then the vector drawn from the tail of first vector to the head of the last.
This method allows also us to add several vec-
tors at once. a + b + cc
b
a
Should it happen that vectors add together forming a loop, so that the end point is the same as
the initial point, then the vector sum is 0. Thus for example if A,B,C are any three points in
space−−→AB +
−−→BC +
−→CA = 0.
We can also add two vectors u and v geometrically by drawing them from the same point and
completing a parallelogram with the two vectors as adjacent sides. The diagonal vector drawn
from the common tail to the common head point is then the vector u+ v.
From the parallelogram method of vector addition we see that u+ v = v + u.
The opposing diagonal, drawn towards v, is the vector v − u.
The unit coordinate vectors.
Vectors of length one unit are called unit vec-
tors. The unit vectors parallel to the positive x
and y axis in the plane are labelled i and j.
In three dimensional space we add a further
unit vector, k , parallel to the z axis.
Any vector r in space can be written as a com-
bination of multiples of i, j and k. The coeffi -
cients of i, j and k are called its rectangular
components.
r = (x, y, z) = xi+ yj+ zk
i
k
j
x
z
y
The magnitudes of vectors given in component form.
Using Pythagoras’theorem it is an easy matter to find the lengths of vectors:
In three dimensions where we have v = ai+ bj+ ck, then ‖v‖ = ‖ai+ bj+ ck‖ =√a2 + b2 + c2.
Example: ‖i− j+ k‖ =√
(1)2 + (−1)2 + (1)2 =√
3
In two dimensions the length of v = ai+ bj is given by |v| = ‖ai+ bj‖ =√a2 + b2.
ENG1091 Mathematics for Engineering page 2
The Scalar or “Dot”Product
In the previous section we saw how vectors can be added/subtracted together, and we saw how
to multiply them by scalars. The question naturally arises: is it possible to multiply two vectors
together?
There are two types of vector multiplication that are generally useful-the scalar or dot product
and the vector or cross product. Now for a word of warning. Many of the rules we take for
granted in ordinary arithmetic don’t hold when it comes to vector multiplication. When we look
at the vector cross product later this lecture we will see that a×b 6= b×a.We will also see thatthere is no such thing as vector division-vectors don’t have reciprocals! Of course we don’t just
multiply vectors for fun-we do it because it has useful applications.
First, consider the scalar product. One modern use of the scalar product is the projection of
a 3D image on a 2D screen and to do it in such a way as to convince the viewer that he/she is
looking at a 3D image.
Given two vectors a and b then we define their scalar or ‘dot’product as
a · b = (‖a‖ ‖b‖ cos θ)
where θ is the angle between the two vectors.
Note that a · b is a scalar quantity-it is not a vector.
Historically the reason that the scalar product was studied is that in physics the work done by
a force F in moving an object a displacement d is the dot product of force with displacement,
i.e. W = F · d.
From the definition we immediately get the following:
(i) a · a = ‖a‖2 (because the angle between a vector a and itself is 0.)
(ii) If a ⊥ b then a · b = 0
The dot products of the unit vectors i, j and k.
Given the definition above we see thati · j = j · k = k · i = 0
and i · i = j · j = k · k = 1
Properties of the Dot Product
(i) a · b = b · a the dot product is commutative
(ii) λa · b = a · λb =λ (a · b) , for any scalar λ
(iii) a · (b+ c) = a · b+ a · c the dot product is distributive
Notice that the expression a · (b · c) has absolutely no meaning because it is attempting toform a dot product of vector a with the scalar b · c.
The expression a (b · c) has a meaning though it is better written as (b · c)a. The expression(b · c)a means to multiply vector a by the scalar b · c, resulting in a vector having the same oropposite direction as a and of length: = |b · c| ‖a‖ .
ENG1091 Mathematics for Engineering page 3
Notice how we can use the distributive law to simplify the dot product of two vectors given in
component form: Let a = a1i+ a2j+ a3k, and b = b1i+ b2j+ b3k then
a · b = (a1i+ a2j+ a3k) · (b1i+ b2j+ b3k)
= a1i · (b1i+ b2j+ b3k) + a2j · (b1i+ b2j+ b3k) + a3k · (b1i+ b2j+ b3k)
= a1i · b1i+ a2j · b2j+ a3k · b3k (since i · j = j · k = i · k = 0)
= a1b1 + a2b2 + a3b3 (since i · i = j · j = k · k = 1)
This gives a computational formula for evaluating a · b = a1b1 + a2b2 + a3b3
Example: This next example should convince you that there is no such thing as being able to
‘cancel’out common vectors from a dot product.
Let a = 2i− j+ 4k, b = −i+ 2k, and c = 3i. Show a · b = a · c. Comment.
a · b = (2) (−1) + (−1) (0) + (4) (2) = 6 and a · c = (2) (3) + (−1) (0) + (4) (0) = 6.
Observe that b 6= c.
We conclude it is not possible to cancel out vectors (even non-zero vectors) from a dot product
like we can in ordinary arithmetic.
As a geometrical application we use the dot product to find the angle between two vectors:
cos θ =a · b‖a‖ ‖b‖ .
Example: Find the angle between the main diagonal of a cube and the diagonal of a face which
it meets:
This angle will be the same regardless of the size of the cube so lets assume the cube has a side
length equal to 1.
Then the face diagonal a is i+ k and the main
diagonal b is i+ j+ k.
Now a · b = (1) (1) + (0) (1) + (1) (1) = 2 and
‖a‖ =√
(1)2 + (1)2 =√
2 and
‖b‖ =√
(1)2 + (1)2 + (1)2 =√
3 giving
cos θ =2√2√
3from which θ = 35.26
θ
The dot product provides a very easy way of telling when two vectors are perpendicular.
If a · b = 0 then θ = 90o and we write a ⊥ b.
Example: Show that the points P (2, 1,−3) , Q (4, 2,−5) and R (3, 3,−1) are the vertices of a
right angled triangle.−−→PQ =
−−→OQ−−−→OP = (4i+ 2j− 5k)− (2i+ j− 3k) = 2i+ j− 2k
−→PR =
−−→OR−−−→OP = (3i+ 3j− k)− (2i+ j− 3k) = i+ 2j+ 2k
−−→QR =
−−→OR−−−→OQ = (3i+ 3j− k)− (4i+ 2j− 5k) = −i+ j+ 4k,
and from these it is clear that−−→PQ · −→PR = 0 so we conclude the triangle is right angled at P.
ENG1091 Mathematics for Engineering page 4
The Vector or “Cross”Product
This is a way of ‘multiplying’ two vectors to-
gether which results in a vector. Given two
vectors a = a1i+ a2j+ a3k, and
b = b1i+ b2j+ b3k then we define their ‘vector’
or ‘cross’product as
a× b = (‖a‖ ‖b‖ sin θ)n
where θ is the angle between the two vectors,
and n is the unit vector perpendicular to both
a and b, in a right-hand rule direction:
Note: (i) a× b = −b× a(ii) If θ = 0o then a× b = 0
(iii) If θ = 90o then ‖a× b‖ = ‖a‖ ‖b‖
The cross products of the unit coordinate vectors i, j and k.
Given the definition above we see that
i× j = k
j× k = i
k× i = j
and i× i = j× j = k× k = 0
Properties of the Cross Product
(i) a× b = − (b× a) cross product is anti-commutative
(ii) λa× b = a× λb =λ (a× b) , for any scalar λ
(iii) a× (b+ c) = a× b+ a× c cross product is distributive
(iv) a× (b× c) 6= (a× b)× c (in general) non-associativity of the cross product
So if a = a1i+ a2j+ a3k, and b = b1i+ b2j+ b3k then
a× b = (a1i+ a2j+ a3k)× (b1i+ b2j+ b3k)
= (a1b1) i× i+ (a1b2) i× j+ (a1b3) i× k
+ (a2b1) j× i+ (a2b2) j× j+ (a2b3) j× k
+ (a3b1)k× i+ (a3b2)k× j+ (a3b3)k× k
ENG1091 Mathematics for Engineering page 5
continuing:
a× b = (a2b3 − a3b2) i− (a1b3 − a3b1) j+ (a1b2 − a2b1)k
=
∣∣∣∣∣ a2 a3
b2 b3
∣∣∣∣∣ i−∣∣∣∣∣ a1 a3
b1 b3
∣∣∣∣∣ j+∣∣∣∣∣ a1 a2
b1 b2
∣∣∣∣∣knote the ‘− ’in the j term
=
∣∣∣∣∣∣∣∣i j k
a1 a2 a3
b1 b2 b3
∣∣∣∣∣∣∣∣A geometrical application of the cross-product:
Two vectors a and b, if drawn from the same point, define a parallelogram:
Now we can determine the area of the parallelogram by breaking it up into two identical triangles.
Area = 2× 12base× perpendicular height
A = ‖a‖ × ‖b‖ sin θ
= ‖a× b‖
Examples
(a) Let P,Q,R be the points P (2, 1,−3) , Q (3, 4, 7) and R (1,−2, 3). Find the area of the
parallelogram which has PQ and PR as adjacent sides.−−→PQ =
−−→OQ−−−→OP = (3i+ 4j+ 7k)− (2i+ j− 3k) = i+ 3j+ 10k
−→PR =
−−→OR−−−→OP = (i− 2j+ 3k)− (2i+ j− 3k) = −i− 3j+ 6k
So−−→PQ×−→PR =
∣∣∣∣∣∣∣∣i j k
1 3 10
−1 −3 6
∣∣∣∣∣∣∣∣ = i
∣∣∣∣∣ 3 10
−3 6
∣∣∣∣∣− j∣∣∣∣∣ 1 10
−1 6
∣∣∣∣∣+ k
∣∣∣∣∣ 1 3
−1 −3
∣∣∣∣∣= ((3) (6)− (−3) (10)) i− ((1) (6)− (−1) (10)) j+ ((1) (−3)− (−1) (3))k
= 48i− 16j
Hence Area =∥∥∥−−→PQ×−→PR∥∥∥ =
√(48)2 + (−16)2 = 16
√32 + 1 = 16
√10.
(b) Find area 4QPR = 12
∥∥∥−−→PQ×−→PR∥∥∥ = 8√
10.
ENG1091 Mathematics for Engineering page 6
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Vectors
Lecture 2 lines in 3DText Reference: §4.3.1
1. Revision of straight lines in two dimensional space
We are all quite familiar with the two-dimensional representation of a line as y = mx+ b, (called
its Cartesian equation) where m is the slope and b is the y-intercept. Students should also be
familiar with the point-slope equation of a straight line:
(y − y0) = m(x− x0) (1)
Given any two points (x1, y1) and (x2, y2) in the x-y plane, we can readily get the equation of
the line passing through these two points by finding the slope m = (y1−y2)(x1−x2) , and using this value
in the equation (1) above. The basic equation of a straight line is unique up to a scalar factor,
regardless of which point is chosen as (x0, y0) .
It would be natural to try to extend the equation of line in 2D space to 3D space. Perhaps one
might consider z = m1x + m2y + b. Unfortunately, this does not work, indeed, we will see in a
future lecture that this is actually the Cartesian equation of a plane in three-dimensional space.
2. Equations of straight lines in three dimensional space
In three-dimensional space, the concept of a slope is not so easily defined. Instead of slope, a
straight line will have an orientation associated with it that can be represented as a vector. The
line is then fully defined by a point on the line, say A, and an orientation vector, say v. Note
that the magnitude of the orientation vector doesn’t actually matter, as long as we travel in the
right direction, we should stay on the line. Working in Cartesian coordinates, we can define the
point A = (a, b, c), by its position vector and v as a vector with components (p, q, r). Then the
position vector−−→OP of any point P on the line is given by
−−→OP =
−→OA +
−→AP where
−→AP = tv for
some scalar t. We can define the equation of a line r(t) as:
r(t) =−−→OP =
−→OA+ tv,
i.e. (x, y, z) = (a, b, c) + t(p, q, r)
This is the vector equation of a line. The variable t, which can take on any real value, is known
as the parametric variable. Breaking this equation up into the three components we obtain the
parametric equations of a straight line:
x(t) = a+ pt,
y(t) = b+ qt,
z(t) = c+ rt.
ENG1091 Mathematics for Engineering page 7
(Note: students may have actually been introduced to parametric variables when learning
trigonometry. A circle of radius a centred at the origin can be represented by the paramet-
ric equations x = a cos(t) and y = a sin(t), where t can represent the angle from the x-axis.)
If we are given two-points, say A (x1, y1, z1) and B (x2, y2, z2), then the line between these two
points can be readily found by defining the orientation (direction) vector as the vector from A
to B.
Example1: Define the (vector and parametric) equation of the line between the points A (2, 3, 4)
and B (1, 1, 1) .
−−→AB = (i+ j+ k)− (2i+ 3j+ 4k) = −i− 2j− 3k = v
Equation of line:−−→OP =
−→OA+ tv = (2i+ 3j+ 4k) + t (−i− 2j− 3k)
= (2− t) i+ (3− 2t) j+ (4− 3t)k
Parametrically:
x(t) = 2− t,
y(t) = 3− 2t,
z(t) = 4− 3t.
Notice how the parametric variable works. If t = 0, we are at one point, A (2, 3, 4) , and if t = 1
we are at the other point, B (1, 1, 1) .
Example 2: From the previous example, find the value of t that defines the point (0,−1,−2) .
Again, any value of t defines some point on the line. The value t = 1/2 defines the mid-point of
AB.
Equate x values: solve 2− t = 0 from which t = 2. If this value of t gives matching y and z values
we know the point (0,−1,−2) is on the line. Otherwise the point lies off the line.
With t = 2, y(2) = 3 − 4 = −1, and z(2) = 4 − 6 = −2. Therefore we conclude the point
(0,−1,−2) is on the line.
With t = 1/2, x(12) = 2 − 1
2 = 32 , y(1
2) = 3 − 1 = 2, and z(12) = 4 − 3
2 = 52 ; so
(32 , 2,
52
)is the
midpoint of AB.
Also note, however, that the equation of a line is not unique. The line between the points (2, 3, 4)
and (0,−1,−2) is equivalent to the equation found in the first example, but the equation looks
different: v = ((−j− 2k)− (2i+ 3j+ 4k)) = −2i− 4j− 6k
−−→OP =
−→OA+tv = (2i+ 3j+ 4k)+t ((−j− 2k)− (2i+ 3j+ 4k)) = (2− 2t) i+(3− 4t) j+(4− 6t)k
So
x(t) = 2− 2t,
y(t) = 3− 4t,
z(t) = 4− 6t.
The equation looks different but is it really?
ENG1091 Mathematics for Engineering page 8
Finally note that the equation of a line can be manipulated to eliminate the parametric variable,
t. In this form the equation of the line is:
x− ap
=y − bq
=z − cr
This is sometimes called the algebraic equation of a straight line. Students should note that
given this form of the equation of a line, we can immediately read off a point on the line and its
orientation vector.
Example 3: Given the relationx+ 2
1=
y
−2=z − 3
2
find any two points on the line.
Solution: By examining the general from in the previous equation we see x = −2, y = 0, z = 3
is one such point (equate each numerator to zero).
Now of course the choice of zero is completely arbitrary; we can of course equate each fraction
to 1 (or any real number)
we do this: x+ 2
1= 1 giving x = −1
y
−2= 1 giving y = −2
z − 3
2= 1 giving z = 5
Thus the point (−1,−2, 5) is also on the line.
Importantly, a direction vector for the line can also be read off namely: v = i− 2j+ 2k. This
choice of v is unique up to scalar multiplication, (i.e. the only other direction vectors for this
line are non-zero scalar multiples of i− 2j+ 2k).
We have a problem if the orientation vector is parallel to any of the axes. In such a case p, q
or r would be equal to zero. For that reason it is best to initially work with the parametric
representation and then find the algebraic form.
After understanding the basic principles of lines, more sophisticated problems can be attempted.
Example 4: Find the minimum distance between the point B (1, 2, 3) and the line defined by
x+ 2
1=
y
−2=z − 3
2
Which point on the line is closest to the point B (1, 2, 3)?
Solution: A point on the line is A (−2, 0, 3) and a direction vector for the line is v = i−2j+2k.
The point B (1, 2, 3) is not on the line. (Check this.)
The shortest distance between the point B and the line is
d =∥∥∥−−→AB∥∥∥ sin θ
=
∥∥∥−−→AB × v∥∥∥‖v‖ ; (draw a diagram)
ENG1091 Mathematics for Engineering page 9
Now−−→AB = (i+ 2j+ 3k)− (−2i+ 3k) = 3i+ 2j and
∥∥∥−−→AB × v∥∥∥ =
∣∣∣∣∣∣∣∣i j k
3 2 0
1 −2 2
∣∣∣∣∣∣∣∣ = 4i− 6j− 8k.
The shortest distance is thusd =
‖4i− 6j− 8k‖‖i− 2j+ 2k‖
=
√16 + 36 + 64
3
=2√
29
3.
The closest point to B
Solution
Converting the equation of the line into parametric form we have:
x = −2 + t y = −2t z = 3 + 2t
So a general point on the line is P (−2 + t,−2t, 3 + 2t)
Hence−−→BP =
−−→OP −−−→OB = (−3 + t) i+ (−2− 2t) j+ 2tk
[Key step!!] The closest point P on the line must satisfy−−→BP · v = 0.
Now
−−→BP · v = (−3 + t) (1) + (−2− 2t) (−2) + 2t (2)
= 1 + 9t
= 0 when t = −19 .
Hence the closest point is P (−2 + t,−2t, 3 + 2t) when t = −19
The closest point is(−21
9 ,29 , 3−
29
)=(−19
9 ,29 ,
259
)
ENG1091 Mathematics for Engineering page 10
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Vectors
Lecture 3 planes in 3DText Reference: §4.3.2
1. Planes in three-dimensional space
When defining a straight line in three-dimensional space, we needed a point on the line and an
orientation vector.
To define a plane in three-dimensional space, we need a point in the plane and a normal vector,
to the plane. Here n is the normal to the plane. For our immediate purpose, the magnitude of
n is not important, only its direction.
So let’s assume that we have some point on the plane which we label A (a, b, c) and we have a
normal vector n = pi+ qj+ rk. We take a general point on the plane P (x, y, z) . Now the vector−→AP lies in the plane and hence is normal to n.
Thus−→AP · n = 0.
This equation is the Cartesian equation of the plane
Explicitly this becomes (x− a) p+(y − b) q+(z − c) r = 0, which can be simplified to the general
form:
Ax+By + Cz = D.
Example 1: Find the equation of the plane that contains the point (2, 2, 3) and is normal to
the vector 〈−1, 1, 2〉 .
Solution:−→AP = 〈x, y, z〉 − 〈2, 2, 3〉 = 〈x− 2, y − 2, z − 3〉
−→AP · n = 〈x− 2, y − 2, z − 3〉 · 〈−1, 1, 2〉 = −x− 6 + y + 2z
Hence the equation of the plane is −x− 6 + y + 2z = 0 or −x+ y + 2z = 6
Example 2: Find the equation of the plane going through the points (−1, 0, 4) , (2, 5, 0) ,
(2, 2,−1) .
Solution: Label the points A (−1, 0, 4) , B (2, 5, 0) ,and C (2, 2,−1) . A normal vector n is given
by n =−−→AB ×−→AC.
Now−−→AB = 〈2, 5, 0〉 − 〈−1, 0, 4〉 = 〈3, 5,−4〉 and −→AC = 〈2, 2,−1〉 − 〈−1, 0, 4〉 = 〈3, 2,−5〉 .
Thus n =
∣∣∣∣∣∣∣∣i j k
3 5 −4
3 2 −5
∣∣∣∣∣∣∣∣ = i
∣∣∣∣∣ 5 −4
2 −5
∣∣∣∣∣− j∣∣∣∣∣ 3 −4
3 −5
∣∣∣∣∣+ k
∣∣∣∣∣ 3 5
3 2
∣∣∣∣∣ = −17i+ 3j− 9k.
Now−→AP = 〈x, y, z〉 − 〈−1, 0, 4〉 = 〈x+ 1, y, z − 4〉 and
−→AP · n = 〈x+ 1, y, z − 4〉 · 〈−17, 3,−9〉 = 0
that is −17x− 17 + 3y − 9z + 36 = 0
giving the equation of the plane as 17x− 3y + 9z = 19.
ENG1091 Mathematics for Engineering page 11
We should check that all three points satisfy the plane’s equation:
A (−1, 0, 4) : 17x− 3y + 9z = −17 + 36 = 19 X
B (2, 5, 0) : 17x− 3y + 9z = 34− 15 = 19 X
C (2, 2,−1) : 17x− 3y + 9z = 34− 6− 9 = 19. X
There are two observations that can be made. Firstly, the equation of a plane in three-dimensional
space is unique (up to multiplication by a scalar constant). Secondly, parallel planes have the
same normal vector and hence will only differ by the constant D.
Example 3: Find the minimum distance between the parallel planes 2x + 3y − z = 6 and
2x+ 3y − z = 0.
Let P (x1, y1, z1) be any point in the plane 2x+ 3y − z = 6 and
Q (x2, y2, z2) be any point in the plane 2x+ 3y − z = 0. [Notice that the equations of the planes
are arranged so that they have identical coeffi cients. Rearrange the equations if necessary-this
is important for what comes next.]
The distance between two parallel planes with normal n is then (diagram)
d =∥∥∥−−→PQ∥∥∥ cos θ =
−−→PQ · n‖n‖
=
(−−→OQ−−−→OP
)· n
‖n‖
=
−−→OQ · n−−−→OP · n
‖n‖
however−−→OQ · n = 2x2 + 3y2 − z2 = 0 and similarly
−−→OP · n = 2x1 + 3y1 − z1 = 6.
Thus (and taking absolute value since we seek a distance):
d =
∣∣∣∣∣∣ 0− 6√22 + 32 + (−1)2
∣∣∣∣∣∣ =6√14
2. Lines and Planes
Combining the knowledge of lines, planes and basic vector operations allows for a wide range of
problems to be addressed in three-dimensional space. For example, we can find:
· the minimum distance from a point to a plane,
· the minimum distance from a point to line,
· the angle between two intersecting planes,
· the minimum distance between two non-intersecting lines.
Example 4: Find the line defined by the intersection of the planes −x+y+z = 2 and x+2y = 4
and the angle of intersection.
Solution: A direction vector of the line of intersection is easily found: it is normal to both
−i+ j+ k and i+ 2j and hence could be obtained using the cross product. To find the equation
ENG1091 Mathematics for Engineering page 12
of the line of intersection is best done using Gauss elimination (next lecture).
A direction vector is
∣∣∣∣∣∣∣∣i j k
−1 1 1
1 2 0
∣∣∣∣∣∣∣∣ = −2i + j − 3k. (Of course any non-zero scalar multiple of
this is also a direction vector.)
The angle between two planes is defined as being the angle between its normals (diagram).
(−i+ j+ k) · (i+ 2j) = −1 + 2 = 1
‖(−i+ j+ k)‖ =√
(−1)2 + 12 + 12 =√
3 and ‖(i+ 2j)‖ =√
5
The angle θ between the planes is then given by cos θ = 1√3√
5, hence θ = 75.04 .
3. Parametric representation of a plane
Recall that straight lines have parametric equations giving x, y, z as function of one parametric
variable (usually t). Planes have parametric equations where x, y, z are given as functions of two
parametric variables (usually u and v).
Suppose we know a point P0 (a, b, c) in the plane and two non-parallel direction vectors
w1 = pi+ qj+ rk, and w2 = li+mj+ nk also in the plane: (diagram):
w1
r(x,y,z)vw2
w2
O
uw1P0
Let r = xi + yj + zk denote the position vector of an general point P (x, y, z) in the plane, so
that r =−−→OP0 + uw1 + vw2 where u, v are any scalars (parameters).
This gives r (x, y, z) = 〈a, b, c〉+ u 〈p, q, r〉+ v 〈l,m, n〉 and hence
x (u, v) = a+ pu+ lv,
y (u, v) = b+ qu+mv,
z (u, v) = c+ ru+ nv.
Theses 3 equations are the parametric equations of a plane. The fact that two parameters (u
and v) are needed to describe it indicates that a plane is a 2 dimensional surface.
In more advanced mathematics (i.e. 2nd level maths), it will be imperative to represent surfaces
parametrically.
ENG1091 Mathematics for Engineering page 13
Example 5: Find a parametric representation of the plane going through the points (−1, 0, 4) , (2, 5, 0)
and (2, 2,−1) .
Solution: label the points P (−1, 0, 4) , Q (2, 5, 0) and R (2, 2,−1) .
Now a choice for w1 is−−→PQ = 〈2, 5, 0〉 − 〈−1, 0, 4〉 = 〈3, 5,−4〉
and a choice for w2 is−→PR = 〈2, 2,−1〉 − 〈−1, 0, 4〉 = 〈3, 2,−5〉 .
Check that these are non-parallel X. (Otherwise the three points are collinear and the ques-tion cannot be answered properly-there will be an infinite number of planes.)
In vector form the parametric equations are
r = (−1, 0, 4) + u (3, 5,−4) + v (3, 2,−5)
= (−1 + 3u+ 3v, 5u+ 2v, 4− 4u− 5v)
Hence
x (u, v) = −1 + 3u+ 3v,
y (u, v) = 5u+ 2v,
z (u, v) = 4− 4u− 5v.
ENG1091 Mathematics for Engineering page 14
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Systems of Linear Equations
Lecture 4 · echelon form · Gauss eliminationText Reference: §5.5
Our object in this lecture is to solve a system of equations like
2x+ y + z + w = 4
4x+ y + 3z + 2w = 7
−2x+ z − w = 9.
Such a system is called linear because each of the equations on the left hand side is a linear
function of the unknown variables x, y, z and w. Simple linear systems of 2 or 3 variables
are commonly encountered in secondary school and is instructive to view an example before
discussing a more general procedure.
Suppose we wish to solve a system like
x+ 2y = 3 (1)
2x− 3y = −8 (2)
One way to proceed is to multiply equation 1 by 2 and subtract this from equation 2:
x+ 2y = 3 (1)
−7y = −14 (2(a))
The reason why this is effective is that one of the variables is eliminated. Equation (2a) is now
easily solved giving y = 2, and substituting this into equation 1 we find x = −1. Geometrically,
the equations x+ 2y = 3 and 2x− 3y = −8 represent two straight lines in the x− y plane whichintersect at the point (−1, 2).
The important point is that both of the systemsx+ 2y = 3
2x− 3y = −8and
x+ 2y = 3
−7y = −14have identical
solutions. Think about the operations we could perform on the two original equations.
We could
• interchange the two equations
• multiply either equation by any number we choose except zero, and
• add a multiple of one equation to the other.
Now performing any of these operations without thinking is not guaranteed to be effective but
at least we are assured that the resulting system of equations has an identical set of solutions.
Notice that the names of the variables is irrelevant: solvingx+ 2y = 3
2x− 3y = −8is exactly the same
as solving the systemu+ 2v = 3
2u− 3v = −8, only the coeffi cients are important.
1. The first step in solving a linear system is to write the system in augmented matrix form.
This is a way of writing the system using only the coeffi cients.
ENG1091 Mathematics for Engineering page 15
For example we write the system
2x+ y + z + w − 4 = 0
4x+ y + 3z + 2w = 7
−2x+ z − w = 9
as
2 1 1 1
4 1 3 2
−2 0 1 −1
∣∣∣∣∣∣∣∣4
7
9
.Notice each equation is written as a single row and that coeffi cients belonging to the same variable
are written directly underneath each other. (Equation 3, which appears to have no y, has in fact
a y−coeffi cient of zero.) Each constant term must be placed on the right hand side of the ‘equals’sign (the ‘−4’becomes +4 on the right hand side of equation 1) and the vertical partition is
used to separate the left hand side from the right hand side. (Think of it as replacing all of the
equals’signs.)
Example: Write the system
r + s + 2t = 0
2r − 3t = 1
6s − 5t = 0
in augmented matrix form.
Solution:
1 1 2
2 0 −3
0 6 −5
∣∣∣∣∣∣∣∣0
1
0
.2. Gaussian elimination
Gaussian elimination is a systematic method of solving linear equations by first reducing the
corresponding system into an equivalent system, called row echelon form, where the unknowns
can be calculated by back substitution.
Example: Given the system
r + s + 2t = 0
s − 3t = 1
− 5t = 5
find solutions to each of the variables
using back substitution.
Solution: t = 5−5 = −1 s = 1 + 3t
= −2
r = −2t− s= 2 + 2
= 4The system of equations in the last example has the augmented matrix
1 1 2
0 1 −3
0 0 −5
∣∣∣∣∣∣∣∣0
1
5
andwhich is one that is already in row echelon form. We saw how easy it is to find solutions of
systems in this form.
Definition: A matrix is in row echelon form when
• the leading (non-zero) coeffi cient of each row (called the pivot entry) has zeros below
it, and
• the pivot entries of following rows are located in columns further to the right.
• any rows which have no pivot (and therefore consist entirely of zeros) must come last.
ENG1091 Mathematics for Engineering page 16
Example: Given the following partitioned matrices, choose those which are in row echelon form:
A.
1 1 2
1 1 13
0 0 1
∣∣∣∣∣∣∣∣0
1
5
B.
1 0 2 0
0 1 −3 0
0 0 0 1
∣∣∣∣∣∣∣∣2
1
10
C.
1 0 0
0 1 0
0 0 1
∣∣∣∣∣∣∣∣0
1
5
no yes yes
D.
2 1 2
0 3 −3
0 0 2
∣∣∣∣∣∣∣∣0
6
5
E.
1 1 2
0 3 13
0 0 0
∣∣∣∣∣∣∣∣0
1
5
F.
1 0 0 0
0 1 1 0
1 0 0 1
∣∣∣∣∣∣∣∣0
0
0
yes yes no
G.
1 2 0 1 −3 1
0 0 0 1 2 −3
0 0 0 0 0 1
∣∣∣∣∣∣∣∣0
1
5
yes
To obtain the equivalent row echelon form of a system we apply a sequence of the three elementary
row operations on the augmented matrix. As discussed above these row operations do not change
the solution set of the corresponding system of linear equations.
The three elementary row operations are:
• Interchanging two rows
• Multiplying a row by a non-zero scalar
• Adding to one row a multiple of another
2. Row echelon forms
To reduce a matrix row echelon form systematically we follow these steps:
1. Locate the left-most column that doesn’t consist entirely of zeros.
2. Ensure that the top entry of this column is a non-zero entry. If necessary, interchange top
row with another row to achieve this.
3. Multiply this top row by the appropriate constant so that the first non-zero entry of this
row is 1. This entry is the pivot for that column. (It is not absolutely necessary that the
value of each pivot be 1 but this is certainly the most convenient value to have. As an
alternative to multiplying each row by a constant we can add/subtract multiples of other
rows to obtain a 1.)
4. Add a suitable multiple of this first row to each row below, so that all entries below this
pivot are 0.
5. Consider the submatrix obtained by removing the top row, and apply to this matrix steps
1 to 4.
ENG1091 Mathematics for Engineering page 17
Repeat steps 1-5 until the next submatrix under consideration has no rows left.
