Lecture notes for Mathematics
1E2 - MATH19682
Lecturer:
SergeiR.
Fedotov
Lecturer:
C. Assier
Email: [email protected]
These notes and other ressources are available on Blackboard
These notes have been written by Prof. J.S.B. Gajjar
Semester 2, 2014
1
Contents
1 Integration
1.1
1.2
1.3
1.4
1.5
Fundamentals . . . . . . . . . . . . . . . .
Elementary indenite integrals . . . .
Denite Integrals . . . . . . . . . . . . .
Fundamental Theorem of Calulus . .
Properties of integrals . . . . . . . . .
1.5.1 Basi properties . . . . . . . . . . .
1.5.2 Integration by parts . . . . . . .
1.5.3 Substitution . . . . . . . . . . . . .
1.6 Innite Limits. . . . . . . . . . . . . . . .
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4
4
5
6
6
7
7
7
8
9
2 Harder Integration Tehniques
10
3 Appliations of Integration
12
2.1 Some standard integrals . . . . . . . . . . . . . . . . . . . . . . .
3.1 Calulation of Area . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1.1 Area bounded by the x-axis . . . . . . . . . . . . . . . . .
3.1.2 Area bounded by the y -axis . . . . . . . . . . . . . . . . .
3.1.3 Area in polar oordinates . . . . . . . . . . . . . . . . . .
3.1.4 Area under a urve given parametrially . . . . . . . . . .
3.2 Ar-length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.1 Ar length in Cartesian Coordinates . . . . . . . . . . . .
3.2.2 Ar-length in polar oordinates . . . . . . . . . . . . . . .
3.2.3 Ar-length of a urve given parametrially in polar oordinates. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Surfae area and volume . . . . . . . . . . . . . . . . . . . . . . .
3.3.1 Volume of solid of revolution . . . . . . . . . . . . . . . . .
3.3.2 Surfae area . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4 Mean and rms values . . . . . . . . . . . . . . . . . . . . . . . . .
3.5 Centroids and entre of mass . . . . . . . . . . . . . . . . . . . . .
4 Sequenes and Series
4.1 Finite and Innite Series . . .
4.1.1 Arithmeti Progression
4.2 Geometri progressions . . . .
4.3 Innite Series . . . . . . . . .
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5 Taylor and Malaurin Series
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5.1 Taylor's theorem . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
10
12
12
12
13
13
14
14
15
16
17
17
17
18
18
20
20
21
22
22
25
25
6
Funtions of more than one variable, partial derivatives
7
Line integrals and multiple integrals
8
6.1
6.2
6.3
6.4
Partial dierentiation
Chain Rule . . . . .
Small hanges . . . .
Gradient . . . . . . .
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7.1 Path Independene . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 Multiple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3 Non-retangular regions . . . . . . . . . . . . . . . . . . . . . . .
Examples
8.1
8.2
8.3
8.4
Examples 1
Examples 2
Examples 3
Examples 4
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28
29
30
32
33
36
38
41
42
44
44
46
49
51
1
1.1
Integration
Fundamentals
The proess of integration an be thought as being the reverse operation to differentiation. Given a funtion f (x) the indenite integral F (x) of f (x) is that
funtion whih when dierentiated gives f (x) ie
dF (x)
= f (x).
dx
The notation we use for the indenite integral is
F (x) =
Z
x
f (t) dt.
Note that f (t) here is alled the integrand and t is regarded as a dummy variable
and we talk of integrating with respet to t. In plae of t one an use other
dummy variables s for instane.
Another form for the indenite integral is
F (x) =
Z
f (x) dx.
Integration is not an abstrat mathematial idea. It arises in real appliations.
For example suppose you need to design an instrument whih is going to alulate
the distane s travelled in a given time t when you are know the speed v(t). We
know that speed and distane are related by
ds
= v(t),
dt
and hene to nd the speed we need to work out the integral
s(t) =
Z
v(t) dt + C,
and our instrument needs to be able to be designed to do this. This is just one
example. There are many appliations involving integration one an nd in all
branhes of engineering.
Example If f (x) = 2x then
F (x) =
where C is an arbitrary onstant.
Z
2x dx = x2 + C
4
1.3 Denite Integrals
Suppose
F (x) =
Then the denite integral
Z
x
f (t) dt.
b
a
In the integral
Z
f (t) dt = F (b) − F (a).
Z
b
f (t) dt
a
the numbers a and b are alled the lower and upper limits respetively. We will
also use the notation
[F (x)]ba = F (b) − F (a).
Example
F (x) =
Z
0
2
Z
x
2t dt = x2 + C,
2t dt = [22 + C − 02 − C] = 22 − 02 = [x2 ]20 = 4.
1.4 Fundamental Theorem of Calulus
f (x) is ontinuous on the interval a ≤ x ≤ b then the integral F (x) =
Suppose
Rx
f (t) dt is also ontinuous for a < x < b and
a
dF (x)
= f (x).
dx
The denite integral ab f (x) dx an be regarded as the area under the urve
y = f (x) from x = a to x = b. If we split the interval [a, b] into N equally sized
intervals of width dx = (b−a)/(N −1) with xk = a+(k −1)dx, k = 1, 2, . . . , N ,
then
Z
R
b
f (x) dx = lim
a
N →∞
N
−1
X
k=1
dx ∗ f (ξk )
where xk−1 ≤ ξk ≤ xk . Note that f (ξk ) gives the average height of the retangle
whose width is dx, so f (ξk ) ∗ dx gives
the area of the thin retangular strip
Rb
between xk−1 and xk . The integral a f (x) dx tends to the area under the urve
from x = a to x = b as we let the the number of intervals beomes large (or
equivalently by shrinking the width dx of the interval to zero).
6
f (ξk )
f (x)
ξk
dx
f (ξk )
a
b
dx
x
Figure 1: Denite integral as area under urve.
1.5
1.5.1
Properties of integrals
Basi properties
The following properties are easy to prove.
Z
(f (x) + g(x)) dx =
Suppose k is a onstant. Then
Z
Z
f (x) dx +
kf (x) dx = k
Z
Z
g(x) dx.
f (x) dx.
Interhanging limits hanges the sign so that
Z
a
1.5.2
b
f (x) dx = −
Z
a
f (x) dx.
b
Integration by parts
Now
df (x)
dg(x)
d
[f (x)g(x)] =
g(x) + f (x)
.
dx
dx
dx
So integrating both sides with respet to x
Z
Z
Z
d
df (x)
dg(x)
[f (x)g(x)] dx =
g(x) dx + f (x)
dx.
dx
dx
dx
7
Hene
f (x)g(x) =
df (x)
g(x) dx +
dx
Z
dg(x)
dx.
dx
Z
f (x)
Z
df (x)
g(x) dx.
dx
We an rearrange this to get the integration by parts formula
Z
dg(x)
f (x)
dx = f (x)g(x) −
dx
Example
Use integration by parts to evaluate
cos x and dierentiate the x. So set
dg
= cos x
dx
R
−→ g(x) = sin x.
Also
df
= 1.
dx
f (x) = x,
Using the above formula gives
Z
1.5.3
x cos x dx = x sin x −
Z
x cos x dx. Here we an integrate the
1. sin x dx = x sin x + cos x + C.
Substitution
Note that
d
[f (g(x))] = f ′ (g(x))g ′ (x) = f ′ (z)g ′ (x)
dx
where we have put z = g(x). If we integrate both sides we obtain
Z
Z
d
d
[f (g(x))] dx =
[f (z)] dx
dx
dx
Z
df dz
dx = f (z) + C = f (g(x)) + C.
=
dz dx
This gives a means by whih one an use substitutions to perform ertain integrals. First put
h(z) =
Suppose want to evaluate
f (z) =
Z
df (z)
.
dz
h(z) dz.
The above formula suggests that we an put z = g(x) and then
f (z) =
Z
h(z) dz =
8
Z
h(g(x))
dz
dx.
dx
This is the way we an handle substitutions. Notie that we just replae
dz =
and
z
by
g(x)
dz
by
dz
dx,
dx
in the integrand.
Example
Consider
Put
z = 2 tan x.
Z
Then
z2
1
dz.
