This article is dedicated to Lisa Sim, daughter of my former colleague, Dr Simon Sim who spoke to me about her interest and passion for mathematics. She told me she wants to be like Albert Einstein. May she be blessed with academic talents.
Mathematics too, besides medicine and all branches of sciences, has been my passion since
university days. I have read mathematics in my undergraduate. So, I thought I
should also write an article on mathematics especially about calculus.
In fact, I
have already written one here on:
Tuesday,
August 6, 2024
Mathematics:
As Tough as Granite Extremely Hard to Smash
https://scientificlogic.blogspot.com/search?q=mathematics
I suppose
mathematic is not for everyone!
But before I
proceed further, I must apologise for any error here because it was extremely
difficult for me to write mathematical symbols using Microsoft Word as this is
the only application I have. Microsoft can only type alphabets like abcdef...z,
and numerical like 123456
So, what I did was to use Google to search for mathematical symbols, copy and paste them as I go along. This was exceedingly hard for me to do this for chains of equations. The sizes may not even fit, let alone typing out exponential and bases. It took me almost the entire day to write to sort out this problem. For instance, in calculus we often use the mathematical symbol for integration that looks like an elongated "S". Microsoft Word hasn't got that symbol. We can get it from Google, copy and paste it of course. But often we need to integrate within a certain limit, and this should be written at the top and bottom of the integral sign. But Microsoft does not have that facility to do that. So I need to write it separately in parenthesis besides the integral sign. This is very troublesome and time consuming.
Hence, I may make mistakes attempting to type mathematical equations using Microsoft Word. Kindly overlook errors I may have made. Thank you for the understanding. Having understood the problem we can now proceed.
What is Calculus?
Calculus is
advanced mathematics as an extension of algebra.
I'll start
with the Introduction to Calculus, and from there, we can proceed section by
section.
I.
Introduction to Calculus
Calculus is a branch of mathematics that focuses on the study of change. It allows us to describe how things change (differential calculus) and how quantities accumulate (integral calculus). Calculus is fundamental in understanding rates of change in fields like physics, engineering, economics, biology, and more.
The Role of
Limits
At its core,
calculus is based on the concept of a limit. A limit describes the value a
function approaches as the input gets closer to a particular point. Limits help
us understand instantaneous rates of change and areas under curves.
Differential
vs. Integral Calculus
Differential
calculus concerns itself with the rate at which quantities change. The
derivative is a tool for finding this rate.
Integral
calculus is about accumulation. It helps find areas under curves and can
represent total quantities when given a rate of change.
II. Limits and
Continuity
Concept of a
Limit
1. Intuitive
definition of limits
2. Formal
definition using epsilon-delta
3. Basic
properties of limits
4. Finding
limits: Direct substitution, factoring, rationalization
Continuity:
Definition of
continuity
Types of
discontinuities (removable, jump, and infinite discontinuities)
III.
Differential Calculus
1.
Introduction to Derivatives
Concept of the
derivative as a rate of change
Definition of
the derivative (as the limit of difference quotients)
Geometric
interpretation: Slope of a tangent line
2. Basic
Differentiation Rules
Power rule,
product rule, quotient rule
Chain rule
3. Derivatives
of Common Functions
Polynomials,
exponentials, trigonometric functions, and logarithms
4.
Higher-Order Derivatives
Second
derivative, acceleration in physics
5.
Applications of Derivatives
Finding
tangent lines
Optimization
(maxima and minima)
Related rates
Motion
problems (velocity and acceleration)
IV. Integral
Calculus
1. Antiderivatives
and Indefinite Integrals
Definition of
an antiderivative
Basic
integration rules (power rule for integrals)
Integration of
common functions
Integration by
substitution
2. Definite
Integrals
Riemann sums
and the area under a curve
Fundamental
Theorem of Calculus
Evaluating
definite integrals
3. Applications
of Integrals
Area between
curves
Volume of
solids of revolution (disc/washer method)
Applications
in physics: Work, mass, and centre of mass
Accumulation
problems
V. Advanced
Topics (Optional but Useful)
1. Techniques
of Integration
Integration by
parts
Trigonometric
integrals
Partial
fraction decomposition
2. Improper
Integrals
Convergence of
improper integrals
3.
Multivariable Calculus
Partial
derivatives
Double and
triple integrals
Limits and
Continuity
1. The Concept
of a Limit
In calculus, a
limit describes the value that a function approaches as the input (or variable)
gets closer to a specific point. Limits are foundational to defining
derivatives and integrals.
