Leaving Cert Applied Maths - Higher Level - J.P. McCarthy ...

Leaving Cert Applied Maths 50

So when the object is moving a friction force of magnitude µmg opposes the motion. What if there is no
horizontal force on the object? Friction only exists when there is a force in the opposite direction, and in its
absence the object stays at rest.
If there is a force on the object which is less than or equal to µR then the particle will not move: the force
of friction will be strong enough to oppose this force (motion) but once the force (motion) is opposed, the
friction doesn’t get any stronger.
Finally if there is a force on the object which is great than µR then the friction force takes on the constant
value µR. In this sense µR is called the limiting friction:

Figure 5.3: A graph of the friction force, F , vs an applied force, T . Note when the applied force is less than
the limiting friction, µR, the friction force equals the applied force. When the applied force exceeds the
limiting friction, the friction force takes the constant value µR.

5.1.10 Definition

The angle of friction, λ, is the angle between R and R + F:

From the diagram it is clear that

F
tan λ = = µ

R

Chapter 6

Differential Equations

6.1 Introduction

We are familiar with equations in the set of real numbers: (6.1)
x2 − 3x + 2 = 0

We have techniques of algebra and analysis that tell us that the solutions to this equation are 2, 1 ∈ R. A
differential equation is another creature altogether. Consider now briefly the derivative of a function. The
derivative of a function, f (x) is a function whose value at x0 is the slope of the tangent to the curve at
(x0, f (x0)):

25 25 123
20 20
15 15
10 10
5 5

2 15 1 2 3 2 15

A differential equation is an equation whose solution is a function rather than a number, and the terms
of the equation will be functions and derivatives. For example:

dy = √ (6.2)
2x 1 − y2
dx

√
asks for which function y = f (x), is the slope of the tangent at (x, y) given by 2x 1 − y2?

6.2 Differentiation, Integration and the Fundamental Theorem of
Calculus

In this section the derivative and definite integral are introduced in their correct setting. At Leaving Cert
level, integration is introduced merely as the inverse of differentiation. This approach simplifies things but
greater understanding comes out of a proper treatment. Historically integration was developed separately
of differentiation and the link between them was later discovered. The link between them; namely that
integration is indeed the inverse of differentiation, is known as the Fundamental Theorem of Calculus. The
topics of coordinate geometry, limits and functions should be studied in more depth prior to a thorough
study calculus, but this brief chapter is merely intended as an exposition to aid understanding.

51

Leaving Cert Applied Maths 52

6.2.1 Differentiation

In the figure below, the line from a to b is called a secant line.

Figure 6.1: Secant Line.

Introduce the idea of slope. The slope of a line is something intuitive. A steep hill has a greater slope
than a gentle rolling hill. The slope of the secant line is simply the ratio of how much the line travels
vertically as the line travels horizontally. Denote slope by m:

m = |ac| . (6.3)
|bc|

What about the slope of the curve? From a to b it is continuously changing. Maybe at one point its slope
is equal to that of the secant but that doesn’t tell much. It could be estimated, however, using a ruler the
slope at any point. It would be the tangent, as shown:

Figure 6.2: Tangent Line

The above line is the slope of the curve at t. Construct a secant line:
Now with respect to analytic geometry, with a function f (x), the slope of this secant is given by:

m = f (x + h) − f (x)
(x + h) − x

⇒ m = f (x + h) − f (x)
h

It is apparent that the secant line has a slope that is close, in value, to that of the tangent line. Let h
become smaller and smaller:

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Figure 6.3: Secant line and Tangent line

Figure 6.4: Secant line approaching slope of Tangent line

The slope of the secant line is almost identical to that of our tangent. Let h → 0. Of course, if h = 0
there is no secant. But if h got so close to 0 as doesn’t matter then there would be a secant and hence a slope.
With respect to analytic geometry, with a function f (x), the slope of this secant, which is indistinguishable
from that of the tangent to the point, is given by:

m = lim f (x + h) − f (x) . (6.4)

h→0 h

This m is the derivative of f(x). This gives the slope of the curve at every point on the curve.

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Notation
There are two significant branches of notation used to denote the derivative of f . The difference is just
notation. They are different ways of writing down the same thing. This section is included to fight against
simple misunderstandings.