Example: Reduce the following matrix to row echelon form:
0 0 −2 0 12
3 6 −15 9 42
2 4 −5 6 −1
Solution:
0 0 −2 0 12
3 6 −15 9 42
2 4 −5 6 −1
(
13
)R2 → R2
R1 ↔ R2
1 2 −5 3 14
0 0 −2 0 12
2 4 −5 6 −1
R3 − 2R1 → R3
1 2 −5 3 14
0 0 −2 0 12
0 0 5 0 −29
(−1
2
)R2 → R2
1 2 −5 3 14
0 0 1 0 −6
0 0 5 0 −29
R3 − 5R2 → R3
1 2 −5 3 14
0 0 1 0 −6
0 0 0 0 1
← row echelon form
Exercise: Find a row echelon form of the matrix
1 0 −1 0
2 1 0 8
0 1 −2 0
1 −1 −2 −6
Solution:
1 0 −1 0
2 1 0 8
0 1 −2 0
1 −1 −2 −6
R2 − 2R1 → R2
R4− R1 → R4
1 0 −1 0
0 1 2 8
0 1 −2 0
0 −1 −1 −6
R3− R2 → R3
R4+ R3 → R4
1 0 −1 0
0 1 2 8
0 0 −4 −8
0 0 1 2
−14R3 →R3
1 0 −1 0
0 1 2 8
0 0 1 2
0 0 1 2
R4−R3 →R4
1 0 −1 0
0 1 2 8
0 0 1 2
0 0 0 0
← row echelon form
ENG1091 Mathematics for Engineering page 18
3. Solving a system using Gaussian elimination: To solve the system
x + 3y + 2z = 1
2x + 7y + 3z = 2
−3x − 10y − 6z = −5
1. we write the augmented matrix:
1 3 2
2 7 3
−3 −10 −6
∣∣∣∣∣∣∣∣1
2
−5
2. by performing appropriate row operations we find an equivalent row echelon form:
R2 − 2R1 → R2
R3 + 3R1 → R3
1 3 2
0 1 −1
0 −1 0
∣∣∣∣∣∣∣∣1
0
−2
R3+R2 →R3
1 3 2
0 1 −1
0 0 −1
∣∣∣∣∣∣∣∣1
0
−2
(−1)R3 →R3
1 3 2
0 1 −1
0 0 1
∣∣∣∣∣∣∣∣1
0
2
3. Use back substitution to find the values of the unknowns, in this case:
z = 2, y = z = 2 and x = 1− 2z − 3y = 1− 4− 6 = −9
So the three planes intersect in a single point: x = −9, y = 2, z = 2.
Note: The pivot in a column does not need to be equal to 1 − any non-zero number would do.
Exercise:
Solve the systems:
(a) −2a − 2b + 3c = 1
2a − 2b + c = 1
a + b − c = −3
ANS: Solution is: a = −52 , b = −11
2 , c = −5
(b) r + s + 2t = 0
2r + 4s − 3t = 1
3r + 6s − 5t = 0
ANS: Solution is: r = −17, s = 11, t = 3
Example: Find a vector equation for the line which forms the solution set of x+ y − z = 3
2x+ y + 2z = 1
(You will recall an example similar to this at the end of lecture 3.)
Writing the augmented matrix of this system and taking the system to row echelon form:[1 1 −1
2 1 2
∣∣∣∣∣ 3
1
]R2 − 2R1 →R2
[1 1 −1
0 −1 4
∣∣∣∣∣ 3
−5
](−1)R2 →R2
[1 1 −1
0 1 −4
∣∣∣∣∣ 3
5
]Here is a system of equations with an infinite solution set.
ENG1091 Mathematics for Engineering page 19
Notice that the pivot entries correspond to variables x and y.[1 1 −1
0 1 −4
∣∣∣∣∣ 3
5
]
The non-pivot variable, z, is said to be free and is set equal to a parameter t.
Let z = t
y − 4z = 5 hence y = 5 + 4z = 5 + 4t
x+ y − z = 3 hence x = 3 + z − y = −2− 3t
....parametric form
The solution can be written in vector form as
(x, y, z) = (−2− 3t, 5 + 4t, t) = (−2, 5, 0) + t (−3, 4, 1)
or in algebraic form:x+ 2
−3=y − 5
4=z − 0
1.
This shows that the solution is a straight line passing through the point (−2, 5, 0) and with
direction vector−3i+ 4j+ k.
Example: (from the previous lecture) Find an equation of the line of intersection of the
planes −x+ y + z = 2 and x+ 2y = 4.
Augmented matrix:
[−1 1 1
1 2 0
∣∣∣∣∣ 2
4
]
R2 + R1 → R2
[−1 1 1
0 3 1
∣∣∣∣∣ 2
6
](now in echelon form)
z is free, y = 2− 13z, x = −2 + z + y = 2
3z
set z = 3t, y = 2 − t, x = 2t and hence (x, y, z) = (0, 2, 0) + t (2,−1, 3) . (Compare with the
direction vector found in that example.)
ENG1091 Mathematics for Engineering page 20
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Consistency of Linear Equations
Lecture 5 · no solution case · infinite solution caseText Reference: §5.6
The equation systems given in the last lecture were rather special in the sense that they all had
solutions.
An example of this is the equation systemx+ 2y = 3
2x− 3y = −8, which consists of two straight lines
intersecting in the point (−1, 2).
But of course straight lines do not always intersect. The equation systemx+ 2y = 3
2x+ 4y = 1represents
two parallel straight lines and has no solution.
The question is how can we do this systematically:
How do we use Gauss elimination to recognise when a system of
equations has no solution.
Notice what happens when we employ Gauss elimination to solve the system of equations likex+ 2y = 3
2x+ 4y = 1
Augmented matrix:
[1 2
2 4
∣∣∣∣∣ 3
1
]
Converting to row echelon form: (one step only),
[1 2
0 0
∣∣∣∣∣ 3
−5
].
Notice that in the last row all entries left of the partition are zero, and that there is a non zero
number to the right of the partition. Since it is impossible for 0x + 0y = −5 we know that the
system has no solution.
Definition: A linear system of equations without solution is called inconsistent.
Now of course the previous example didn’t need Gauss elimination to demonstrate its inconsis-
tency. However, a system of 3 equations in 3 unknowns (represented by three planes in space) is
rather more complex. A 3× 3 system of equations will be inconsistent if either
• the three planes are parallel
• two planes are parallel and are intersected by the third,
• neither of the planes is parallel but each pair of planes intersects in a line parallel to theothers.
Geometrically the situation for higher dimensions (>3 unknowns) is even more complex still but
algebraically very easy to sort out provided we apply Gauss elimination.
ENG1091 Mathematics for Engineering page 21
The great advantage of Gauss elimination is that it takes the guess work out of equation manip-
ulation by being systematic. We can tell whether equations are inconsistent or not by using the
following very simple test:
When the augmented matrix corresponding to a system of inconsistent equations
is converted into a row echelon form, there will be at least one row where
all entries left of the partition are zero and there is a non-zero entry to the right
of the partition.
To put it another way, the row echelon form of an inconsistent linear system will have a row of
type[
0 0 0 · · · 0∣∣∣ ∗ ] where ∗ is some non-zero number.
Moreover, the test is completely diagnostic: if no such row exists then the equation systemmust
have solutions.
Example: The following partitioned matrices are row echelon forms corresponding to various
systems of linear equations. Which linear systems are inconsistent?
A.
1 1 2
0 1 13
0 0 0
∣∣∣∣∣∣∣∣0
1
5
B.
1 0 2 0
0 1 −3 0
0 0 0 0
∣∣∣∣∣∣∣∣2
1
0
C.
1 1 −3
0 2 1
0 0 0
0 0 0
∣∣∣∣∣∣∣∣∣∣∣
0
4
1
0
D.
2 1 2
0 0 0
0 0 0
∣∣∣∣∣∣∣∣1
0
0
E.
1 0 2
0 2 0
0 0 0
∣∣∣∣∣∣∣∣0
1
0
F.
1 0 0 0
0 1 1 0
0 0 0 1
∣∣∣∣∣∣∣∣0
0
0
G.
1 2 0 1 −3 1
0 0 0 1 2 −3
0 0 0 0 0 0
∣∣∣∣∣∣∣∣0
1
5
Example: Show that the following system of equations is inconsistent by forming its augmented
matrix and then using row operations convert it to a matrix in row echelon form:
x + 2z = 1
y − z = 0
x + y + z = 2
Solution
Augmented matrix:
1 0 2
0 1 −1
1 1 1
∣∣∣∣∣∣∣∣1
0
2
(not in echelon from)
R3 − R1 → R3
1 0 2
0 1 −1
0 1 −1
∣∣∣∣∣∣∣∣1
0
1
R3 − R2 → R3
1 0 2
0 1 −1
0 0 0
∣∣∣∣∣∣∣∣1
0
1
The shaded row indicates inconsistency.
ENG1091 Mathematics for Engineering page 22
What is the geometric interpretation of this inconsistent system?
Answer: Since none of the three planes are parallel (why?) we conclude that each pair of planes
intersects in a line parallel to the others.
[Examine the normal vectors (1, 0, 2) , (0, 1,−1) , (1, 1, 1) . Since no two of these is parallel neither
is there a pair of parallel planes.]
A system of linear equations that does not have solutions is said to be inconsistent, so obviously
a consistent system is one that does have solutions.
Now we encounter a remarkable fact: either a consistent linear system has a unique solution
(exactly one solution for each of the unknowns) or else it possesses infinitely many! To put it
another way, if a linear system of equations is known to have two different solutions (say) then
that system must have infinitely many solutions.
2. Systems with infinitely many solutions:
The augmented matrix of the system
x − 3y + z = 1
2x − 6y + 3z = 4
−x + 3y = 1
reduces to the following equivalent row-echelon form:
Working: Augmented matrix:
1 −3 1
2 −6 3
−1 3 0
∣∣∣∣∣∣∣∣1
4
1
R2 − 2R1 → R2
R3 + R1 → R3
1 −3 1
0 0 1
0 0 1
∣∣∣∣∣∣∣∣1
2
2
R3 − R2 → R3
1 −3 1
0 0 1
0 0 0
∣∣∣∣∣∣∣∣1
2
0
row echelon form:
1 −3 1
0 0 1
0 0 0
∣∣∣∣∣∣∣∣1
2
0
The echelon form matrix gives us all the information concerning the original system. First of all
we notice there is no row of the type[
0 0 0 · · · 0∣∣∣ ∗ ] where ∗ is non-zero, so we know
that the system has solutions.
The third row is entirely zero and in effect is totally redundant. We ignore rows that consist
entirely of zeros.
From 2nd row we have z = 2.
Solving the first row for x we have x = 1− z + 3y = −1 + 3y (since z = 2).
So z = 2 and x = −1 + 3y where the choice for y is completely arbitrary. There are infinitely
many solutions, one for each value of y.
It is customary to assign a parameter to the free variable y. We can then write the solution set
ENG1091 Mathematics for Engineering page 23
as y = t, x = −1 + 3t, z = 2, where t is arbitrary.
What is the graphical interpretation of this consistent system?
Answer: The three planes x − 3y + z = 1, 2x − 6y + 3z = 4, and −x + 3y = 1 intersect
in a straight line in 3D space. This line has a vector equation (x, y, z) = (−1 + 3t, t, 2) =
(−1, 0, 2)+t (3, 1, 0) , and therefore passes through the point (−1, 0, 2) and points in the direction
of the vector 3i+ j+ 0k.
Example: Solve the 3× 4 system of linear equations:
2x+ y + z + w = 4
4x+ y + 3z + 2w = 7
−2x+ 2y + z − w = 9
Solution:
We write the system in augmented matrix form and use elementary row operations to convert
the system to an equivalent one in echelon form. (Gauss elimination.)
Augmented matrix:
[A | b] =
2 1 1 1
4 1 3 2
−2 2 1 −1
∣∣∣∣∣∣∣∣4
7
9
2 1 1 1
4 1 3 2
−2 2 1 −1
∣∣∣∣∣∣∣∣4
7
9
R2 − 2R1 →R2
2 1 1 1
0 −1 1 0
−2 2 1 −1
∣∣∣∣∣∣∣∣4
−1
9
R3+ R1 →R3
2 1 1 1
0 −1 1 0
0 3 2 0
∣∣∣∣∣∣∣∣4
−1
13
R3+ 3R2 →R3
2 1 1 1
0 −1 1 0
0 0 5 0
∣∣∣∣∣∣∣∣4
−1
10
This time the pivot variables are x, y and z (since the pivot entries occur in columns 1,2, and 3,
corresponding to the variables x, y, z).
The free variable is w.
w = free = t (say)
from row 3: 5z = 10 ∴ z = 2
from row 2: −y + z = −1 ∴ y = z + 1 = 3
from row 1: 2x+ y + z + w = 4 ∴ x = 2− 12z −
12y −
12w = −1
2 −12 t
Writing the solutions in vector form:
〈x, y, z, w〉 =⟨−1
2 −12 t, 3, 2, t
⟩=⟨−1
2 , 3, 2, 0⟩
+ t⟨−1
2 , 0, 0, 1⟩.
ENG1091 Mathematics for Engineering page 24
Exercise:
The row echelon form of a system with unknowns r, s, t, and u, is
1 1 0 1
0 0 1 1
0 0 0 0
0 0 0 0
∣∣∣∣∣∣∣∣∣∣∣
1
1
0
0
Describe the solutions of the system.
ANS: infinite number of solutions with s and u free t = 1− u, r = 1− t− s∴ (r, s, t, u) = (1− t− s, s, 1− u, u) where s, u are arbitrary.
Exercises: Solve the following systems of linear equations:
(a)
x − y − 2z = 3
x + 2y − z = 0
2x − y + z = 5
x − y − z = 3
ANS: unique solution x = 2, y = −1, z = 0
(b)
x + y + z = 2
x − y + z = 1
2x + 2z = 4
ANS: no solution
(c)
−a + b + c + 2d + e = 0
a − c + d − e = 1
2b + c − d − 2e = −1
ANS: infinite solution set where d and e are free.
Solving for a, b, c we get a = −2 + 6d+ 3e, b = 1− 3d, c = −3 + 7d+ 2e
i.e. (a, b, c, d, e) = (−2 + 6d+ 3e, 1− 3d,−3 + 7d+ 2e, d, e)
= (−2, 1,−3, 0, 0) + (6d,−3d, 7d, d, 0) + (3e, 0, 2e, 0, e)
= (−2, 1,−3, 0, 0) + d (6,−3, 7, 1, 0) + e (3, 0, 2, 0, 1)
showing that the solution set is a plane in 5D space
ENG1091 Mathematics for Engineering page 25
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Matrices
Lecture 6 · matrices · matrix arithmeticText Reference: §5.1-5.2
A matrix is a rectangular arrangement of numbers or variables, which can be either real or
complex, enclosed in square brackets. It is usual to denote matrices using capital letters. For
example:
A =
−2 3
4 5 0
7 π√
2 −1
−0.18 7 20 −78
, B =
[1 2 3
4 5
]is not a matrix
A matrix has rows, running left to right, and columns running form top to bottom.
The matrix A has three rows and four columns and consists of 12 entries.
• A matrix with m rows and n columns is called a m × n matrix; the matrix A in the
example is a 3× 4 matrix.
• The position of each entry is determined by the column and row numbers. We use
subindices to indicate this, for example,
a24 is the entry in row 2 , column 4. In matrix A, a24 = −1.
a13 is the entry in row 1 , column 3. In matrix A, a13 = 5.
We use the notation A = [aij ] to indicate that A is a matrix (hence the square brackets) whose
entries are generically indicated as aij . The notation A = [aij ]m×n means that A is an m × nmatrix.
Some special matrices
1. A 1× n matrix is a row matrix or row vector, e.g.[
1 2 4 3]is a 1× 4 row vector.
2. An m× 1 matrix is a column matrix or column vector; e.g.
x =
1
2
3
is a 3 × 1 column vector. Column matrices are usually identified with ordinary vectors andfor this reason it is common to use lower case boldface letters to denote them.
3. A matrix with the same number of rows and columns is called a square matrix; e.g.[1 3
2 4
]is a 2× 2 matrix
4. A zero or null matrix contains all zero entries. That is 0 = [0ij ] where 0ij = 0 for all i and
j.
Operations with matrices
ENG1091 Mathematics for Engineering page 26
1. Addition and subtraction
Addition and subtraction are possible only between matrices of the same order. These
are performed by adding or subtracting the corresponding entries respectively.
Example: 1 −1
3 5
−4 8
+
7 12
6 1
−3 5
=
8 11
9 6
−7 13
The addition of matrices is commutative i.e. A+B = B +A.
2. Multiplication by scalars
Given a matrix A and a number k, the multiplication of A by the scalar k and is obtained
by multiplying each entry of A by k.
For example let k = 3 and A =
1 −1
3 5
−4 8
, then 3A = 3
1 −1
3 5
−4 8
=
3 −3
9 15
−12 24
Note that subtraction can be expressed in terms of a scalar product (k = −1) and an
addition: A−B = A+ (−B)
For any matrix A, A−A = 0.
3. Multiplication
Two matrices A and B can be multiplied together only when the number of columns in
A equals the number of rows in B. To find the ij entry in the product AB we multiply
the entries along the ith row of A pairwise with entries on the jth column of B and then
add:
A =
1 −1
3 5
−4 8
, B =
[1 −1 3
2 4 −2
], C =
1
2
3
(a)
AB =
1 −1
3 5
−4 8
[
1 −1 3
2 4 −2
]=
1× 1 +−1× 2 1×−1 +−1× 4 1× 3 +−1×−2
3× 1 + 5× 2 3×−1 + 5× 4 3× 3 + 5×−2
−4× 1 + 8× 2 −4×−1 + 8× 4 −4× 3 + 8×−2
=
−1 −5 5
13 17 −1
12 36 −28
(b) AC is not defined
(c)
BA =
[1 −1 3
2 4 −2
]1 −1
3 5
−4 8
=
[1× 1 +−1× 3 + 3×−4 1×−1 +−1× 5 + 3× 8
2× 1 + 4× 3 +−2×−4 2×−1 + 4× 5 +−2× 8
]
=
[−14 18
22 2
]
ENG1091 Mathematics for Engineering page 27
This example demonstrates something very important: matrix multiplication is not usually
commutative, i.e. AB 6= BA in general.
In fact AB and BA need not be of the same order, or even if one product AB is defined,
the other product, BA, need not be.
In general if A = [aij ]m×p and B = [bij ]p×n then AB is defined and AB = C = [cij ]m×n
where cij = ai1b1j + ai2b2j + · · ·+ aipbpj =p∑
k=1
aikbkj .
To illustrate: . . . . . . . . . . . .
. . . . . . . . . . . .
. . . cij . . . . . .
. . . . . . . . . . . .
m×n
=
. . . . . . . . . . . .
. . . . . . . . . . . .
ai1 ai2 . . . aip
. . . . . . . . . . . .
m×p
. . . b1j . . . . . .
. . . b2j . . . . . .
. . . . . . . . . . . .
. . . bpj . . . . . .
p×n
=
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . ai1b1j + ai2b2j + · · ·+ aipbpj . . . . . .
. . . . . . . . . . . .
So cij = ai1b1j + ai2b2j + · · ·+ aipbpj =
p∑k=1
aikbkj
4. Examples:[2 3
−1 5
]2× 2
[−1 2
−2 3
]2× 2 = 2× 2
[1 −2 3
−4 5 6
] −1 1 3
0 2 −1
3 5 4
2× 3 3× 3 = 2× 3
=
[2 (−1) + 3 (−2) 2 (2) + 3 (3)
−1 (−1) + 5 (−2) −1 (2) + 5 (3)
]=
[−1 + 9 1− 4 + 15 3 + 2 + 12
4 + 18 −4 + 10 + 30 −12− 5 + 24
]
=
[−8 13
−9 13
]=
[8 12 17
22 36 7
]
ENG1091 Mathematics for Engineering page 28
5. Important:
• We stress again that to be able to perform the matrix product AB there is a size
restriction:
the number of columns in A (the matrix on the left) must equal the number of rows
in B (the second matrix in the product). We then say that AB is defined.
• If A is a m× p matrix, and B is a p× n matrix, then AB is a m× n matrix.
6. Properties of matrix multiplication
If A,B, and C are matrices of appropriate sizes, and k is a scalar then:
• A(B + C) = AB +AC
• (B + C)A = BA+ CA
• (AB)C = A(BC)
• k(AB) = (kA)B = A (kB)
• AB 6= BA in general.
Exercises
1. Find the following product of matrices[2 1
−3 5
]×[
3 −1
−2 4
]=
[4 2
−19 23
]
2. The product in the reverse order, although possible, leads to a different matrix:[3 −1
−2 4
]×[
2 1
−3 5
]=
[9 −2
−16 18
]
3. Given
A =
[1 3
−1 2
], B =
0
7
8
, C =
[2 4 6
8 10 12
], D =
9 8 7 6
5 4 3 2
1 0 9 8
determine which of the following are defined and give their sizes (orders).
(a) AB not defined
(b) AC 2× 3
(c) CD 2× 4
(d) AD not defined
(e) DC not defined
(f) CB 2× 1
(g) BC not defined
(h) (AC)D 2× 4
(i) A(CD) 2× 4
ENG1091 Mathematics for Engineering page 29
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Matrices
Lecture 7 · transpose · matrix inversesText Reference: §5.4
The transpose of a matrix
The transpose of a matrix is obtained by interchanging its rows and columns. That is, the entries
of the ith row become the entries of the ith column.
So, if A is a m× n matrix, then its transpose, denoted AT , is a n×m matrix.
Example:
Let A be the 3× 2 matrix
1 3
2 4
5 8
.
Then AT is the 2× 3 matrix: AT =
1 3
2 4
5 8
T
=
[1 2 5
3 4 8
].
Transpose of a product
If A and B are such that AB is defined then (AB)T = BTAT .
In words: the transpose of a matrix product is equal to the product of individual transposes
taken in the reverse order.
First we illustrate this with an example.
Let
A =
[1 −2
3 0
], B =
[2 −1 4
0 1 3
]we have
AB =
[1 −2
3 0
][2 −1 4
0 1 3
]=
[2 −3 −2
6 −3 12
]so that
(AB)T =
[2 −3 −2
6 −3 12
]T=
2 6
−3 −3
−2 12
On the other hand
BTAT =
2 0
−1 1
4 3
[
1 3
−2 0
]=
2 6
−3 −3
−2 12
Now we give a proof to show why (AB)T = BTAT is always true.
Let A = [aij ]m×p and B = [bij ]p×n.
Then AB is an m× n matrix and for any i = 1, ...,m and j = 1, ..., n we have
(AB)ij = ai1b1j + ai2b2j + · · ·+ aipbpj =p∑
k=1
aikbkj .
ENG1091 Mathematics for Engineering page 30
Now the (i, j) entry of (AB)T is the the (j, i) entry of (AB); this is found by swapping
i’s and j’s in the formula for (AB)i,j :
(AB)Ti,j =
p∑k=1
ajkbki = aj1b1i + ai2b2i + · · ·+ ajpbpi
= b1iaj1 + b2iaj2 + · · ·+ bpiajp =
p∑k=1
bkiajk
= the sum of products found by multiplying, term by term, the
ith row of BT with the jth column of AT and this is the (i, j)
entry of BTAT
Special matrices
Some types of matrices that are particularly important are given below. This list is not exhaus-
tive.
• A symmetric matrix is one which is equal to its transpose: e.g.
1 2 3
2 4 5
3 5 6
• Diagonal matrices are square matrices where any non-zero entries occur on the main diag-
onal: e.g.
1 0 0
0 2 0
0 0 0
• Identity matrices are square matrices where the main diagonal entries are all 1’s. For
example I2 =
[1 0
0 1
], I3 =
1 0 0
0 1 0
0 0 1
.If the size of the identity matrix can be understood from the context, or is irrelevant, the
symbol I is used.
If A is a square matrix and I is the identity matrix the same size as A, then AI = IA = A.
Identity matrices play a role analogous to the number 1 in ordinary arithmetic.
The inverse of a matrix
Definition: The inverse of a square n× n matrix A is an n× n matrix B, (if one exists),such that AB = BA = I where I is the n× n identity matrix.Note: If such a B exists it is unique and we write it as A−1.
Warning: A−1 does not mean 1A .
If A has an inverse, then we say that matrix A is invertible or non-singular.
We can calculate A−1 by forming the augmented matrix consisting of A and I, the identity
matrix. We then apply a systematic sequence of row operations until, if possible, A becomes I.
In the process, (in other words apply exactly the same sequence of row operations to I), I will
have become A−1.
ENG1091 Mathematics for Engineering page 31
Schematically: [A | I] row operations−→
[I | A−1
]Recall that elementary row operations are:
1. Interchanging two rows.
2. Multiplying a row by a non-zero scalar.
3. Adding to one row a multiple of another.
STAGE 1: Forward elimination process
1. Let C = [A | I] .
2.If there is a stage where C has a column consisting entirely of zeros, we stop immediately:
A has no inverse.
3. Ensure that the top left entry of C is a non-zero entry, which we will label as a. (If necessary,
interchange the top row with another row to achieve this.)
4. Multiply this row by 1a so that the first non-zero entry of this row is 1. This entry is the
pivot for that column. (Alternatively this can sometimes be affected by row interchange.)
5. Add a suitable multiple of this first row to the rows below row so that all entries in the
column below the pivot become 0.
If there is a stage where there the sub-matrix of C left of the partition has a row consisting
entirely of zeros, we stop immediately: the matrix A has no inverse.
6. Consider the submatrix of C found by removing its 1st row and 1st column, regard this
as a new matrix C. Repeat steps 2-6 until the next submatrix under consideration has no
rows left.
7. Provided the algorithm has not been exited at steps 2 or 5 the full matrix is now in echelon
form. The pivots are all 1 and located on the main diagonal of the matrix left of the
partition.
STAGE 2: Backward elimination process
1. Notice that all pivots are 1 and are located on the main diagonal of the matrix left of the
partition. Locate the row containing the right-most pivot, (which must be in the bottom
row).
2. Add suitable multiples of this row to the rows above so that all entries in the column above
become 0.
3. Locate the next pivot by moving up the diagonal and repeat steps 2 and 3.
4. This procedure is repeated until the top left pivot is reached, at which point the full matrix
is[I | A−1
].
ENG1091 Mathematics for Engineering page 32
Examples:
Find inverses of the following (if they exist).
1. A =
0 1 1
109
0√
2 −4
0 3 π
, A has a column of zeros and hence no inverse.
2. A =
1 1 −1
1 1 0
1 1 1
We form [A | I] = C =
1 1 −1
1 1 0
1 1 1
∣∣∣∣∣∣∣∣1 0 0
0 1 0
0 0 1
Step 1We note that C has a pivot in the top left entry and that this pivot is 1. Steps 3-4
Subtract row 1 from row 2:
1 1 −1
0 0 1
1 1 1
∣∣∣∣∣∣∣∣1 0 0
−1 1 0
0 0 1
Subtract row 1 from row 3:
1 1 −1
0 0 1
0 0 2
∣∣∣∣∣∣∣∣1 0 0
−1 1 0
−1 0 1
. Step 5 is now complete.Step 7. We apply the algorithm again to the submatrix of C found by deleting its 1st row and
column (shaded)
1 1 −1
0 0 1
0 0 2
∣∣∣∣∣∣∣∣1 0 0
−1 1 0
−1 0 1
but since this new matrix has a column of zeros, we conclude the matrix
1 1 −1
1 1 0
1 1 1
has noinverse. (Exiting the algorithm at step 2.)
ENG1091 Mathematics for Engineering page 33
3. Find the inverse of A =
2 7 1
1 4 −1
1 3 0
Solution: [A | I] =
2 7 1
1 4 −1
1 3 0
∣∣∣∣∣∣∣∣1 0 0
0 1 0
0 0 1
R1 ↔ R2
1 4 −1
2 7 1
1 3 0
∣∣∣∣∣∣∣∣0 1 0
1 0 0
0 0 1
R2 − 2R1 → R2
1 4 −1
0 −1 3
1 3 0
∣∣∣∣∣∣∣∣0 1 0
1 −2 0
0 0 1
R3 − R1 → R3
1 4 −1
0 −1 3
0 −1 1
∣∣∣∣∣∣∣∣0 1 0
1 −2 0
0 −1 1
(−1)R2 → R2
1 4 −1
0 1 −3
0 −1 1
∣∣∣∣∣∣∣∣0 1 0
−1 2 0
0 −1 1
R2 + R3 → R3
1 4 −1
0 1 −3
0 0 −2
∣∣∣∣∣∣∣∣0 1 0
−1 2 0
−1 1 1
(−1
2
)R3 → R3
1 4 −1
0 1 −3
0 0 1
∣∣∣∣∣∣∣∣0 1 0
−1 2 012 −1
2 −12
R2 + 3R3 → R2
1 4 −1
0 1 0
0 0 1
∣∣∣∣∣∣∣∣0 1 012
12 −3
212 −1
2 −12
R1 + R3 → R1
1 4 0
0 1 0
0 0 1
∣∣∣∣∣∣∣∣12
12 −1
212
12 −3
212 −1
2 −12
R1 − 4R2 → R1
1 0 0
0 1 0
0 0 1
∣∣∣∣∣∣∣∣−3
2 −32
112
12
12 −3
212 −1
2 −12
=[I | A−1
].
Hence A−1 =
−3
2 −32
112
12
12 −3
212 −1
2 −12
= 12
−3 −3 11
1 1 −3
1 −1 −1
Check: 12
−3 −3 11
1 1 −3
1 −1 −1
2 7 1
1 4 −1
1 3 0
= 12
−6− 3 + 11 −21− 12 + 33 −3 + 3 + 0
2 + 1− 3 7 + 4− 9 3− 3 + 0
2− 1− 1 7− 4− 3 1 + 1 + 0
= 12
2 0 0
0 2 0
0 0 2
=
1 0 0
0 1 0
0 0 1
.
Strictly speaking we should also check that
2 7 1
1 4 −1
1 3 0
−3
2 −32
112
12
12 −3
212 −1
2 −12
=
1 0 0
0 1 0
0 0 1
,however it is a known fact for matrices that a left inverse is also a right inverse (and vice versa),
so a one sided check is suffi cient.
ENG1091 Mathematics for Engineering page 34
Inverses of 2× 2 matrices
Example: find the inverse of the matrix
[2 4
1 3
].
Solution: [A | I] =
[2 4
1 3
∣∣∣∣∣ 1 0
0 1
]12R1 → R1
[1 2
1 3
∣∣∣∣∣ 12 0
0 1
]R2 − R1 → R2
[1 2
0 1
∣∣∣∣∣ 12 0
−12 1
]
R1 −2R2 → R1
[1 0
0 1
∣∣∣∣∣ 32 −2
−12 1
]. Hence
[2 4
1 3
]−1
=
[32 −2
−12 1
].
However there is a simple formula for the inverse of 2× 2 matrices.
The inverse of a 2× 2 matrix:
Let A =
[a b
c d
], and suppose ad−bc 6= 0. Then A is invertible and A−1 =
1
ad− bc
[d −b−c a
].
The number ad− bc is called the determinant of A and is denoted by∣∣∣∣∣ a b
c d
∣∣∣∣∣ or det (A) .
The determinant of any square matrix A is also defined (see next lecture) and this number
determines whether or not A is invertible:
A square matrix A is invertible if and only if its determinant is non-zero..