Formal
Definition:
lim f (x) = L
(x → c)
This means as
x approaches c, f (x) approaches L.
Example 1:
Calculating a Simple Limit
Let’s evaluate
the limit:
lim x → 2 (x2 +
3x)
Step 1: Direct
substitution—just plug x = 2 into the function.
f (2) = 22 +
3 (2) = 4 + 6 = 10
Thus, lim x → 2 (x2 + 3 x) = 10.
Example 2:
Finding Limits Involving a Factorable Expression
Evaluate:
lim x → 3 (x2 −
9) / (x – 3)
Step 1:
If we
substitute x = 3 directly, the expression becomes:
(32 −
9) / (3 – 3) = 0 / 0 (indeterminate form)
To resolve
this, we factor the numerator:
(x−3).(x+3) /
(x – 3 )
Step 2:
Cancel out the
common factor (x−3):
lim x→3 (x
+ 3) = 6
Thus, the
limit is 6.
2. Continuity
A function f
(x) is said to be continuous at a point x = c if:
1. f (c) is
defined.
2. lim x → c, f (x) exists.
3. lim x → c, f (x) = f (c).
Example of
Continuity:
Consider the
function:
f (x) = (x2 -1)
/ (x – 1) for ≠ 1, f (1) = 2
Is this
function continuous at x = 1?
Step 1: Find
limx→1 (x2 –
1) / (x – 1).
Factoring the
numerator:
(x−1). (x+1) /
x – 1 = x + 1
So:
lim x→1 (x + 1) = 2
Since limx→1 f(x) = f (1) = 2, the function is continuous at
x = 1.
III.
Differential Calculus
1.
Derivatives: Definition and Interpretation
The derivative
represents the rate of change or the slope of the tangent line to a curve at
any point. It is the limit of the average rate of change as the interval
becomes infinitesimally small.
Formal
Definition of the Derivative:
f′(x) =
lim h → 0 [f (x + h) – f (x)] / h
Example 1:
Derivative of a Polynomial Function
Find the
derivative of f (x) = x2.
Step 1: Use
the definition of the derivative:
f′(x) = lim h → 0 [(x
+ h)2 – x2] / h
Expand (x + h)2:
f′ (x) =
lim h→ 0 [(x 2+ 2 x h) +
h2 – x2 ] / h
= lim h→ 0 (2xh
+ h2) / h
Cancel h:
f′(x) =
lim h→0 (2 x +
h) = 2 x
Thus, the
derivative of x2 is 2x.
Example 2:
Derivative of a Trigonometric Function
Find the
derivative of f(x) = sin(x).
Using the
standard differentiation rules:
f′(x) = cos(x)
2. Basic
Differentiation Rules
1. Power Rule:
d / dx (xn) = nx n−1
Example: d / x
(x3) = 3 x2.
2. Product
Rule:
d / dx
[f (x) g (x)]
= f’ (x) g (x)
+ f (x) g’ (x)
3. Quotient
Rule:
d / dx [(f (x)
/ g (x)] = [f’ (x) g (x) – f (x) g’ (x) / g’ (x)] / g (x)2
4. Chain
Rule:
d / dx f (g (x) = f′(g (x) )g′ (x)
Applications
of Derivatives
Optimization:
Find the
maximum or minimum of a function.
Example: Suppose
the profit P(x) for a company is given by P(x) = − 2x2 +
4x + 100, where x represents the number of products sold. To maximize
profit, we find the critical points.
Step
1: Find the derivative:
P′(x) = − 4x +
4
Step
2: Set the derivative equal to zero to find critical points:
−4x + 4 =
0 ⇒ x = 1
Step
3: Use the second derivative test to confirm if this is a maximum:
P′′(x) = −4
(negative, so it’s a maximum).
Thus, the
company maximizes profit by selling 1 unit of the product.
IV. Integral
Calculus
1. Indefinite
Integrals
The reverse of
differentiation, or antiderivatives, gives us the original function from its
derivative.
Example:
Find ∫2x
dx.
∫2x dx =
x2 + C
where C is the
constant of integration.
2. Definite
Integrals:
Used to
calculate areas under curves.
Example:
Find the area
under the curve f (x) = x2 from x = 0 to x = 2.
∫ x2 dx = [x 3/
3] = 8/3 – 0 (limit
between 0 – 2)
= 8/3
Thus, the area
is 8 / 3
Applications
of Calculus:
1. Engineering:
Calculus is used in designing systems where changing parameters (e.g., stress,
pressure, electrical currents) are modelled.