+4
dz
= 2 sec2 x,
dx
and the integral beomes
Z
1
dz =
z2 + 4
Z
1
2 sec2 x dx.
4 + 4 tan2 x
Now make use of the identity
sec2 x = 1 + tan2 x
and the integral redues to
Z
2 sec2 x
dx =
4 sec2 x
Z
1
1
1
z
dx = x + C = tan−1 + C.
2
2
2
2
1.6 Innite Limits.
The integral
Z
b
f (x) dx
a
an have one or both limits equal to
Z
∞.
∞
Consider for example
e−x dx.
0
To evaluate this we just integrate as normal and evaluate the answer at the upper
limit
x = b with b → ∞. Thus
Z ∞
Z b
−x
e dx = lim
e−x dx = lim [−e−x ]b0 = lim [1 − e−b ] = 1.
0
b→∞
0
b→∞
9
b→∞
2
Harder Integration Tehniques
Integration an be a bit of an art sometimes although one an identify useful
substitutions and triks whih an help in some ases.
Below we have ollated
some brief notes on the type of substitutions whih work for the type of integrand
being disussed. It is a useful exerise to work through the details and onvine
yourself that the substitution does work and redues the integral to one of the
more familiar types.
2.1
1.
Some standard integrals
Z
dx
x 2 + a2
with
a 6= 0.
Use
x = a tan θ
and
dx
x2
= a sec2 θ = a(1 + 2 )
dθ
a
to get
Z
2.
Z
x2
dx
,
− a2
dx
=
2
x + a2
with
Z
a 6= 0.
a sec2 θ
1
1
x
dθ = θ + C = tan−1 + C.
2
2
a sec θ
a
a
a
Use partial frations to write
·
¸
1
1
1
1
=
−
x 2 − a2
2a x − a x + a
and then
Z
dx
=
2
x − a2
=
3.
Z
√
dx
,
− x2
a2
with
Z
·
¸
1
1
1
dx
−
2a x − a x + a
1
[log |x − a| − log |x + a|] + C.
2a
a 6= 0.
Use
x = a sin θ
and then
r
dx
x2 √
= a cos θ = a 1 − 2 = a2 − x2 .
dθ
a
Then
Z
dx
√
=
a2 − x2
Z √ 2
a − x2
x
√
dθ = θ + C = sin−1 + C
a
a2 − x2
10
4.
Z
√
dx
. Use x = a sinh θ and then
a2 + x 2
r
x2 √
dx
= a cosh θ = a 1 + 2 = a2 + x2 .
dθ
a
Hene
Z
dx
√
=
2
a + x2
Z √ 2
a + x2
x
√
dθ = θ + C = sinh−1 + C.
2
2
a
a +x
R
dx
. Use x = a cosh θ and then
x 2 − a2
5.
Z
√
6.
Z
ax2
√ dx
x2 −a2
= cosh−1
x
a
+ C.
dx
with a 6= 0. First omplete the square so that
+ bx + c
ax2 + bx + c = a(x2 +
b
b2
bx
) + c = a(x + )2 + c −
a
2a
4a
and then use the substitution z = x + 2ab to redue to one of standard type.
dx
with a 6= 0. Complete the square and then use the sub+ bx + c
stitution z = x + 2ab to redue to one of standard type.
Z
dx + e
8.
dx. We try and express the numerator as a multiple of the
2
ax + bx + c
7.
Z
√
ax2
derivative of the denominator. Thus
d
bd
(2ax + b)
e − 2a
dx + e
2a
=
+
.
ax2 + bx + c
ax2 + bx + c ax2 + bx + c
The rst term on the right hand side an be integrated (or use a substitution
z = ax2 +bx+c) and the seond term is of a type we have already disussed.
9.
P (x)
dx where a 6= 0 and P (x) is a polynomial.
+ bx + c
Z
ax2
Z
dx
√
with a 6= 0. Use the substitution z = 1/(x + d).
(x + d) ax2 + bx + c
P (x)
If P (x) is of degree greater than 1 we do long division to obtain Q(x)
=
S(x)
R(x) + ax2 +bx+c where S(x) is at most of degree 1. This is now of a form
introdued earlier.
10.
This redues the integral to a type we have already studied.
11
3
Appliations of Integration
In this setion we will disuss a number of dierent appliations whih involve
integrals. For further pratie you should attempt the spei exerises in Examples 1 and 2 given at the end of the booklet.
3.1
Calulation of Area
3.1.1
x-axis
Area bounded by the
The area A bounded by the urve y = f (x) between x = a and x = b and the
x−axis, see Fig.2, is given by
A=
b
Z
y dx =
b
Z
f (x) dx.
a
a
Note that a negative result implies a net portion of the area is below the x−axis.
y = f (x)
y
a
b
x
Figure 2: The area trapped by y = f (x), x = a, x = b and the x−axis.
Calulate the area between x = 0 and x = 2, bounded by the x−axis
and the urve y = ex − 1.
The required area is
Example
Z
0
3.1.2
2
(ex − 1) dx = [ex − x]20 = e2 − 2 − e0 + 0 = e2 − 3.
y -axis
Area bounded by the
The area A bounded by the urve x = g(y) between y = a and y = b and the y−
axis, see Fig. 3 is given by
A=
Z
b
x dy =
a
Z
a
12
b
g(y) dy.
y
b
a
x = g(y)
x
and the
y−axis.
Note a negative result implies a net portion is to the left of the
y−axis.
Figure 3: The area trapped by
3.1.3
x = g(y), y = a, y = b
Area in polar oordinates
A of the region bounded by the urve r = f (θ) and
In polar oordinates the area
θ = θ 0 , θ = θ1 ,
the rays
see Fig. 4 is
Z
1
A=
2
θ1
θ0
1
r dθ =
2
2
Z
θ1
[f (θ)]2 dθ.
θ0
y
r = f (θ)
θ1
θ0
Figure 4: The area trapped by
r = f (θ)
and
Example Calulate the area trapped by the urve
x
θ = θ0 , θ = θ 1 .
r = θ
between
θ = 0
and
θ = π/4.
The area
A
is given by
A=
Z
0
3.1.4
π
4
¸π
·
1 3 4
1 2
π3
θ dθ = θ
.
=
2
6
384
0
Area under a urve given parametrially
A bounded by a urve given parametrially as y = f (t),
x = a and x = b and the x−axis is
Z b
Z t1
dg
A=
y dx =
f (t) dt
dt
a
t0
Area
13
x = g(t) between
where t0 , t1 are the points orresponding to
a = g(t0 ),
y = f (t),
and
b = g(t1 ),
see Fig. 5.
x = g(t)
y
a
t = t0
b
t = t1
Figure 5: The area trapped by
Example Calulate the area between
urve
y = sin t, x = cos t.
x = 0 = cos t
Now when
t=
we have
π
. Hene the required area
3
A=
Z
π
3
π
2
A
A=
3.2
3.2.1
π
3
π
2
is
x = 1/2,
π
and when
2
Z
π
3
π
2
the
x =
1
2
x−
axis and the
= cos t
− sin2 t dt.
2 sin2 t = 1 − cos 2t
to get
"
√ #
·
¸ π3
1 π
1 1
1
3
(cos 2t − 1) dt =
sin 2t − t =
+
.
2
2 2
2 6
4
π
2
Ar-length
Ar length in Cartesian Coordinates
Note that a small element of length
or also
t =
y = f (t), x = g(t).
and
d
sin t (cos t) dt =
dx
We an make use of the identity
Z
x=0
x
ds,
see Fig. 6 is given by
p
ds = dx2 + dy 2 =
ds =
p
dx2 + dy 2 =
14
r
1+(
s
1+(
dy 2
) dx,
dx
dx 2
) dy.
dy
we have
y
y
y = f (x)
y = f (x)
ds
dx
x
x
(a)
(b)
Figure 8: Curve y = f (x) rotated about x−axis to form a volume of revolution.
3.3
3.3.1
Surfae area and volume
Volume of solid of revolution
Consider a solid generated by rotating y = f (x) about the x−axis between x = a
and x = b, see Fig. 8.