Using matrix methods to solve linear systems of equations
Consider the 3× 3 linear system:
2x1 + 7x2 + x3
x1 + 4x2 − x3
x1 + 3x2
=
=
=
1
4
5
which can also be written in matrix form
2 7 1
1 4 −1
1 3 0
x1
x2
x3
=
1
4
5
.Any n× n linear system can be written in the form Ax = b, where x and b are column vectors
(matrices).
If A is invertible we can multiply on the left by A−1 and so obtain the unknown
matrix x :
Ax = b
A−1Ax = A−1b
Ix = A−1b
giving x = A−1b
This method is somewhat more restrictive than Gaussian elimination. It only works
for n × n systems and either produces a unique solution (when det (A) 6= 0) but is
ENG1091 Mathematics for Engineering page 35
incapable of distinguishing between the no solution or infinite solution cases which
occur when det (A) = 0.
The main advantage to using matrix inverse method occurs when working with mul-
tiple equations with the same set of coeffi cients.
Example:
Solve: (a)
2x1 + 7x2 + x3 = 1
x1 + 4x2 − x3 = 4
x1 + 3x2 = 5
and (b)
2x1 + 7x2 + x3 = −2
x1 + 4x2 − x3 = 4
x1 + 3x2 = 6
In (a) we have
x1
x2
x3
=
2 7 1
1 4 −1
1 3 0
−1
1
4
5
, and in (b)
x1
x2
x3
=
2 7 1
1 4 −1
1 3 0
−1
−2
4
6
.
Now
2 7 1
1 4 −1
1 3 0
−1
= 12
−3 −3 11
1 1 −3
1 −1 −1
(shown above),
giving the solution to (a):
x1
x2
x3
= 12
−3 −3 11
1 1 −3
1 −1 −1
1
4
5
=
20
−5
−4
and to (b):
x1
x2
x3
= 12
−3 −3 11
1 1 −3
1 −1 −1
−2
4
6
=
30
−8
−6
.Exercise: Solve the following system of equations using matrix inversion followed by matrix
multiplication:2x+ 3y = 7
4x+ y = 3
In matrix form:
[2 3
4 1
][x
y
]=
[7
3
]
If exists
[2 3
4 1
]−1
we may write
[2 3
4 1
]−1 [2 3
4 1
][x
y
]=
[2 3
4 1
]−1 [7
3
]
Now
∣∣∣∣∣ 2 3
4 1
∣∣∣∣∣ = 2− 12 6= 0 so
[2 3
4 1
]−1
exists.
The formula for the inverse of a 2×2matrix gives
[2 3
4 1
]−1
= 12−12
[1 −3
−4 2
]=
[− 1
10310
25 −1
5
]
So
[1 0
0 1
][x
y
]=
[− 1
10310
25 −1
5
][7
3
]=
[15115
]; giving x = 1/5 and y = 11/5
ENG1091 Mathematics for Engineering page 36
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Determinants
Lecture 8 · Determinants · Cramer’s RuleText Reference: §5.3
Determinants
The determinant of a 2× 2 matrix A =
[a b
c d
]is defined by detA =
∣∣∣∣∣ a b
c d
∣∣∣∣∣ = ad− bc.
As we noted in the previous lecture determinants are used to determine whether a square matrix
is invertible or not.
Determinants can also be used to solve n× n systems of linear equations using a rule known asCramer’s rule.
For example the system :2x + 3y = 5
7x + 11y = 13has the solution:
x =
∣∣∣∣∣ 5 3
13 11
∣∣∣∣∣∣∣∣∣∣ 2 3
7 11
∣∣∣∣∣, y =
∣∣∣∣∣ 2 5
7 13
∣∣∣∣∣∣∣∣∣∣ 2 3
7 11
∣∣∣∣∣The denominator is always the determinant of coeffi cients and the determinant on the top line
replacing the column containing the coeffi cients of the variable in question with the numbers on
the right hand side.
Evaluating the determinants gives
x =55− 39
22− 21, y =
26− 35
22− 21
i.e.
x = 16, y = −9.
Cramer’s rule works provided determinant of coeffi cients (the denominator) is non-zero,
when it is zero Cramer’s rule fails, and the system of equations has either no solutions or
infinitely many.
Determinants of larger matrices
The determinant is a number we assign to any square matrix. It plays an important role in
finding the inverse of a matrix, solving systems of equations, multiplication of vectors, finding
areas of triangles, etc.
To find the determinant of larger matrices we need to know about cofactors. A cofactor of a
particular entry in a matrix is the (smaller) determinant consisting of those elements which
remain if we removed the row and column belonging to that entry, together with a sign, + or −,depending on where the entry is located.
Example In the matrix A =
−1 3 7
4 2 −2
5 6 9
the cofactor of the (2, 3) entry, namely −2, is the
ENG1091 Mathematics for Engineering page 37
signed determinant −∣∣∣∣∣ −1 3
5 6
∣∣∣∣∣ .
The minus sign comes from the sign matrix:
+ − + . . .
− + − . . .
+ − + . . ....
......
and the minor determinant∣∣∣∣∣ −1 3
5 6
∣∣∣∣∣ is obtained by removing row 2 and column 3We refer to the cofactor of the (i, j) entry as Cij .
In the example above, C23 = −∣∣∣∣∣ −1 3
5 6
∣∣∣∣∣ = − (−6− 15) = 21.
Example Find, but do not evaluate, C41 in the matrix
−1 4 3 −7
2 −3 9 1
1 8 −6 −1
−1 2 1 10
.
This is clearly −
∣∣∣∣∣∣∣∣4 3 −7
−3 9 1
8 −6 −1
∣∣∣∣∣∣∣∣ .Note that the ‘−’comes from the position not the sign of the entry.
How to evaluate determinants.
1. Choose any row or column.
2. For each position in the selected row or column, calculate the corresponding cofactor.
3. Form the product of each cofactor with the corresponding entry. The determinant is the
sum of these products.
Example Find detA =
∣∣∣∣∣∣∣∣−1 3 7
4 2 −2
5 6 9
∣∣∣∣∣∣∣∣ .We choose to expand along the second row.
detA = 4×−∣∣∣∣∣ 3 7
6 9
∣∣∣∣∣+ 2×∣∣∣∣∣ −1 7
5 9
∣∣∣∣∣− 2×−∣∣∣∣∣ −1 3
5 6
∣∣∣∣∣= 4×− (27− 42) + 2× (−9− 35) + 2 (−6− 15) = −70.
If we chose instead to expand along the 1st column the answer is the same.
detA = −1×∣∣∣∣∣ 2 −2
6 9
∣∣∣∣∣+ 4×−∣∣∣∣∣ 3 7
6 9
∣∣∣∣∣+ 5×∣∣∣∣∣ 3 7
2 −2
∣∣∣∣∣ = −30 + 4× 15 + 5×−20 = −70
ENG1091 Mathematics for Engineering page 38
Which row or column? It is a remarkable fact that the answer is independent of the row and
column we choose.
As a practical consideration we would do well to choose that row/column that has the greatest
number of zeros.
Example: Find the 4× 4 determinant
∣∣∣∣∣∣∣∣∣∣∣
−1 4 3 −7
0 −3 9 1
0 0 −6 −1
0 0 0 10
∣∣∣∣∣∣∣∣∣∣∣.
An obvious choice is to expand along the 1st column:∣∣∣∣∣∣∣∣∣∣∣
−1 4 3 −7
0 −3 9 1
0 0 −6 −1
0 0 0 10
∣∣∣∣∣∣∣∣∣∣∣= −1
∣∣∣∣∣∣∣∣−3 9 1
0 −6 −1
0 0 10
∣∣∣∣∣∣∣∣+ other terms all zero.
And again: −1
∣∣∣∣∣∣∣∣−3 9 1
0 −6 −1
0 0 10
∣∣∣∣∣∣∣∣ = −1×−3
∣∣∣∣∣ −6 −1
0 10
∣∣∣∣∣+ other terms all zero
= −1×−3× (−6× 10− 0) = −1×−3×−6× 10 = −180
This is a general rule:the determinant of a triangular matrix (either upper or lower) is the product of the entries
along the main diagonal.
Exercise: Find detB where B =
−2 1 1
1 3 3
10 5 2
.
detB =
∣∣∣∣∣∣∣∣−2 1 1
1 3 3
10 5 2
∣∣∣∣∣∣∣∣(notice the different bracketing which distinguishes
a matrix from its determinant)
= − (1)
∣∣∣∣∣ 1 1
5 2
∣∣∣∣∣+ (3)
∣∣∣∣∣ −2 1
10 2
∣∣∣∣∣− (3)
∣∣∣∣∣ −2 1
10 5
∣∣∣∣∣(expanding along row 2)= − (−3) + 3 · −14− 3 · −20
= 3− 42 + 60
= 21.
ENG1091 Mathematics for Engineering page 39
Properties of Determinants
To illustrate the following properties of determinants we will work with an arbitrary 3×3 matrix
A =
a1 a2 a3
b1 b2 b3
c1 c2 c3
. We stress that the all the following properties are true regardless of size.1. Transpose property: det (A) = det
(AT)
∣∣∣∣∣∣∣∣a1 a2 a3
b1 b2 b3
c1 c2 c3
∣∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣∣a1 b1 c1
a2 b2 c2
a3 b3 c3
∣∣∣∣∣∣∣∣2. Scaling property: to multiply a determinant by a number we multiply any row or column
by that number.
k
∣∣∣∣∣∣∣∣a1 a2 a3
b1 b2 b3
c1 c2 c3
∣∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣∣a1 a2 ka3
b1 b2 kb3
c1 c2 kc3
∣∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣∣ka1 ka2 ka3
b1 b2 b3
c1 c2 c3
∣∣∣∣∣∣∣∣ (we may pick any row or column)
Hence det (kA) = det
ka1 a2 a3
b1 b2 b3
c1 c2 c3
=
∣∣∣∣∣∣∣∣ka1 ka2 ka3
kb1 kb2 kb3
kc1 kc2 kc3
∣∣∣∣∣∣∣∣ = k3
∣∣∣∣∣∣∣∣a1 a2 a3
b1 b2 b3
c1 c2 c3
∣∣∣∣∣∣∣∣If A is n× n then det (kA) = kn det (A)
3. Interchange property: Swapping any two rows, or two columns, changes the sign of the
determinant
e.g.
∣∣∣∣∣∣∣∣a1 a2 a3
b1 b2 b3
c1 c2 c3
∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣a1 a2 a3
c1 c2 c3
b1 b2 b3
∣∣∣∣∣∣∣∣Hence a matrix with two identical rows or columns has determinant = zero∣∣∣∣∣∣∣∣
a1 a2 a3
a1 a2 a3
c1 c2 c3
∣∣∣∣∣∣∣∣ = 0.
4. Elimination property: Adding a multiple of a row to another row does not alter the
value of a determinant. Similarly for columns.
e.g.
∣∣∣∣∣∣∣∣a1 a2 a3
b1 + kc1 b2 + kc2 b3 + kc3
c1 c2 c3
∣∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣∣a1 a2 a3
b1 b2 b3
c1 c2 c3
∣∣∣∣∣∣∣∣ .5. Matrix multiplication property: Let A and B be square matrices of the same size (both
n× n), then
det (AB) = detAdetB
ENG1091 Mathematics for Engineering page 40
Special case: if A is invertible then AA−1 = I so that det(AA−1
)= detAdetA−1 = det I = 1.
In particular detA 6= 0, and
det(A−1
)=
1
detA.
It can be shown that the condition detA 6= 0 is also suffi cient for invertibility, i.e.
A is invertible if and only if detA 6= 0.
Application: Simplify the 3× 3 Vandemonde determinant:∣∣∣∣∣∣∣∣1 a a2
1 b b2
1 c c2
∣∣∣∣∣∣∣∣=
∣∣∣∣∣∣∣∣1 a a2
0 b− a b2 − a2
1 c c2
∣∣∣∣∣∣∣∣ R2− R1 → R2
=
∣∣∣∣∣∣∣∣1 a a2
0 b− a b2 − a2
0 c− a c2 − a2
∣∣∣∣∣∣∣∣ R3− R1 → R3
= (b− a) (c− a)
∣∣∣∣∣∣∣∣1 a a2
0 1 b+ a
0 1 c+ a
∣∣∣∣∣∣∣∣taking out the common factors of (b− a) from row 2
and (c− a) from row 3
= (b− a) (c− a)
∣∣∣∣∣∣∣∣1 a a2
0 1 b+ a
0 0 c− b
∣∣∣∣∣∣∣∣ R3− R1 → R2
= (b− a) (c− a) (c− b) multiplying down the main diagonal to evaluate the determinant
ENG1091 Mathematics for Engineering page 41
Cramer’s Rule
Example: Solve the system of equations below using Cramer’s rule:
x+ 2y + z = 1
2x− 3y + 7z = −4
−x+ y − 3z = −1
Solution:
x =
∣∣∣∣∣∣∣∣1 2 1
−4 −3 7
−1 1 −3
∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 2 1
2 −3 7
−1 1 −3
∣∣∣∣∣∣∣∣, y =
∣∣∣∣∣∣∣∣1 1 1
2 −4 7
−1 −1 −3
∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 2 1
2 −3 7
−1 1 −3
∣∣∣∣∣∣∣∣, z =
∣∣∣∣∣∣∣∣1 2 1
2 −3 −4
−1 1 −1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 2 1
2 −3 7
−1 1 −3
∣∣∣∣∣∣∣∣.
∣∣∣∣∣∣∣∣1 2 1
−4 −3 7
−1 1 −3
∣∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣∣1 2 1
0 5 11
0 3 −2
∣∣∣∣∣∣∣∣ =
∣∣∣∣∣ 5 11
3 −2
∣∣∣∣∣ = −43
∣∣∣∣∣∣∣∣1 1 1
2 −4 7
−1 −1 −3
∣∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣∣1 1 1
0 −6 5
0 0 −2
∣∣∣∣∣∣∣∣ = 12
∣∣∣∣∣∣∣∣1 2 1
2 −3 −4
−1 1 −1
∣∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣∣1 2 1
0 −7 −6
0 3 0
∣∣∣∣∣∣∣∣ = 18
∣∣∣∣∣∣∣∣1 2 1
2 −3 7
−1 1 −3
∣∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣∣1 2 1
0 −7 5
0 3 −2
∣∣∣∣∣∣∣∣ = 14− 15 = −1
giving x =−43
−1= 43 y =
12
−1= −12 z =
18
−1= −18
ENG1091 Mathematics for Engineering page 42
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Eigenvalues & Eigenvectors
Lecture 9&10 EigenvaluesText Reference: §5.7
1. Homogeneous systems of equations
Generally speaking, when the determinant of an n× n system of equations is zero, we can only
deduce that the system has no solutions or infinitely many.
A homogeneous system of equations, introduced earlier (i.e. Ax = 0), always has the trivial
solution x = 0. If the determinant of coeffi cients of a homogeneous system is zero the system
must have infinitely many solutions.
Example 1: Let
A =
a 1 3
2 2 −1
−2 a 1
,find the values of a, such that Ax = 0 has non-trivial solutions.
2. Eigenvalues and eigenvectors
Definitions: Let A be an n×n matrix and x be an n×1 vector. Any scalar λ satisfying Ax = λx
for some non-zero x is called an eigenvalue of A. The corresponding non-zero vectors x for
which Ax = λx are called the eigenvectors of A corresponding to λ.
To quote from the textbook, “such problems arise naturally in many branches of engineering.
For example, in vibrations the eigenvalues and eigenvectors describe the frequency and mode of
vibration respectively, while in mechanics they represent principal stresses and the principal axes
of stress in bodies subject to external forces.”
Example 2: Show that x =
[1
1
]is an eigenvector of A =
[2 1
1 2
]corresponding to the
eigenvalue λ = 3.
This is straightforward matrix arithmetic :[2 1
1 2
][1
1
]=
[3
3
]= 3
[1
1
]
Note also that if we multiply of the sides of this equation by the scalar t we get[2 1
1 2
][t
t
]=
[3t
3t
]= 3
[t
t
]
This demonstrates that any non-zero multiple of an eigenvector corresponding to λ is also an
eigenvector.
ENG1091 Mathematics for Engineering page 43
3. Finding eigenvalues
It is tempting to rewrite the equation Ax = λx as (A−λ)x = 0, but this cannot possibly be
correct why?
We write instead: (A−λI)x = 0, this is a homogeneous system of equations. Now we know the
trivial solution x = 0 is always available, but we are interested only in the non-zero solutions
(called eigenvectors). This is the requirement that a homogeneous system has infinite number of
solutions which happens precisely when
det (A− λI) = 0
Now det (A− λI) is a polynomial of degree n. This will have n (possibly complex) linear factors,
and so det (A− λI) = (−1)n (λ− λ1) (λ− λ2) · · · (λ− λn) ; the roots of this polynomial are the
eigenvalues λ1, λ2, . . . , λn.
Notes
1. It is possible for the roots of the polynomial to be repeated so we should not assume the
λ1, λ2, . . . , λn are necessarily distinct.
2. In this course we only deal with the case where the matrix A has real entries. When this is
so, the polynomial det (A− λI) has real coeffi cients and if any of its roots are imaginary these
will occur as conjugate pairs.
Example 3: Find the eigenvalues of
A =
[0 −1
1 0
]
Solution: The characteristic polynomial: det (A− λI) =
∣∣∣∣∣ −λ −1
1 −λ
∣∣∣∣∣ = λ2 + 1 = 0 for λ = ±i.
Example 4: Find the eigenvalues of
A =
1 1 −2
−1 2 1
0 1 −1
Solution: A− λI =
1 1 −2
−1 2 1
0 1 −1
− λ
1 0 0
0 1 0
0 0 1
=
1− λ 1 −2
−1 2− λ 1
0 1 −1− λ
The characteristic polynomial: det (A− λI) =
∣∣∣∣∣∣∣∣1− λ 1 −2
−1 2− λ 1
0 1 −1− λ
∣∣∣∣∣∣∣∣= −
∣∣∣∣∣∣∣∣−1 2− λ 1
1− λ 1 −2
0 1 −1− λ
∣∣∣∣∣∣∣∣ R1 ↔R2
ENG1091 Mathematics for Engineering page 44
= −
∣∣∣∣∣∣∣∣−1 2− λ 1
0 (1− λ) (2− λ) + 1 −1− λ0 1 −1− λ
∣∣∣∣∣∣∣∣ R2 + (1− λ)R1 →R2
= − (−1)
∣∣∣∣∣ (1− λ) (2− λ) + 1 −1− λ1 −1− λ
∣∣∣∣∣ expanding along col1= (−1− λ)
∣∣∣∣∣ (1− λ) (2− λ) + 1 1
1 1
∣∣∣∣∣ factoring col2= (−1− λ) [(1− λ) (2− λ) + 1− 1] = (−1− λ) (1− λ) (2− λ) . Hence eigenvalues: λ = 1, 2,−1
4. Finding the eigenvectors of unique eigenvalues
Having solved the nth order polynomial (characteristic equation) for the n roots (eigenvalues),
it still remains to find the corresponding eigenvectors. For the moment let’s assume that the
eigenvalues are distinct (non-repeated.)
Let ej be an eigenvector that corresponds to the eigenvalue λj so that Aej = λjej
Example 3 (again): Find the eigenvectors of
A =
[0 −1
1 0
]
Solution: Consider the eigenvalue λ = i :[0 −1
1 0
][x1
x2
]= i
[x1
x2
]is equivalent to
[−i −1
1 −i
][x1
x2
]=
[0
0
]The 2× 2 case always leads to two identical equations, in this example x1 = ix2 and −ix1 = x2.
Thus the eigenvectors are t (i, 1) where t 6= 0.
Now λ = −i :[0 −1
1 0
][x1
x2
]= i
[x1
x2
]is equivalent to
[i −1
1 i
][x1
x2
]=
[0
0
]hence x1 = −ix2
giving eigenvectors t (−i, 1) where t 6= 0.
Example 4 (again): Find the eigenvector corresponding to the dominant eigenvalue of
A =
1 1 −2
−1 2 1
0 1 −1
Solution: The dominant eigenvalue is λ = 2.
Solving Ax = 2x :1 1 −2
−1 2 1
0 1 −1
x1
x2
x3
= 2
x1
x2
x3
and this is equivalent to the homogeneous system−x1 + x2 − 2x3 = 0
−x1 + x3 = 0
x2 − 3x3 = 0
The ‘augmented’matrix is (we don’t need to include the column of zeros):
ENG1091 Mathematics for Engineering page 45
−1 1 −2
−1 0 1
0 1 −3
(−1)R1 →R1
1 −1 2
−1 0 1
0 1 −3
R1+R2 →R2
1 −1 2
0 −1 3
0 1 −3
R3+R2 →R3
1 −1 2
0 −1 3
0 0 0
The non-pivot variable x3 is chosen free, so x3 = t
x2 = 3x3 = 3t from row 2, and x1 = x2 − 2x3 = t from row 1, giving (x1, x2, x3) = t (1, 3, 1)
hence an eigenvector corresponding to λ = 2 is 〈1, 3, 1〉 .
Check
1 1 −2
−1 2 1
0 1 −1
1
3
1
=
2
6
2
= 2
1
3
1
.Note that the eigenvectors are only unique up to multiplication by scalars.
Also, while the 0 vector is always a solution to the system (A−λI)x = 0 the eigenvectors are
the non-zero solutions.
The vector 0 is never an eigenvector.
5. Finding the eigenvectors of repeated eigenvalues
If the eigenvalues of the matrix A are distinct, then it can be shown that the corresponding
eigenvectors are linearly independent. If, however, the eigenvalues are repeated, it may not be
possible to find n linearly independent eigenvectors. (By repeated roots of the characteristic
equation, we mean that two or more of the eigenvalues are the same.)
Example 5: Find the eigenvalues of the matrix
A =
0 0 1
0 1 2
0 0 1
.
Solution: Characteristic polynomial is det (A− λI) =
∣∣∣∣∣∣∣∣−λ 0 1
0 1− λ 2
0 0 1− λ
∣∣∣∣∣∣∣∣ = −λ (1− λ)2
which has roots λ = 0 and λ = 1 (multiplicity = 2)
It is not clear how many independent eigenvectors exist when λ = 1. The eigenvalue has a
multiplicity of 2, but that doesn’t assure us that there will be two independent eigenvectors.
Example 6: From the matrix in the previous example, find the eigenvector(s) corresponding to
each eigenvalue.
Solution: Eigenvectors for λ = 1 :
A− λI = A− I =
−1 0 1
0 0 2
0 0 0
which is in echelon form.ENG1091 Mathematics for Engineering page 46
Solving
−1 0 1
0 0 2
0 0 0
x1
x2
x3
=
0
0
0
we have x3 = 0, x2 = t and −x1 +x3 = 0 so x1 = 0 also.
Thus the eigenvectors for λ = 1 are
x1
x2
x3
=
0
t
0
= t
0
1
0
, that is, the non-zero multiplesof the vector x = (0, 1, 0) .
Eigenvectors for λ = 0 :
A− λI = A =
0 0 1
0 1 2
0 0 1
R1 ↔ R2
0 1 2
0 0 1
0 0 1
R3− R2 → R3
0 1 2
0 0 1
0 0 0
(now in ech-elon form)
Solving
0 1 2
0 0 1
0 0 0
x1
x2
x3
=
0
0
0
we have x3 = 0, x2 + 2x3 = 0 (and hence x2 = 0) while
x1 is free and we set x1 = t.
Thus the eigenvectors for λ = 0 are
x1
x2
x3
=
t
0
0
= t
1
0
0
, that is, the non-zero multiplesof the vector x = (1, 0, 0) .
Consider the next example.
Example 7: Find the eigenvalues and corresponding eigenvectors for the matrix
A =
0 0 0
0 1 0
1 0 1
Solution: Characteristic polynomial is det (A− λI) =
∣∣∣∣∣∣∣∣−λ 0 0
0 1− λ 0
1 0 1− λ
∣∣∣∣∣∣∣∣ = −λ (1− λ)2
which has roots λ = 0 and λ = 1 (multiplicity = 2)
The eigenvectors:
λ = 1 :
A− λI = A− I =
−1 0 0
0 0 0
1 0 0
R3+ R1 → R3
−1 0 0
0 0 0
0 0 0
which is in echelon form.
Solving
−1 0 0
0 0 0
0 0 0
x1
x2
x3
=
0
0
0
we have −x1 = 0, x2 and x3 are free so we set x2 = s
and x3 = t.
ENG1091 Mathematics for Engineering page 47
Thus the eigenvectors for λ = 1 are
x1
x2
x3
=
0
s
t
= s
0
1
0
+ t
0
0
1
, that is, the sums ofnon-zero multiples of (0, 1, 0) and (0, 0, 1) .
In this example we do have 2 independent eigenvectors for λ = 1.
λ = 0 :
A− λI = A =
0 0 0
0 1 0
1 0 1
R3 ↔ R1
1 0 1
0 1 0
0 0 0
which is in echelon form.
Solving
1 0 1
0 1 0
0 0 0
x1
x2
x3
=
0
0
0
we have x2 = 0, and x1 + x3 = 0 with x3 as free.
We set x3 = t, giving
x1
x2
x3
=
−t0
t
= t
−1
0
1
that is, the non-zero multiples of thevector x = (−1, 0, 1) .
ENG1091 Mathematics for Engineering page 48
6. Properties of eigenvalues
Let us assume that we have an n × n matrix A with the eigenvalues λ1, λ2, λ3, . . . λn. (Not
necessarily distinct.)
Property 1: The sum of the eigenvalues of A is equal to the sum of the elements of the diagonal
of A.n∑1
λi = λ1 + λ2 + ...+ λn =n∑1
aii
(The right-hand summation is known as the trace of A.)
Property 2: The product of the eigenvalues of A is equal to the determinant of A.
n∐1
λi = λ1 · λ2 · ... · λn = det(A)
Property 3: If A−1 exists, each of the eigenvalues of A must be non-zero, (use Property 2).
The eigenvalues of A−1 are:1
λ1,
1
λ2, . . . ,
1
λn
Property 4: The eigenvalues of the transpose matrix AT are the same as those of A.
Property 5: If k is a scalar then the eigenvalues of kA are kλ1, kλ2, kλ3, . . . , kλn.
Property 6: If k is a scalar and I is the n × n identity matrix then eigenvalues of A ± kI arerespectively
λ1 ± k, λ2 ± k, λ3 ± k, . . . λn ± k.
Property 7: If k is a positive integer then the eigenvalues of Ak are
λk1, λk2, λ
k3, . . . , λ
kn
ENG1091 Mathematics for Engineering page 49
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Further Calculus
Lecture 11 · Implicit differentiation · Logarithmic differentiationText Reference: §8.3.14&8.4
Functions such as f(x) = x sinx express f(x) explicitly in terms of x. Expressions of the form
x2 + y2 = 4, or 2y + x = 11, define an implicit relationship between x and y. In these cases, we
can still determine the rate of change of one variable with respect to the other by the technique
known as implicit differentiation. As an illustration, consider the equation
x2 + y2 = 4.
The graph of this equation is the circle radius 2 centre (0, 0). This graph obviously is not the
graph of a function since, for instance, there are two points on the graph whose x-coordinate is
1. (Functions must satisfy the vertical line test.)
We denote the slope (gradient) of the curve at the point(1,√
3)by
dy
dx|(1,√
3)
In general, the slope (gradient) at the point (a, b) is denoted by
dy
dx|(a,b)
In a small vicinity of the point (a, b), the curve looks like the graph of a function. [That is, on
this part of the curve, y = g(x) for some function g(x).] We say that this function is defined
implicitly by the equation.
We obtain a formula fordy
dxby differentiating both sides of the equation with respect to x
while treating y as a function of x.
1. Finddy
dxin the relation x2 + y2 = 4. (This simple example is used mainly to illustrate the
general technique of implicit differentiation.)
Now we differentiate both sides of the equation with respect to x.
The first term x2 has the derivative 2x as usual. We need to think of the second term y2 as
having the form [g(x)]2 . To differentiate we apply the chain rule:
d
dx[g(x)]2 = 2 [g(x)]1 g′(x)
or, equivalentlyd
dx
(y2)
= 2ydy
dx.
The right hand side of the original equation, the derivative of the constant function 4 is zero.
Thus implicit differentiation of x2 + y2 = 4 yields
2x+ 2ydy
dx= 0
ENG1091 Mathematics for Engineering page 50
solving for dydx , we have
2ydy
dx= −2x
dy
dx= −2x
2y= −x
y.
notice that this slope (gradient) formula involves y as well as x. This reflects the fact that the
slope (gradient) of the circle at a point depends on the y-coordinate as well as the x-coordinate.
At the point(1,√
3)the slope (gradient) is
dy
dx|(1,√
3) = −xy|(1,√
3)
= − 1√3
At the point(1,−√
3)the slope (gradient) is
dy
dx|(1,−
√3) = −x
y|(1,−
√3)
=1√3
When y = 0 the slope (gradient) is undefined indicating that the tangent is vertical at these
points.
2. The equation 5x2− 6xy+ 5y2 = 16 defines an ellipse graphed below, find a formula fordy
dxin
this relation.
3 2 1 1 2 3
3
2
1
1
2
3
x
y
Differentiate both sides with respect to x, taking care with the product 6xy and the chain rule
on 5y2 :10x−
(6 · y + 6x · dydx
)+ d
dx
(5y2)
= 0 using the product rule
10x− 6y − 6x dydx + ddx
(5y2) dydx = 0 using the chain rule
10x− 6y − 6x dydx + 10y dydx = 0
Grouping: 10x− 6y + dydx (10y − 6x) = 0
Now solve for dydx :
ANS:dy
dx=
6y − 10x
10y − 6x=
3y − 5x
5y − 3x. (Observe also the faint lines 3y − 5x = 0 where the curve is
horizontal and 5y − 3x = 0 where the curve is vertical.)
ENG1091 Mathematics for Engineering page 51
3. Find a formula fordy
dxin the relation x3 + y3 = 6xy
5 4 3 2 1 1 2 3 4 5
5
4
3
2
1
1
2
3
4
5
x
y
Differentiate both sides with respect to x, taking care with the chain rules on x3 and y3 :
x3 + y3 = 6xy
3x2 + ddx
(y3)
= 6 · y + 6x · dydx3x2 + d
dx
(y3) dydx = 6y + 6x dydx
3x2 + 3y2 dydx = 6y + 6x dydx
dydx
(3y2 − 6x
)= 6y − 3x2
ANS:dy
dx=
6y − 3x2
3y2 − 6x
=2y − x2
y2 − 2x
4. Problem type: Find the point(s) on the curve (x− 1)2 + (y + 2)2 = 4
where the gradient of the tangent is 1.
Differentiate both sides with respect to x :
2(x− 1) + 2(y + 2) dydx = 0
so dydx = −
(x−1y+2
)This is equal 1 where − (x− 1) = y + 2,
i.e. along the line y = −x− 1 (Sketch)
Intersection with the curve (x− 1)2 + (y + 2)2 = 4
(x− 1)2 + (−x+ 1)2 = 4
(x− 1)2 + (−1)2 (x− 1)2 = 4
2(x− 1)2 = 4 or (x− 1)2 = 2
so (x− 1) = ±√
2
giving x = 1 +√
2 or x = 1−√
2
The points of intersection are:(1 +√
2,−2−√
2)and
(1−√
2,−2 +√
2)
2 1 1 2 3 4
5
4
3
2
1
1
x
y
ENG1091 Mathematics for Engineering page 52
Logarithmic Differentiation
The calculation of derivatives of complicated functions involving products, quotients, or powers
can often be simplified by taking logarithms. The method is called logarithmic differentiation.