Newtonian Notation
The function is denoted f (x) and the graph is the set of points (x, f (x)):

fx

x

Figure 6.5: Newtonian notation for functions

The derivative of f (x) is denoted f ′(x). Other names for the derivative of f (x) include:
• the differentiation of f (x)

• the derived function for f (x)

• the slope of the tangent at (x, f (x))

• the gradient

• df
dx

Leibniz Notation
The function is denoted y = f (x) (e.g. y = x2); and the graph is the set of points (x, y):

y

x

Figure 6.6: Leibniz notation for functions (6.5)
In this notation, y is equivalent to f (x). However, the notation for the derivative of y is:

dy
.

dx

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It must be understood that if y = f (x); then

f ′(x) ≡ dy (6.6)
,
dx

and there is no notion of canceling the ds; it is just a notation. It is an illuminating one because if the second
graph of figure 6.4 is magnified about the secant:

Figure 6.7: Leibniz notation for the derivative

If dy is associated with a small variation in y ∼ f (x+h)−f (x); and dx associated with a small variation
in x ∼ h; then dy/dx makes sense.

To reiterate if y = f (x); then dy/dx is the same thing as:
• the derivative of f (x)
• the differentiation of f (x)
• the derived function for f (x)
• the slope of the tangent at (x, f (x))
• the gradient
• f ′(x)

So to emphasise f (x) and y are freely interchangeable; as are f ′(x) and dy/dx.

6.2.2 Integration

What is the area of the shaded region under the curve f (x)?

Start by subdividing the region into n strips S1, S2, ..., Sn of equal width as Figure 6.8.
The width of the interval [a, b] is b − a so the width of each of the n strips is

∆x = b−a .

n

Approximate the ith strip Si by a rectangle with width ∆xi and height f (xi), which is the value of f at the
right endpoint. Then the area of the ith rectangle is f (xi) ∆xi:

The area of the original shaded region is approximated by the sum of these rectangles:

A ≈ f (x1) ∆x + f (x2) ∆x + · · · + f (xn) ∆x. (6.7)

This approximation becomes better and better as the number of strips increases, that is, as n → ∞. Therefore
the area of the shaded region is given by the limit of the sum of the areas of approximating rectangles:

A = lim [f (x1) ∆x + f (x2) ∆x + · · · f (xn) ∆x]. (6.8)

n→∞

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Figure 6.8:

Definition

If f (x) is a function defined in [a, b] and xi, ∆x are as defined above, then the definite integral of f from a

to b is ∫ b ∑n

f (x) dx = lim f (xi) ∆x. (6.9)

a n→∞ i=1

∫∑
So an integral is an infinite sum. Associate ∼ limn→∞ n. Again dx is associated with a small variation

in x ∼ ∆x.

6.2.3 Fundamental Theorem of Calculus

Rough Version

If f is a function with derivative f ′ then
∫b
f ′(x) dx = f (b) − f (a)

a

Proof
Identical to that of Theorem 2.2.1 replacing s(t) by f (x) and v(t) by f ′(x).

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The Indefinite Integral

In Leaving Cert where integrals are defined as the inverse of derivatives, an indefinite integral defines inte-
gration.

Definition (6.10)

If f (x) is a function and it differential with respect to x is f ′(x), then
∫
f ′(x) dx = f (x) + c

where c is called the constant of integration.

Note the constant of integration. It’s inclusion is vital because if f (x) is a function with derivative f ′(x)
then f (x) + c also has derivative f ′(x) as:

d df d
(f (x) + c) = + c,
dx dx dx

=f ′(x) =0

⇒ d (f (x) + c) = f ′(x).
dx

Geometrically a curve f (x) with slope f ′(x) has the same slope as a curve that is shifted upwards; f (x) + c.

Note that the constant of integration can be disregarded for the indefinite integral. Suppose the integrand
is f ′(x) and the anti-derivative is f (x) + c. Then:

∫b
f ′(x) dx = (f (b) + c) − (f (a) + c),

a ∫b
⇒ f ′(x) dx = f (b) − f (a).

a

The cs cancel!