The volume δV of the solid element of thikness δx obtained by rotating
y = f (x) about the x−axis an be alulated by assuming this is a thin ylinder
of radius y = f (x). The volume of this element is then given by
δV = πy 2 dx.
Integrating we obtain
V =
Z
b
Z
2
πy dx =
a
3.3.2
b
π[f (x)]2 dx.
a
Surfae area
The surfae area of the element δA is alulated by assuming that this is part of
one. For a setion of a irular one, see g. 9, the surfae area of the urved
portion is
S = 2πrl
where l is the length of the slanted edge and r is the average radius of the setion.
In our ase onsidering the element as part of a irular one, we have l = ds and
r = y + dy
. Then the surfae area of this small element
2
δA ≈ 2πy ds
to leading order (after negleting dyds terms) and so integration gives
A=
Z
x=b
x=a
2πy ds =
Z
b
s
f (x) 1 +
a
17
µ
df
dx
¶2
dx.
r1
l
r2
r = 21 (r1 + r2 )
Figure 9: Surfae area of a setion of a irular one is S = 2πrl.
3.4
Mean and rms values
Consider a signal y = f (t) for a ≤ t ≤ b. The mean value ȳ of the signal is
dened by
Z
ȳ =
b
1
(b − a)
f (t) dt.
a
The root mean square (rms) value ȳrms value is dened as
ȳrms =
3.5
½
1
(b − a)
Z
a
b
¾ 21
[f (t)] dt .
2
Centroids and entre of mass
The entroid (x̄, ȳ ) of a lamina with shape given by the area A trapped by the
urve y = f (x), the x−axis and the setions x = a and x = b is dened by
1
x̄ =
A
Z
b
1
ȳ =
2A
xf (x) dx,
a
Note that
A=
Z
b
Z
b
Z
b
yf (x) dx.
a
f (x) dx.
a
If the lamina is rotated about the x−axis to generate a solid with volume V .
The entre of mass of the solid has oordinates given by (X, 0) where
π
X=
V
x[f (x)]2 dx.
a
18
Here
V =
Z
b
2
πy dx =
Z
a
a
19
b
π[f (x)]2 dx.
4
Sequenes and Series
In this setion we will disuss sequenes, nite and innite series and this will
lead into Taylor and Malaurin Series for funtions. Additional exerises for you
to try are given in Examples 3 and 4 at the bak of this booklet.
Denition A sequene an is rule by whih we alulate a number (real or omplex) from a set of integer values.
Example: an = n1 generates a1 = 1, a10 = 1/10,
Example: an =
1
n(n+1)
gives
1
a1 = ,
2
1
a2 = ,
6
a5 =
1
.
30
We will often be interested in what happens to the sequene for large values
of n- this is alled taking the limit for large n, or as n → ∞.
Example: If an = 1 + n1 , then
a1 = 2,
a10 = 1 + 0.1 = 1.1,
a100 = 1.01,
a1000 = 1.001
and we see that as n beomes very large ie n → ∞ the sequene an → 1.
Example: If an = rn then if −1 < r < 1, an → 0 as n → ∞. If r = 1 then
an = 1 for all values of n. If r > 1 then an beomes indenitely large, ie we say
an → ∞ (read this as an tends to innity).
4.1 Finite and Innite Series
We an take a sum of a nite number of terms to form another sequene. Thus
dene
S N = a1 + . . . aN =
N
X
an .
n=1
Here SN is alled the N th partial sum.
Example: Suppose an = n then
SN = 1 + 2 + . . . N =
N (N + 1)
.
2
[For why the nal answer is as shown read the next setion℄
20
4.1.1
Arithmeti Progression
Suppose we start with a sequene with a0 = r and the rule is add d to the last
term to get the next term. Then this forms an arithmeti progression with
a1 = r,
a1 = r + d,
a2 = r + 2d,
an = r + (n − 1)d.
To nd the sum of an arithmeti progression let
SN = a1 + a2 + · · · + aN −1
SN = r + (r + d) + · · · + r + (N − 1)d.
(4.1)
SN = r + (N − 1)d + r + (N − 2)d + · · · + r.
(4.2)
We an write this bakwards as well to get
Hene adding the last two expressions (4.1,4.2) we see that
2SN = [r + (r + (N − 1)d] + [r + d + (r + (N − 2)d] + . . . [r + (N − 1)d + r].
Eah grouped term adds to 2r + (N − 1)d and there are N terms. Hene
2SN = [2r + (N − 1)d]N,
and so
SN =
N
(2r + (N − 1)d).
2
Notie that the sum of an arithmeti series an also be written as SN = (half the
sum of the rst and last terms times the total number of terms.
Suppose we take r = a and d = 1 then this gives the sequene an = n
and so sum of the rst N terms is given by
Example:
SN =
N (N + 1)
.
2
Find the sum of all the numbers from 1 to 100.
Here r = 1 and d = 1. We have
Examples:
S100 = 1 + 2 + . . . 100,
S100 = 100 + 99 + . . . 1.
Hene
2S = (101) ∗ 100 = 10100 =⇒ S100 = 5050.
21
4.2 Geometri progressions
Suppose we take a sequene suh that eah suessive term is r times the previous
term, and the rst term is a. Then
a1 = a,
a3 = ar2 ,
a2 = ar,
. . . , an = arn−1 .
Suh a sequene forms a geometri progression. To nd the sum to N terms we
an write
SN = a + ar + ar2 + · · · + arN −1 .
Hene if we multiply by r we obtain
rSN = ar + ar2 + · · · + arN .
Subtrating the last two expressions gives
SN (r − 1) = arN − a = a(rN − 1).
Hene
SN =
a(rN − 1)
.
r−1
Suppose you invest ¿100 in a bank whih give ompound interest rate
of 5% eah year. Find how muh money is in your aount after 10 years, assuming
that the interest rate remains the same and no money has been withdrawn.
Here a = 100 and at the end of the year to get the new sum we add the
interest, ie the new total at the end of the year is
Example:
a+
5
a = 1.05a.
100
We need to multiply the new sum eah year by 1.05 and so we have a geometri
progression with r = 1.05. The total sum after 10 years is
a10 = ar10 = 100 ∗ 1.0510 = £162.89.
4.3 Innite Series
Suppose we have a partial sum SN to N terms of a sequene an and we investigate
what happens as N beomes very large. If SN remains nite as N → ∞ and
approahes the value S we say that the sum of the innite series
a1 + · · · + a n + · · · =
22
∞
X
n=1
an = S.
In this ase the series is said to onverge. If SN does not approah a nite value
as N → ∞ the series is said to diverge.
Example
Consider an =
1
n(n+1)
an =
for n ≥ 1. We see that
1
1
1
= −
,
n(n + 1)
n n+1
and hene the partial sum to N terms is
S N = a1 + a2 + . . . aN
1
1 1
1
1
SN = (1 − ) + ( − ) + · · · + ( −
).
2
2 3
N
N +1
Notie that there is a lot of anellation and we are left with
SN = 1 −
1
.
N +1
We an see that as N → ∞ then SN → 1 and so
1=
∞
X
n=1
1
1 1
1
= + +
+ ....
n(n + 1)
2 6 12
Consider the sum to N terms of a geometri progression with rst
term a and ommon ratio r. We have
Example
SN = a
(rN − 1)
.
r−1
We an see that if −1 < r < 1 then rN → 0 as N goes to innity. So then
SN → S =
and
a + ar + ar2 + · · · + arn−1 + · · · =
If |r| ≥ 1 the series diverges.
Example
.
1+
a
,
1−r
∞
X
arn−1 =
n=0
1
1
1 1
+ + ··· + n + ··· =
2 4
2
1−
1
2
a
,
1−r
|r| < 1.
= 2.
Suppose we take a = 1 and r = x then
1 + x + x2 + . . . xn + · · · =
23
1
,
1−x
|x| < 1.
(4.3)
The latter (4.3) is an example of a power series. Notie that if we put r = −x
we then obtain
1 − x + x2 − x3 + . . . (−1)n xn + · · · =
1
,
1+x
(4.4)
|x| < 1.
Both the above power series only onverge when |x| < 1.