Review of logarithm laws: If x, y, a > 0, and n ∈ R, then
1. loga(xy) = loga(x) + loga(y)
2. loga(xy ) = loga(x)− loga(y)
3. loga(xn) = n loga(x)
4. loga(1) = 0
5. loga(a) = 1
Steps in Logarithmic Differentiation
1. Take logarithms of both sides of an equation y = f(x).
2. Differentiate implicitly with respect to x.
3. Solve the resulting equation for y′.
Example 1 Differentiate y =x4√x2 + 1
(2x+ 3)6
Take loge of both sides:
ln y = ln(x4√x2 + 1
)− ln
((2x+ 3)6
)= ln
(x4)
+ ln√x2 + 1− 6 ln(2x+ 3)
= 4 ln (x) + 12 ln
(x2 + 1
)− 6 ln(2x+ 3)
Now differentiate both sides implicitly with respect to x :
1
y
dy
dx=
4
x+
1
2· 2x
x2 + 1− 6 · 1
2x+ 3· 2
=4
x+
x
x2 + 1− 12
2x+ 3
Now solve fordy
dx:
dy
dx= y
(4
x+
x
x2 + 1− 12
2x+ 3
)=x4√x2 + 1
(2x+ 3)6
(4
x+
x
x2 + 1− 12
2x+ 3
)
ENG1091 Mathematics for Engineering page 53
Example 2 Find the derivative of y = 3x
Take loge of both sides:
ln y = ln (3x)
= x ln 3
Now differentiate both sides implicitly with respect to x :
1
y
dy
dx= ln 3
Now solve fordy
dx:
dy
dx= y ln 3
= 3x ln 3
Hence the derivative of an exponential function is a constant multiple of the exponential function.
Example 3 Find the derivative of y =(x− 1)3(2x+ 5)5
(3− 4x)2(1 + x2)3
Take loge of both sides:
ln y = ln[(x− 1)3(2x+ 5)5
]− ln
[(3− 4x)2(1 + x2)3
]= 3 ln (x− 1) + 5 ln (2x+ 5)− 2 ln(3− 4x)− 3 ln(1 + x2)
Now differentiate both sides implicitly with respect to x :
1
y
dy
dx=
3
x− 1+ 5 · 2
2x+ 5− 2 · −4
3− 4x− 3 · 2x
1 + x2
=3
x− 1+
10
2x+ 5+
8
3− 4x− 6x
1 + x2
Now solve fordy
dx:
dy
dx= y
(3
x− 1+
10
2x+ 5+
8
3− 4x− 6x
1 + x2
)=
(x− 1)3(2x+ 5)5
(3− 4x)2(1 + x2)3
(3
x− 1+
10
2x+ 5+
8
3− 4x− 6x
1 + x2
)Example 4 (Problem type): Determine the location and nature of any stationary points of
y = xx and hence sketch the graph of y = xx.
Find both dydx and
d2ydx2
implicitly:
Take loge x of both sides of the equation y = xx:
loge y = x loge x,
now differentiate implicitly, with a product rule on the right:
(1/y)dy
dx= loge x+ 1
sody
dx= y (loge x+ 1) = xx (loge x+ 1) .
Notice thatdy
dx= 0 at the point where loge x + 1 = 0 and only there, giving the coordinates of
ENG1091 Mathematics for Engineering page 54
the stationary point as(e−1,
(e−1)e−1)
=(e−1, e−e
−1).
For the second derivative we differentiate the equation dydx = y (loge x+ 1) implicitly:
d2y
dx2=dy
dx(loge x+ 1) + y · d
dx(loge x+ 1) (product rule)
=dy
dx(loge x+ 1) + y
(1
x
)= xx (loge x+ 1)2 + xx
(1
x
)= xx (loge x+ 1)2 + xx−1.
> 0 when x = e−1 (in fact d2ydx2
> 0 for all x > 0). Therefore the point(e−1, e−e
−1)
is a local minimum point.
Sketch:
x
y
ENG1091 Mathematics for Engineering page 55
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Further Calculus
Lecture 12 ·: Hyperbolic functions · IdentitiesText Reference: §2.7, 8.3.12
1. Definitions: Trig functions are often called ‘circular’functions because (cos t, sin t) lies on
the curve x2 + y2 = 1, (i.e. the unit circle).
Hyperbolic functions have a very similar relationship with the hyperbola x2 − y2 = 1, with the
point (cosh t, sinh t) lying on the right branch of this curve.
x x
y y
0 Q
x
x
2
2
2
2
y
y
+
0
=1
P cos t, sin t( ) P cosh t, sinh t( )
=1
The Hyperbolic functions arise from certain combinations of exponential functions, and occur
frequently in applications of mathematics. For example, the shape of a hanging wire (a catenary
curve) is described by a ‘cosh’expression.
Definitions: coshx =ex + e−x
2pron. ‘cosh’
sinhx =ex − e−x
2pron. ‘shine’
tanhx =sinhx
coshx=ex − e−xex + e−x
pron. ‘tanch’
The reciprocal functions can also be defined:
sechx =1
coshx=
2
ex + e−xpron. ‘sech’as in fetch
cschx =1
sinhx=
2
ex − e−x pron. ‘cosech’as in go-fetch
cothx =1
tanhx=
coshx
sinhx=ex + e−x
ex − e−x pron. ‘coth’as in goth
ENG1091 Mathematics for Engineering page 56
2. Graphs:
5 4 3 2 1 1 2 3 4 5
1
1
2
3
4
5
x
y
y = coshx, y = 12ex, y = 1
2e−x
5 4 3 2 1 1 2 3 4 5
5
4
3
2
1
1
2
3
4
5
x
y
y = sinhx, y = 12ex, y = −1
2e−x
ENG1091 Mathematics for Engineering page 57
3 2 1 1 2 3
2
1
1
2
x
y
y = tanhx, y = 1, y = −1
5 4 3 2 1 1 2 3 4 5
5
4
3
2
1
1
2
3
4
5
x
y
y = cschx, y = sinhx
ENG1091 Mathematics for Engineering page 58
3. Hyperbolic identities
Hyperbolic identities hold in similar ways to the trig identities; some of these include
a) cosh2 x− sinh2 x = 1
b) 1− tanh2 x = sech2 x
c) sinh(x+ y) = sinhx cosh y + coshx sinh y
Osborn’s rule: In general, to obtain the formula for hyperbolic functions from the analogous
identity for circular functions, replace each circular function by the corresponding hyperbolic
function and change the sign of every product (or implied product) of two sines.
Examples
cos2 x+ sin2 x = 1 becomes cosh2 x− sinh2 x = 1
cos 2x =
cos2 x− sin2 x
2 cos2 x− 1
1− 2 sin2 x
cosh 2x =
cosh2 x+ sinh2 x
2 cosh2 x− 1
1 + 2 sinh2 x
sin 2x = 2 sinx cosx sinh 2x = 2 sinhx coshx
1 + tan2 x = sec2 x 1− tanh2 x = sech2 xbecause tan2 x = sin2 x
cos2 xis an
implied product of two sines
4. Relationship between circular and hyperbolic functions
Eulers’formula provides the link:
Since (i) eiθ = cos θ + i sin θ and (ii) e−iθ = cos (−θ) + i sin (−θ) = cos θ − i sin θ
adding: 2 cos θ = eiθ + e−iθ from which cos θ = cosh (iθ)
and subtracting (ii) from (i) gives sin θ =eiθ − e−iθ
2i=
sinh (iθ)
i
Then cosh iθ = cos θ
sinh iθ = i sinθ
cos iθ = cosh θ
sin iθ = i sinh θ
These relationships provide the justification for Osborn’s rule.
5. Derivatives of Hyperbolic Functions
These are easily found using differentiation of exponentials ( again note the similarities with trig
function derivatives).
d
dx(sinhx) = coshx (apply definition and use derivatives of ex and e−x)
d
dx(coshx) = sinhx (apply definition and use derivatives of ex and e−x)
ENG1091 Mathematics for Engineering page 59
d
dx(tanhx) =
cosh2 x− sinh2 x
cosh2 x= sech2 x (apply definition and quotient rule, use identity
cosh2 x− sinh2 x = 1)
d
dx(cschx) = − coshx
cosh2 x− 1= − cschx cothx (apply definition and quotient rule, use identity
cosh2 x− sinh2 x = 1)
d
dx(sechx) = − sinhx
cosh2 x= − sechx tanhx (apply definition and quotient rule, use identity
cosh2 x− sinh2 x = 1)
d
dx(cothx) = −sinh2 x− cosh2 x
sinh2 x= − csch2 x (apply definition and quotient rule, use identity
cosh2 x− sinh2 x = 1)
Examples
1. Find the derivative of f(x) =√
coshx.
f ′(x) = 12 (coshx)−1/2 d
dx (coshx) (applying the chain rule)
= 12 (coshx)−1/2 (sinhx) (applying the rule obtained 4 (ii))
=
(sinhx
2√
coshx
)2. Find the derivative of f(x) = cosh
√x.
f ′(x) = sinh (√x) d
dx (√x) (applying the chain rule)
= (sinh (√x)) 1
2 (x)−1/2
=
(sinh (
√x)
2√x
)3. Find f ′(x) if f(x) = sinhx+ coshx.
f ′(x) = coshx+ sinhx = f(x)
Notice also that f(0) = 1
There is only one function f which is equal its derivative and which satisfies f (0) = 1. Namely
f (x) = ex. It must be therefore that coshx+ sinhx = ex.
ENG1091 Mathematics for Engineering page 60
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Further Calculus
Lecture 13 · Inverse Hyperbolic functions · Log form · DerivativesText Reference: §2.7.5
Since the hyperbolic sine, sinh, and the hyperbolic tangent, tanh, are one-to-one, their inverses
are fully defined without needing to consider domain restrictions. For the hyperbolic cosine,
cosh, which is not one-to-one, we use a restricted domain of [0, ∞) to define its inverse. Thisfunction will be called the principal branch of coshx.
Since the hyperbolic functions are defined in terms of the exp function, their inverses can be
defined in terms of natural logarithms. This definition is often called the logarithmic form of the
inverse.
Examples
1. Find the logarithmic form, domain and range of the inverse of the function f(x) = coshx, x ≥ 0
Sketch the graph of y = f(x) and its inverse on the same axes.
The domain of f is given and its range is easily determined. From this we can write down the
domain and range of the inverse.
f f−1
domain: [0,∞) [1,∞)
range: [1,∞) [0,∞)
deriving the log form of cosh−1 x: y = cosh−1 x so cosh y = x
x = 12 (ey + e−y)
2x = ey + e−y
2xey = e2y + 1
0 = e2y − 2xey + 1
ey =2x±
√4x2 − 4
2(quadratic formula)
ey =2x+ 2
√x2 − 1
2
(the negative root is ignored
since the range of cosh−1 is
[0,∞) and ∴ ey ≥ 1)
ey = x+√x2 − 1
y = cosh−1 x = ln(x+√x2 − 1
)
ENG1091 Mathematics for Engineering page 61
(restricted) coshx : domain: [0,∞) cosh−1 x : domain: [1,∞)
range: [1,∞) range: [0,∞)
log form: ln(x+√x2 − 1)
1 1 2 3 4 5
1
1
2
3
4
5
x
y
coshx and cosh−1 x
The derivatives of the inverse functions can be found by differentiating the logarithmic form, or
by implicit differentiation.
For f(x) = cosh−1 x, the derivative is found as follows:
For f(x) = sinh−1 x, the derivative is found as follows:
ENG1091 Mathematics for Engineering page 62
Function Properties Log Form Derivative
sinh−1 x
Domain: Range:
R R
4 2 2 4
4
2
2
4
x
y
ln(x+√x2 + 1)
1√1 + x2
cosh−1 x
Domain: Range:
[1,∞) [0,∞)
1 1 2 3 4 51
1
2
3
4
5
x
y
ln(x+√x2 − 1)
1√x2 − 1
tanh−1 x
Domain: Range:
(−1, 1) R
2 1 1 2
2
1
1
2
x
y
1
2ln(
1 + x
1− x)1
1− x2
csch−1 x
Domain: Range:
R\ 0 R\ 0
4 2 2 4
4
2
2
4
x
y
ln
(1
x+
√1
x2+ 1
)−1
|x|√x2 + 1
ENG1091 Mathematics for Engineering page 63
sech−1 x
Domain: Range:
(0, 1) [0,∞)
1 1 2 3 4 51
1
2
3
4
5
x
y
ln
(1
x+
√1
x2− 1
)−1
x√
1− x2
coth−1 x
Domain: Range:
R\ [−1, 1] R\ 0
4 2 2 4
4
2
2
4
x
y
1
2ln
(1 + x
x− 1
)1
1− x2
Examples
1. Show that f(x) = tanh−1 x is always increasing.
d
dx
(tanh−1 x
)=
1
1− x2
and 1− x2 is positive on the domain of tanh−1, namely (−1, 1) . Thus
d
dx
(tanh−1 x
)=
1
1− x2> 0
and hence tanh−1 x is always increasing.
2. Find the derivative of tanh−1 (sinx).
d
dxtanh−1 (sinx) =
1
1− sin2 xcosx
=1
cos2 xcosx using 1− sin2 x = cos2 x
= secx
3. Evaluate∫ 1
0
dx√1 + x2
.
∫ 1
0
dx√1 + x2
=[sinh−1 x
]10
= sinh−1 (1)− sinh−1 0
= sinh−1 (1) = loge
(1 +√
2)
using the log form of sinh−1
ENG1091 Mathematics for Engineering page 64
4. Find∫ 3
0
dx√9 + 4x2
.
∫ 3
0
dx√9 + 4x2
=
∫ 3
0
dx
3√
1 + 4x2
9
=
∫ 3
0
dx
3
√1 +
(2x3
)2=
1
3
[3
2sinh−1
(2x
3
)]3
0
=1
2sinh−1 (2)− 0
=1
2loge
(2 +√
5)
using the log form of sinh−1
5. Find f ′(x) when f(x) = sinh−1(x2).
f ′(x) =1√
1 + (x2)2· 2x
=2x√
1 + x4
6. Find the derivative of sinh−1 (tanx). Comment in light of Q.2
d
dxsinh−1 (tanx) =
1√1 + tan2 x
sec2 x
=1
secxsec2 x using 1 + tan2 x = sec2 x
= secx
=d
dxtanh−1 (sinx) from example 2
Notice also sinh−1 (tan (0)) = 0 and tanh−1 (sin 0) = 0.
Provided sinh−1 (tan (x)) and tanh−1 (sinx) are defined on the same interval, and one which
includes x = 0 (for example(−π
2 ,π2
)), we must conclude that they are the same function
on this interval.
ENG1091 Mathematics for Engineering page 65
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Further Calculus
Lecture 14 · Integration by parts · use of complex exponentialText Reference: §8.8.3
Differentiation techniques are usually fairly routine, following set rules and patterns. This is
not the case for antidifferentiation, where it can be far more challenging to find the appropriate
technique; some careful thinking must often be done to find the antiderivative. Sometimes an
antiderivative can’t be found in terms of elementary functions! Remember that all antiderivatives
can be checked by differentiation, be prepared to have a go (even guess the answer?) and check
it back through by differentiation.
e.g. Guess the answer for∫x cosxdx
The formula for Integration by Parts arises from the Product Rule for Differentiation:
d
dx(u(x)v(x)) =
du
dxv(x) + u(x)
dv
dxso∫ [
du
dxv(x) + u(x)
dv
dx
]dx = u(x)v(x)∫
du
dxvdx+
∫udv
dxdx = uv∫
du
dxvdx = uv −
∫udv
dxdx
This effectively means that we are replacing the problem of finding∫u′(x)v(x)dx with the (eas-
ier?) problem of finding∫u(x)v′(x)dx. To use this rule effectively, we have to be careful in
choosing u and v. There are no general rules for choosing u and v, but the purpose is to obtain
a simpler integral.
Integration by parts is often used when integrating a product (but not always) and is usually
the second technique we would think to employ. (The method of substitution, covered in the
ENG1090 and specialist maths syllabus, being the first.)
Examples:
•∫x cosxdx
∫x cosxdx =
∫xd
dx(sinx) dx
= x sinx−∫
1 · sinxdx
= x sinx+ cosx+ c
•∫xexdx
∫xexdx =
∫xd
dx(ex) dx
= xex −∫
1 · exdx
= xex − ex + c
ENG1091 Mathematics for Engineering page 66
•∫x lnxdx ∫
x lnxdx =
∫lnx
d
dx
(1
2x2
)dx
=1
2x2 lnx−
∫1
x·(
1
2x2
)dx
=1
2x2 lnx− 1
2
∫xdx
=1
2x2 lnx− 1
4x2 + c
•∫
lnxdx∫
lnxdx =
∫lnx
d
dx(x) dx
= x lnx−∫
1
x· (x) dx
=1
2x lnx−
∫dx
= x lnx− x+ c
•∫x2exdx ∫
x2exdx =
∫x2 d
dx(ex) dx
= x2ex −∫
2x · exdx
= x2ex − 2
∫xexdx
now integrate by parts again or use the
result from example 2 above
= x2ex − 2 (xex + ex) + c
•∫
tan−1 xdx∫
tan−1 xdx =
∫tan−1 x
d
dx(x) dx
= x tan−1 x−∫xd
dx
(tan−1 x
)dx
= x tan−1 x−∫
x
1 + x2dx
= x tan−1 x− 1
2
∫2x
1 + x2dx
= x tan−1 x− 1
2loge
(1 + x2
)+ c
ENG1091 Mathematics for Engineering page 67
•∫ex cosxdx
∫ex cosxdx =
∫ex
d
dx(sinx) dx
= ex sinx−∫
d
dx(ex) sinxdx
= ex sinx−∫ex sinxdx
= ex sinx+
∫ex
d
dx(cosx) dx
= ex sinx+ ex cosx−∫
(cosx)d
dx(ex) dx
= ex sinx+ ex cosx−∫
(cosx) exdx
2
∫ex cosxdx = ex sinx+ ex cosx
Hence:
∫ex cosxdx =
1
2ex (cosx+ sinx) + c.
Complex numbers are extremely useful in obtaining integrals of the type∫eax cos bxdx or∫
eax sin bxdx, and are usually much quicker than integration by parts.
To better understand these examples we are going to use two important facts about complex
numbers:
1
a+ ib=
1
a+ ib× a− iba− ib and eibx = cos bx+ i sin bx
=a− iba2 + b2
so that e(a+ib)x = e(a+ib)x
= eax · eibx
= eax (cos bx+ i sin bx)
Examples
•∫ex cosxdx
∫ex cosxdx = Re
∫ex (cosx+ i sinx) dx
= Re
∫ex · eixdx
= Re
∫ex+ixdx
= Re
∫ex(1+i)dx
Now
∫ex(1+i)dx =
1
1 + iex+ix
=1
1 + iex (cosx+ i sinx)
=1− i
2ex (cosx+ i sinx) .
Taking the real part:
∫ex cosxdx =
1
2ex (cosx+ sinx)
Hence:
∫ex cosxdx =
1
2ex (cosx+ sinx) + c.
ENG1091 Mathematics for Engineering page 68
•∫e−x sin 2xdx
∫e−x sin 2xdx = Im
∫e−x (cos (2x) + i sin (2x)) dx
= Im
∫e−x · ei2xdx
= Im
∫e−x+2ixdx
= Im
∫ex(−1+2i)dx
Now
∫ex(−1+2i)dx =
1
−1 + 2iex(−1+2i)
=1
−1 + 2ie−x (cos 2x+ i sin 2x)
=−1− 2i
5e−x (cos 2x+ i sin 2x)
Taking the imaginary part
∫e−x sin 2xdx = −2
5e−x cos 2x− 1
5e−x sin 2x+ c
•∫e3x cosxdx
∫e3x cosxdx = Re
∫e3x (cosx+ i sinx) dx
= Re
∫e3x · eixdx
= Re
∫ex(3+i)dx
Now
∫ex(3+i)dx =
1
3 + ie3x (cosx+ i sinx)
=3− i10
e3x (cosx+ i sinx)
Taking the real part:
∫e3x cosxdx =
1
10e3x (3 cosx+ sinx) + c
ENG1091 Mathematics for Engineering page 69
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Limits of Functions
Lecture 15 · Limit properties · ‘Squeeze’Principle · l’Hopitals Rule
Definition: We write
limx→a
f (x) = L
and say “the limit of f (x) , as x approaches a equals L” if we can make the values of f (x)
arbitrarily close to L (as close to L as we like) by taking x to be suffi ciently close to a (on either
side of a) but not equal to a.
Notice the phrase “but x 6= a”in the definition. This means in finding the the limit of f (x) as x
approaches a, we are not interested in the value of the function at x = a. In fact f (x) may not
even be defined when x = a. The only thing that maters is how f (x) is defined near a.
Illustrative example: Use a calculator and guess the value of
limx→0
sinx
x.
The limit laws.
The limit laws are listed below. Essentially they allow ‘common sense’manipulation of limit
expressions, following normal algebraic operations, e.g. the limit of a sum is the same as the sum
of its limits. It is important to note that these laws can only be applied when the combining
functions have an existing limit.
Suppose that c is a constant and the limits limx→a f(x) and limx→a g(x) exist. Then
1. limx→a
[f(x) + g(x)] = limx→a
f(x) + limx→a
g(x)
2. limx→a
[f(x)− g(x)] = limx→a
f(x)− limx→a
g(x)
3. limx→a
[cf(x)] = c limx→a
f(x)
4. limx→a
[f(x)× g(x)] = limx→a
f(x)× limx→a
g(x)
5. limx→a
f(x)
g(x)=
limx→a f(x)
limx→a g(x)if lim
x→ag(x) 6= 0.
6. To evaluate limits we will make frequent use of the continuous function rule: :
Suppose limx→a
g (x) = b and f is continuous at b.
Then limx→a
f(g (x)) = f(
limx→a
g (x))
= f (b)
To make effective use of rule 6 we will take it as known that the elementary functions (poly-
nomial, exponential, logarithmic, trigonometric and hyperbolic functions) are continuous
on their respective domains.
Examples: (examples 1-4 are evaluated using the limit laws above)
ENG1091 Mathematics for Engineering page 70
1. Evaluate limx→1
x2(6x+ 3)(2x− 7)
(x3 + 4)(x+ 17).
limx→1
x2(6x+ 3)(2x− 7)
(x3 + 4)(x+ 17)=
limx→1 x2(6x+ 3)(2x− 7)
limx→1(x3 + 4)(x+ 17)by rule 5 since lim
x→1(x3 + 4)(x+ 17) 6= 0
=limx→1 1(9)(−5)
limx→1(5)(18)
= −1
2.
In example 1 we could have found the limit by merely substituting in the value x = 1. If
we could always evaluate limits by doing this the concept of a limit would be superfluous.
However the notion of a limit of a function f (x) as x → a is most useful when f (x) is
undefined at x = a.
2. Find limx→1
1x − x1x − 1
limx→1
1x − x1x − 1
= limx→1
1− x2
1− x
= limx→1
(1− x) (1 + x)
1− x= lim
x→1(1− x)
= 0
3. Find limx→0
1x+4 −
14
x
limx→0
1x+4 −
14
x
= limx→0
4− (x+ 4)
x (x+ 4) 4
= limx→0
−xx (x+ 4) 4
= limx→0
−xx (x+ 4) 4
= limx→0
−1
(x+ 4) 4
= −1÷ limx→0
(x+ 4) 4
= − 1
16
ENG1091 Mathematics for Engineering page 71
4. Find limt→0
√2− t−
√2
t.
limt→0
√2− t−
√2
t= lim
t→0
√2− t−
√2
t×√
2− t+√
2√2− t+
√2
= limt→0
2− t− 2
t(√
2− t+√
2)
= limt→0
−tt(√
2− t+√
2)
= limt→0
−1(√2− t+
√2)
= −1÷ limt→0
(√2− t+
√2)
= − 1
2√
2
The last four examples demonstrate the use of algebra in evaluating limits. However in
evaluating most limits the use of algebra alone will not be suffi cient. The next technique
we introduce is much more powerful than algebraic methods.
ENG1091 Mathematics for Engineering page 72
Indeterminate forms and L’Hopital’s rule
Applying the limit techniques (particularly direct substitution) discussed earlier can often lead
to ‘meaningless’expressions of the type 00 or
∞∞ . These are called indeterminate forms, since
they have not correctly determined the true limit value.
However, if we ‘zoom in’around x = a for 2 functions f and g, such that f(a) = g(a) = 0 we
can see that the value of f(x)g(x) ≈
f ′(x)g′(x) .
f
g
x
y
0 a
This forms the basis of a rule known as L’Hopital’s Rule: Suppose f and g are differentiable,
with f(a) = g(a) = 0. If f ′ and g′ are continuous (but g′(x) 6= 0), then
limx→a
f(x)
g(x)= lim
x→af ′(x)
g′(x).
This rule can be applied for two-sided and one-sided limits, approaching a fixed value a or ±∞,which give the indeterminate form
0
0or∞∞ . To reduce expressions to a meaningful term, it may
be necessary to apply L’Hopital’s Rule two or more times.
Examples
limx→0
sin 2x
x limx→0
sin 2x
xis of the form ‘
0
0’so that L’Hopital’s rule may be applied:
= limx→0
2 cos (2x)
1
= limx→0
2
1= 2
limx→∞
lnx
x limx→∞
lnx
xis of the form ‘
∞∞’so that L’Hopital’s rule may be applied:
= limx→∞
(1x
)1
= limx→∞
1
x= 0
ENG1091 Mathematics for Engineering page 73
limx→0
x− sinx
x3
limx→0
x− sinx
x3is of the form ‘
0
0’so that L’Hopital’s rule may be applied:
= limx→0
1− cosx
3x2is of the form ‘
0
0’so that L’Hopital’s rule may be applied again
= limx→0
sinx
6xis of the form ‘
0
0’
= limx→0
cosx
6
=1
6
There are other types of indeterminate forms, involving combinations of 0 and ∞, dealt with asfollows:
Indeterminate Product 0. ∞
If limx→a
f(x)g(x) = 0 · ∞ re-arrange f(x)g(x) to f(x)1/g(x) , then apply L’H Rule.
limx→0+
x lnx limx→0+
x lnx is of the form ‘0 · ∞’so that some rearrangement is necessary
= limx→0
lnx
(1/x)is now of the form ‘
∞∞’so that L’Hopital’s rule may be applied
= limx→0
x−1
−x−2
= limx→0
x2
x
= limx→0
x
1= 0
Indeterminate Difference ∞−∞
If limx→a[f(x) − g(x)] = ∞ − ∞, convert the expression to a single fraction, using commondenominators, factorisation, or rationalisation, to produce a 0
0 or∞∞ form. Then apply L’H Rule.
Examples
limx→0
[1
x− 1
sinx]
limx→0
[ 1x −
1sinx ] is of the form ‘∞−∞’so that some rearrangement is necessary
= limx→0
sinx−xx sinx is now of the form ‘
0
0’so that L’Hopital’s rule may be applied
= limx→0
cosx−1sinx+x cosx a ‘
0
0’form’
= limx→0
sinxcosx−x sinx+cosx applying L’Hopital’s rule
= limx→0 sinxlimx→0(cosx−x sinx+cosx) applying rule (5)
= limx→0
02
= 0
ENG1091 Mathematics for Engineering page 74
Indeterminate Powers 00, 1∞, ∞0.
For these indeterminate forms, begin with y = f(x)g(x).
Take the natural logarithm of both sides, to give ln y = g(x) ln f(x).
The indeterminate product 0.∞ which results is then re-arranged as above.
Lastly, if limx→a ln y has been found to be L, limx→a y = eL.
Examples
limx→0+
xx = a ‘00’form
suppose for the moment that limx→0+
xx exists, so let limx→0+
xx = L
then ln
(limx→0+
xx)
= lnL
since ln is continuous ln
(limx→0+
xx)
=
(limx→0+
ln (xx)
)so lim
x→0+ln (xx) = lnL
now limx→0+
x ln (x) = 0 from the example above hence
lnL = 0 giving L = e0 = 1
Hence limx→0+
xx = 1
limx→0
(1 +
2
x
)x= a ‘∞0’form
suppose for the moment that limx→0
(1 +2
x)x exists, so let lim
x→0(1 +
2
x)x = L
then ln
(limx→0
(1 +2
x)x)
= lnL
since ln is continuous ln
(limx→0
(1 +2
x)x)
=
(limx→0
ln(1 +2
x)x)
so(
limx→0
ln(1 +2
x)x)
= lnL
now limx→0
ln
(1 +
2
x
)x= lim
x→0x ln(1 +
2
x) = ....a ‘0 · ∞’form
= limx→0
ln(1 + 2x)
1/x= ....a
∞∞ form
= limx→0
1/(1 + 2x) ·
(−2/x2
)−1/x2
applying L’Hopital’s rule
= limx→0−2
1
(1 + 2x)
= −2
lnL = −2 giving L = e−2.
Hence limx→0
(1 +
2
x
)x= e−2
ENG1091 Mathematics for Engineering page 75
The squeeze theorem
If a function g(x) is ‘trapped’between 2 other functions f and h such that f(x) ≤ g(x) ≤ h(x),
and limx→a
f(x) = limx→a
h(x) = L, then limx→0
g(x) = L.
L
a
fg
h
x
We can sometimes use this to evaluate limits of expressions where Limit Laws cannot successfully
be applied:
Example:
Show that
limx→0
x sin1
x= 0.
This limit cannot be found by finding limx→0
x and limx→0
sin1
xand multiplying them together since
limx→0
sin1
xdoes not exist.
graph of y = sin(
1x
)We know that −1 ≤ sin
1
x≤ 1, so we can introduce a ‘squeeze’situation by using
− |x| ≤ x sin1
x≤ |x|
ENG1091 Mathematics for Engineering page 76
graph of y = x sin(
1x
)Now lim
x→0|x| = 0 and limx→0− |x| = 0, so we have lim
x→0x sin
1
x= 0.
Example: A very common limit encountered by engineering students is
limx→∞
e−x sinx
Solution: We know that −1 ≤ sinx ≤ 1, so we can form the ‘squeeze’using
−e−x ≤ e−x sinx ≤ e−x
Now limx→∞
e−x = 0 and similarly limx→∞
−e−x = 0 hence limx→∞
e−x sinx = 0.
ENG1091 Mathematics for Engineering page 77
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Further Calculus
Lecture 16 · Improper integralsText Reference: §9.2
Example: What is wrong with the following calculation:∫ 1
−1
1
x2dx =
[−x−1
]1−1
= −1− 1 = −2?
ANS: The function f (x) = 1x2is always positive. The definite integral of a positive function can
never be negative. (Definite integrals give the ‘signed’area between a curve and the x−axis.For a curve which is always positive this signed area must also be positive.)
We have applied the Fundamental Theorem of Calculus in circumstances where we were
not entitled to do so.