6.2.4 Conclusion

Finding the derivative of a function f at x is finding the slope of the tangent to the curve at x. Integration

meanwhile measures the area between two points x = a and x = b. The Fundamental Theorem of Calculus

states however that differentiation and integration are intimately related; that is given a function f :

d ∫

dx f (x) dx = f (x),
d
∫

dx f (x) dx = f (x) + c.

i.e. differentiation and integration are essentially inverse

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(6.11)
6.3 Differentiation (6.12)
(6.13)
6.3.1 Proposition
(6.14)
d xn = nxn−1 (6.15)
dx (6.16)
d (6.17)

sin x = cos x
dx
d cos x = − sin x
dx

Proof
By using (6.4):

f ′(x) = lim f (x + h) − f (x)
h→0 h

6.3.2 Proposition

(D1) The Sum/ Difference Rule
If u(x) and v(x) are two functions and if

f (x) = u(x) ± v(x), then

df = du ± dv .
dx dx dx

(D2) The Product Rule
If u(x) and v(x) are two functions and if

f (x) = u(x).v(x), then

df dv du
= u. + v. .

dx dx dx

(D3) The Quotient Rule
If u(x) and v(x) are two functions and if

u(x)
f (x) = , then

v(x)

df = v. du − u. dv .
dx dx dx

v2

(D4) The Scalar Rule
If g(x) is a function and if f (x) = k g(x), where k is a constant (k ∈ R) then

df dg
= k. .

dx dx

Proof
By using (6.4).

Leaving Cert Applied Maths 59

6.3.3 Corollary

d (anxn + an−1xn−1 + · · · + a1x + a0) = nanxn−1 + (n − 1)an−1xn−2 + · · · + 2a2x + a1 (6.18)
dx (6.19)

d tan x = sec2 x
dx

6.3.4 The Chain Rule

The Chain Rule allows us to differentiate a composite function. If u(x) and v(x) are functions and f (x) =

u(v(x)) = u ◦ v(x) then:

df du dv (6.20)
=.

dx dv dx

6.3.5 Inverting

Ordinarily in LC maths, y = f (x). That is y is a function of x. However, for some functions y = f (x) there
is an inverse: x = f −1(y). For such functions, y = f (x), x = g(y) may be written; where gf (x) = x. Then
x can be differentiated with respect to y. Suppose

dx
= u(y)

dy

some function of y. Then

dy 1
=,

dx dx/dy
⇒ dy = 1 .

dx u(y)

Now substitute f (x) for y in u(x). This method of inverting is used to differentiate both ln x and the
inverse trigonometric functions sin−1 x and tan−1 x - whose inverses are, respectively, x = ey, x = sin y and
x = tan y.

6.3.6 Corollary

d sin−1(x) = √ 1 (6.21)
dx 1 − x2 (6.22)

d tan−1(x) = 1
dx 1 + x2

6.3.7 The Exponential Function & The Natural Logarithm

The Exponential Function

The Exponential Function may be defined in many different ways. Possibly the best way to think of it at

LC level is to consider the following:

As it turns out there is a very special number, e ≈ 2.718 given by:

( 1 )n ∑∞ 1
e := lim 1 + = (6.23)
n→∞ n i!
i=0

such that the function ex and its derivative are equal; that is:

d ex = ex
dx

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10 10
8 8
6 6
4 4
2 2

0.0 0.5 1.0 1.5 2.0 0.0 0.5 1.0 1.5 2.0

Figure 6.9: A plot of the functions ax and their derivatives for a = 1, 2, 3, 4. Observe the similarity between
ax and dax/dx.

By the Fundamental Theorem of Calculus: (6.24)

∫
ex dx = ex + c

Remark

Later it will be seen in Further Calculus and Series that functions have a power series expansion. The power

series expansion of ex is:

ex = 1 + x + x2 + x3 + . . . (6.25)
1! 2! 3!

The general term in this series is: xn
tn = n!

By the Sum Rule, the differentiation of a sum is the sum of the derivatives. Hence consider

d = n xn−1 = n(n − 1)(n n xn−1 = xn−1 = tn−1.
dx tn n! − 2) · · · (2)(1) (n − 1)!