Suppose we integrate the series (4.4) for 1/(1 + x) term by term (and assume
that this is allowed) then we obtain
Z
dx
x2 x3
xn
= ln |(1 + x)| + C = x −
+
+ · · · + (−1)n+1 + . . . .
1+x
2
3
n
Setting x = 0 on both sides gives C = 0 and thus
∞
X
xn
ln(1 + x) =
(−1)n+1 ,
n
n=1
|x| < 1.
One an show that the above series also onverges for x = 1. From this we an
dedue that
∞
X
1+x
x3 x5
x2n+1
x2n+1
ln |
| = 2(x +
+
+ ...
+ ...) = 2
1−x
3
5
(2n + 1)
(2n + 1)
n=1
|x| < 1.
Next onsider the series (4.4) for 1/(1 + x) and replae x by x2 to get
1
= 1 − x2 + x4 + · · · + (−1)n x2n + . . . .
1 + x2
Next integrate this term by term to nd
Z
2n+1
x3 x5
dx
−1
n x
=
tan
x
+
C
=
x
−
+
+
·
·
·
+
(−1)
+ ....
1 + x2
3
5
2n + 1
Again evaluating both sides for x = 0 gives C = 0 and thus
tan−1 x =
∞
X
n=1
(−1)n
x2n−1
x3 x5
x2n−1
= x − + + · · · + (−1)n
+...,
(2n − 1)
3
5
2n − 1
|x| < 1.
In general a power series is a series of the form
∞
X
n=0
an x n = a0 + a1 x + a2 x 2 + · · · + a n x n + . . . .
where the an are onstants. If the power series onverges for x = R(6= 0) then
the series will onverge for all |x| < R. If the series diverges for x = R then the
series will diverge for |x| > R. The maximum value of R for whih the series
onverges is alled the radius of onvergene.
24
5
Taylor and Malaurin Series
Suppose we have a funtion f (x) whih is dierentiable N times in an interval.
Then we dene a Taylor Polynomial Pn (x) or order n, (n = 0, 1, . . . , N ) about
x = a by the formula
Pn (x) = f (a)+f ′ (a)(x−a)+
f ′′ (a)
(x − a)k
(x − a)n
(x−a)2 +· · ·+f (k) (a)
+· · ·+f (n) (a)
.
2
k!
n!
(5.1)
Consider the funtion f (x) = ex . We an obtain the Taylor polynomial for ex or order 3 about x = 0 as follows.
Example:
f (x) = ex ,
f ′ (x) = ex ,
f ′′ (x) = ex ,
f ′′′ (x) = ex ,
f (0) = e0 = 1,
f ′ (0) = 1,
f ′′ (0) = 1,
f ′′′ (0) = 1.
So
P3 (x) = 1 + x +
x2 x3
+ .
2
6
Consider the funtion f (x) = sin x. We will obtain the Taylor polynomial for sin x of order 3 about x = 0.
Example:
f (x) = sin x ,
f ′ (x) = cos x ,
f ′′ (x) = − sin x,
f ′′′ (x) = − cos x,
f (0) = sin 0 = 0,
f ′ (0) = 1,
f ′′ (0) = 0,
f ′′′ (0) = −1.
So
P3 (x) = x −
5.1
x3
.
6
Taylor's theorem
Suppose a funtion f (x) is dierentiable n times in some inteval ontaining a and
x. Then under ertain onditions Taylor's theorem states that we an write
f (x) = Pn (x) + Rn (x)
where
Pn (x) = f (a)+f ′ (a)(x−a)+
(x − a)k
(x − a)n
f ′′ (a)
(x−a)2 +· · ·+f (k) (a)
+· · ·+f (n) (a)
,
2
k!
n!
25
is the
term
nth order Taylor polynomial for f (x) about x = a and Rn (x) is a remainder
(and the error in approximating f (x) by Pn (x)) whih is given by
Rn (x) =
for some number
When
Rn (x)
obtain the
f (n+1) (c)
(x − a)n+1
(n + 1)!
a < c < x.
n→∞
goes to zero as
∞
X
f (n) (a)
n!
n=0
The
x
inside some interval then we
Taylor Series for the funtion f (x) about x = a given by
f (x) =
a = 0.
for all
(x − a)n .
Malaurin Series is a speial ase of the Taylor series when we take
Example if we take the funtion f (x) = ex then
ex = 1 + x +
x2 x3
xn
+
+ ...
+ Rn (x)
2
6
n!
where
Rn (x) =
We an show that
Rn (x) → 0
n→∞
as
x
e =
ec
xn+1 .
(n + 1)!
∞
X
xn
n=0
n!
for all
,
x
for all
and hene
x.
Other important funtions have the following Taylor series expansions:
Taylor series of elementary funtions
1. Geometri series
∞
X
1
=
zn = 1 + z + z2 + . . . ,
1−z
n=0
2.
|z| < 1.
(5.2)
Exponential series
z
e =
∞
X
zn
n=0
This onverges for all
n!
=1+z+
z.
26
z2 z3
+
+ ....
2
6
(5.3)
3.
Sine series
sin z =
∞
X
(−1)n
n=0
This onverges for all
4.
z3
z5
z 2n+1
=z−
+
+ ...
(2n + 1)!
6
120
z.
Cosine series
∞
X
z 2n
z2 z4
cos z =
(−1)
=1−
+
+ ...
(2n)!
2
24
n=0
This onverges for all
5.
n
Hyperboli funtions
∞
X
n=0
This onverges for all
This onverges for al
z 2n+1
z3
z5
=z+
+
+ ...
(2n + 1)!
6
120
(5.6)
z.
cosh z =
z.
∞
X
z2 z4
z 2n
=1+
+
+ ...
(2n)!
2
24
n=0
(5.7)
Logarithm
ln(1 + z) =
∞
X
(−1)n+1
n=1
ln(1 − z) = −
ln
7.
(5.5)
z.
sinh z =
6.
(5.4)
µ
1+z
1−z
¶
zn
z2 z3
=z−
+
+ ...,
n
2
3
∞
X
zn
n=1
n
= −z −
z2 z3
−
+ ...,
2
3
∞
X
z 2n−1
z3
=2
= 2(z +
+ . . . ),
(2n − 1)
3
n=1
|z| < 1.
|z| < 1.
|z| < 1.
(5.8)
(5.9)
(5.10)
Binomial expansion
−m
(1+z)
=
¶
∞ µ
X
−m
n=0
n
z n = 1−mz+
m(m + 1) 2 m(m + 1)(m + 2) 3
z −
z +. . . ,
2!
3!
(5.11)
27
|z| < 1.
6
Funtions of more than one variable,
partial
derivatives
So far we have met funtions of just one variable say y = f (x). However in many
appliations we have funtions of two or more variables.
Example The power P
dissipated by a resistor is given by
P (r, v) =
v2
r
where r is the resistane and v is the instantaneous voltage.
Example Energy losses E due to urrents are given by
E(ω, B) = kω 2 B 2
where k is a onstant, ω is the frequeny and B the magneti strength.
Example The speed v of a partile satises
s(x, y, z) =
p
u2 (x, y, z) + v 2 (x, y, z) + w2 (x, y, z)
where u, v, w are the veloity omponents and (x, y, z) the partile position.
In the rst two examples P and E are funtions of two variables and in the
last example s is a funtion of three variables. Funtions of one variable an
be thought of as urves in the plane, whilst funtions of two variables an be
visualized as a surfae in three-dimensional spae.
1
0
1.0
-1
0.5
-2
-1.0
0.0 y
-0.5
0.0
-0.5
x
0.5
1.0
-1.0
Figure 10: A plot of the surfae generated by the funtion z = x2 − 2y 2 .
28
6.1 Partial dierentiation
Consider a funtion of two variables say z = f (x, y). We are often interested
in rates of hange of the funtion in dierent diretions and need to alulate
derivatives. For example the rst derivative of z with respet to x is written as
∂z
∂x
and alled the partial derivative of z with respet to x. Mathematially this is
alulated by looking at small hanges in the x diretion keeping y xed, ie
∂z
f (x + h, y) − f (x, y)
≈
∂x
h
with h approahing zero. Similarly, dierentiation in y is written as
∂z
∂y
and
∂z
f (x, y + k) − f (x, y)
= lim
.