The Fundamental Theorem of Calculus which enables us to evaluate a definite integral by
taking an antiderivative of the integrand requires that the integrand be continuous over a finite
domain of integration [a, b].
The function 1x2is not continuous on the domain [−1, 1]. (In fact of course it is not even defined
on [−1, 1] .)
If we break the integral up we obtain
∫ 1
−1
1
x2dx =
∫ 1
0
1
x2dx+
∫ 0
−1
1
x2dx
However this introduces a new problem. The integrands in both these integrals are not Riemann
integrable because they are not bounded. (The function 1x2is unbounded near x = 0.)
We can extend the theory of Riemann integration by introducing ‘improper integrals’.
There are two types of improper integrals:
• an expression like∫ ∞
1exdx is improper because the domain of integration, in this
case [1,∞) , is not bounded;
• expressions like∫ 1
0
1
x2dx where the range of the integrand is unbounded on the
interval of integration. (In this case the function1
x2is unbounded on [0, 1] .)
When the domain of integration is not finite we have a Type 1 improper integral.
When the integrand is unbounded at a particular point, but continuous elsewhere, we have
a Type 2 improper integral.
Type 1: Infinite intervals
For these integrals, we are attempting to find the area of an ‘infinite space’. To do this, we
evaluate the definite integral over a finite interval, and investigate the limit of the integral as the
interval is extended.
ENG1091 Mathematics for Engineering page 78
Example∫ ∞1
1
x2dx [geometrically this is the area under the curve y =
1
x2to the right of x = 1].
∫ ∞1
1
x2dx = lim
t→∞
∫ t
1
1
x2dx (diagram)
= limt→∞
[−x−1
]t1
= limt→∞
(−1
t
)+ 1
= 1.
We say that∫ ∞
1
1
x2dx is convergent.
We use the following definitions to evaluate these integrals:
To define∫ ∞a
f(x)dx we require two things, (i) that∫ t
af(x)dx exists for every number t ≥ a
(ii) that the limt→∞
∫ t
af(x)dx exists and is finite.
We then say that∫ t
af(x)dx converges.
Provided these two conditions are satisfied we define∫ ∞a
f(x)dx = limt→∞
∫ t
af(x)dx.
A similar statement can be made regarding the definition of∫ a
−∞f(x)dx.
The integral∫ ∞−∞
f(x)dx is also considered a type I integral, we define
∫ ∞−∞
f(x)dx =
∫ 0
−∞f(x)dx+
∫ ∞0
f(x)dx
provided the two improper integrals on the right are convergent independently.
Note: In each of these cases, if the integral exists, we say that the improper integral is convergent
and that the limit becomes the value of the improper integral. If the limit fails to exist, the
improper integral is divergent.
ExampleDetermine if
∫ ∞0
e−2xdx is convergent or divergent.
∫ ∞0
e−2xdx = limt→∞
∫ t
0e−2xdx
= limt→∞
[−1
2e−2x
]t0
= limt→∞
(−1
2e−2t
)+
1
2
=1
2. (The integral is convergent.)
ENG1091 Mathematics for Engineering page 79
ExampleDetermine if
∫ ∞1
1
xdx is convergent or divergent.
∫ ∞1
1
xdx = lim
t→∞
∫ t
1
1
xdx
= limt→∞
[loge x]t1
= limt→∞
(loge t)− 0
=∞ since loge x is an unbounded function as x→∞.
This means the integral∫ ∞
1
1
xdx diverges.
ExampleFor what values of p is
∫ ∞1
1
xpdx convergent?
The case where p = 1 was considered in the previous example, so we know that∫ ∞
1
1
xpdx
diverges when p = 1.
Now consider what happens if p 6= 1 :∫ ∞1
1
xpdx = lim
t→∞
∫ t
1
1
xpdx
= limt→∞
[1
1− px−p+1
]t1
provided p 6= 1
= limt→∞
(1
1− pt−p+1
)− 1
1− p
Now1
1− pt−p+1 →∞ as t→∞ if p < 1, while
1
1− pt−p+1 → 0 as t→∞ if p > 1.
We conclude∫ ∞
1
1
xpdx converges if p > 1 and diverges if p ≤ 1.
Example
Evaluate∫ ∞−∞
1
1 + x2dx or else explain why the integral diverges.
If∫ ∞−∞
1
1 + x2dx is to converge we require the (independent) convergence of both
∫ ∞0
1
1 + x2dx
and∫ 0
−∞
1
1 + x2dx.
Now∫ ∞
0
1
1 + x2dx = lim
t→∞
∫ t
0
1
1 + x2dx While
∫ 0
−∞
1
1 + x2dx = lim
t→−∞
∫ 0
t
1
1 + x2dx
= limt→∞
[tan−1 x
]t0
= limt→−∞
[tan−1 x
]0t
= limt→∞
(tan−1 t
)− tan−1 (0) = 0− lim
t→−∞
(tan−1 t
)=π
2− 0 = 0−− π
2
=π
2. =
π
2.
ENG1091 Mathematics for Engineering page 80
So both∫ ∞
0
1
1 + x2dx and
∫ 0
−∞
1
1 + x2dx converge,
and we have∫ ∞−∞
1
1 + x2dx =
∫ ∞0
1
1 + x2dx+
∫ 0
−∞
1
1 + x2dx = π.
Type 2 - integrand unbounded at a single point
Suppose f is a function continuous on [a, b) but is not bounded at x = b, that is, limx→b− f (x) =
∞ or −∞.
Provided limt→b−
∫ t
af(x)dx exists, we define
∫ b
af(x)dx = lim
t→b−
∫ t
af(x)dx
The analogous definition can be made when f is not bounded at a :
Suppose f is a function continuous on (a, b] but is not bounded at x = a, that is, limx→a+ f (x) =
∞ or −∞.
Provided limt→a+
∫ b
tf(x)dx exists, we define
∫ b
af(x)dx = lim
t→a+
∫ b
tf(x)dx
Now we see why we have the apparent contradiction in the example:∫ 1
−1
1
x2dx.
The integral∫ 1
−1
1
x2dx is undefined because neither
∫ 1
0
1
x2dx nor
∫ 0
−1
1
x2dx exists.
(The failure of just one of these limits to exist results in the integral being undefined.)
∫ 1
0
1
x2dx = lim
t→0+
∫ 1
t
1
x2dx
= limt→0+
[−x−1
]1t
= limt→0+
(−1 +
1
t
)=∞. (The integral is divergent.)
Similarly∫ 0
−1
1
x2dx diverges.
ENG1091 Mathematics for Engineering page 81
Example
Is the area under the curve y = 1√xfrom x = 0 to x = 1 finite? If so, what is it?
Solution: The area, if it exists, is given by∫ 1
0
1√xdx. This integral is improper since the
integrand is unbounded at x = 0.
Now∫ 1
0
1√xdx = lim
t→0+
∫ 1
t
1√xdx
= limt→0+
[2x1/2
]1
t
= limt→0+
(2− 2
√t)
= 2.
The area under the curve is finite and is equal 2 sq. units.
Examples: Evaluate each of the following when they exist and explain the situation otherwise:
Find∫ 1
0
1√1− x2
dx
∫ 1
0
1√1− x2
dx = limt→1−
∫ t
0
1√1− x2
dx
= limt→1−
[sin−1 x
]t0
= limt→1−
(sin−1 t− 0
)= sin−1 (1)
= π/2
Find∫ e
0lnxdx
∫ e
0lnxdx = lim
t→0+
∫ e
tlnxdx diagram:
= limt→0+
[x lnx− x]et (see lecture 14:∫
lnxdx = x lnx− x),
= e ln e− e− limt→0
(t ln t− t)
= e− e− 0 since limt→0
(t ln t) = 0,
= 0
ENG1091 Mathematics for Engineering page 82
The Comparison Test for Improper Integrals allows us to discuss the convergence of an
improper integral without evaluating it directly, by comparing it to a known or easier integral.
If f and g are continuous functions, where f(x) ≥ g(x) ≥ 0, then
1.∫ ∞a
g(x)dx is convergent if∫ ∞a
f(x)dx is convergent.
2.∫ ∞a
f(x)dx is divergent if∫ ∞a
g(x)dx is divergent.
Example
Show that∫ ∞
1e−x
2dx is convergent. (This integral cannot be evaluated by elementary means
since the antiderivative of e−x2is not an elementary function).
Solution: We compare the integrand e−x2with e−x.
Since x2 ≥ x for all x ≥ 1 we have1
ex2≤ 1
exi.e. e−x
2 ≤ e−x for all x > 1 (in fact e−x2approaches
0 at a very much faster rate than does e−x).
So, using the comparison test,∫ ∞
1e−x
2dx converges if we can show
∫ ∞1
e−xdx converges.
∫ ∞1
e−xdx = limt→∞
∫ t
1e−xdx
= limt→∞
[−e−x
]t1
= limt→∞
(−e−t
)+ e−1.
Now limt→∞
(−e−t
)exists, in fact it is zero, and hence
∫ ∞1
e−xdx converges to the value e−1.
Thus∫ ∞
1e−x
2dx also converges. Its value (whatever it might be) is a number less than e−1.
ENG1091 Mathematics for Engineering page 83
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Further Calculus
Lecture 17 · slab and washer methods · shell methodText Reference: §8.9.1
Volumes of solids
Most regular solids have a ‘formula’to use to calculate their volume
e.g. volume (sphere) =43πr
3,
volume (cone) =13πr
2h etc.
Where do these formulae come from, and how do we find volumes of other solids?
A shape is positioned along co-ordinate axes
and a representative slice is used for the cross-
sectional area.
The width of this slice is taken as ∆x and thus
the volume of a typical slice is
∆V = A(x)∆x.
V =
∫ b
aA(x)dx
Example 1 Find the volume of a sphere of radius r with centre at the origin.
∆V = A(x)∆x
= π [y (x)]2 ∆x
So that
V =
∫ r
−rπ(√
r2 − x2)2dx
= 2π
∫ r
0
(r2 − x2
)dx since
√r2 − x2 is an even function
= 2π
[r2x− 1
3x3
]r0
= 2π
(r3 − 1
3r3
)=
4
3πr3
Slab method:
The sphere is an example of a solid of revolution. These are formed when a region (in this
case the region bounded by the x -axis and the upper half of the circle centred at the origin and
ENG1091 Mathematics for Engineering page 84
of radius r) of the Cartesian plane is rotated about the x -axis. The cross-sectional area of a
typical slice is then in the shape of a disk, and being circular has area
A = πr2 = πf(x)2, where f(x) = height of each slice above the x -axis and therefore the radius
of each slab.
Thus, for a volume of a solid of revolution bounded by the x-axis, y = f(x), x = a and x = b, we
have
V =
b∫a
π [f(x)]2 dx
Washer method
The volume formed by rotation around the x -axis of an area between 2 curves can often be
determined by using the washer method. For this we use
V =
∫ b
aπ[f (x)2 − g(x)2
]dx
The shape created will be a washer, sitting perpendicular to the x -axis.
Example 2 Find the volume of the solid formed when the region bounded by y = x and y = x2
is rotated through 2π radians about the x -axis.
∆V = A(x)∆x
= π(
[f (x)]2 − [g(x)]2)
∆x
V =
∫ b
aπ[f (x)2 − g(x)2
]dx
= π
∫ 1
0
(x2 −
(x2)2)
dx
= π
[1
3x3 − 1
5x5
]1
0
= π
(1
3− 1
5
)=
2π
15
Example 3 Find the volume of the solid formed when the region bounded by y = x and y = x2
is rotated through 2π radians about the y-axis.
∆V = A(y)∆y
= π(
[x2 (y)]2 − [x1(y)]2)
∆y
V = π
∫ 1
0
[(√y)2 − y2
]dy the y terminals are y = 0 and y = 1
the outer radius x2 is y = x2 or x2 (y) =√y
and the inner radius is x1 (y) = y
= π
∫ 1
0
(y − y2
)dy
=
[1
2y2 − 1
3y3
]1
0
= π
(1
2− 1
3
)=π
6
ENG1091 Mathematics for Engineering page 85
Shell method:
In finding the volume of a solid of revolution which has been rotated about the y-axis, it may
sometimes be more useful to find the volume using cylindrical (hollow) shells, where the shells
will be thin with axis the y-axis.
We use the fact that the shell opens to give a
flat rectangular solid, where
∆V = length× height× thickness
= 2πx · f(x) · dx
to arrive at the expression for the total volume
of the solid of revolution
V =
∫ b
a2π × (shell radius)× (shell height) dx
=
∫ b
a2πxf(x)dx
To use the Shell Method:
1. Draw the diagram, including a line to represent the radius perpendicular to the axis of
revolution.
2. Find the limits of integration, along the required axis of revolution.
3. Integrate the product 2 π (shell radius) (shell height) to give the total volume.
Example 4 Find the volume of the solid obtained by rotating about the y-axis the region
bounded by y = x(x− 1)2 and y = 0. (To attempt this example using the washer method would
be almost impossible.)
∆V = 2πx · x(x− 1)2∆x
ENG1091 Mathematics for Engineering page 86
V = 2π
∫ 1
0x2 (x− 1)2 dx
= 2π
∫ 1
0x4 − 2x3 + x2dx
= 2π
[1
5x5 − 1
2x4 +
1
3x3
]1
0
= 2π
(1
5− 1
2+
1
3
)
=π
15
1.0 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
0.5
0.4
0.3
0.2
0.1
0.1
0.2
0.3
0.4
0.5
x
y
Example 5 (Example 3 again but this time via shell method.): Find the volume of the solid
formed when the region bounded by y = x, and y = x2 is rotated through 2π radians about the
y-axis.
∆V = 2πx(x− x2
)∆x
V = 2π
∫ 1
0x(x− x2
)dx
= 2π
[1
3x3 − 1
4x4
]1
0
= 2π · 1
12
=π
6same as that obtained previously
The answers obtained by either method are identical, but the shell method avoids the use of
squaring.
Example 6 Find the volume of the solid generated when the region bounded by y = 1x , y = 0,
x =1 and x = 10 is rotated about the y-axis, using cylindrical shells.
∆V = 2πx
(1
x− 0
)∆x
= 2π∆x
V = 2π
∫ 10
11dx
= 2π · 9
= 18π
The next example shows that the shell method can also be used to find volumes of revolution
about the x axis.
ENG1091 Mathematics for Engineering page 87
Example 7 The region bounded by y =√x, the x -axis, and the line x = 4 is revolved about
the x -axis to generate a solid. Find its volume using shells.
∆V = 2πy (4− x) ∆y
= 2πy(4− y2
)∆y
V = 2π
∫ 2
0y(4− y2
)dy note use of y values as terminals
= 2π
[2y2 − 1
4y4
]2
0
= 8π
Here the shell method is more complicated than the washer method: V = π∫ 4
0 (√x)
2dx = 8π.
ENG1091 Mathematics for Engineering page 88
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Sequences and Series
Lecture 18 · sequences · limits of sequencesText Reference: §7.1,7.2,7.5
1. Definition: An infinite sequence is a special kind of function whose domain is a set of
integers extending from some starting integer (usually 1) and then continuing indefinitely.
The sequence a1, a2, a3, a4, ... is the ordered list of function values of a function a wherea (n) = an at each positive integer n. We usually specify a sequence by giving its general
term, the formula for an.
2. Examples:
(a) an =
(−1
2
)n=
−1
2,1
4,−1
8,
1
16, ...
(b) an =
n− 1
n=
0,
1
2,2
3,3
4, ...
(c) an = (−1)n−1 = 1,−1, 1,−1, ...
(d) an =
(n2
2n
)=
1
2, 1,
9
8, 1,
25
32,
9
16, ...
(e) an =
(cos nπ2n
)=
0,−1
2, 0,
1
4, 0,−1
6, ...
(f) an =
(1 +
1
n
)n=
2,
(3
2
)2
,
(4
3
)3
,
(5
4
)4
, ...
.
3. Definition: An infinite sequence has a limit L if the terms of the sequence tend to that
limit. This is all very well but it doesn’t say very much. A real (or complex) number L
is the limit of a sequence an if for any number ε > 0 there is a number N such that all
terms of the sequence beyond N are within ε of L. Consult the picture on page 439 of your
text for a visual illustration of this definition. When an infinite sequence an has a limitL we write
limn→∞
an = L.
We are not going to use this definition in any formal sense because we are going to establish
convergence or divergence of sequences using the limit theorems which follow. However it
is important to bear in mind that the proofs of these theorems depend ultimately on this
definition.
Not all sequences have limits and those that do are said to be convergent to their limit.
If a sequence has no limit we say it diverges.
Many people have a false idea of a limit as a number which the terms of the sequence ‘get
closer to’somehow. Notice example (e) above which has the limit 0. Notice also that it is
not true to say that successive terms are getting closer to zero, in fact each non-zero term
is farther away from zero than its predecessor, which of course is exactly zero.
ENG1091 Mathematics for Engineering page 89
4. Examples:
(a) an =
(−1
2
)n=
−1
2,1
4,−1
8,
1
16, ...
converges to 0.
(b) an =n− 1
n=
0,
1
2,2
3,3
4, ...
converges to 1.
(c) an = (−1)n−1 = 1,−1, 1,−1, ... diverges since it oscillates indefinitely between −1
and 1.
(d) an =
(n2
2n
)=
1
2, 1,
9
8, 1,
25
32,
9
16, ...
converges to 0.
(e) an =
(cos nπ2n
)=
0,−1
2, 0,
1
4, 0,−1
6, ...
converges to 0.
(f) an =
(1 +
1
n
)n=
2,
(3
2
)2
,
(4
3
)3
,
(5
4
)4
, ...
, converges to e.
(g) an = n = 1, 2, 3, 4, ... , diverges since an →∞, we also say that an is unbounded.
5. Demonstrating divergence. Showing that a particular sequence diverges can in many
ways be more problematic.
(a) If we can show that the sequence is unbounded the sequence diverges. A sequence
an is unbounded if for all numbers M > 0 we may find an n such that |an| > M.
However, please remember that many bounded sequences are also divergent.
(b) If a sequence appears to have two or more different ‘limits’the sequence diverges. It
may happen, for example, that the sequence of odd terms of a converges to a limit
which is different to the limit of the sequence of even terms. This behaviour is apparent
in the example (c) above.
(c) Many divergent sequences behave like the divergent sequence an = sin (n) . The range
of this sequence is dense in the set [−1, 1] which means we can pick any number in
[−1, 1] and specify any positive distance we like, then there exists an n such that
sin (n) is as close as we please to our chosen number.
ENG1091 Mathematics for Engineering page 90
6. Sequence theorems
Suppose that c and p are constants and (unless stated otherwise) the limits limn→∞ an
and limn→∞ bn exist. Then
(a) limn→∞
[an + bn] = limn→∞
an + limn→∞
bn
(b) limn→∞
[an − bn] = limn→∞
an − limn→∞
bn
(c) limn→∞
[can] = c limn→∞
an
(d) limn→∞
[an × bn] = limn→∞
an × limn→∞
bn
(i) if limn→∞ bn 6= 0 then limn→∞anbn
= limn→∞ anlimn→∞ bn
,
(ii) if an is a bounded sequence and bn is unbounded then limn→∞
anbn
= 0. (It is
not necessary that limn→∞ an exists.)
(e) limn→∞
[anp] = [ lim
n→∞an]p
Part (f) is really a special case of the Continuous function theorem which says
that
if f is a continuous function then limn→∞
[f (an)] = f(
limn→∞
an
).
(f) limn→∞
c = c
(g) limn→∞
cn = 0 if |c| < 1 and divergent otherwise.
7. The following examples illustrate how the various properties listed above can be used to
establish convergence of sequences and find their limits.
(a) an = n diverges since an is unbounded.
(b) an = 1n converges to 0. Rather obvious but a special case of rule (e)ii.
(c) an =n2 − 3n+ 1
2n2 + 1
Write an =n2 − 3n+ 1
2n2 + 1
=n2(1− 3
n + 1n2
)n2(2 + 1
n2
)=
(1− 3
n + 1n2
)(2 + 1
n2
)
So limn→∞
an =limn→∞
(1− 3
n + 1n2
)limn→∞
(2 + 1
n2
) (apply rule (e))
=(1− 0 + 0)
(2 + 0)(apply rules (a),e(ii))
=1
2
ENG1091 Mathematics for Engineering page 91
(d) an =2n2 + 3n+ 1
n3 + 1
Write an =2n2 + 3n+ 1
n3 + 1
=n2(2 + 3
n + 1n2
)n3(1 + 1
n3
)=
1(2 + 3
n + 1n2
)n(1 + 1
n2
)
So limn→∞
an = limn→∞
(1
n
)×
limn→∞(2 + 3
n + 1n2
)limn→∞
(1 + 1
n2
) (apply rules (d,e))
= 0× 2
= 0
(e) an =√n+ 1−
√n
an =
√n+ 1−
√n
1×√n+ 1 +
√n√
n+ 1 +√n(a trick that often works with difference of sq. roots)
=n+ 1− n√n+ 1 +
√n
=1√
n+ 1 +√n
So limn→∞
an = limn→∞
1√n+ 1 +
√n
= 0 (since the sequences√n+ 1,
√n are unbounded)
Exercises Find the limits of the following sequences if they exist, or if they are divergent explain
why.
1. an =√n2 + 2n− n ANS: convergent: limn→∞ an = 1.
2. an =n2 − 4
n+ 5ANS: divergent: an = n2−4
n+5 is not bounded.
3. an = ln (n+ 1)− ln (2n− 1) ANS: convergent: limn→∞ an = ln 12 = − ln 2.
ENG1091 Mathematics for Engineering page 92
An important sequence
Show limn→∞
(1 +
x
n
)n= ex.
Use L’Hopitals rule but first we need to change it from a sequence limit to a function of a
continuous variable.
Consider instead limx→∞(1 + a
x
)x= L (a) .
Then lnL (a) = ln(limx→∞
(1 + a
x
)x)= limx→∞ ln
(1 + a
x
)x= limx→∞ x ln
(1 + a
x
)=∞ · 0
= limx→∞ln(1+ a
x)1/x = ·0
0
= limx→∞
(−ax2
)(1+ a
x)÷(−1x2
)applying L’Hopitals rule
= limx→∞a
(1+ ax)
= a
so L (a) = ea hence limx→∞(1 + a
x
)x= ea.
We conclude that the sequence limit also exists and limn→∞(1 + x
n
)n= ex.
Note that the existence of the function limit implies the existence of the corresponding sequence
limit but not vice versa.
For example limn→∞
sin (2πn) = 0 but limx→∞
sin (2πx) does not exist.
ENG1091 Mathematics for Engineering page 93
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Sequences and Series
Lecture 19&20 · series · geometric series · convergenceText Reference: §7.6
Many students find this lecture very diffi cult and the material covered here is quite sparse. An
excellent account of this material can be found in the first chapter of Mathematical Methods in
the Physical Sciences by Mary L Boas. (Available in the Hargrave library.)
1. An infinite series is a formal sum of infinitely many terms; for example a1 + a2 + a3 + a4 + ...
is a series formed by adding the terms of the sequence an . This series is also denoted∞∑n=1
an.
∞∑n=1
an = a1 + a2 + a3 + a4 + ...
Examples:
1.∞∑n=1
(−1
2
)n= −1
2+
1
4− 1
8+
1
16− ....
2.∞∑n=1
n− 1
n= 0 +
1
2+
2
3+
3
4+ ...
3.∞∑n=1
(−1)n−1 = 1− 1 + 1− 1 + ...
4.∞∑n=1
(n2
2n
)=
1
2+ 1 +
9
8+ 1 +
25
32+
9
16+ ...
5.∞∑n=1
n = 1 + 2 + 3 + 4 + ...
6.∞∑n=2
(1
lnn
)=
1
ln 2+
1
ln 3+
1
ln 4+ ...
To every series∞∑n=1
an there is an associated sequence called the sequence of partial sums sn
whose nth term is the sum of the first n terms of the series:
s1 = a1
s2 = a1 + a2
s4 = a1 + a2 + a3
s4 = a1 + a2 + a3 + a4
...
sn =
n∑k=1
ak
...
Definition: We say that the series∞∑n=1
an converges to the sum s if the sequence of partial
sums sn , where sn =
n∑k=1
ak, converges to s. If this is the case we write∞∑n=1
an = s.
If the sequence of partial sums is a divergent sequence then the series∞∑n=1
an is said to diverge.
Recall what it means for a sequence sn to converge. Given any ε > 0 there exists N such that
|sn − L| < ε for all n > N. In particular the distance between any two terms sn and sn+1 must
ENG1091 Mathematics for Engineering page 94
be less than 2ε whenever n > N. To see this:
|sn+1 − sn| = |sn+1 − L+ L− sn|
≤ |sn+1 − L|+ |L− sn| by triangle inequality
< ε+ ε whenever n > N
But |sn+1 − sn| = |an+1| so the sequence an converges to zero. Thus we have the followingnecessary condition for convergence.
Theorem: The infinite series∑∞
n=1 an converges only if the parent sequence an converges tozero.
Example: Discuss the convergence or divergence of the series∞∑n=1
n− 1
n+ 1.
We have limn→∞
an = limn→∞
n− 1
n+ 1
= 1. Since this is not zero the series∞∑n=1
n− 1
n+ 1diverges.
Important note: The test limn→∞
an = 0 is a condition necessary for convergence; it is not
suffi cient.
Later on we show that the series∞∑n=1
1
nis a divergent series despite the fact that lim
n→∞1
n= 0.
2. Geometric series
A series of the form
a+ ar + ar2 + ....
where a 6= 0 is called a geometric series. The number a is its first term and the number r is
called the common ratio since it is the value of the ratio of any term to its predecessor.
Repeating decimals are infinite geometric series, e.g.
0.12 = 0.12121212... =12
100+
12
10, 000+
12
1, 000, 000+ ...; r =
1
100
Finding an explicit formula for sn for a geometric series is easy:
sn = a+ ar + ar2 + ....+ arn−1, (1)
and
rsn = ar + ar2 + ar3 + ....+ arn (2)
e.g. (1)− e.g. (2):
hence
sn =a (1− rn)
1− r
• For |r| < 1 we have limn→∞ rn = 0 and so the geometric series converges to
∞∑n=1
arn−1 =a
1− r .
ENG1091 Mathematics for Engineering page 95
• For r > 1 the sequencearn−1
is unbounded and so the geometric series diverges.
• For r = 1, and a 6= 0 we have the divergent constant series a+ a+ a+ .... and for r = −1
we have the series a − a + a − a + .... which alternates between a and 0, and hence also
diverges.
Exercise Use the formula∞∑n=1
arn−1 =a
1− r to find the fraction equivalent of the repeating
decimal 0.12.
0.12 = 0.121212...
Exercises: Discuss the convergence or divergence of each of the following series:
1. Use partial fractions to show1
n (n+ 1)=
1
n− 1
n+ 1. Use this to find a formula for its nth
partial sum sn. Hence show∞∑n=1
1
n (n+ 1)converges by finding its limit.
The nth partial sum is sn =n∑k=1
1
n (n+ 1)=
(1
1− 1
2
)+
(1
2− 1
3
)+ ...+
(1
n− 1
n+ 1
)= 1− 1
n+ 1
Hence∞∑n=1
1
n (n+ 1)= lim
n→∞
(1− 1
n+ 1
)= 1.
2.∞∑n=1
n− 1√n2 + 1
.
ENG1091 Mathematics for Engineering page 96
Tests for Series Convergence
The convergence or divergence of the geometric series was determined by finding a formula for
the sequence of partial sums sn . This is not always possible for more general series and hencethe need to establish some tests which are suffi cient to determine convergence or divergence.
For now we deal exclusively with positive series, that is series of the type∞∑n=1
an where an ≥ 0
for all n.
1. Integral Test.
Example: Determine the convergence or divergence of the series∞∑n=1
1
n2.
Notice that all of the terms of the series are positive.
The essential idea of the integral test is that the series∞∑n=1
1
n2and the improper integral
∫ ∞1
1
x2dx
either both converge, or both diverge (to ∞).
Now a quick calculation shows∫ ∞
1
1
x2dx converges:
Notice that∞∑n=2
1
n2<
∫ ∞1
1
x2dx (diagram) so that
∞∑n=1
1
n2< 1 +
∫ ∞1
1
x2dx
Since an =1
n2is always positive, the sequence of partial sums is increasing (since sn+1 − sn =
an+1 > 0).
Therefore the series is bounded above by 1 +
∫ ∞1
1
x2dx.
An increasing sequence sn that is bounded above converges, hence the series∞∑n=1
1
n2converges.
Example: Determine the convergence or divergence of the series∞∑n=1
1
n.
Notice once again that all of the terms of the series are positive. This time the corresponding
improper integral is∫ ∞
1
1
xdx which diverges (to ∞).
Calculation:
Notice that∞∑n=1
1
n>
∫ ∞1
1
xdx (diagram).
Hence∞∑n=1
1
n>
∫ ∞1
1
xdx is unbounded, and therefore
∞∑n=1
1
nis also unbounded and therefore
diverges.
Note: the divergent series∞∑n=1
1
nis called the harmonic series. It is rather special because it is
an example of a series that diverges and yet whose parent sequence, an = 1n , converges to zero.
ENG1091 Mathematics for Engineering page 97
Example (p-series): The series class∞∑n=1
1
np. are known collectively as p−series . By comparing
with the corresponding integral∫ ∞
1
1
xpdx a quick calculation shows:
∞∑n=1
1
npdiverges for p ≤ 1 and
∞∑n=1
1
npconverges for p > 1.
2. The comparison test
The integral test works by comparing an infinite series with the corresponding improper integral.
Why not compare two series? This then is the comparison test.
Example The series∑∞
n=11
n2+1≤∑∞
n=11n2because 1
n2+1≤ 1
n2for all n. We know
∑∞n=1
1n2
converges and since it dominates∑∞
n=11
n2+1this series must also converge. (Once again the fact
that∑∞
n=11
n2+1and
∑∞n=1
1n2are both series of positive terms is crucial here.)
The precise statement of the comparison test is as follows:
Let∑∞
n=1 an and∑∞
n=1 bn both be series of positive terms and that the
convergence or divergence of∑∞
n=1 bn is known.
Showing convergence: If∑∞
n=1 bn converges and an ≤ bn for all n, then∑∞
n=1 an
converges.
Showing divergence: If∑∞
n=1 bn diverges and an ≥ bn for all n, then∑∞
n=1 an diverges.
Warning: When using the comparison test it is important to get the inequalities the
correct way about and avoid using too coarse a comparison.
For example, it is true that 1n2+1
≤ 1n for all n and that
∑∞n=1
1n diverges. What can we say
about the behaviour of∑∞
n=11
n2+1on the basis of this comparison? Absolutely nothing!
Exercises: Discuss the convergence or divergence of each of the following series:
1.∞∑n=2
1
n lnn. [Compare with the integral
∫ ∞2
1
x lnxdx.]
2.∞∑n=1
en cos2 n
πn. [Compare with the geometric series
∞∑n=1
en
πn]
3.∞∑n=1
√n− 1
n2 + 1. [Compare with the p-series
∞∑n=1
1
n3/2]
4.∞∑n=1
n− 1
2n (n+ 1). [Compare with the geometric series
∞∑n=1
1
2n]
ENG1091 Mathematics for Engineering page 98
The Ratio Test
Recall that the infinite geometric series∑∞
n=1 arn−1 = a + ar + ar2 + ... converges for r < 1
and diverges for r > 1, where the common ratio r is the ratio of two consecutive terms of the
geometric sequence, i.e. r = an+1an
.