⇒ d ex = 1 + x + x2 + x3 + . . .
dx 1! 2! 3!
⇒ d ex = ex.
dx

The Natural Logarithm

The natural logarithm is the inverse of the exponential function. Suppose ea = b; then ln b = a. So ex and

ln are inverse in the sense:

ex = y ⇔ ln y = x (6.26)

Figure 6.10: The exponential and natural logarithm functions are inverse

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From our tables we can see that d1
ln x =
(6.27)
dx x

However we can derive this as follows. Suppose

y = ln x

⇒ ey = x
⇒ ey dy = 1
dx dx

⇒ dy = 1 = 1
dx ey x

Obviously then ∫∫

1 dx ≡ dx = ln |x| + c (6.28)
x x

The absolute value sign is there because ln is only defined for strictly positive numbers. Note if we try to

use the formula: ∫ xn dx = xn+1 + c
n+1
(6.29)

for

∫∫
1
dx = x−1 dx
x
∫
⇒ 1 x0
dx = + c
x0

which is undefined.

6.4 Integration (6.30)

From the Fundamental Theorem of Calculus
∫
f ′(x) dx = f (x) + c

Thus: f (x) ∫
Also because f (x)
xn (n ≠ −1)
xn+1
cos x n+1
sin x
ex sin x + c
sec2 x − cos x + c

1 ex + c
x tan x + c

ln |x| + c

d
sin nx = n cos nx , and

dx

d cos nx = −n sin nx
∫ dx
sin nx
⇒ cos nx dx = + c and
n
∫ sin nx dx = − cos nx + c
⇒ n

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Also, let a > 0; d1 x 1 1 1 (6.31)
Also dx a arctan a = a 1 + x2/a2 . a (6.32)

6.4.1 Proposition ⇒ d 1 x1 1 1
arctan = .
dx a a a¡ (a2 + x2)/ a 2 a¡

⇒ d 1 arctan x = a2 1
dx a a + x2
∫
1 1x
⇒ a2 + x2 dx = a arctan a + c

d x √ 1 1
arcsin =
dx a 1 − x2/a2 a

⇒ d arcsin x = √ 1 1

dx a (a2 − x2)/a2 a
√

⇒ d x  a 2 ¨√1a¨2
arcsin = − x2
dx∫ a a2

⇒ √1 x
dx = arcsin
a2 − x2 a

• ∫ ∫∫
• (f (x) ± g(x)) dx = f (x) dx ± g(x) dx
Proof
∫∫
k f (x) dx = k f (x) dx , where k ∈ R

Remarks

There is no direct analogue of the product, quotient nor chain rule for integration. Although the substitution
method below is like a chain rule for integrals.

Leaving Cert Applied Maths 63
(6.33)
6.4.2 The Substitution Method for Evaluating Integrals

∫∫
f (g(x))g′(x) dx = f (u) du

where u = g(x)

Proof (Non-Examinable)

u = g(x)

⇒ du = g′(x)
dx

⇒ dx = 1
du g′(x)

⇒ dx = du
g′(x)

So
∫∫∫
f (g(x))g′(x)dx = f (u)¨g′(¨x¨)¨g′d(¨ux¨) = f (u)du

LIATE
If we cannot see a g(x), g′(x) pattern we can use the LIATE rule. Choose u according to the most complicated
expression in the following hierarchy:

L ogarithms

I nverses (inverse sine, tan)

A lgebraic (polynomials in x)

T rigonometric

E xponential
In general this works well (also works for Integration by Parts).

6.5 Second Derivatives

In a sense, differentiation is a function on the set of functions1, C(R):

D : C(R) → C(R) , f → f ′(x) (6.34)

So the derivative of a function is not a number: but another function. If we differentiate f ′(x) we will get

another function:

D(f ′(x)) = f ′′(x) (6.35)

This f ′′(x) is the second derivative of f (x). In Leibniz Notation:

d2f ≡ f ′′(x) (6.36)
dx2

6.6 Laws of Logs

In this chapter the particular integral: ∫
dx = ln |x| + c
x (6.37)

will frequently arise. It would be wise therefore to state here the definition and laws of logs:

1strictly the set of differentiable functions but we won’t worry about these technicalities

Leaving Cert Applied Maths 64
(6.38)
6.6.1 Definition

Let a, x ∈ R; a, x > 0.

loga x = p ⇔ ap = x.