∂y k→0
k
The notation limk→0 should be read as taking the limit of the quantity for k
approahing zero. Here we treat x as xed and look at small hanges in the y
diretion.
Example Consider z = xy + sin(x2 + y). Then
∂z
= y + 2x cos(x2 + y),
∂x
∂z
= x + cos(x2 + y).
∂y
Notie that in dierentiating with respet to x we keep y xed and in dierentiating with respet to y we keep x xed.
Higher derivatives an be alulated in a similar way. Thus
Also
¢
∂ 2z
∂ ¡
2
=
y
+
2x
cos(x
+
y)
= 2 cos(x2 + y) − 4x2 sin(x2 + y).
∂x2
∂x
¢
∂ 2z
∂ ¡
2
=
x
+
cos(x
+
y)
= − sin(x2 + y).
2
∂y
∂y
We an also ompute mixed derivatives suh as
µ ¶
∂ ∂z
∂2z
,
=
∂x∂y
∂x ∂y
=
¢
∂ ¡
x − cos(x2 + y) = 1 − 2x sin(x2 + y).
∂x
29
Also
∂
∂2z
=
∂y∂x
∂y
=
We see that
µ
∂z
∂x
¶
,
¡
¢
∂
= y + 2x cos(x2 + y) = 1 − 2x sin(x2 + y).
∂y
∂2z
∂2z
=
.
∂x∂y
∂y∂x
A more ompat notation that is often used
is fx whih means
the partial
∂2f
∂2f
derivative with respet to x. Also fxx means ∂x2 and fxy means ∂x∂y .
Note that if fxy and fyx are ontinuous at a point (a, b) then fxy = fyx at
(a, b).
6.2 Chain Rule
Consider a funtion z = f (x, y) of two variables and x and y are themselves
funtions of u and v . Then z is also a funtion of u and v , say F (u, v) where
F (u, v) = f (x(u, v), y(u, v)).
Let us alulate the partial derivative of z with respet to u. Now
F (u + δu, v) − F (u, v)
∂z
≈
.
∂u
δu
But the point (u + δu, v) will orrespond to a point (x + δx, y + δy) and (u, v) to
a point (x, y). Hene
F (u + δu, v) − F (u, v)
f (x + δx, y + δy) − f (x, y)
=
,
δu
δu
¶
¶
µ
µ
f (x + δx, y + δy) − f (x, y + δy) δx
f (x, y + δy) − f (x, y) δy
=
+
.
δx
δu
δy
δu
If we let δu go to zero we obtain
∂z ∂x ∂z ∂y
∂f ∂x ∂f ∂y
∂z
=
+
=
+
.
∂u
∂x ∂u ∂y ∂u
∂x ∂u ∂y ∂u
Similarly
∂z
∂z ∂x ∂z ∂y
∂f ∂x ∂f ∂y
=
+
=
+
.
∂v
∂x ∂v ∂y ∂v
∂x ∂v
∂y ∂v
30
This is alled the hain
rule for partial dierentiation.
Example Suppose x = u + v, y = u − v and z = x2 + y2.
and
∂z
∂v
using the hain rule.
We an alulate
∂z
∂u
∂z
∂z ∂x ∂z ∂y
=
+
.
∂u
∂x ∂u ∂y ∂u
Now
and so
Similarly
∂z
= 2x + 2y = 2(x + y) = 4u.
∂u
∂z ∂x ∂z ∂y
∂z
=
+
,
∂v
∂x ∂v ∂y ∂v
n and
Hene
∂y
= 1,
∂u
∂z
= 2y,
∂y
∂x
= 1,
∂u
∂z
= 2x,
∂x
∂y
= −1.
∂v
∂x
= 1,
∂v
∂z
= 2x − 2y = 2(x − y) = 4v.
∂v
We an also alulate
∂z
∂u
and
∂z
∂v
diretly. Sine
z = x2 + y 2 = (u + v)2 + (u − v)2 = u2 + 2uv + v 2 + u2 − 2uv − v 2 = 2(u2 − v 2 ).
Thus
and
∂z
= 4u,
∂u
∂z
= −4v.
∂v
Example
Suppose z = f (x, y) and x = r cos θ,
y = r sin θ. Then
∂z
∂z ∂x ∂z ∂y
=
+
.
∂r
∂x ∂r ∂y ∂r
Now
∂x
= cos θ,
∂r
∂y
= sin θ,
∂r
31
and so
∂z
∂z
∂z
= cos θ
+ sin θ ,
∂r
∂x
∂y
= cos θ
Also
∂f
∂f
x ∂f
y ∂f
+ sin θ
=
+
.
∂x
∂y
r ∂x r ∂y
∂z
∂z ∂x ∂z ∂y
=
+
.
∂θ
∂x ∂θ ∂y ∂θ
Now
∂x
= −r sin θ,
∂θ
and so
∂y
= r cos θ,
∂θ
∂z
∂z
∂z
= −r sin θ
+ r cos θ ,
∂r
∂x
∂y
= −r sin θ
∂f
∂f
∂f
∂f
+ r cos θ
= −y
+x .
∂x
∂y
∂x
∂y
Example Suppose that z = f (x, y) and x = x(u), y = y(u).
hain rule gives
Then applying the
∂z
∂f ∂x ∂f ∂y
=
+
.
∂u
∂x ∂u ∂y ∂u
But sine now z is just a funtion of u, ie of one variable we an write the partial
derivative as the total derivative ie,
∂f ∂x ∂f ∂y
dz
=
+
.
du
∂x ∂u ∂y ∂u
Note that x = x(u), y = y(u) are the parametri equations of a urve in the x − y
plane, and so the total derivative above gives the rate of hange along this urve.
6.3
Small hanges
Suppose that we have z = f (x, y) and we have small hanges dx and dy in x and
y . What is the onsequent hange in z ? Now
dz = f (x + dx, y + dy) − f (x, y)
µ
¶
µ
¶
f (x + dx, y + dy) − f (x, y + dy)
f (x, y + dy) − f (x, y)
=
dx +
dy.
dx
dy
Hene for small dx, dy we obtain
dz =
∂f
∂f
dx +
dy.
∂x
∂y
32
In general if z = f (x1 , x2 , . . . , xn ) is a funtion of n independent variables
then
n
dz =
X ∂f
∂f
∂f
∂f
dx1 +
dx2 + . . .
dxn =
dxk ,
∂x1
∂x2
∂xn
∂xk
k=1
is alled the total dierential.
Example Consider the power P
dissipated in a resistor given by
P =
v2
,
r
where v is the instantaneous voltage and r is the resistane. Calulate the maximum perentage error in alulating P if r = 2 ± 0.01Ohms and v = 6 ± .1volts.
We have
dP =
and
∂P
∂P
dr +
dv,
∂r
∂v
v2
∂P
= − 2,
∂r
r
Hene
dP = −
and
2
− v2 dr +
dP
= r v2
P
r
∂P
2v
= .
∂v
r
2v
v2
dr + dv,
2
r
r
2v
dv
r
2
1
= − dr + dv.
r
v
So the maximum possible positive error in P is given by taking dr negative and
dv positive and is
dP
1
2
= ∗ (.01) + (0.1) = 0.0833.
P
2
6
The maximum possible perentage error in P is 8.33% . The minimum perentage
value is given by taking dr positive and dv negative with
1
2
dP
= − ∗ (.01) − (0.1) = −0.0833.
P
2
6
The minimum perentage error is -8.33%.
6.4
Gradient
Consider the funtion the funtion of two variables W = f (x, y). From earlier we
have the total dierential
dW =
∂f
∂f
dx +
dx.
∂x
∂y
33
Then for a small hange along a urve element ds we have
dW
∂f dx ∂f dy
≈
+
.
ds
∂x ds ∂y ds
From geometry, see g. 11, we see that
dx
= cos θ,
ds
PSfrag replaements ds
dy
= sin θ
ds
dy
θ
dx
Figure 11: Element of length along a urve and ds = ds t = ds (cos θ i + sin θ j) .