The ratio test for convergence of a series is a generalisation of this to other types of series.
Ratio Test: Suppose we have a series∞∑n=1
an where an > 0 for all n, and for which limn→∞
an+1
aneither exists or is infinite.
Let ρ = limn→∞
an+1
an.
• If ρ < 1 then∞∑n=1
an converges. (As a consequence we get limn→∞
an = 0.)
• If ρ > 1 then limn→∞
an =∞ and∞∑n=1
an diverges.
• If ρ = 1, then the ratio test fails as the series may converge, or diverge to ∞.
Notice that this test could also be used to test for convergence of a geometric series since in this
case limn→∞an+1an
= an+1an
= r, a constant.
Examples
1.∞∑n=1
1
n2
(ρ = 1 and therefore ratio test fails, but we know this series converges by earlier tests)
2.∞∑n=1
2n
n!
(ρ = 0, series converges by ratio test)
ENG1091 Mathematics for Engineering page 99
3.∞∑n=1
n100
2n
(ρ = 12 , series converges by ratio test)
4.∞∑n=1
n!
nn
(ρ = 1e , series converges by ratio test)
5. Use the ratio test to show the series∞∑n=1
ne−nconverges.
ENG1091 Mathematics for Engineering page 100
Absolute and Conditional convergence
All of the series in the previous section were series of positive terms. We can now remove
this restriction and allow arbitrary terms an. We can obtain a series of positive terms from an
arbitrary series by replacing all the terms with their absolute values.
Definition: The series∞∑n=1
an is said to be absolutely convergent if the series∞∑n=1
|an| con-
verges.
Absolute convergence Theorem: If a series converges absolutely then the series converges.
Thus the tests for series of positive terms can be used to determine the convergence of any series
converges by it showing converges absolutely.
Example: Show the series∞∑n=1
(−1)n
n2converges absolutely.
However the absolute convergence test (if we call it that) is a suffi cient condition for convergence,
but it is not a necessary condition. Many series may fail to be absolutely convergent and yet are
convergent just the same. We call such series conditionally convergent.
Example: The series∞∑n=1
(−1)n
ndoes not converge absolutely because if we replace all the terms
by their absolute values we get the divergent harmonic series.∞∑n=1
1
n.
However the alternating harmonic series∞∑n=1
(−1)n
nconverges (conditionally) as we will show.
We cannot use any of the tests previously discussed to show that the series∑∞
n=1(−1)n
n converges
as these tests apply only to series of positive terms. Generally speaking, to demonstrate conver-
gence where the convergence is not absolute is usually quite diffi cult. We will discuss but one of
ENG1091 Mathematics for Engineering page 101
many tests that do the job; this test is very easily applied but is quite restrictive as it can only
be used on special types of series.
The Alternating series test. Suppose we have a series of the form∞∑n=1
(−1)n an where the
sequence an satisfies:
(i) an ≥ 0, for all n
(ii) limn→∞ an = 0 and
(iii) an+1 ≤ an for all n.
Then the series∞∑n=1
(−1)n an converges.
Example: The series∞∑n=1
(−1)n
n.
(i) The series is of the required form with an = 1n . Clearly an > 0 for all n.
(ii) limn→∞1n = 0,
(iii) an − an+1 = 1n −
1n+1 = 1
n(n+1) > 0 for all n and hence an+1 ≤ an.
The three parts of the alternating series test are satisfied and we deduce that∞∑n=1
(−1)n
nconverges.
Example: The series∞∑n=2
cosnπ
loge n.
(i) Since cosnπ = (−1)n the series is of the required form with an = 1loge n
. Since loge n > 0
for all n ≥ 2, we have an > 0.
(ii) Also, limn→∞1
loge n= 0,
(iii) To show an − an+1 = 1loge n
− 1loge(n+1) > 0 for all n, is a little more awkward than that
for the previous example and one way of doing this is to show the function 1/ loge (x) is
decreasing for all x ≥ 2. This is easy using calculus:
The function 1/ loge (x) has derivative:
d
dx(loge (x))−1 = −1 (loge (x))−2 × d
dxloge (x)
= − 1
x (loge (x))2
This is clearly negative, and hence 1/ loge (x) is a decreasing function. Thus1
loge n− 1
loge(n+1) > 0 for all n ≥ 2.
All three parts of the alternating series test are satisfied and we deduce that∞∑n=2
cosnπ
loge nconverges.
ENG1091 Mathematics for Engineering page 102
The alternating series test is quite restrictive as it cannot be used to show the conditional
convergence of series whose terms do not strictly alternate in sign.
For example, the series∑∞
n=1sinnn is also convergent conditionally, but its terms do not strictly al-
ternate in sign. A more general test for conditional convergence (and which works for∑∞
n=1sinnn )
is Dirichlet’s test but will not be examined in this course.
ENG1091 Mathematics for Engineering page 103
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Sequences and Series
Lecture 21 Taylor’s theoremText Reference: §9.4.1-9.9.4.2
The pragmatic reason for spending all this time on sequences and series was to get to Taylor
series. The idea of the Taylor series is to approximate a function with a power series. This series
can then be used to find values of the original function in an effi cient manor. Calculators and
computers regularly use Taylor series expansions for more sophisticated functions.
To begin with, let’s construct an approximation to the function f(x) at the point a, given the
value of the function at the point and it’s slope. If we don’t really know anything about the
shape of the function, then we will stick with the basic approximation of a straight line.
f (x) ≈ f (a) + (x− a) f ′ (a)
If we are also given the second derivative evaluated at the point x = a, then we have an extra
constraint. Instead of a straight line, we can approximate f(x) with a parabola.
f (x) ≈ f (a) + (x− a) f ′ (a) +(x− a)2
2!f ′′ (a)
Example: Given f (2) = −1, f ′ (2) = 0, f ′′ (2) = −1, find the 2nd order polynomial approxima-
tion to f(x) about x = 2.
The Taylor polynomial of degree 2 is
f (x) ≈ f (a) + (x− a) f ′ (a) +(x− a)2
2!f ′′ (a)
and with a = 2 becomes
f (x) ≈ f (2) + (x− 2) f ′ (2) +(x− 2)2
2!f ′′ (2)
= −1− 1
2(x− 2)2 .
This can readily be extended to higher order polynomials.
Example: Given, f (0) = 2, f ′ (0) = −1, f ′′ (0) = 3 and f ′′′ (0) = 1 find the 3rd order polynomial
approximation to f(x) about x = 0.
Now
f (x) ≈ f (a) + (x− a) f ′ (a) +(x− a)2
2!f ′′ (a) +
(x− a)3
3!f ′′′ (a)
with a = 0 this becomes:
f (x) ≈ f (0) + xf ′ (0) +(x)2
2!f ′′ (0) +
(x)3
3!f ′′′ (0)
= 2− x+3
2x2 +
1
6x3
ENG1091 Mathematics for Engineering page 104
Taylor polynomials are approximations of the function f.
If we replace the notion of polynomial with infinite series the ‘approximate equals’sign can be
replaced by equality [under appropriate conditions]:
f (x) = f (a) + (x− a) f ′ (a) +(x− a)2
2!f ′′ (a) + · · ·+ (x− a)n
n!f (n) (a) + ... (21.1)
This expression is known as the Taylor series of f.
[One of the conditions for equality in eq. (2.1) is that the Taylor series converges. The Taylor
series of some functions converge for all x, while others (typically) converge only on some interval
of the real line. Equality can only apply for those x for which the series converges.]
Example: Find the Taylor series for f(x) = ex about x = 1.
f (x) = ex f (1) = e
f ′ (x) = ex f ′ (1) = e
f ′′ (x) = ex f ′′ (1) = e
f ′′′ (x) = ex f ′′′ (1) = e
f (4) (x) = ex f (4) (1) = e
· · · · · ·Therefore the Taylor series of f (x) = ex is
f (a) + (x− a) f ′ (a) +(x− a)2
2!f ′′ (a) + · · ·+ (x− a)n
n!f (n) (a) + · · ·
= e+ e (x− 1) +(e
2
)(x− 1)2 +
( e3!
)(x− 1)3 + ...
In the instance when the expansion is about the point x = 0, the Taylor series is then
called a Maclaurin series.
f (x) = f (0) + xf ′ (0) +x2
2!f ′′ (0) + · · ·+ xn
n!f (n) (0) + · · ·
Example: Find the Maclaurin series for f(x) = ln (1 + x) about x = 0.
f (x) = ln (1 + x) f (0) = 0
f ′ (x) =1
1 + xf ′ (0) = 1
f ′′ (x) =−1
(1 + x)2 f ′′ (0) = −1
f ′′′ (x) =2
(1 + x)3 f ′′′ (0) = 2
f ′′′ (x) = − 2 · 3(1 + x)4 = − 3!
(1 + x)4 f (4) (0) = −3!
· · · · · ·Therefore the Maclaurin series of f (x) = ln (1 + x) is
f (0) + (x) f ′ (0) +(x)2
2!f ′′ (0) + · · ·+ (x)n
n!f (n) (0) + · · ·
= 0 + 1 (x)− x2
2+
2! · x3
3!− 3! · x4
4!+ ...
= x− x2
2+x3
3− x4
4+ ...
ENG1091 Mathematics for Engineering page 105
Example: Find the Maclaurin series for f(x) = cos (x) about x = 0.
f (x) = cosx f (0) = 1
f ′ (x) = − sinx f ′ (0) = 0
f ′′ (x) = − cosx f ′′ (0) = −1
f ′′′ (x) = sinx f ′′′ (0) = 0
f ′′′ (x) = cosx f (4) (0) = 1
· · · · · ·Therefore the Maclaurin series of f (x) = cosx is
f (0) + (x) f ′ (0) +(x)2
2!f ′′ (0) + · · ·+ (x)n
n!f (n) (0) + · · ·
= 1− 0 (x)− x2
2+ 0
x3
3!+x4
4!+ ...
= 1− x2
2!+x4
4!− x6
6!+ ...
It can be rather tedious finding Taylor or Maclaurin series from scratch each time. Using a known
Taylor series it is possible to find the Taylor series of other related functions by substitution, or
by integrating or differentiating term by term.
Example: Find the Maclaurin series expansion to f(x) =1
1 + x, given the expansion of f(x) =
ln (1 + x) from the earlier example.
We differentiate the Maclaurin series for f(x) = ln (1 + x)
The Maclaurin series for f(x) =1
1 + xis then
(x− x2
2+x3
3− x4
4+ ...
)′= 1− x+ x2 − x3 + ...
(This is an infinite geometric series with common ratio r = −x.)
ENG1091 Mathematics for Engineering page 106
Example: Find the Maclaurin series expansion to f(x) =
∫coshxdx.
First we find the Maclaurin series expansion to coshx :
f (x) = coshx f (0) = 1
f ′ (x) = sinhx f ′ (0) = 0
f ′′ (x) = coshx f ′′ (0) = 1
f ′′′ (x) = sinhx f ′′′ (0) = 0
f ′′′ (x) = coshx f (4) (0) = 1
· · · · · ·Giving
coshx = 1 +x2
2!+x4
4!+x6
6!+ ...
Note that the Maclaurin series for cosx can also be obtained by the identity cosx = cosh (ix) .
Now integrating term by term we obtain∫coshxdx =
∫1 +
x2
2!+x4
4!+x6
6!+ ...dx
= x+x3
3!+x5
5!+x7
7!+ ...+ C
with C = 0 we obtain the Maclaurin series expansion to sinhx = x+x3
3!+x5
5!+x7
7!+ ....
(This may be obtained directly of course from the Taylor series formula.)
ENG1091 Mathematics for Engineering page 107
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Multivariable Calculus
Lecture 22&23 · partial derivatives · directional derivatives · chain ruleText Reference: §9.6.1-9.6.5
1. Functions of several variables
Throughout our discussions on differentiation and integration we have examined functions with
only one independent variable. Yet we can think of any number of examples in engineering in
which a quantity is defined by two or more independent variables. The volume of a cylinder is a
function of the height of the cylinder and the radius of its base:
V = πr2h
The density of ocean water is a function of its temperature and salinity: density:
ρ = ρ (T, σ)
For the moment let us focus on functions with two independent variables, x and y. For further
convenience, we can assume that x and y are our familiar Cartesian coordinates. Given an
arbitrary function of our two independent variable, z = f(x, y), it is possible to view the variable
z as the height above the x-y plane. This function of two variables is thus a three-dimensional
surface above the x-y plane, which, unfortunately, is very diffi cult to graph on a piece of paper. In
graphing f(x, y), it is common to draw lines of constant height z (i.e. contours). Such diagrams
are completely analogous to contour maps used in bushwalking and mountaineering.
It is worth the time to graph a few simple functions to help with future lectures.
Consider the contour maps/surface plots for the functions below:
z1 =√
16− x2 − y2 z2 = 16− x2 − y2
ENG1091 Mathematics for Engineering page 108
z3 = 2x− y3
and
z4 = cos (x) cos (y) (not examinable)
It is worth noting that the function f (x, y) is often called a scalar field in vector calculus. Also,
we can readily extend this material to three dimensions and beyond; only it isn’t simple to draw
such functions on paper.
ENG1091 Mathematics for Engineering page 109
2. Partial differentiation: The aim of this section is to extend some of the principles of
basic calculus to functions with multiple independent variables. We begin with differentiation.
Thinking back to one independent variable, if f is a function of a single variable, x say, then we
define the derivative of f with respect to x as
df
dx= lim
∆x→0
f (x+ ∆x)− f (x)
∆x
Now if f is a function of two independent variables, x and y, then we can define the derivative
of f with respect to each of these variables as follows
∂f
∂x= lim
∆x→0
(∆f
∆x
)y=const
= lim∆x→0
(f(x+ ∆x, y)− f(x, y)
∆x
)y=const
(1)
In this operation we treat y as a constant. It is basically ignored. Note the special notation used
for the partial derivative. Note that∂f
∂xand
df
dx
have different meanings in multivariable calculus, so we need to be careful. The partial derivative
with respect to y is similarly defined as
∂f
∂y= lim
∆y→0
(∆f
∆y
)x=const
= lim∆y→0
(f(x, y + ∆y)− f(x, y)
∆y
)x=const
(2)
where x is held constant throughout.
The basic concepts of differentiation (e.g. the product rule,quotient rule, associative and distrib-
utive properties) extend across to higher dimensions as expected.
ENG1091 Mathematics for Engineering page 110
Returning to our visualisation of z = f (x, y) as representing a height or a 3-D surface, then the
partial derivative∂z
∂x
represents the change in height in the x direction or the slope of the surface in the x direction.
Example: Find both partial derivatives of
f (x, y) = sin (xy) + x2 + x/y
∂f
∂x= cos (xy) · y + 2x+ 1/y
∂f
∂y= cos (xy) · x− xy−2
= y cos (xy) + 2x+ 1/y = x cos (xy)− xy−2
Example: Given
f(x, y) = sin(xy) + x2 + x/y,
find both ∂f∂x and
∂f∂y at the point (π, 1) .
∂f
∂x|(π,1)= cos (π) + 2π + 1
∂f
∂y|(π,1)= π cos (π)− π
= 2π = −2π
As the text notes, partial differentiation can readily be extending to instances of more than two
independent variables.
Example (from text): Given
f (x, y, z) = xyz2 + 3xy − z
find∂f
∂x,∂f
∂yand
∂f
∂z.
∂f
∂x= yz2 + 3y
∂f
∂y= xz2 + 3x
∂f
∂z= 2xyz − 1
Suppose we want to evaluate the partial derivative at a specified point. That is, we want to
quantify the slope given a choice of x and y. Just as in one dimension, we must take the derivative
first before plugging in the variable. Note that since y is held constant in calculating∂f∂x ,, it doesn’t
really matter when we substitute in the given value of y.
ENG1091 Mathematics for Engineering page 111
3. The gradient and directional derivatives
Staying in Cartesian coordinates, it is natural to extend the partial derivatives to include a
direction. That is, we can turn them into a vector. Assuming that ∂f∂x points in the direction of
x and ∂f∂y points in the direction of y, then we call define the gradient of the field f(x, y) as
∇f(x, y) =∂f
∂xi+
∂f
∂yj (3)
where i and j are the unit vectors in the direction of x and y, respectively. The gradient of the
field f is often abbreviated as ‘gradf’and given the notation ∇f .
Example: Given the scalar field
f(x, y) =√
16− x2 − y2,
calculate ∇f . Sketch these vectors on the contour map of f(x, y).
Solution:
∇f(x, y) =∂f
∂xi+
∂f
∂yj =
1
2
(16− x2 − y2
)−1/2 · −2xi+1
2
(16− x2 − y2
)−1/2 · −2yj
=−1√
16− x2 − y2(xi+ yj)
Note that the gradient vector is always perpendicular to a level curve at a given point and
points towards the direction of increasing function value.
The previous example revealed a noteworthy point about the gradient. At all points the vectors
of the gradient are at right angles to the contour lines. In this two-dimensional, Cartesian
coordinate picture, the gradient points us in the direction of greatest change of our scalar field
f (x, y). Going back to our analogy of f (x, y) representing the contours of height on a map, the
gradient of f (x, y) gives us a vector that tells us the direction of the maximum slope and its
magnitude.
Example: Given the scalar field f(x, y) = xy, draw the contour field, calculate ∇f and sketchthe gradient vectors over the contour lines.
∇f = ∂f∂x i+ ∂f
∂y j = yi+ xj
5 4 3 2 1 1 2 3 4 5
5
4
3
2
1
1
2
3
4
5
x
y
ENG1091 Mathematics for Engineering page 112
Please note that the gradient can readily be extended to higher dimensions.
Example: Given the scalar field
f(x, y, z) = z + (x2 + y2)
calculate ∇f . Sketch a level surface f(x, y, z) = k for some suitable value of k and plot ∇f ata point on this surface. (The graphic illustrates the case k = 1, i.e. the surface z + (x2 + y2).)
42
4
4 2
3
2
z0
y
00
1
2
3
x2
1
42
Example: Given the scalar field f(x, y, z) = xyz2 + 3xy − z calculate∇f.
∇f =∂f
∂xi+
∂f
∂yj+
∂f
∂zk =
(yz2 + 3y
)i+(xz2 + 3x
)j+ (2xyz − 1)k
Directional derivative
We’ve seen that ∇f is a vector that tells us the direction and magnitude of the rate of changeof the scalar field f (x, y). We can also use ∇f to find the rate of change of the scalar field f (x,y) in some arbitrary direction. This is known as the directional derivative. Specifically, if we are
given a scalar field f (x , y) and a specified orientation to follow, say
v = vxi+ vyj
the unit vector having same direction as v is v =v
‖v‖ where ‖v‖ =√v2x + v2
y ;
then the directional derivative Dvf is defined as
Dvf = ∇f ·(v
‖v‖
)(4)
Example: Given the scalar field f(x, y) = xy, find the directional derivative in the direction of
v = 3i+ 4j
at the points ((1, 1), (1,−1) , and (−4, 3).
v = 3i+ 4j so that ‖v‖ =√
(3)2 + (4)2 = 5 and hence v = 35 i+ 4
5 j.
ENG1091 Mathematics for Engineering page 113
∇f = ∂f∂x i+ ∂f
∂y j = yi+ xj
Hence Dvf (x, y) = ∇f · v = 35y + 4
5x.
Dvf (1, 1) = 75 , Dvf (1,−1) = 1
5 , Dvf (−4, 3) = −75 .
The definition of the directional derivative presented here is different, in notation, than that
presented in the text. One would find that the definitions are identical in practice since:
v
‖v‖ =vxi+ vyj√v2x + v2
y
=
vx√v2x + v2
y
i+
vy√v2x + v2
y
j = cos(α)i+ sin(α)j (5)
where α is the angle that the vector v makes with the x axis. Using the dot product, eq.(4)
becomes:
∇f ·(v
‖v‖
)=
(∂f
∂xi+
∂f
∂yj
)· (cos(α)i+ sin(α)j)
=∂f
∂xcosα+
∂f
∂ysinα (6)
Equation (6) is the definition of directional derivative (of functions of two variables) given in the
text.
The vector definition presented in these notes is, in general, far more widely used in mathematics
and engineering as it can readily be extended to other coordinate systems and higher dimensions.
4. The chain rule
In one dimension the chain rule was employed when f (x ) and x (t). In such a case,
df
dt=df
dx× dx
dt.
When moving to multiple dimensions, the basic concept is extended.
Suppose that we have z = f (x, y) and that x (s, t) and y(s, t). Here we have f as a function of
two variables, and each of these variables, in turn is a function of two variables. In this case we
may find an expression for the change in f with respect to s and t.
∂z
∂s=∂f
∂x
∂x
∂s+∂f
∂y
∂y
∂s
and∂z
∂t=∂f
∂x
∂x
∂t+∂f
∂y
∂y
∂t
As the text notes, a good example of this is when undertaking a coordinate transformation. If
a function is defined in Cartesian coordinates, and we wish to change over to polar coordinates
(r, θ) then we need to recall the relations
x = r cos θ, and y = r sin θ.
In calculating the partial derivatives, one can either completely change coordinate systems first,
and then compute the partial derivatives, or apply the chain rule.
Example: Given the function z = sin(xy) is defined in for a Cartesian coordinate system, find
the partial derivatives∂z
∂r, and
∂z
∂θ.
ENG1091 Mathematics for Engineering page 114
∂z
∂x= y cos (xy)
∂z
∂y= x cos (xy)
From x = r cos θ, and y = r sin θ we have:
∂x
∂r= cos θ
∂x
∂θ= −r sin θ
∂y
∂r= sin θ
∂y
∂θ= r cos θ
Now
∂z
∂r=
∂z
∂x
∂x
∂r+∂z
∂y
∂y
∂r
= y cos (xy) cos θ + x cos (xy) sin θ
= 2r cos θ sin θ cos (xy)
= r cos (xy) sin (2θ)
∂z
∂θ=
∂z
∂x
∂x
∂θ+∂z
∂y
∂y
∂θ
= y cos (xy) · −r sin θ + x cos (xy) r cos θ
= cos (xy)(r2 cos2 θ − r2 sin2 θ
)= r2 cos (xy) cos (2θ)
Suppose now we have z = f(x, y) and that x and y are functions of a single variable t. Here we
might think of x and y being our Cartesian coordinates again, but these values are functions of
the time t. (Thus x (t) and y(t) define a path of some particle as it moves in the x -y plane.)
We can then define a derivative of z with regards to t as follows:
dz
dt=∂f
∂x
dx
dt+∂f
∂y
dy
dt
ENG1091 Mathematics for Engineering page 115
Example: Given z (x, y) = x2y − y lnx− 2x with the further relations x (t) = t2 and
y (t) = cos (t).
Finddz
dtand evaluate it at the time t = π.
dz
dt=∂z
∂x
dx
dt+∂z
∂y
dy
dt
∂
∂x
(x2y − y lnx− 2x
)= 2xy − y
x− 2
∂
∂y
(x2y − y lnx− 2x
)= x2 − lnx
dx
dt= 2t
= 2π when t = π
dy
dt= − sin t
= 0 when t = π
dz
dt=
∂z
∂x
dx
dt+∂z
∂y
dy
dt
=(
2xy − y
x− 2)
2π + 0
=
(−2π2 +
1
π2− 2
)2π substituting x = π2 and y = −1 when t = π
= −4π3 +2
π− 4π
ENG1091 Mathematics for Engineering page 116
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Multivariable Calculus
Lecture 24 higher derivatives · total differential · exact differentialText Reference: §9.6.7
1. Higher order derivatives
We can extend the partial differentials to higher order derivatives. Given the function f (x, y),
we could create four second order derivatives.
∂
∂x
(∂f
∂x
)=∂2f
∂x2= fxx (1)
∂
∂y
(∂f
∂y
)=∂2f
∂y2= fyy (2)
∂
∂y
(∂f
∂x
)=
∂2f
∂y∂x=
∂
∂y(fx) = fxy (3)
∂
∂x
(∂f
∂y
)=
∂2f
∂x∂y=
∂
∂x(fy) = fyx (4)
Please note the order of the notation in these equations. The partial derivative within the
brackets is the first operation, so in equation (3) the partial derivative with respect to x is first
undertaken, and then with respect to y. Also note that there are functions when equations (3)
and (4) are NOT equal. However, for the purposes of this course, we will neglect these special
cases and assume that order of differentiation can readily be swapped. I.e., we will assume that
∂
∂y
(∂f
∂x
)=
∂
∂x
(∂f
∂y
).
Example: Find given
f (x, y) = x3y3 + sin (y)
find fxx, fyy, fxy and fyx.
fx = ∂∂x
(x3y3 + sin (y)
)fy = ∂
∂y
(x3y3 + sin (y)
)= 3x2y3 = 3x3y2 + cos y
fxx = ∂∂x
(3x2y3
)= 6xy3
fxy = ∂∂y
(3x2y3
)= 9x2y2
fyx = ∂∂x
(3x3y2 + cos y
)= 9x2y2
fyy = ∂∂y
(3x3y2 + cos y
)= 6x3y − sin y
ENG1091 Mathematics for Engineering page 117
Extending this work to higher order derivatives, and/or functions of more than two independent
variables is straightforward.
2. The total differential and small errors
Suppose we are given a function z = f(x, y), and we wish to appreciate the change in z given a
small change in x and y.
∆z = f(x+ ∆x, y + ∆y)− f(x, y)
This can readily be manipulated to
∆z = f(x+ ∆x, y + ∆y)− f(x+ ∆x, y) + f(x+ ∆x, y)− f(x, y) ≈ ∂f
∂x∆x+
∂f
∂y∆y (5)
If we turn the change of independent variables into a vector
dr = (∆xi+ ∆yj)
then the total differential can be written succinctly as ∇f · dr.
In the limiting case of ∆x→ 0 and ∆y → 0 we can define total differential as
dz =∂f
∂xdx+
∂f
dydy (6)
with dz ≈ ∆z.
Example (from text): Find the total differential for the function z(x, y) = x2y3.
dz =∂f
∂xdx+
∂f
dydy
=
The text notes that the concept of the total differential is commonly used in setting error esti-
mates given some uncertainty in the independent variables. The relative error is defined as
∣∣∣∣duu∣∣∣∣ ≈ ∣∣∣∣∆uu
∣∣∣∣
ENG1091 Mathematics for Engineering page 118
Example (from text): Find the relative error of the volume of a circular cylinder given the
radius r = 3± 0.01 and the height h = 5± 0.005.
3 Exact differentials
In the previous topic, we started with a well-defined function z = f(x, y) and developed the
total differential in equation (24.6). The idea now is to start with something in the form of the
right-hand side of equation (24.6) and see if it is, indeed, an exact differential. Assume we have
P (x, y)dx+Q(x, y)dy. (7)
Does there exist a function z = f(x, y) such that
dz = P (x, y)dx+Q(x, y)dy?
For this to hold we need
P (x, y) =∂f
∂xand Q(x, y) =
∂f
∂y(8)
Assuming that f(x, y) has second order derivatives we have
∂P
∂y=
∂
∂y
(∂f
∂x
)=
∂2f
∂y∂xand
∂Q
∂x=
∂
∂x
(∂f
∂y
)=
∂2f
∂x∂y
so if the mixed partial derivatives are equal we have
∂P
∂y=∂Q
∂x.
Provided this equation is satisfied our original expression (24.7) may be considered an exact
differential. Note that this test does not tell us how to recover the original function f(x, y). This
must be done through integrating both parts of (24.8) to find a common function.
Example: Verify the expression
(2x+ 2y)dx+ (2x+ 1/y)dy
is an exact differential and recover the function defined by it.
ENG1091 Mathematics for Engineering page 119
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Multivariable Calculus
Lecture 25 Taylor’s theorem in two dimensions · OptimisationText Reference: §9.7
4. Taylor’s theorem in two dimensions
Taylor series can readily be extended to functions of two (or more variables). For a function of
two independent variables, f (x, y), we can make an extension around the point (a, b) as follows.
f(a+ h, b+ k) = f(a, b) +1
1!
(h∂
∂x+ k
∂
∂y
)f(x, y)(a,b) +
1
2!
(h∂
∂x+ k
∂
∂y
)2f(x, y)|(a,b) +
. . .+1
n!
(h∂
∂x+ k
∂
∂y
)nf(x, y)|(a,b) + ... (25.1)
Some new notation has been introduced here(h∂
∂x+ k
∂
∂y
)rf(x, y)|(a,b) ≡
hr
∂r
∂xr+
(r
1
)hr−1k
∂r
∂xr−1∂y+ . . .+
(r
s
)hr−sks
∂r
∂xr−s∂ys+ . . .
+
(r
r − 1
)hkr−1
∂r
∂x∂yr−1+ kr
∂r
∂yr
f(x, y)|(a,b)
For example:(h∂
∂x+ k
∂
∂y
)3f(x, y)|(a,b) = h3
∂3f
∂x3|(a,b) +3h2k
∂3f
∂y∂x3|(a,b) +3hk2
∂3f
∂y2∂x|(a,b) +k3
∂3f
∂y3|(a,b)
= (x− a)3 fxxx (a, b) + 3 (x− a)2 (y − b) fyxx (a, b) + 3 (x− a) (y − b)2 fyyx (a, b) + (y − b)3 fyyy (a, b)
Here we have assumed that all of the nth order partial derivatives exist and are continuous in
some domain close to the point (a, b).
Example: Up to second order, find the Taylor series expansion to the function ln(xy) about the
point (1, 1) .
ENG1091 Mathematics for Engineering page 120
Please note that the first order Taylor approximation to f (x, y) is
T (x, y) = f(a, b) +1
1!
((x− a)
∂
∂x+ (y − b) ∂
∂y
)f(x, y)|(a,b)
= f(a, b) +1
1!((x− a) fx (a, b) + (y − b) fy (a, b))
The equation
z = f(a, b) + ((x− a) fx (a, b) + (y − b) fy (a, b))
is the equation of the tangent plane in 3-D to the surface z = f (x, y) at the point (a, b, f((a, b)) .
This is analogous to earlier work with functions of one independent variable, f(x), in which the
first order Taylor series approximation returned the tangent line.
5. Optimisation of unconstrained functions
We’ve learned that the local extrema of a continuous function of one independent variable f(x)
occur at critical points where the derivative f ′(x) is equal to zero. If the derivative is equal to
zero, then we can have a local minimum, maximum or point of inflection. We then used the
second derivative to, hopefully, tell help us classify the extrema. We wish to extend this work to
a function of two independent variables, f(x, y).
Using the Taylor series expansion just presented, we see that in the neighbourhood of the point
(a, b) the change in f (x, y) is
∆f = f(a+h, b+k)−f(a, b) =
(h∂
∂x+ k
∂
∂y
)f(x, y)|(a,b)+
1
2!
(h∂
∂x+ k
∂
∂y
)2f(x, y)|(a,b)+ . . .