2.Example:
Calculating torque in machines using derivatives.
3.Physics:
Derivatives are used to find velocity and acceleration from displacement
functions.
Examples:
1.
Newton’s second law, F=ma, involves calculating acceleration as the derivative
of velocity.
2. Research
& Technology: In optimization problems, calculus helps to minimize cost
functions or maximize outputs.
3. Machine
learning algorithms often use gradient descent, which relies on derivatives to
find the optimal solution.
V. Integral
Calculus: Volume of a Solid of Revolution
The volume of
a solid can be calculated when a region is rotated about an axis. This is
typically done using the method of disk/washer or cylindrical shells.
Example:
Finding the Volume of a Solid of Revolution (Disk Method)
Let’s consider
the function:
f(x) =√ x
and rotate the
region under the curve from x = 0 to x = 4 around the x-axis.
Step-by-Step
Solution:
1. Set up the
formula for the volume using the disk method: The volume is given by the
formula:
V = Ï€∫ [f(x)] 2 dx (limits between a and
b)
In this case,
we’re rotating the curve y = x around
the x-axis, so f(x) = x, and
the limits are a = 0 and b = 4.
2. Substitute
the function into the formula:
V = Ï€ ∫ (x)2 dx
(between the limits of a = 0 and b = 4)
3. Simplify
the integral:
V =
Ï€ ∫ x dx
(limits between 0 – 4)
4. Evaluate
the integral:
V = π [2 2 / 2] (between 0 and 4)
= 42 /2
– 02/2
V = π 16 / 2] = 8π
Thus, the
volume of the solid is:
V = 8Ï€
(approximately 25.13 cubic units).
VI.
Applications to Motion: Velocity and Acceleration
In physics, we
use calculus to describe the motion of objects, where velocity is the rate of
change of position, and acceleration is the rate of change of velocity.
1. Example:
Finding Velocity from Position
Suppose the
position of an object moving along a straight line is given by the function:
s(t) = 3t2+2t
where s(t) is
the position in meters, and t is the time in seconds. Find the velocity of the
object after t=3 seconds.
Step-by-Step
Solution:
1.Find the
velocity function by taking the derivative of the position function:
v(t) = dt /
d [3t2+2t]
Using the
power rule:
v(t) = 6t + 2
Substitute t =
3 into the velocity function:
v (3) = 6(3) +
2 = 18+2 = 20 meters per second.
Thus, the
velocity of the object at t=3 seconds is 20 m/s.
2. Example:
Finding Acceleration from Velocity
Now, let’s
find the acceleration of the object from the previous example. The velocity
function is:
v(t) = 6t + 2
We find the
acceleration by taking the derivative of the velocity function.
Step-by-Step
Solution:
Find the
acceleration function by taking the derivative of v (t):
a(t) = d/ dt [6t+2]
a(t) = 6
Thus, the
acceleration is a constant 6m/s2, meaning the object is accelerating at a
steady rate of 6m/s2.
Changing
Parameters: Motion Under Acceleration
When an object
undergoes constant acceleration, we can use integration to find the position or
velocity over time.
Example:
Motion Under Constant Acceleration
Suppose an
object starts from rest and accelerates at a constant rate of 4m/s2. How far
does it travel in 5 seconds?
Step-by-Step
Solution:
1. Given:
Initial
velocity v0 = 0
(starts from rest).
Acceleration a
= 4m /s2.
2. Find
the velocity as a function of time using integration:
Since
acceleration a = dv/dt = 4,
integrate to find the velocity:
v(t)
= ∫4dt = 4t + C
Since the
object starts from rest (v (0) = 0), we know C = 0, so:
v(t) = 4t
3. Find
the position as a function of time by integrating the velocity function:
s(t)
= ∫v(t) dt = ∫4tdt = 2t2 + D
Since the
object starts at the origin (s (0) = 0), we know D = 0, so:
s(t) = 2t2
4. Substitute
t = 5 into the position function:
s (5) = 2(52)
= 2 (25) = 50 meters.
Thus, the object travels 50 meters in 5 seconds.
Summary of Key
Concepts So Far:
1. Volume
of a Solid of Revolution:
Use the disk
method to find the volume when rotating a curve around an axis.
Key formula:
V = Ï€∫[f(x)]2 dx. (within the limits of a
and b)
2. Velocity
and Acceleration:
Velocity is
the derivative of the position function.
Acceleration
is the derivative of the velocity function.
Key formula
for position under constant acceleration: s(t) =1/2 at2 + v0t +s0.
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