Hene for ds going to zero we obtain
dW
∂f
∂f
= cos θ
+ sin θ
= t.∇f,
ds
∂x
∂y
where
∇f = (
∂f ∂f
∂f
∂f
,
)=
i+
j
∂x ∂y
∂x
∂y
is alled the gradient vetor of the funtion W = f (x, y). The derivative
gives the rate of hange of W in the diretion
t = (cos θ, sin θ) = cos θi + sin θj,
and i, j are the unit vetors in the x and y diretions respetively.
Notie that if dW
= 0 then ∇f is normal to the urve t.
ds
Example The surfaes of onstant eletri potential satisfy
φ(x, y) = c
for dierent c. If we onsider a small hange along this surfae then
dφ = φ(x + dx, y + dy) − φ(x, y) = 0 = ∇φ.t.
34
dW
ds
∇φ
PSfrag replaements
φ=c
Figure 12: Equipotentials φ(x, y) = x2 + y 2 = c with the gradient vetor ∇φ.
So the gradient ∇φ is perpendiular to the surfaes of onstant potential. In
eletrostatis the eletri eld E is given by
E = −∇φ.
Example Suppose φ(x, y) = x2 + y2. Then the equipotentials, ie urves of on-
stant φ satisfy
φ(x, y) = c = x2 + y 2 .
√
For c > 0 these are irles with radius c. Next
∇φ =
∂φ
∂φ
i+
j = 2xi + 2yj.
∂x
∂y
But if P is the point with oordinates (x, y) then
OP = r = xi + yj
and so
∇φ = 2r
and is in the diretion of the radial vetor from the origin, g. 12.
Example Suppose φ(x, y) = x2 + sin(xy) then
∇φ = (2x + y cos(xy))i + x cos(xy)j) .
We an generalize the onept of a gradient to funtions of more than two
variables. In three-dimensions the gradient vetor for the funtion f (x, y, z) is
dened as
¶
µ
∇f =
∂f ∂f ∂f
,
,
∂x ∂y ∂z
35
.
7
Line integrals and multiple integrals
Suppose y = f (x) is a single-valued real ontinuous funtion of x in some interval
a < x < b and is represented by the urve C , see g. 13. We an onstrut the
line integral along the urve C of some funtion P (x, y) as
I=
Z
P (x, y) dx,
C
or
J=
Z
P (x, y) dy.
C
The rst of these is just
Z
Z
P (x, y) dx =
x=b
P (x, f (x)) dx.
x=a
C
Notie that the line integral implies a spei diretion from the point where
x = a to the point where x = b and so
Z
−C
P (x, y) dx = −
Z
P (x, y)dx.
C
y
PSfrag replaements
C
b
a
x
Figure 13: Integral along a urve C .
Now sine
dy = f ′ (x) dx
the seond line integral an be written as
J=
Z
P (x, y) dy =
Z
x=b
P (x, f (x))f ′ (x) dx.
x=a
C
In many appliations in engineering and physis, the urve C an be losed,
see g. 14.
Example Ampere's Law states that the urrent passing through a losed urve
C is related to the magneti eld B by
I
I
µ0 I = B.ds = (B1 dx + B2 dy + B3 dz)
C
C
36
where B = (B1 , B2 , B3 ) in artesian oordinates and µ0 is a onstant.
Example In uid mehanis the irulation K around a losed urve C is given
by
K=
I
U.ds =
C
I
(u dx + v dy + w dz),
C
where U = (u, v, w) is the uid veloity.
B
C
PSfrag replaements
Figure 14: Current indued by a magneti eld B passing through a iruit C .
Example Find
Z
(x2 + y 2 ) dx
C
along the urves C given by (a) x + y = 1 from x = 0 to x = 1; (b) the urve
given parametrially by y = t, x = 2t2 from (x, y) = (0, 0) to (x, y) = (2, 1).
(a) Figure 15(a) shows the line y = 1 − x and so
I=
Z
2
2
(x + y ) dx =
1
Z
0
C
(x2 + (1 − x)2 ) dx = [
x3 (1 − x)3 1 2
−
]0 = .
3
3
3
(b) The urve C is given by y = t, x = 2t2 , see g. 15(b), and so dx = 4tdt.
Also when (x, y) = 0, t = 0 and when (x, y) = (2, 1), t = 1. Hene
I=
Z
2
2
(x + y ) dx =
1
(4t4 + t2 )4t dt.
0
C
We nd
Z
·
16 6
t + t4
I=
6
¸1
0
Example The magneti eld B at a point P
by
B=
=
11
.
3
from a long straight wire is given
x
−y
i+ 2
j.
2
+y
x + y2
x2
37
PSfrag replaements
y
y
PSfrag replaements
1
1
x+y =1
x
1
0
x = 2t2
y=t
O
(a)
x
2
(b)
Figure 15: (a) C : x + y = 1, (b) C :
x = 2t2 ,
y = t.
Calulate the urrent passing through the iruit given by x = a cos θ, y =
a sin θ, 0 ≤ θ ≤ 2π . We have
µo I =
I
(B1 dx + B2 dy),
C
and x = a cos θ, y = a sin θ. So
dx = −a sin θ,
dy = a cos θ.
This gives
µ0 I =
Z
0
2π
a2 cos2 θ
a2 sin2 θ
dθ
+
dθ =
a2
a2
Hene
I=
7.1
Z
2π
dθ = 2π.
0
2π
.
µ0
Path Independene
It should be noted that the line integral usually depends not only on the endpoints, but also on the urve over whih the integral is taken. In this setion we
examine ases when the value of the integral does not depend on the urve C .
Suppose we alulate the integral
I=
Z
P (x, y)dx + Q(x, y)dy,
C
along some path C . If we an nd a funtion φ suh that
∂φ
= P (x, y),
∂x
38
∂φ
= Q(x, y)
∂y
then
P (x, y)dx + Q(x, y)dy =
Hene
I=
Z
Z
P dx + Qdy =
C
C
∂φ
∂φ
dx +
dy = dφ.
∂x
∂y
dφ = (φ(B) − φ(A)),
where B, A are the endpoints of the path C and φ(B) denotes the value of φ(x, y)
at the point B .
If suh a φ an be found then the above shows that the integral I is independent of the path taken from A to B , see g. 16.
y
PSfrag replaements y1
B
A
y0
y0
x
x1
x0
Figure 16: If the integral is path independent, then
on the values of φ at the endpoints A and B .
R
If C is a losed path then in this ase
Z
(
C
∂φ
∂φ
dx +
dy) = 0.
∂x
∂y
Example Consider
I=
Here we see that if
Z
(x dx + y dy).
C
∂φ
= x, ⇒ φ =
∂x
∂φ
= y, ⇒ φ =
∂y
x2
+ g(y),
2
y2
+ h(x).
2
Comparing the two expressions suggests that
1
φ(x, y) = (x2 + y 2 ) + C.
2
39
C
P dx + Qdy depends only
So I is independent of the path and
I = [φ(x, y)]B
A = φ(x1 , y1 ) − φ(x0 , y0 ),
where B = (x1 , y1 ), A = (x0 , y0 ).
A neessary and suient ondition to hek whether an integral of the form
Z
P (x, y)dx + Q(x, y)dy
C
is path independent is that
∂Q
∂P
=
.
∂y
∂x
Example Consider
Z
C
(x2 dx − 2xy dy),
where C onsists of staight line segments joining the points O = (0, 0), A = (1, 0)
and B = (0, 1), see g. 17.
y
PSfrag replaements
B
A
O
x
Figure 17: Sketh of losed urve OAB for example.
Here P (x, y) = x2 ,
Q(x, y) = −2xy and we see that
∂P
∂Q
= 0 6=
= −2y.
∂y
∂x
So we need to ompute the integral along the given path.
On OA we have y = 0 so dy = 0 and thus
IOA =
Z
1
x2 dx =
0
x3
1
= .
3
3
On AB we have x + y = 1 and so dy = −dx and we nd
Z
AB
=
Z
1
0
2
(x dx + 2x(1 − x)dx) = −
Z
1
¸1
· 3
2
x
2
=− .
(−x + 2x) dx = − − + x
3
3
0
2
0
40
Example Consider
Z
2
3
Z
Z
3
x3 x
+
=
3
3
¸3
1
2
2
(x + y ) dx dy =
2
0
·
7.3
2
·
y3
x y+
3
2
= 10 −
¸1
dx =
Z
2
0
3
1
[x2 + ] dx
3
10
20
= .