∆f must be either strictly negative or positive for an extrema. Notice that the first term on the
right-hand side depends linearly on h and k. Since these values can be either positive or negative,
the first partial derivatives∂f
∂xand
∂f
∂y
must be zero for ∆f to be strictly positive or negative. This is a necessary condition, which
then leaves our difference depending on the second order partial derivatives. Since we are only
interested in very small values of h and k, we can ignore the higher order partial derivatives, as
these will involve terms like h3, which is much less than h2. Ultimately we require
∆f ≈ 1
2
(h2fxx (a, b) + 2hkfxy (a, b) + k2fyy (a, b)
)(25.2)
to be either positive or negative. This expression can be manipulated as follows
fxx (a, b) ∆f ≈ 1
2
(h2 (fxx (a, b))2 + 2hkfxx (a, b) fxy (a, b) + k2fxx (a, b) fyy (a, b)
)ENG1091 Mathematics for Engineering page 121
Complete the square on the first two terms:
=1
2
[((hfxx (a, b)) + kfxy (a, b))2 − k2 (fxy (a, b))2 + k2fxx (a, b) fyy (a, b)
]=
1
2
[((hfxx (a, b)) + kfxy (a, b))2 + k2
(fxx (a, b) fyy (a, b)− (fxy (a, b))2
)]First, in order for ∆f to be strictly positive in the neighbourhood of a stationary point we require
both∂2f
∂x2and
∂2f
∂x2∂2f
∂y2−(∂2f
∂x∂y
)2(25.3)
be positive. This is thus a requirement for a local minimum.
The results are summarised in the folowing theorem:
Let (a, b) be an interior point of the domain for the function f and suppose that the first and
second partial derivatives of f exist and are continuous on some circular disk with (a, b) as its
centre and contained in the domain of f. Assume that (a, b) is a critical point of f, so that
fx(a, b) = fy(a, b) = 0. Define
∆ =
∣∣∣∣∣ fxx(a, b) fxy(a, b)
fyx(a, b) fyy(a, b)
∣∣∣∣∣ = fxx(a, b)fyy(a, b)− (fxy(a, b))
Then:
1. If ∆ > 0 and fxx(a, b) < 0 or fyy(a, b) < 0, then (a, b) is a local maximum.
2. If ∆ > 0 and fxx(a, b) > 0 or fyy(a, b) > 0, then (a, b) is a local minimum.
3. If ∆ < 0, then (a, b) is a saddle point.
4. If ∆ = 0, then this test is inconclusive.
A saddle point, as the name suggests, is a point on the domain of f (x, y) where a minimum is
approached in one direction, but a maximum is approached from a different direction.
Example 1: Verify that the point (2,−1) is a local maximum for the function f(x, y) = 1 −(x− 2)2 − (y + 1)2
Solution: fx = −2 (x− 2) · 1 and fy = − (y + 1) · 1 and these are zero when x = 2 and when
y = −1. Hence there is a single stationary point of (2,−1) .
To determine the nature of the stationary point we evaluate
D (x, y) =
∣∣∣∣∣ fxx fxy
fyx fyy
∣∣∣∣∣ =
∣∣∣∣∣ −2 0
0 −1
∣∣∣∣∣ = 2 > 0
so (2,−1) is either a local minimum or a local maximum.
Since fxx = −2 < 0 we have that (2,−1) is a local maximum point and that the local maximum
value of f is f(2,−1) = 1.
ENG1091 Mathematics for Engineering page 122
Example 2: Find the critical points of the function f(x, y) = x2 − 5xy + 3y2 + 13y. Determine
the nature of each stationary point.
Solution:
fx = 2x− 5y and fy = −5x+ 6y + 13 and these are zero when
2x− 5y = 0
−5x+ 6y = −13.
Using Cramer’s rule gives
x =
∣∣∣∣∣ 0 −5
−13 6
∣∣∣∣∣∣∣∣∣∣ 2 −5
−5 6
∣∣∣∣∣= 5 and y =
∣∣∣∣∣ 2 0
−5 −13
∣∣∣∣∣∣∣∣∣∣ 2 −5
−5 6
∣∣∣∣∣= 2
So there is one stationary point: (5, 2) .
Its nature:
D (x, y) =
∣∣∣∣∣ fxx fxy
fyx fyy
∣∣∣∣∣ =
∣∣∣∣∣ 2 −5
−5 6
∣∣∣∣∣ = 12− 25 < 0
so (5, 2) is a saddle point.
Example 3: Show that the function f(x, y) = x3−3xy+y3 has two stationary (critical) points.
Find the second order partial derivatives of f and evaluate the determinant
D (x, y) =
∣∣∣∣∣ fxx fxy
fyx fyy
∣∣∣∣∣at each stationary point. Hence determine the nature of each stationary point.
If the function f has local maximum or minimum values find these.
Solution:
fx = 3x2 − 3y and fx = 0 when x2 = y and fy = 3y2 − 3x and fy = 0 when y2 = x.
Thus the only critical points occur when x4 = x, i.e. when
x4 − x = x (x− 1)(x2 + x+ 1
)= 0, namely at x = 0 and x = 1.
Hence f has two critical points, (0, 0) and (1, 1) .
The nature of the two critical points:
fxx = 6x, fxy = fyx = −3 and fyy = 6y.
Hence D (x, y) =
∣∣∣∣∣ 6x −3
−3 6y
∣∣∣∣∣ = 36xy − 9.
Now D (0, 0) < 0 indicating (0, 0) is a saddle point of f.
On the other hand D (1, 1) > 0 and fxx (1, 1) > 0 indicating that f has a local minimum at
(1, 1) , and its minimum value is f (1, 1) = −1.
ENG1091 Mathematics for Engineering page 123
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Multivariable Calculus
Lecture 26 Taylor’s theorem in two dimensions · OptimisationText Reference: §9.7
6. Optimisation of constrained functions (the technique of Lagrange multipliers)
In this section we wish to explore the optimisation of a function of several independent variables,
given a constraint. In two dimensions this is often straightforward. For example, suppose we
wanted to find the maximum of the function
f(x, y) = 1− (x− 2)2 − (y + 1)2
subject to the constraint g(x, y) = x/y = 3. Visually, this could be done by drawing the contour
map of f (x, y) and then drawing the hyperbola x = 3y over the top of the contours, on the
say sheet of paper. The maximum contour value along the hyperbola is the solution we want
to find. Mathematically, we could attack this problem by simple substitution. The constraint is
equivalent to saying that x = 3y so the original function becomes
f(y) = 1− ((3y)− 2)2 − (y + 1)2 = −10y2 + 10y − 4
The extreme for this can readily be found by solving
f ′(y) = −20y + 10 = 0
solution: y = 1/2
substituting into x = 3y we get x = 3/2
Thus the point (3/2, 1/2) should be the maximum (or minimum) point to the function
f(x, y) = 1− (x− 2)2 − (y + 1)2
subject to the constraint g(x, y) = x/y = 3. We get the value f(3/2, 1/2) = −1.5 as a solution
to the original problem.
Example: Find the extrema for the function
f(x, y) = 12xy − 4x2y − 3xy2
given the constraint x+ 2y = 2.
12xy − 4x2y − 3xy2 = 12 (2− 2y) y − 4 (2− 2y)2 y − 3 (2− 2y) y2
= 8y + 2y2 − 10y3
d
dy
(8y + 2y2 − 10y3
)= 8 + 4y − 30y2
= 0 when y =1
15+
1
15
√61, or y =
1
15− 1
15
√61
ENG1091 Mathematics for Engineering page 124
substituting into x = 2− 2y
= 2− 2
(1
15+
1
15
√61
)=
28
15− 2
15
√61
or x = 2− 2
(1
15− 1
15
√61
)=
28
15+
2
15
√61
Suppose, now that we are working with functions of three independent variables. Namely suppose
we wish to find the extrema of the function f (x, y, z ) subject to the constraint
g(x, y, z) = 0. (26.1)
Sometimes we can manipulate the constraint and substitute it into the original function and
lower the number of independent variables.
For example, consider the function
f(x, y, z) = x2 + xy + xz + y2z2,
and the constraint
g(x, y, z) = 2x2 + 3y − z = 2,
then we could define z = 2x2+ 3y− 2 and substitute this into f to leave it with two independent
variables,
f(x, y) = x3 + xy + x(2x2 + 3y − 2) + y2(2x2 + 3y − 2)2
We are then back to optimising a function of two independent variables and we could approach
the problem as was done in the previous section.
Please note however that this can be very tedious. We can actually manipulate this problem
to present it in a manner that is usually easier to solve. Consider the constraint (26.1). This,
in general, represents a surface in 3-D space. We will define small motions along this surface
as ds = (dx, dy, dz). Without any loss of generality, we can consider this to be a vector in the
3-D Cartesian space. Since g(x, y, z) is constrained to be zero, we know that motion along this
surface won’t change the value of g(x, y, z) :
dg = ∇g · ds =∂g
∂xdx+
∂g
∂ydy +
∂g
∂zdz = 0
Now assume that we are at the point that conditional stationary point that actually both satisfies
the constraint and optimises f(x, y, z) under this constraint. Then small motions along the
surface will also require
df = ∇f · ds =∂f
∂xdx+
∂f
∂ydy +
∂f
∂zdz = 0
Using our basic understanding of the vector dot product we know that both ∇f and ∇g isperpendicular to ds. Thus they may be expressed as a linear combination of one another.
∇f − λ∇g =
(∂f
∂x,∂f
∂y,∂f
∂z
)− λ
(∂g
∂x,∂g
∂y,∂g
∂z
)= (0, 0, 0) (26.2)
ENG1091 Mathematics for Engineering page 125
Here λ is basically another unknown variable. At this point in time, some students might be
asking what the advantage in all of this is. We have moved from our initial optimisation problem
with three unknowns (x, y and z ) to a system with four equations [(26.1) and the three of (26.2)]
and four unknowns (x, y, z and λ). Experience tells us that this new approach is often easier to
solve than the original problem. Please note that the variable λ is called the Lagrange multiplier
and the function
φ(x, y, z) = f(x, y, z)− λg(x, y, z)
is called the auxiliary function.
Example: Find the extrema of the function
f(x, y, z) = x2 + y2 + z2
subject to the constraint
g(z, y, z) = x2 + 2y2 − z2 − 1 = 0
Solution: The three Lagrange multiplier equations can be written:
∇f = (2x, 2y, 2z) = λ∇g = λ (2x, 4y,−2z)
The first equation 2x = λ2x gives λ = 1 or x = 0
If λ = 1 (x is arbitrary) then the second component gives 2y = 4y hence y = 0; and the third
component 2z = −2z gives z = 0.
Solving the constraint equation x2 + 2y2 − z2 − 1 = 0 with y = z = 0 gives x = ±1.
Using the equation 2y = λ4y we have λ = 1/2.
If λ = 1/2, then y can be arbitrary and equations 1 and 3 give x = z = 0. The constraint equation
x2 + 2y2 − z2 − 1 = 0 with x = z = 0 gives y = ±1/√
2.
Using the equation 2z = λ (−2z) we have λ = −1.
If λ = −1, then z can be arbitrary and equations 1 and 2 give x = y = 0. The constraint equation
x2 + 2y2 − z2 − 1 = 0 becomes −z2 = 1 which has no solution.
There are thus the 4 constrained extreme points (±1, 0, 0) with f (x, y, z) = 1 and(0,±1/
√2, 0)
with f (x, y, z) = 1/2.
ENG1091 Mathematics for Engineering page 126
Example: Find the extrema of the function f(x, y, z) = xyz subject to the constraint
g(x, y, z) = x2 + y2 + z2 = 1.
Solution: The three Lagrange multiplier equations can be written:
∇f = (yz, xz, xy) = λ∇g = λ (2x, 2y, 2z)
λ =yz
2x=xz
2y=xy
2z
y2 = x2; z2 = y2; x2 = z2
x2 + y2 + z2 = 1 so 3x2 = 1⇒ x = ± 1√3
we have y = ± 1√3
; z = ± 1√3
so eight points:
(± 1√
3,± 1√
3,± 1√
3
).
Example: Use the method of Lagrange multipliers to find the maximum possible volume of a
cone inscribed in a sphere of radius a.
Solution: Let the cone have height h and radius r.
The function to be maximised is V = 13πr
2h.
The fact that the cone is inscribed in the sphere leads to the constraint:
a2 = r2 + (h− a)2 = g (r, h) .
This time there are two Lagrange multiplier equations:
∇V =
(2
3πrh,
1
3πr2)
= λ∇g = λ (2r, 2 (h− a))
so λ =23πrh
2r=
13πr
2
2 (h− a)
hence2h
r=
r
h− a and hence 2h2 − 2ah+ h2 − 2ah+ a2 = a2
3h2 − 4ah = 0 and hence h (3h− 4a) = 0⇒ h =4a
3(or h = 0)
From r2 + (h− a)2 = a2 we get
r2 = a2 − (h− a)2 = a2 −(a
3
)2=
8
9a2
r =2√
2
3a
Vmax =1
3πr2h =
32π
81a3
ENG1091 Mathematics for Engineering page 127
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Ordinary Differential Equations
Lecture 27 · introduction · classificationText Reference: §10.1-10.5
1. Introduction and definition
The derivatives of y(x ) have been further expressed by the notation
dy
dx,d2y
dx2,d3y
dx3
(or more concisely as y′, y′′, y′′′) for the first, second and third order derivatives.
Equations (or physical relationships) involving derivatives are known as differential equations.
Examples:dy
dx= 5y + 2 (27.1)
y′′′ + y cosx = 0 (27.2)
d2s
dt2+ t
ds
dt+ t2s = −t (27.3)
xx2 + t2x = ln t (27.4)
Although these examples have no particular physical relevance, there are many simple examples
of relevant differential equations. In basic calculus, the exponential function was commonly
defined through the differential equation
dN
dt= λN
and was used to model ideal population growth.
2. Examples of Engineering Applications:
Ordinary Differential Equations (or ODEs) also have a number of basic engineering applications.
For example, Newtonian physics requires that the forces applied to it define the rate of change
of momentum of a body. For simple gravity
mdv
dt= −mg or
d2s
dt2= −g
ord2s
dt2= −g
where g defines gravity, s is height, v is velocity and t is time. If a drag is considered, then the
equation becomes
mdv
dt= −mg + bv2
ord2z
dt2= −g +
b
m
(dz
dt
)2where b is a constant. The dynamics of a spring can readily be modelled with an ODE. Here
the resistance force is not gravity or drag, but rather it is proportional to the displacement.
ENG1091 Mathematics for Engineering page 128
Consider a basic problem in thermodynamics with the heating (or cooling) of a body to room
temperature. The rate of change of the temperature of the body T b is proportional to the
temperature difference between the body and room temperature (Tb − Tr). Specifically the
governing equation isdTbdt
= −α(Tb − Tr)
Another classical example models an electrical circuit involving a resistor, an inductor and a
capacitor. If we define the inductance as L, the capacitance as C and the resistance as R, then
the current i(t) of the LCR circuit can be modeled as
Ld2i
dt2+R
di
dt+
1
Ci = 0
3. Classification of ODEs
The notation y(x ) has commonly been used to define y as a dependent function of the inde-
pendent variable x. It is common to use x or t as the dependent variable to signify position or
time.
Given a differential equation, if the dependent variable is a function of only one independent
variable, then the differential equation will be classified as an ordinary differential equation
or sometimes ODEs. All of the examples discussed so far have been of ordinary differential
equations.
In multivariable calculus the function y (also called the dependent variable) might be a function
of two or more independent variables. (For example y might be a function of the displacement
x and the time t, we write y(x, t)). The derivatives are partial derivatives:(∂y
∂xand
∂y
∂t
)Equations involving partial derivatives are logically referred to as partial differential equa-
tions (or PDEs) and will be covered in 2nd level engineering maths. PDEs are commonly used
to study fluid dynamics, heat flow and other engineering applications.
Differential equations will be further classified by their order, which is the degree of the highest
derivative that appears in the differential equation. Example 27.1 is a first order, ordinary
differential equation. Example 27.2 is a 3rd order ODE. Examples 27.3 and 27.4 are both 2nd
order ODEs.
Another important differential equation type are linear ODEs. We define linear differential
equations as those in which the dependent variable terms and their derivatives do not occur as
products, raised to powers (other than one) or in non-linear functions. Otherwise the differential
equation is said to be non-linear.
Examples 27.1, 27.2 and 27.3 are linear while example 27.4 is non-linear because of the xx2 term.
A very important classification of a differential equation is whether it is homogenous or not. A
homogeneous differential equation is one in which every term involves either the dependent
variable or derivatives of the dependent variable. Otherwise it is said to be non-homogeneous.
ENG1091 Mathematics for Engineering page 129
A non-homogeneous equation will have one or more terms that are either constant or that contain
the independent variable only. Equation 27.1 is non-homogeneous because of the ‘2’on the right
hand side. Such a term will be referred to in these notes as the ‘inhomogeneous term’.
It is common to write differential equations with all terms involving the dependent variable
(including derivatives) on the left-hand side of the equation and any remaining terms on the right.
Thus if the right-hand side of the differential equation is zero, it is classified to be homogeneous.
Example 27.2 is homogeneous while the remaining examples are non-homogeneous. Note that
homogenous equations will always have the trivial solution y(x ) = 0, while non-homogenous
equations will not.
In summary, example 27.1 is a first order, linear, non-homogeneous ordinary differential equation.
Example 27.2 is a third order, linear, homogeneous ordinary differential equation. 27.3 is second
order, linear, non-homogeneous ordinary differential equation, and example 27.4 is a second
order, non-linear, non-homogeneous ordinary differential equation.
4. Solving differential equations
Ideally a solution of an ODE would be an explicit representation of the independent variable
y(x ). Sometimes an analytic solution of an ODE may be found, but only in an implicit form,
e.g. H (x, y) = 0, and sometimes no analytic solution to an ODE is possible.
For example, the exponential function
N(t) = αeλt
is an explicit solution to the simple ODE
dN
dt= λN
where α is an arbitrary constant.
As a second example, consider the 2nd order linear ODE
x+ x = t
By inspection we can see that x(t) = t is a solution to the ODE since the second derivative of
x(t) would be zero. A more general solution, however, would be x(t) = A sin(t) + B cos(t) + t,
where A and B are arbitrary constants.
x(t) = A sin(t) +B cos(t) + t
x(t) = A cos(t)−B sin(t)
x(t) = −A sin(t)−B cos(t)
so substituting in these values,
x+ x = [−A sin(t)−B cos(t)] + [A sin(t) +B cos(t) + t] = t
We define the general solution of an ODE as one that contains the arbitrary constants and
retains the maximum degrees of freedom possible. As demonstrated in the first example, the
ENG1091 Mathematics for Engineering page 130
solution to our first order ODE has one degree of freedom in its solution. The solution of the
second order ODE has two degrees of freedom in the solution. Both of these examples are linear
ODEs, and an nth order linear ODE will have n degrees of freedom in its general solution.
If the solution of an ODE contains no free constants, then we say that the solution is a particular
solution. Typically a particular solution is found by placing additional constraints on the ODE
that define the arbitrary constants. For example, the ODE
dN
dt= λN
could be further constrained by the condition when t = 0, N (t) = 5. So the solution would have
to then be N(t) = 5eλt.
In the second example, the 2nd order linear ODE requires two constraints to fully define the
arbitrary constants. These two constraints could be at different points in the domain (e.g. x (0)
= 4 and x (10) = -2) or all the constraints could be given at the same point in the domain (e.g.
x (0) = 4 and x(0) = 3.) The first set of constraints is called boundary conditions and the
later is called initial conditions. The definition typically reflects the physical nature of the
physical problem. As there is only one constraint for first order linear ODEs, it doesn’t really
matter what you call it (but it is common to refer to the single constraint as the initial condition.)
The statement of an ODE with the boundary (initial) conditions is commonly called a boundary
(initial) value problem.
The ODE x− 4x = 4t will allow a general solution of
x(t) = Ae2t +Be−2t − t
(Use substitution to verify this.) While the initial value problem
x− 4x = 4t, x(0) = 0, x(0) = 4
requires the particular solution of
x(t) = e2t − e−2t − t
(Again, this can be verified through substitution.) In the coming lectures we will learn a number
of techniques for finding analytic solutions to a select set of ODEs. When analytic solutions
are not possible, one may be interested in employing a graphical approach (for 1st order ODEs)
and/or numerical techniques for higher order problems.
5. Graphical interpretation of first order ODEs
Let us initially assume that we have a simple 1st order ODE that we can write in the form
dy
dx= F (x, y)
with no initial condition specified. These slopes can then be drawn and produce what is known
as a direction field.
ENG1091 Mathematics for Engineering page 131
4 2 2 4
4
2
2
4
Slope field for dydx = y
4 2 2 4
4
2
2
4
Slope field for dydx = −x
y
4 2 2 4
4
2
2
4
Slope field for dydx = x
y
4 2 2 4
4
2
2
4
Slope field for dydx = y
x
Given an initial condition, the solution can be mapped out graphically. This is known as a
solution curve. Different initial conditions will normally lead to different solutions. Simply
plotting a few arbitrary solution curves will produce a family of solution curves. In a preview
to a later lecture, this graphical technique is the basis of many common numerical techniques for
solving ODEs.
ENG1091 Mathematics for Engineering page 132
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Ordinary Differential Equations
Lecture 28 · separable first order ODEsText Reference: §10.5
1. Separable equations
A number of techniques may be used to find analytic solutions of various ODEs. Perhaps the
most simple approach would for ODEs that are separable. By this we mean that the basic ODE
can be re-written with all components of the dependent variable on one side of the equation, and
all components of the independent variable on the other.
Example 1:dy
dx= xy can be rewritten to
dy
y= xdx
Both the left and right hand side of the equation can be readily integrated:∫dy
y=
∫xdx
which leads to
ln y =x2
2+ c
This can be further manipulated to
y(x) = c1ex2/2
One can readily verify by substitution that this is the general solution to the original 1st order
linear ODE.
Example 2: Find the solution to the ODE
dy
dx=−xy
and verify that the solution does solve the ODE.
In general, the technique for separation of variables requires that the ODE be of the form
dx
dt=h(t)
f(x)(28.1)
which can be rewritten to f(x)dx = h(t)dt and that both integrals may be solved with F (x) =∫f(x)dx and H(t) =
∫h(t)dt.
Then the general solution of the separable ODE will be
F (x)−H(t) = c. (28.2)
Note that not all ODEs are separable.
Moreover, even if a 1st order ODE is separable, that does not mean that the components can be
integrated to get a neat analytic solution.
ENG1091 Mathematics for Engineering page 133
Example 3:
y′ = ex+y.
Example 4: Find the solution to the ODE
exdy
dx− 2y = 1
and verify that the solution does solve the ODE.
ENG1091 Mathematics for Engineering page 134
2. Substitution
Just as when we learned basic integration, simple substitutions may sometimes be able to trans-
form the given ODE into a separable 1st order ODE. The standard example of this pertains to
ODEs of the form:dx
dt= f
(xt
)Here we can make the substitution w = x/t or x = tw.
Example (from text): Solve the ODE
t2dx
dt= x2 + xt
Write the DE as a function ofx
t:dx
dt=(xt
)2+(xt
)now use the substitution x = tw so that
dx
dt= w + t
dw
dtdx
dt=(xt
)2+(xt
)becomes w + t
dw
dt= w2 + w
so that tdw
dt= w2
now separate:dw
w2= tdt
integrate:∫dw
w2=
∫dt
t
−w−1 = ln t+ C
hencet
x= − ln t+ C
giving x (t) =t
C − ln t.
ENG1091 Mathematics for Engineering page 135
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Ordinary Differential Equations
Lecture 29 · first order linear ODEsText Reference: §10.5.9
1. First order linear ODEs
As an initial point we will consider a homogeneous 1st order linear ODE,
dx
dt+ p (t)x = q (t) (29.1)
When we have a linear d.e. in this form we multiply both sides of the equation by an integrating
factor g (t) that will make the LHS of the ODE the derivative of a product.
g (t)dx
dt+ g (t) p (t)x = g (t) q (t)
The integrating factor is
g (t) = e∫p(t)dt (29.2)
Example: Find the integrating factor and solve the ODE
dy
dx+ xy = 0
[This equation could also be solved by separating the variables.]
Integrating factor: g (x) = e∫xdx = e
12x2
Multiply both sides: e12x2 dy
dx+ xe
12x2y = 0
Combine the LHS into a single derivative:d
dx
(e12x2y)
= 0
Integrate both sides: e12x2y = c
y = ce−12x2
2. Non-homogenous first order linear ODEs
Example: Solve the following ODE by finding the integrating factor.
dy
dx− y
x= 2
Integrating factor: g (x) = e∫− 1xdx = e− lnx =
1
x
Multiply both sides:1
x
dy
dx− 1
x2y =
2
x
Combine the LHS into a single derivative:
d
dx
(1
xy
)=
2
x
Integrate both sides:y
x=
∫2
xdx = 2 lnx+ c
so y = 2x lnx+ cx
ENG1091 Mathematics for Engineering page 136
Example: Find the integrating factor and solve the initial value problem
tdx
dt+ x = t2 with x(2) = 1/3.
Rewrite in standard form:dx
dt+
1
tx = t
Integrating factor: g (t) = e∫1tdt = eln t = t
Multiply both sides: tdx
dt+ x = t2
Combine the LHS into a single derivative:d
dt(tx) = t2
Integrate both sides
tx =1
3t3 + C
so x (t) =1
3t2 +
C
t
Now use the initial condition:
x (2) =4
3+C
2=
1
3, which gives C = −2
Hence x (t) =1
3t2 − 2
t.
Example: Find the integrating factor and solve the initial value problem
dx
dt+ 5x− t = e−2t, x(−1) = 0.
Rewrite in standard form:dx
dt+ 5x = e−2t + t
Integrating factor: g (t) = e∫5dt = e5t
Multiply both sides: e5tdx
dt+ 5e5tx = e3t + te5t
Combine the LHS into a single derivative:d
dt
(e5tx
)= e3t + te5t
Integrate both sides (note the integration by parts):
e5tx =
∫e3t + te5tdt
=1
3e3t +
1
5
∫td
dt
(e5t)dt
=1
3e3t +
1
5
(te5t)− 1
5
∫ (e5t)dt
=1
3e3t +
1
5
(te5t)− 1
25e5t + C
so x (t) =1
3e−2t +
1
5t− 1
25+ Ce−5t
Now use the initial condition:
x (−1) =1
3e2 − 6
25+ Ce5 = 0, which gives C = −0.015...
Hence x (t) =1
3e−2t +
1
5t− 1
25− 0.015e−5t
ENG1091 Mathematics for Engineering page 137
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Systems of Differential Equations
Lecture 30&31 · Homogeneous linear systems
Systems of differential equations (time continuous dynamical systems)
Consider the linear differential equation system
x = 2x− y
y = x+ y
To solve this system we need to find both x and y as explicit functions of t.
Now the first step in solving such a system is to write it in matrix form:[dxdtdydt
]=
[2 −1
1 1
][x
y
]
Which in vector form can be written:
dx
dt= Ax where x =
[x
y
], and A =
[2 −1
1 1
]. (30.1)
This is a first order homogeneous system.
Such systems arise frequently in engineering applica-
tions. As an example let us consider the mechanical
system consisting of two masses on two springs as
shown in the diagram.
The displacements y1 (t) and y2 (t) are the displace-
ments of the two masses from their equilibrium posi-
tions when the whole system is at rest.
The upper mass is connected to two springs and
Hooke’s law gives an upward spring force of −3y1 and
a downward spring force of 2 (y2 − y1) since the dis-placement of the lower spring is (y2 − y1) from its
equilibrium.
The lower mass experiences an upward spring force of
−2 (y2 − y1) .
So..y1 = −3y1 + 2 (y2 − y1) = −5y1 + 2y2..y2 = −2 (y2 − y1)
.
This is a 2nd order system.
ENG1091 Mathematics for Engineering page 138
To convert it to first order we let x1 = y1 x2 =·y1 x3 = y2 x4 =
·y2
=·x1 =
·x3
and then
·x1 =
·y1 = x2
·x2 =
··y1 = −5x1 + 2x3
·x3 =
·y2 = x4
·x4 =
··y2 = 2x1 − 2x3
Written in matrix form this is·x1·x2·x3·x4
=
0 1 0 0
−5 0 2 0
0 0 0 1
2 0 −2 0
x1
x2
x3
x4
.
We wont solve the system in this particular example but the general solution takes a remarkably
simple form provided we know the eigenvalues and eigenvectors of the matrix A.
The solution first order homogeneous systems.
Suppose we have a general first order homogeneous linear system of d.e.s:
dx
dt= Ax (30.2)
where x =
x1 (t)...
xn (t)
and A is an n× n constant matrix.If A has n linearly independent eigenvectors v1,v2, . . . ,vn corresponding to the eigenvalues
λ1, λ2, . . . , λn then the general solution to (30.2) is
x = c1eλ1tv1 + c2e
λ2tv2 + . . .+ cneλntvn (30.3)
Proof: (part)
Using x = c1eλ1tv1 + c2e
λ2tv2 + . . .+ cneλntvn we have
dx
dt= c1λ1e
λ1tv1 + c2λ2eλ2tv2 + . . .+ cnλne
λntvn.
Now, remembering that each v1,v2, . . . ,vn is an eigenvector so that Av1 = λ1v1,
Av2 = λ2v2, . . . , and Avn = λnvn we have
dx
dt= c1λ1e
λ1tv1 + c2λ2eλ2tv2 + . . .+ cnλne
λntvn
= c1eλ1tAv1 + c2e
λ2tAv2 + . . .+ cneλntAvn
= A(c1e
λ1tv1 + c2eλ2tv2 + . . .+ cne
λntvn
)= Ax.
ENG1091 Mathematics for Engineering page 139
ExamplesSolve the system:
dx1dt
= x1 + x2,dx2dt
= 4x1 − 2x2
subject to the initial conditions: x1 (0) = 1, x2 (0) = 6.
Solution: First write system in matrix form:
[dx1dtdx1dt
]=
[1 1
4 −2
][x1
x2
].
The matrix
[1 1
4 −2
]has eigenvalues−3, 2 corresponding to eigenvectors
[1
−4
],
[1
1
]respectively.
Show this:
The characteristic polynomial is det (A− λI) =
∣∣∣∣∣ 1− λ 1
4 −2− λ
∣∣∣∣∣= (1− λ) (−2− λ)− 4
= −2 + λ+ λ2 − 4
= λ2 + λ− 6
= (λ+ 3) (λ− 2)
and hence the eigenvalues are λ = −3 and λ = 2.
Now for the eigenvectors:
For λ = −3 we solve
[1 1
4 −2
][x1
x2
]= −3
[x1
x2
](equivalently,
[4 1
4 1
][x1
x2
]=
[0
0
])
hence 4x1 + x2 = 0 yielding eigenvectors of the form s
[1
−4
]for s 6= 0.
For λ = 2 we solve
[1 1
4 −2
][x1
x2
]= 2
[x1
x2
](equivalently,
[−1 1
4 −4
][x1
x2
]=
[0
0
])
hence x1 = x2 yielding eigenvectors of the form s
[1
1
]for s 6= 0.
Hence the general solution of the system is
[x1 (t)
x2 (t)
]= c1e
−3t
[1
−4
]+ c2e
2t
[1
1
]
Now we need to find c1 and c2, to do this solve c1
[1
−4
]+ c2
[1
1
]=
[1
6
]giving c1 = −1
and c2 = 2.