6.6.2 Proposition

Let a, b, x, y ∈ R; a, b, x, y > 0, n ∈ R.

(L1)

loga x + loga y = loga xy

(L2) ()

loga x − loga y = loga x
y

(L3)

n loga x = loga xn

(L4)

loga x = logb x
logb a

Proof

(L1) Let loga x = p and loga y = q:
⇒ ap = x , and aq = y
⇒ xy = apaq
⇒ xy = ap+q

⇒ loga xy = p + q = loga x + loga y

(L2) Let loga x = p and loga y = q:

⇒ ap = x , and aq = y

x ap
⇒ y = aq

⇒ x = ap−q
y
()
x
⇒ loga y = p − q = loga x − loga y

(L3) Let loga x = p

⇒ ap = x
⇒ xn = apn = anp
⇒ loga xn = np = n loga x

(L4) Let loga x = p:

⇒ x = ap

⇒ logb x = logb ap

⇒ logb x = (loga x)(logb a)

⇒ loga x = logb x
logb a

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6.7 First Order Separable Differential Equation

A first order separable differential equation has the form: (6.39)
dy

f (x)g(y) = p(x)q(y)
dx

The key is to put all the x-terms on one side and the y-terms on the other:

g(y) p(x)
dy = dx

q(y) f (x)

Integrate both sides:

∫∫
g(y) p(x)
dy = dx
q(y) f (x)

⇒ F (y) + c1 = G(x) + c2

⇒ F (y) = G(x) + (c2 − c1)

:=c

Now solve for y:

y = F −1 (G(x) + c) = H(x) + C (6.40)

Now this answer is not unique due to the constant of integration term. A boundary condition will be given:

y(x0) = y0 (6.41)

Plugging this in:

y0 = H(x0) + C
⇒ C = y0 − H(x0)
⇒ y = H(x) + y0 − H(x, 0)

6.8 Second Order Separable Differential Equation

A second order separable differential equation has the form:

( dy ) d2y ()
dy
f dx dx2 = g dx (6.42)

Let (6.43)
(6.44)
dy
v=

dx
⇒ f (v) dv = g (v)

dx

⇒ f (v) dv = dx
g(v)
∫ f (v) ∫
⇒ dv = dx
g(v)

⇒ F (v) = x + c

⇒ v = F −1(x + C1) = H(x) + C1

But this is just the first order separable differential equation:

dy
dx = H(x) + C1
∫ ⇒ dy =∫ (H(x) + C1) dx

⇒ dy = (H(x) + C1) dx

⇒ y = P (x) + C1x + C2

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Again this answer is not unique due to the constant of integration terms. A boundary condition will
be given:

y(x0) = y0 (6.45)
dy (6.46)
dx (x1) = y1

Plugging this in (6.43):

y1 = H(x1) + C1
⇒ C1 = y1 − H(x1)

Plugging this into (6.44):

y0 = P (x0) + (y1 − H(x1))x0 + C2
⇒ C2 = y0 − P (x0) − (y1 − H(x1))x0
⇒ y = P (x) + (y1 − H(x1))x + y0 − P (x0) − (y1 − H(x1))x0

6.9 Second Order Separable Differential Equation requiring the
Chain Rule

A second order separable differential equation requiring the chain rule has the form:

d2y (6.47)
f (y) dx2 = g(y)

Let

dy
v=

dx

⇒ d2y = dv = dv · dy = dv
dx2 dx dy dx v

dy

⇒ dv = g(y)
f (y)v
dy

which is a first order separable differential equation and so solvable as before for:

v = F (y) + C1
Again there will be boundary conditions:

y(x0) = y0 (6.48)
dy (6.49)
dx (y1) = v1

the second of which will allow us to calculate C1. Now we solve:

v = F (y) + C1

⇒ dy = F (y) + C1
dx ∫
∫
dy
⇒ = dx
F (y) + C1

⇒ P (y) = Q(x) + C2

which with the first boundary condition will determine y uniquely when we solve for y.