3
3
Non-retangular regions
Suppose next we have a non-retangular region R. Here we may use the same
approah as before and subdivide the region into smaller elements and sum over
the smaller elements and take the limit as the the size of the elements approahed
zero. If the region is suh that it is bounded by the urves
tL (x) ≤ y ≤ tU (x) for a ≤ x ≤ b
see g. 19(a) then we an use vertial strips to divide the region. The integral is
then
Z Z
Z
Z
x=b
tU (x)
f (x, y) dx dy =
f (x, y) dy dx.
x=a
R
tL (x)
If the region is instead bounded by the urves, see g. 19(b),
sL (y) ≤ x ≤ sR (y) for c ≤ y ≤ d
then if we divide the region into horizontal strips we nd
Z Z
f (x, y) dx dy =
y=d
Z
y=c
R
Example Consider
Z Z
I=
Z
sR (y)
f (x, y) dx dy.
sL (y)
xy dx dy
R
where R is the triangle with verties O = (0, 0), A = (1, 0) and B = (1, 1). If we
divide the region into vertial strips then sine OB is given by y = x we have
I=
Z
x=1
x=0
Z
y=x
2
xy dy dx =
y=0
=
Z
x=1
x=0
Z
0
1
x4
1
dx = .
3
15
42
· 3 ¸x
y
dx,
x
3 0
PSfrag replaements
y
d
y
x = sL (x)
x = sR (x)
dy
y = tU (x)
dx
c
y = tL (x)
b
a
x
x
(a)
(b)
Figure 19: Dividing the region R into (a) vertial strips and (b) horitzontal strips.
However if we divide the region into horizontal strips we obtain
I=
Z
1
y=0
=
Z
0
1
Z
x=1
2
xy dx dy =
x=y
Z
0
1
·
x2 2
y
2
¸1
dy,
y
1 2
1 y3 y5
1 1 1
1
[y − y 4 ] dy = [ − ]10 = ( − ) = .
2
2 3
5
2 3 5
15
43
8
Examples
8.1
Examples 1
Examples 1
1. Sketh the regions desribed in Cartesian oordinates (x, y) by eah of the
following examples and nd the area of eah region.
(a)
(b)
()
(d)
The
The
The
The
region enlosed between the urves y = x and y = x4 .
region enlosed between the urves x2 = 2y and y 2 = 2x.
region between y = sin x, y = ex , x = 0 and x = π/2.
region between y = ln x, y = 1, y = 0 and x − y 2 + 1 = 0.
2. Sketh the regions desribed in Polar oordinates (r, θ) by eah of the following examples and nd the area of eah region.
(a) The region surrounded by the `ardioid' r = 2(1+cos θ) for θ ∈ [0, 2π].
(b) The region between the radial line θ = 0 and the spiral r = θ for
θ ∈ [0, 2π].
3. (a) Use integration to nd the volume and √surfae area of the sphere
formed by rotating the semi-irle y = 4 − x2 of radius 2, about
the x−axis.
(b) Find the volume and surfae area of the shape reated by rotating the
parametri urve x = 2(t − sin t), y = 2(1 − cos t) for 0 ≤ t ≤ 2π
about the x−axis.
4. Sketh the following urves and nd their lengths.
(a) The urve y = 2x3/2 from x = 0 to x = 1.
(b) The parabola y = x2 /2 from x = −1 to x = 3.
() The urve given parametrially by x = 2t2 , y = 1 + 4t with − 14 ≤
t ≤ 1.
5. Find the mean and rms values for the following signals:
(a) y(t) = 1 + 2 sin t with 0 ≤ t ≤ π/2.
(b) y(t) = cos 3t cos t with 0 ≤ t ≤ 2π.
() y(t) = e−t sin t with 0 ≤ t ≤ π .
44
6. The entroid (x̄, ȳ ) of a lamina with shape given by the area A trapped by
the urve y = f (x), the x−axis and the setions x = a and x = b is dened
by
Z
Z
x̄ =
1
A
b
xf (x) dx ȳ =
a
1
2A
b
yf (x) dx.
a
√
(i) Calulate the entroid of the lamina with shape y = x − 1 trapped
between x = 1 and x = 2.
(ii) The lamina is rotated about the x−axis to generate a solid with volume
V . The entre of mass of the solid has oordinates given by (X, 0) where
π
X=
V
Z
b
x[f (x)]2 dx.
a
Calulate the oordinates of the entre of mass for the shape given in (i).
Seleted answers to Examples 1
1. (a) 103 ,
(b) 43 ,
() eπ/2 − 2,
(d) e − 31 .
2. (a) 6π ,
(b) 43 π 3 .
, surfae area =16π .
3. (a) Volume= 32π
3
.
(b) Volume= 40π 2 , surfae area= 256π
3
4. (a)
− 1) ,
√
√
(b) 12 (3 10 + 2 + sinh−1 3 + sinh−1 1).
√
√
() 2 2 + 817 + 2 sinh−1 41 + 2 sinh−1 1.
2
(103/2
27
5. (a) ȳ = 1 + π4 ,
ȳrms =
√1 (3π
π
+ 8)1/2 .
(b) ȳ = 0, ȳrms = 21 .
() ȳ =
1+e−π
,
2π
ȳrms =
6. (x̄, ȳ) = (8/5, 3/8),
√1 (1
8π
X=
− e−2π )1/2 .
5
3
45
8.2
Examples 2
Examples 2
1. Find
Z
1
0
4t3 − 2t2 + 3t − 1
dt.
2t2 + 1
2. The total energy E dissipated in a resistor is given by
E=
Z
∞
P (t) dt,
0
where
V 2 − 2t
e SC
P (t) =
S
is the power disspated in the resistor at time t. You may assume that V, S
and C are onstant. Show that
E=
CV 2
.
2
3. The `Fourier Transform' of a signal sin t is given by the integral
F (p) =
Z
∞
e−pt sin t dt.
0
Use integration by parts to show that
F (p) =
1
.
1 + p2
4. Write out the rst 5 terms of the sequene an dened by:
n+1
, n ≥ 1.
n
n
(b) an = (−1)n 2n! , n ≥ 0.
(a) an =
() an = an−1 + 1, a0 = 1.
(d) an = 2e−an−1 , a0 = 1.
1
(e) an = 2an−1
, a0 = 1.
+1
For parts (a),(d) and (e) what value does an tend to for large n?
5. A digital ounter is set to ount in steps of 2 eah time a button is pressed.
If the ounter starts at 1, what does it show after after the button has
been pressed 100 times. A seond adder iruit adds eah number that is
displayed on the digital ounter and displays the running sum. What does
the adder show after 100 button presses?
46
6. In eah pass through an amplier, the input signal is inreased by 2%. If
the initial signal is 10 db, what is the signal strength after passing through
10 amplier stages.
7. A student has to take out a loan to pay ¿ 9000 in fees and ¿ 5000 to pay for
his food and lodging. His Eletrial Engineering degree ourse lasts for 4
years and the fees and subsistene osts remain the same eah year. At the
end of eah year 3% interest is added to the aumulated amount of loan.
What does he owe after his degree ourse nishes? Beause he is unable
to nd a high paying job, he annot repay anything for at least 10 years.
What does he owe 10 years after he nished his degree?
8. Find the following innite sums:
(a)
∞ µ ¶n
X
1
2
n=0
∞
X
.
µ ¶n
1
.
(−1)
(b)
2
n=0
()
∞
X
n
e−n
n=0
9. Assuming that |r| < 1 and that
∞
X
arn−1 =
n=1
nd
∞
X
n=2
a(n − 1)rn−2 .
Seleted answers to Examples 2
1.
1
4
2.
V 2C
.
2
log 3.
3. F (p) =
a
,
1−r
1
.
1+p2
4.
5. S100 = 10000.
6. 11.951db
47
7.
8. (a) 2
(b) 23
() 1.582
9.
a
.