Show this (Cramer’s rule):
We have c1 + c2 = 1
−4c1 + c2 = 6
hence
c1 =
∣∣∣∣∣ 1 1
6 1
∣∣∣∣∣∣∣∣∣∣ 1 1
−4 1
∣∣∣∣∣and c2 =
∣∣∣∣∣ 1 1
−4 6
∣∣∣∣∣∣∣∣∣∣ 1 1
−4 1
∣∣∣∣∣so c1 =
−5
5= −1 and c2 =
10
5= 2
ENG1091 Mathematics for Engineering page 140
Therefore the solution is x =
[x1 (t)
x2 (t)
]= −e−3t
[1
−4
]+ 2e2t
[1
1
].
Explicitly this gives x1 (t) = −e−3t + 2e2t, x2 (t) = 4e−3t + 2e2t.
Example (repeated eigenvalue)Find the general solution of the system:
dx1dt
= 6x1 + x2,dx2dt
= −x1 + 8x2
Solution: The system in matrix form:
[dx1dtdx1dt
]=
[6 1
−1 8
][x1
x2
].
This time the matrix:
[6 1
−1 8
]has a single (repeated) eigenvalue of 7 corresponding to the
eigenvector
[1
1
].
Show this:
The characteristic polynomial is det (A− λI) =
∣∣∣∣∣ 6− λ 1
−1 8− λ
∣∣∣∣∣= (6− λ) (8− λ) + 1
= λ2 − 14λ+ 49
= (λ− 7) (λ− 7)
and hence there is a single eigenvalue only, namely λ = 7.
Now for the eigenvectors:
For λ = 7 we solve
[6 1
−1 8
][x1
x2
]= 7
[x1
x2
](equivalently,
[−1 1
−1 1
][x1
x2
]=
[0
0
])
hence −x1 + x2 = 0 yielding eigenvectors of the form s
[1
1
]for s 6= 0.
As this matrix only has one independent eigenvector the solution form (30.3) is incomplete.
While x =
[x1 (t)
x2 (t)
]= c1e
7t
[1
1
]is a solution it is only part of the general solution.
The complete general solution cannot be obtained solely through eigenvalue/eigenvector methods.
ENG1091 Mathematics for Engineering page 141
Example (complex eigenvalues)Find the complete general solution the system:
dx1dt
= x1 + x2,dx2dt
= −4x1 + x2
Solution: Write system in matrix form:
[dx1dtdx2dt
]=
[1 1
−4 1
][x1
x2
].
The matrix
[1 1
−4 1
]has eigenvalues 1 + 2i, with corresponding eigenvector
[1
2i
],
and 1− 2i, with corresponding eigenvector
[1
−2i
].
Show this:
The characteristic polynomial is det (A− λI) =
∣∣∣∣∣ 1− λ 1
−4 1− λ
∣∣∣∣∣= (1− λ)2 + 4
= [(1− λ)− 2i] [(1− λ) + 2i]
and hence the eigenvalues are λ = 1± 2i.
Now for the eigenvectors:
For λ = 1+2i we solve
[1 1
−4 1
][x1
x2
]= (1 + 2i)
[x1
x2
](equivalently,
[−2i 1
−4 −2i
][x1
x2
]=
[0
0
])
hence −2ix1 + x2 = 0 yielding eigenvectors of the form s
[1
2i
]for s 6= 0.
For λ = 1−2i we solve
[1 1
−4 1
][x1
x2
]= (1− 2i)
[x1
x2
](equivalently,
[2i 1
−4 2i
][x1
x2
]=
[0
0
])
hence 2ix1 + x2 = 0 yielding eigenvectors of the form s
[1
−2i
]for s 6= 0.
The general solution (30.3) gives x =
[x1 (t)
x2 (t)
]= c1e
(1+2i)t
[1
2i
]+ c2e
(1−2i)t
[1
−2i
].
Now
c1e(1+2i)t
[1
2i
]+ c2e
(1−2i)t
[1
−2i
]= et
[c1 (cos 2t+ i sin 2t)
2ic1 (cos 2t+ i sin 2t)
]+
[c2 (cos 2t− i sin 2t)
−2ic2 (cos 2t− i sin 2t)
]
= et
[(c1 + c2) cos 2t+ i (c1 − c2) sin 2t
2i (c1 − c2) cos 2t− 2 (c1 + c2) sin 2t
]
Setting C1 = c1 + c2, and C2 = i (c1 − c2) (notice that C1 and C2 are real if and only if c1 andc2 are complex conjugates) we obtain
x1 (t) = et (C1 cos 2t+ C2 sin 2t)
x2 (t) = et (2C2 cos 2t− 2C1 sin 2t)
ENG1091 Mathematics for Engineering page 142
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Second Order Differential Equations
Lecture 32&33 · Examples · Homogeneous problemText Reference: §10.8 & 10.9.1
1. Linear higher order ODEs.
Recall in the initial lecture on ODEs that linear differential equations were defined as those
in which the dependent variable or variables and their derivatives do not occur as products,
raised to powers or in non-linear functions.
A general second order linear equation takes the form
p (x)d2y
dx2+ q (x)
d2y
dx2+ r (x) y = f (x) (32.1)
where as the notation implies, p, q, r and f are functions of x only. [Here the independent variable
is x.] If f (x) = 0 the equation is homogeneous:
p (x)d2y
dx2+ q (x)
d2y
dx2+ r (x) y = 0 (32.2)
One feature common to all linear, homogeneous equations is that if y1 (x) and y2 (x) are both
solutions of (32.2) the homogeneous linear differential equation, then so is ay1 (x)+by2 (x), where
a and b are arbitrary constants.
If yp (x) is any solution of the non-homogeneous equation (32.1) and y1 (x) and y2 (x) are both
solutions of (32.2), then for any constants a and b, yp (x)+ay1 (x)+by2 (x) is a solution of (32.1).
We will make use of these facts as we progress.
2. Examples of 2nd order linear ODEs
Gravitation Acceleration
Consider a stone dropped from a tall building. Neglecting air resistance, its acceleration is given
by
a =d2s
dt2= g
where g is gravity. This is a simple 2nd order linear non-homogeneous ODE. The velocity v of
the stone may readily be recovered to
v =ds
dt= v0 + gt
where v0 is the initial velocity.
Its distance from the top of the building s is given by
s = s0 + v0t+ gt2/2
where s0is the initial distance from the top of the building.
Simple Harmonic Motion (Mass on a Spring)
Hooke’s Law: If the spring is stretched (or compressed) s units from its natural length,
ENG1091 Mathematics for Engineering page 143
F = −kswhere k is the spring constant (k > 0).
Now net force = mass × acceleration, so
d2s
dt2= − k
ms = −ω2s
where ω2 =k
m> 0
Here m defines mass. We will find that ω governs the frequency with which the system oscillates.
Assuming that the spring oscillates about the position s = 0, then the solution to this ODE is
s(t) = A cosωt+B sinωt
or by identity s(t) = C sin(ωt+D)
period (time for one complete oscillation) =2π
ω
frequency =1
period=
ω
2π
Example: Verify that equation (32.2) is a solution to equation (32.1).
This type of motion is called simple harmonic motion.
Example: A spring with a mass of 2 kg has natural length 0.5 m.. A force of 12.8 N is required
to maintain it stretched to a length of 0.6 m. If the spring is stretched to a length of 0.6 m and
then released with initial velocity 0, find the position of the mass at any time t.
Damped Oscillations
For a mass on a spring, the frictional force from air resistance increases with the velocity of the
mass. The frictional force is often proportional to the velocity, so we can introduce a damping
term of the form
D dsdt , where D is a constant, called the damping constant, and ds
dt is the velocity. The governing
ODE remains a 2nd order linear ODE with constant coeffi cients.
md2s
dt2= −ks−Dds
dt
or
md2s
dt2+D
ds
dt+ ks = 0
3. Homogeneous 2nd order linear ODEs with constant coeffi cients
For the moment let us focus on homogeneous 2nd order linear ODE with constant coeffi cients.
ad2s
dt2+ b
ds
dt+ cs = 0 (32.3)
where a, b, and c are constants.
The general solution to a 2nd order linear ODE will be a family of functions with two linearly
independent components meaning two arbitrary constants. In the example of the falling body,
ENG1091 Mathematics for Engineering page 144
the initial value problem requires a specification of velocity and position at some point in time.
Since we are examining 2nd order ODEs, we could readily create a boundary value problem
instead of an initial value problem by defining either the velocity or the position at two different
points in time.
Let us assume that a solution of (32.3) has the form s(t) = eλt, for some (as yet unkown value)
of λ.
Example: Verify that equation (32.4) is a solution to (32.3). Identify the constraint that is
placed on λ.
Substituting we obtain: (aλ2 + bλ+ c
)eλt = 0 (32.4)
which requires
aλ2 + bλ+ c = 0. (32.5)
Equation (32.5) is called the auxiliary or characteristic equation, and is a quadratic equation
and has either two real solutions, two imaginary solutions or 1 real solution. These form the
basis of three three cases.
Case 1: b2 − 4ac > 0.
There are two distinct real solutions,
λ1,2 =
(−b±
√b2 − 4ac
2a
)
to the characteristic equation and so the general solution has the form
s (t) = C1eλ1t + C2e
λ2t (32.6)
here C1 and C2 are arbitrary constants. Now eλ1t and eλ2t are two independent solutions of
(32.3) and because (32.6) involves 2 arbitrary constants the solution (32.6) is the full general
solution of (32.3).
ENG1091 Mathematics for Engineering page 145
Example: Solve the ODE
s+ 3s+ 2s = 0
with the constraints of s(0) = −0.5 and s(0) = 3.
Solution:
The d.e. s+ 3s+ 2s = 0 has the characteristic equation
λ2 + 3λ+ 2 = 0
which factorises: (λ+ 2) (λ+ 1) = 0
the equation has two real (unequal) roots λ = −2,−1
hence the general solution consists of linear combinations of the two independent solutions
e−t, and e−2t
i.e. s(t) = C1e−t + C2e
−2t
With the initial conditions s(0) = −0.5 and s(0) = 3 we obtain:
−0.5 = C1 + C2
3 = −C1 − 2C2
giving us C1 = 2 and C2 = −2. 5.
Hence s(t) = 2e−t − 2.5e−2t.
1 2 3 4 5
1
0
1
x
y
Note that the solution passes over the t axis once and approaches it as t approaches infinity.
With different initial conditions, the solution needn’t pass over the axis at all. This case is
sometimes called overdamped.
ENG1091 Mathematics for Engineering page 146
Case 2: b2 − 4ac < 0.
Here there are no real solutions to the characteristic equation, instead there are two complex
conjugate solutions λ1 = p+ qi, λ2 = p− qi where p = −b/2a and
q =√
4ac− b2/2a
The general solution can be written in the form of (32.6) but is usually simplified to
s(t) = ept(C1 cos(qt) + C2 sin(qt)) (32.7)
with the use of Euler’s equation,
eiqt = cos(qt) + i sin(qt).
Exercise: Starting with equation (32.6) derive equation (32.7).
Example: Solve the ODE
s+ 0.4s+ 4.04s = 0
with the constraints of s(0) = 1 and s(0) = −0.2.
Solution: The d.e. s+ 0.4s+ 4.04s = 0 has the characteristic equation
λ2 + 0.4λ+ 4.04 = 0
which has the solutions
λ =−0.4±
√(0.4)2 − 4× 1× 4.04
2using the quadratic formula: λ =
−b±√b2 − 4ac
2a
=−0.4±
√−16
2
=−0.4± 4i
2= −0.2± 2i
The solution eq. (32.7) now becomes
s(t) = e−0.2t(C1 cos (2t) + C2 sin (2t))
This of course is the general solution of s+ 0.4s+ 4.04s = 0.
With the initial conditions s(0) = 1 and s(0) = −0.2. we obtain:
1 = e0(C1 cos (0) + C2 sin (0))
= C1
Nows (t) = −0.2e−0.2t(C1 cos (2t) + C2 sin (2t))
+ e−0.2t(−2C1 sin (2t) + 2C2 cos (2t))
so
−0.2 = s (0) = −0.2C1 + 2C2
ENG1091 Mathematics for Engineering page 147
hence
2C2 = −0.2 + 0.2 = 0
giving the specific solution: s(t) = e−0.2t cos (2t) .
Graph:
1 2 3 4 5 6 7
1.0
0.8
0.6
0.4
0.2
0.0
0.2
0.4
0.6
0.8
1.0
x
y
Note that while the solution is damped and s(t) will approach 0 as t approaches infinity, the
solution oscillates about 0. This is sometimes called and underdamped system.
Case 3: b2 − 4ac = 0.
There is only one distinct real solution (λ = −b/2a), and while s(t) = eλt does satisfy the ODE, it
alone is not the general solution, as we need a second linearly independent component. Another
independent solution has the equation teλt. The general solution of (32.3) is for this case
s (t) = (C1t+ C2) eλt (32.8)
Example: Solve the ODE
s+ 2s+ s = 0
with the constraints of s(0) = 3 and s(0) = 5.
Solution:
The d.e. s+ 2s+ s = 0 has the characteristic equation
λ2 + 2λ+ 1 = 0
which factorises: (λ+ 1)2 = 0
the equation has two equal roots λ = −1,−1
hence the general solution consists of linear combinations of the two independent solutions
e−t, and (note this) te−t
i.e. s(t) = C1e−t + C2te
−t
ENG1091 Mathematics for Engineering page 148
With the initial conditions s(0) = 3 and s(0) = 5 we obtain:
3 = C1e0 + 0
3 = C1
Now s (t) = −C1e−t + C2(e−t − te−t
)so 5 = s (0) = −C1 + C2
hence C2 = 8 giving the specific solution:
s (t) = 3e−t + 8te−t
0 1 2 3 4 5 6 7 8 9 100
1
2
3
4
5
x
y
Case 3 is sometimes called critically damped as it provides the quickest approach to s = 0. This
is similar to case 1, but the damping is just suffi cient to suppress vibrations.
ENG1091 Mathematics for Engineering page 149
Summary: Solutions of
ad2s
dt2+ b
ds
dt+ cs = 0
where a, b, and c are constants.
Roots of aλ2 + bλ+ c = 0 General solution
Two real distinct roots λ1 and λ2 s(t) = C1eλ1t + C2e
λ2t
Repeated (real) root λ s(t) = C1eλt + C2te
λt
Two complex roots p± iq s(t) = ept(C1 cos(qt) + C2 sin(qt))
where C1, C2 are arbitrary constants.
4. Higher order linear ODEs with constant coeffi cients
The text notes that the method of solution developed here is not strictly limited to 2nd order
equations. In particular a homogeneous nth order linear ODE with constant coeffi cients
andns
dtn+ an−1
dn−1s
dtn−1+ · · ·+ a1s+ a0 = 0 (32.9)
where a0 through an are constants, can be solved by assuming a solution in the general form of
s(t) = eλt. As before this will lead to a characteristic equation, which is an nth order polynomial.
anλn + an−1λ
n−1 + · · ·+ a1λ+ a0 = 0 (32.10)
This polynomial will have n roots, which may be some combination of real, repeated and complex
conjugate pairs. As the ODE is linear, we must have n linearly independent components of the
general solution. Please note that it is not trivial to analytically solve a higher order polynomial.
Example: Find the general solution to the ODE
...x − 2x− 5x+ 6x = 0.
Hint: λ = 1 is one solution to the characteristic equation.
Solution: The d.e....x − 2x− 5x+ 6x = 0 = 0 has the characteristic equation
λ3 − 2λ2 − 5λ+ 6 = 0
which factorises to
(λ− 1)(λ2 − λ− 6
)= 0 using the hint
(λ− 1) (λ− 3) (λ+ 2) = 0 completing the factorisation
hence there are three roots
λ = 1, 3,−2
Since these roots are different there are no teλt terms in the solution, and the solution is written:
x (t) = C1et + C2e
3t + C3e−2t
ENG1091 Mathematics for Engineering page 150
Example: Suppose a 4th order linear homogeneous ODE has the characteristic equation
(λ2 + 1)(λ+ 1)2 = 0.
Find the homogeneous equation and find its general solution.
Solution:
The characteristic equation has roots λ = ±i from (λ2 + 1) = 0 and two equal roots λ = −1,−1
from (λ+ 1)2 = 0.
The roots λ = ±i provide the partial solution form (eq. 32.7)
x (t) = e0(C1 cos(t) + C2 sin(t))
= C1 cos(t) + C2 sin(t)
and the two equal roots λ = −1,−1 provide the remaining part of the solution (eq. 32.8):
x (t) = C3e−t + C4te
−t.
Combining: x (t) = C1 cos(t) + C2 sin(t) + C3e−t + C4te
−t
We now write out the original form of the ODE. We need to expand the characteristic equation:
(λ2 + 1)(λ+ 1)2
= (λ2 + 1)(λ2 + 2λ+ 1
)= λ4 + 2λ3 + 2λ2 + 2λ+ 1 = 0
whence we obtain (using dot notation now would be ridiculous)
d4x
dt4+ 2
d3x
dt3+ 2
d2x
dt2+ 2
dx
dt+ x = 0
ENG1091 Mathematics for Engineering page 151
MONASH UNIVERSITY —SCHOOL OF MATHEMATICAL SCIENCES
ENG1091 Second Order Differential Equations
Lecture 34&35 · nonhomogeneous equations · Engineering applicationText Reference: §10.9.2 & 10.10
1. Non-homogeneous 2nd order linear ODEs with constant coeffi cients
Moving from the homogeneous to non-homogeneous 2nd order linear ODE with constant coeffi -
cients means adding a term f(t) to the right-hand side of the equation.
ad2s
dt2+ b
ds
dt+ cs = f (t) (34.1)
where a, b, and c are constants.
The non-homogeneous term f(t) cannot involve the dependent variable s, but it can be non-linear
in the independent variable t. The function f(t) is commonly called the forcing term and this
course will deal with the situation in which f(t) is either a polynomial, exponential or a circular
function.
The general solution s (t) to (34.1) is the sum of the homogeneous solution sc (t) (called the com-
plementary solution) to the homogeneous equation (also called the complementary equation)
and a particular solution, sp (t) .
s (t) = sc (t) + sp (t)
The particular solution thus accounts for the forcing term f(t). The complementary solution
sc(t) will already contain the two independent variables necessary for the general solution.
Much as we had to make a wise guess in finding the solution to homogeneous problem, we will
have to make a wise guess for the nature of the particular solution. Having set the form of the
particular solution it will remain to find the coeffi cient for this term. The technique for this is
commonly called the method of undetermined coeffi cients.
Consider the ODEd2s
dt2− ds
dt− 2s = sin t (34.2)
The solution to the homogeneous equation (called the complementary solution) can readily
be found to be sc (t) = C1e−t + C2e
2t.
A particular solution sp (t) must take the form sp (t) = A cos (t) + B sin (t) where A and B are
undetermined constants which need to be found.
Substituting (34.3) into the original ODE (34.2) allows us to define the coeffi cients A and B into
the differential equation and try to find appropriate values of A and B.
s′p (t) = −A sin (t) +B cos (t) (34.4)
s′′p (t) = −A cos (t)−B sin (t) (34.5)
ENG1091 Mathematics for Engineering page 152
Substituting these equations back into (e.g. 34.2) leads to
−A cos (t)−B sin (t)− (−A sin (t) +B cos (t))− 2 (A cos (t) +B sin (t)) = sin (t)
−3A cos (t)− 3B sin (t) +A sin (t)−B cos (t) = sin (t)
(−3A−B) cos (t) + (A− 3B) sin (t) = sin (t)
This last equation states that (−3A−B) cos (t)+(A− 3B) sin (t) = sin (t) must be true for all t.
This is possible only if A− 3B = 1 and −B − 3A = 0.
A− 3B = 1
−3A−B = 0
A =
∣∣∣∣∣ 1 −3
0 −1
∣∣∣∣∣∣∣∣∣∣ 1 −3
−3 −1
∣∣∣∣∣B =
∣∣∣∣∣ 1 1
−3 0
∣∣∣∣∣∣∣∣∣∣ 1 −3
−3 −1
∣∣∣∣∣=
−1
−10B =
3
−10
Thus A = 0.10 and B = −0.30.
The general solution is now
s (t) = sc (t) + sp (t)
= C1e−t + C2e
2t +A cos (t) +B sin (t) where A = 0.10 and B = −0.30
= C1e−t + C2e
2t + 0.1 cos (t)− 0.3 sin (t)
Series RLC circuits
In an RLC electrical circuit externally driven by E(t).
E
R
L
C
By Kirchoff’s voltage law,
Ld2Q
dt2+R
dQ
dt+
1
CQ = E (t)
ENG1091 Mathematics for Engineering page 153
where Q is the charge on the capacitor at time t.
Differentiate this equation with respect to t (and remember that I =dQ
dt), thus
Ld2I
dt2+R
dI
dt+
1
CI = E′ (t)
Example: Find the charge and current at time t in an RLC circuit if R = 40 Ω, L = 1 H, C =
16× 10−4 F, E(t) = 100 cos (10t) V, and the initial charge and current are both 0.
Solution: Substituting the values for L,R and C we obtain
d2Q
dt2+ 40
dQ
dt+ 625Q = 100 cos (10t)
The homogeneous equation isd2Q
dt2+ 40
dQ
dt+ 625Q = 0
and this has characteristic equation
λ2 + 40λ+ 625 = 0
λ =−40±
√1600− 4× 1× 625
2using the quadratic formula: λ =
−b±√b2 − 4ac
2a
=−40±
√−900
2= −20± 15i
This gives us the complementary function:
Qc (t) = e−20t (C1 cos (15t) + C2 sin (15t))
For the particular solution we try Qp (t) = A cos (10t) +B sin (10t) where A and B are undeter-
mined constants which need to be found.
Q′ (t) = −10A sin (10t) + 10B cos (10t)
Q′′ (t) = −100A cos (10t)− 100B sin (10t)
substituting intod2Q
dt2+ 40
dQ
dt+ 625Q = 100 cos (10t)
we obtain
−100A cos (10t)− 100B sin (10t) + 40 (−10A sin (10t) + 10B cos (10t))
+625 (A cos (10t) +B sin (10t)) = 100 cos (10t)
(525A+ 400B) cos (10t) + (−400A+ 525B) sin (10t) = 100 cos (10t)
ENG1091 Mathematics for Engineering page 154
So
525A+ 400B = 100 or 21A+ 16B = 4
−400A+ 525B = 0 or − 16A+ 21B = 0
A =
∣∣∣∣∣ 4 16
0 21
∣∣∣∣∣∣∣∣∣∣ 21 16
−16 21
∣∣∣∣∣B =
∣∣∣∣∣ 21 4
−16 0
∣∣∣∣∣∣∣∣∣∣ 21 16
−16 21
∣∣∣∣∣=
84
697B =
64
697
Q (t) = Qc (t) +Qp (t)
= e−20t (C1 cos (15t) + C2 sin (15t)) +84
697cos (10t) +
64
697sin (10t)
We now find the values of C1 and C2 given Q (0) = 0 and Q′ (0) = 0
Substituting: Q (0) = 0
C1 +84
697= 0
Finding Q′ (t)
Q′ (t) = −20e−20t (C1 cos (15t) + C2 sin (15t)) + 15e−20t (−C1 sin (15t) + C2 cos (15t))
− 840
697sin (10t) +
640
697cos (10t)
Substituting: Q′ (0) = 0
0 = −20C1 + 15C2 +640
697
Solving
C1 = − 84
697
−20C1 + 15C2 = −640
697
we obtain C1 = − 84
697and C2 = − 464
2091
giving
Q (t) = e−20t(− 84
697cos (15t)− 464
2091sin (15t)
)+
84
697cos (10t) +
64
697sin (10t)
Graphs of Q (t) and Qp (t)
Note that e−20t → 0 as t→∞ and both cos (15t) and sin (15t) are bounded functions.
So for large values of t, Q (t) ≈ Qp (t) , and for this reason, Qp (t) is called the steady state
solution.
ENG1091 Mathematics for Engineering page 155
We have so far considered only one type of externally forcing for our non-homogeneous problem,
namely sinusoidal forcing.
Fortunately we can cover a little more ground than this. Experience tells us that the method
of undetermined coeffi cients can readily be employed when the forcing function is a polynomial
or exponential, in addition to sinusoidal. Note that more complicated forcing may, hopefully, be
readily approximated by some series involving this base functions.
Summary of undetermined coeffi cients
f (t) try sp(t)
acos (kt) +b sin (kt) sp(t) = Acos (kt) +Bsin (kt)
antn + · · ·+ a1t+ a0 sp(t) = Ant
n + · · ·+A1t+A0
ekt sp(t) = Aekt
ekt (acos (ωt) +b sin (ωt)) s(t) = ekt(A cos(ωt) +B sin(ωt))
As the table suggests, if the forcing term f(t) is a polynomial, we anticipate that the particular
solution will be of this form (and degree) too.
Example Find the general solution to the ODE:
d2x
dt2+ 6
dx
dt+ 9x = t2
and this has characteristic equation
λ2 + 6λ+ 9 = 0
λ = −3,−3
xc (t) = C1e−3t + C2e
−3tt
A particular solution xp (t) must take the form xp (t) = at2+bt+c where a, b, c are undetermined
constants which need to be found.
x′p (t) = 2at+ b
x′′p (t) = 2a
Substituting:d2x
dt2+ 6
dx
dt+ 9x = 2a+ 6 (2at+ b) + 9
(at2 + bt+ c
)= 9at2 + (9b+ 12a) t+ 2a+ 9c+ 6b ≡ t2
9a = 1
9b+ 12a = 0
2a+ 9c+ 6b = 0
a =1
9, b = − 4
27, c =
2
27
x (t) = xc (t) + xp (t) = C1e−3t + C2te
−3t + +1
9t2 − 4
27t+
2
27
If the forcing term f(t) is an exponential, we anticipate that the particular solution will be one
also.
ENG1091 Mathematics for Engineering page 156
Example: Find the general solution to the ODE
d2x
dt2+ 5
dx
dt− 6x = e−t
Solution: This has characteristic equation
λ2 + 5λ− 6 = 0
(λ+ 6) (λ− 1) = 0
λ = −6, 1
xc (t) = C1et + C2e
−6t
A particular solution sp (t) must take the form
xp (t) = ae−t
where a is the undetermined constant.
x′p (t) = −ae−t
x′′p (t) = ae−t
d2x
dt2+ 5
dx
dt− 6x = ae−t − 5ae−t − 6ae−t
= −10ae−t ≡ e−t
Clearly
a = − 1
10
x (t) = C1et + C2e
−6t − 1
10e−t
Problems arise if any term of xp (t) is a part of the complementary solution. In such a case
multiply xp (t) by t (or t2 in the case of repeated roots).
Example: Find the general solution to the ODE .
d2x
dt2+ 5
dx
dt− 6x = 7et
Solution: This has characteristic equation
λ2 + 5λ− 6 = 0
λ = −6, 1
xc (t) = C1et + C2e
−6t
A particular solution sp (t) must take the form xp (t) = a(tet)
since et is already part of the complementary solution where a is the undetermined constant.
x′p (t) = a(et + tet
)x′′p (t) = a
(et + et + tet
)d2x
dt2+ 5
dx
dt− 6x = a
(2et + tet
)+ 5a
(et + tet
)− 6a
(tet)
= 0(tet)
+ 7aet ≡ 7et
Clearly a = 1 giving x (t) = C1et + C2e
−6t + tet.
ENG1091 Mathematics for Engineering page 157
If the inhomogeneous term is composed of several functions whose particular solutions can be
individually found then we combine (add) our particular solutions.
Example: Find the general solution to the ODE .
d2x
dt2+ 5
dx
dt− 6x = t+ e−2t sin t
Solution:
Consider the equationd2x
dt2+ 5
dx
dt− 6x = t
A particular solution xp (t) must take the form xp (t) = at+ b
which leads to the particular solution: xp (t) = − 536 −
16 t
While the equationd2x
dt2+ 5
dx
dt− 6x = e−2t sin t
has a particular solution xp (t) of the form: xp (t) = ae−2t cos t+ be−2t sin t
x′p (t) = a(−2e−2t cos t− e−2t sin t
)+ b
(−2e−2t sin t+ e−2t cos t
)= (−2a+ b) e−2t cos t+ (−a− 2b) e−2t sin t
x′′p (t) = (−2a+ b)(−2e−2t cos t− e−2t sin t
)+ (−a− 2b)
(−2e−2t sin t+ e−2t cos t
)= (3a− 4b) e−2t cos t+ (4a+ 3b) e−2t sin t
now substitute into d2xdt2
+ 5dxdt − 6x :
[(3a− 4b) + 5 (−2a+ b)− 6a] e−2t cos t+ [(4a+ 3b) + 5 (−a− 2b)− 6b] e−2t sin t
= (−13a+ b) e−2t cos t+ (−a− 13b) e−2t sin t
so −13a+ b = 0
−a− 13b = 1
from which we obtain: a = − 1170 , b = − 13
170
which leads us to the particular solution:
xp (t) = ae−2t cos t+ be−2t sin t
xp (t) = e−2t(− 1
170cos (t)− 13
170sin (t)
)Now we add particular solutions:
d2x
dt2+ 5
dx
dt− 6x = t+ e−2t sin t so xp (t) = − 5
36− 1
6t+ e−2t
(− 1
170cos (t)− 13
170sin (t)
)The full general solution is
x (t) = xc (t) + xp (t)
= C1et + C2e
−6t − 5
36− 1
6t+ e−2t
(− 1
170cos (t)− 13
170sin (t)
)Note: inhomogenous terms such as ekt sinωt or ekt cosωt can be much better handled using the
complex exponential which is given as an alternative to the above working.
ENG1091 Mathematics for Engineering page 158
Alternative (finding the particular solution of ekt sinωt or ekt cosωt forms) using complex expo-
nentials:
For the particular solution of d2xdt2
+ 5dxdt − 6x = e−2t sin t first we rewrite the equation as
d2xdt2
+ 5dxdt − 6x = e−2t (cos t+ i sin t) = e−2t · eit and we wish to take the imaginary part only.
We try xp (t) = αe(−2+i)t where α is an unknown complex constant.
Now
d2x
dt2+ 5
dx
dt− 6x = α (−2 + i)2 e(−2+i)t + 5α (−2 + i) e(−2+i)t − 6αe(−2+i)t
= α[(−2 + i)2 + 5 (−2 + i)− 6
]e(−2+i)t
= α [(4− 4i− 1)− 10 + 5i− 6] e(−2+i)t notice this step
= α (−13 + i) e(−2+i)t
now introduce the RHS: ≡ e−2t · eit = e(−2+i)t
Now equate coeffi cients:
α (−13 + i) = 1
α = 1−13+i
= 1−13+i ×
−13−i−13−i
= 1132+1
(−13− i)
= 1170 (−13− i)
So for the particular solution of d2xdt2
+ 5dxdt − 6x = e−2t sin t we use the imaginary part of
αe(−2+i)t =(−13170 −
1170 i
)e−2t (cos t+ i sin t) .
The ‘i’term is[(− 1170 i
)(cos t) +
(−13170 i
)(sin t)
]e−2t so the imaginary part of αe(−2+i)t is(
− 1170 cos (t)− 13
170 sin (t))e−2t = xp (t) as before.
ENG1091 Mathematics for Engineering page 159
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