(1 − r)2
48
8.3
Examples 3
Examples 3
1. Use the formulae given in the notes to write out the rst 5 non-zero terms
in the power series expansions about x = 0 expansions for the following
funtions:
(a) e−3x , (b) cos x, () sin 2x, (d)
1
,
1 − x2
(e) (1 + x)−5 , (f) (1 + x2 ) 2 x.
3
2. Find the Taylor polynomial of order N for the following funtions about
the given value of x = a and for the given N .
(a) x sin x,
N = 4.
1
(b) (1 − x2 )1/2 , a = , N = 2.
2
() tanh x, a = 0, N = 3.
(d) e−x ,
2
(e)
x+1
,
x−1
a = π,
a = 1,
a = 0,
N = 3.
N = 4.
3. The power P (t) dissipated in a resistor varies as
P (t) = I 2 (t)R(t)
where I(t) is the urrent and R(t) is the resistane. If the urrent and
resistane vary in time as
2
I(t) = 1 − e−t ,
and R(t) = 5
(t + 1)
,
(t + 2)
use a rst order Taylor polynomial to approximate P (t) about t = 1. Using
any suitable plotting pakage, plot the atual P (t) and ompare with values
from your seond order Taylor polynomial for 0 ≤ t ≤ 2.
4. In a mathematial model of a semiondutor, the diode urrent I at room
temperature is given by
I(V ) = Is (e40V − 1)
where V is the applied voltage and Is is the saturation urrent. If the diode
operates at a voltage Va , nd a seond order Taylor approximation for I(V )
about the operatimg voltage Va .
5. The position x(t) of a partile at time t is given by
√
dx
= x + t + 1,
dt
49
x(t = 0) = 2.
(a) Estimate x(0.1) using a rst order Taylor polynomial.
(b) Estimate x(0.1) using a seond order Taylor polynomial.
Seleted answers to Examples 3
1.
2. (a) P4 (x) = −π(x − π) − (x − π)2 + π6 (x − π)3 + 61 (x − π)4 .
(b) P2 (x) =
√
3
2
−
√1 (x
3
− 21 ) −
4
√
(x
3 3
− 12 )2 .
() P3 (x) = x − x3 .
(d) P3 (x) = e−1 − 2e−1 (x − 1) + e−1 (x − 1)2 + 32 e−1 (x − 1)3 .
(e) P4 (x) = −1 − 2x − 2x2 − 2x3 − 2x4 .
3
3. P1 (t) = P (1) + (t − 1)P ′ (1) = 1.33192 + 3.32258(t − 1).
4. P2 (V ) = Is (e40Va − 1) + 40Is e40Va (V − Va ) + 800Is e40Va (V − Va )2 .
5. (a) x(0.1) = x(0) + (0.1)x′ (0) = 2 + 0.1 ∗ 3 = 2.3.
2
(b) x(0.1) = x(0) + 0.1 ∗ x′ (0) + 0.12 x′′ (0) = 2 + 0.3 + 0.0175 = 2.3175.
50
8.4
Examples 4
Examples 4
1. Find the partial derivatives
ing funtions:
(a)
(b)
()
(d)
∂f ∂f ∂ 2 f ∂ 2 f ∂ 2 f ∂ 2 f
,
,
,
,
,
, for the follow∂x ∂y ∂x2 ∂x∂y ∂y∂x ∂y 2
f (x, y) = 2x2 − xy + y 2 .
f (x, y) = (x + y 2 )3 .
f (x, y) = (3y 2 + x2 + 2xy)1/2 .
f (x, y) = exy cos x.
x
(e) f (x, y) = 2
. Chek that fxx + fyy = 0.
x + y2
2. Find
dz
when
dt
(a) z = x2 + y 2 , x = t2 + 1, y = t − 1.
(b) z = x2 y 1/3 , x = cos t, y = e−t .
3. Find
∂f ∂f
,
, when
∂s ∂t
(a) f (x, y) = x2 − 2y 2 ,
x = st,
(b) f (x, y, z) = x + 2y + z 2 ,
y = s + t.
s
x = , y = s2 + ln t,
t
z = 2s.
4. The total surfae area S of a one of base radius a and height h is given by
√
S = πa2 + πa a2 + h2 .
Calulate the rate of hange of S at a = 1, h = 2 if a is inreasing at the
rate 0.2cms−1 and h is inreasing at the rate 0.1cms−1 .
5. The eletromotive fore E due to a irular loop of wire of radius a rotating
with angular veloity ω in a magneti eld of intensity B inlined at an
angle θ is given by
E = πa2 Bω sin θ.
Suppose a = 10 ± .01cm, B = 2 ± .1Tesla, ω = 10 ± .2 rad/s, θ =
rad. Calulate the value for E with suitable error bounds.
6. Find the gradient ∇f at the given point.
(a) f (x, y) = y − x2 , (1, 1).
√
(b) f (x, y) = x2 − y 2 , (2, 3).
51
π
4
± .1
() f (x, y, z) = x2 + y 2 − 2z 2 , (1, 1, 1).
(d) f (x, y, z) = ex+y cos z, (0, 0, π/6).
7. Find the diretional derivative of f at the point P in the diretion n given
below.
(a) f (x, y, z) = xy + yz + zx, P (1, −1, 2), n = 3i + 6j − 2k.
(b) f (x, y, z) = cos(xy) + eyz + ln(zx), P (1, 1, 1/2), n = i + 2j + 2k.
() f (x, y) = x tan−1 (y/x), P (1, 1), n = i − j.
8. The work W done by the fore F = (F1 , F2 , F3 ) over a urve C is given by
the line integral
W =
Z
F.dr =
C
Find W when
Z
(F1 dx + F2 dy + F3 dz).
C
(a) F = (y − x2 )i + (z − y 2 )j + (x − z 2 )k and C is the urve r = ti + t2 j +
t3 k from (0, 0, 0) to (1, 1, 1).
(b) F = xi + zj + yk and C is the helial urve r = cos t i+sin t j+tk from
0 ≤ t ≤ 2π.
() F = (x − y)i + xj and C is the irle r = cos t i+sin t j from 0 ≤ t ≤ 2π .
9. For the given fores F nd the potential φ so that F = ∇φ and hene
evaluate the integral
Z B
F.dr,
A
(a) F = (yz, xz, xy), A(0, 0, 0), B(2, 3, 1).
(b) F = (2xy, x2 − z 2 , −2yz), A(0, 0, 0), B(1, 2, 3).
() F = (sin y cos x, cos y sin x, 1), A(1, 0, 0), B(0, 0, 1).
10. Sketh the regions of integration and hene evaluate the integrals.
3
Z
Z
1
(4 − y ) dx dy, (b)
Z 01 Z0 y
dx dy.
(d)
(a)
2
Z
8
1
Z
ln y
x+y
e
dx dy, ()
0
Z
0
π
Z
sin x
y dy dx,
0
y2
1/2
11. Sketh the region of integration and evaluate the integral.
Z
(a)
0
2
Z
0
4−x
Z
dy dx (b)
0
4
Z
2
√
x dxdy,
()
Z
0
y
2
Z
ex
y dy dx.
1
Rewrite the above integrals with the order of integration reversed and evaluate your result.
52
seleted answers to Examples 4
1.
2. (a)
(b)
dz
dt
dz
dt
= 2x(2t) + 2y = 2(t2 + 1)2t + 2(t − 1).
= −2xy 1/3 sin t − x2 13 y −2/3 e−t .
3. (a) fs = 2xt − 4y, ft = 2xs − 4y.
(b) fs = 1t + 2(2s) + 4z , ft = − ts2 + 2t .
4. St = 3.22359cm2 s−1 .
5. E = 0.444 ± 0.0844 Volts.
6. (a)
(b)
()
(d)
(1, −2)
√
(4, −2 3)
(2, 2, −4)
√
√
( 3/2, 3/2, −1/2)
7. (a) (4 − 1 + 18)/7 = 3.
(b) 13 (5 − 3 sin(1) + 3e1/2 ).
¤
£
() √12 π4 − 1 .
8. (a) 29
60
(b) 0.
() 2π.
9. (a) 6.
(b) -16.
() 1
10. (a) 3
(b) π4 .
() 1/12.
11. (a) 6.
(b) 4.
() 14 (e4 − 5).
53