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AUTOMOTIVE ENGINEERING MATHEMATICS & SCIENCE  

SECTION 1: MATHEMATICS

Numbers

Natural Numbers (N)

These are the positive integers used for counting.
N = {1,2,3,4,…}

Real Numbers (R)

This set includes all rational and irrational numbers. It encompasses all numbers that can be represented on a number line.

Rational Numbers (Q)

These numbers can be expressed as a fraction p/q, where p and q are integers and q ≠ 0. They include:

Irrational Numbers (I)

These numbers cannot be expressed as a fraction p/q. They have non-terminating, non-repeating decimal expansions. Examples:

Approximation (Rounding Off)

Rounding Off to a Required Decimal Place

Rule:
If the digit following the desired decimal place is 5 or greater, round the preceding digit up. If the digit following the desired decimal place is less than 5, leave the preceding digit as it is.

Examples:

Rounding Off to a Required Significant Figure

Rules:

Examples:

Number System Bases

Conversion Methods

Decimal to Other Bases:

Repeated division: Divide the decimal number by the target base, noting the remainders. The remainders, read in reverse order, form the number in the new base.

Other Bases to Decimal:

Positional notation: Multiply each digit by the base raised to the power of its position (starting from 0 on the right), and then sum the results.

Example Conversions

Let us convert the decimal number 250 to binary, octal, duodecimal, and hexadecimal.

Decimal (250) to Binary

		250 / 2 = 125, remainder 0
		125 / 2 = 62, remainder 1
		62 / 2 = 31, remainder 0
		31 / 2 = 15, remainder 1
		15 / 2 = 7, remainder 1
		7 / 2 = 3, remainder 1
		3 / 2 = 1, remainder 1
		1 / 2 = 0, remainder 1
		Result: 11111010 (binary)

Decimal (250) to Octal

		250 / 8 = 31, remainder 2
		31 / 8 = 3, remainder 7
		3 / 8 = 0, remainder 3
		Result: 372 (octal)

Decimal (250) to Duodecimal

		250 / 12 = 20, remainder 10 (A)
		20 / 12 = 1, remainder 8
		1 / 12 = 0, remainder 1
		Result: 18A (duodecimal)

Decimal (250) to Hexadecimal

		250 / 16 = 15, remainder 10 (A)
		15 / 16 = 0, remainder 15 (F)
		Result: FA (hexadecimal)

Conversions from Other Bases to Decimal

Binary (11111010) to Decimal:

		(1 × 27) + (1 × 26) + (1 × 25) + (1 × 24) + (1 × 23) + (0 × 22) + (1 × 21) + (0 × 20) 
		= 128 + 64 + 32 + 16 + 8 + 0 + 2 + 0 = 250

Octal (372) to Decimal:

		(3 × 82) + (7 × 81) + (2 × 80) = 192 + 56 + 2 = 250

Duodecimal (18A) to Decimal:

		(1 × 122) + (8 × 121) + (10 × 120) = 144 + 96 + 10 = 250

Hexadecimal (FA) to Decimal:

		(15 × 161) + (10 × 160) = 240 + 10 = 250

Example: Convert Decimal 178 to Octal, Binary, and Hexadecimal

1. Decimal (178) to Octal

		178 / 8 = 22, remainder 2
		22 / 8 = 2, remainder 6
		2 / 8 = 0, remainder 2
		Octal result: 262

2. Octal (262) to Binary

Convert each octal digit to 3-bit binary:

		2 = 010
		6 = 110
		2 = 010
		Combine the binary representations: 010 110 010
		Binary result: 010110010 (or 10110010, removing the leading zero)

3. Octal (262) to Hexadecimal

Group the octal digits into groups of four, starting from the right. If needed, add leading zeros:

		0262 (group the numbers as if they were binary numbers, and then use the octal to binary conversion, and group the binary into 4 bit sections.)
		010110010 (binary equivalent)
		0101 1001 0 (group into sets of 4)
		0101 1001 (remove the last zero)
		Convert each 4-bit group to hexadecimal:
		0101 = 5
		1001 = 9
		Hexadecimal result: 59

Example 2: Convert Decimal 350 to Octal, Binary, and Hexadecimal

1. Decimal (350) to Octal

		350 / 8 = 43, remainder 6
		43 / 8 = 5, remainder 3
		5 / 8 = 0, remainder 5
		Octal result: 536

2. Octal (536) to Binary

Convert each octal digit to 3-bit binary:

		5 = 101
		3 = 011
		6 = 110
		Combine the binary representations: 101 011 110
		Binary result: 101011110

3. Octal (536) to Hexadecimal

Group the octal digits into groups of four, starting from the right:

		1010 11110 (binary equivalent of 536)
		1 0101 1110 (add a leading zero, and group into sets of 4)
		Convert each 4-bit group to hexadecimal:
		0001 = 1
		0101 = 5
		1110 = E
		Hexadecimal result: 15E

Fundamental Mathematical Laws

1. Cumulative Laws (Commutative Laws)

Definition: These laws state that the order of operands doesn't affect the result for certain operations.

Addition: a + b = b + a
Multiplication: a × b = b × a

Application:

Example:

Problem: Calculate 12 + 25 + 8.
Solution: Using the commutative law, rearrange: 12 + 8 + 25 = 20 + 25 = 45.

2. Associative Laws

Definition: These laws state that the grouping of operands doesn't affect the result for certain operations.

Addition: (a + b) + c = a + (b + c)
Multiplication: (a × b) × c = a × (b × c)

Application:

Example:

Problem: Calculate 2 × (5 × 9)
Solution: Using the associative law, you can also calculate (2 × 5) × 9 = 10 × 9 = 90. Or you can calculate 2 × 45 = 90.

3. Distributive Law

Definition: This law describes how multiplication distributes over addition (or subtraction).

a × (b + c) = (a × b) + (a × c)
a × (b - c) = (a × b) - (a × c)

Application:

Example:

Problem: Calculate 7 × (10 + 3).
Solution: Using the distributive law, 7 × (10 + 3) = (7 × 10) + (7 × 3) = 70 + 21 = 91.

Problem: Factor 15x + 10.
Solution: 5(3x + 2). 5 was distributed in the original problem.

Combined Application

Let's look at a more complex example that uses all three laws:

Problem: Simplify 4 × (2x + 3) + 6x + 8.
Solution:

Commutative: Order doesn't matter (addition, multiplication).
Associative: Grouping doesn't matter (addition, multiplication).
Distributive: Multiplication spreads over addition/subtraction.

BODMAS

When simplifying arithmetic expressions, it is crucial to follow the correct order of operations to ensure you arrive at the accurate answer. This order is commonly remembered using the acronym BODMAS (or PEMDAS in some regions). Here is a breakdown:

BODMAS stands for:

Key Points:

How to apply BODMAS:

Brackets (Parentheses):

First, simplify any expressions within brackets (or parentheses). If there are nested brackets, start with the innermost ones.

Orders (Exponents):

Next, evaluate any orders (powers or exponents) like squares, cubes, etc.

Division and Multiplication:

Perform division and multiplication from left to right in the order they appear.

Addition and Subtraction:

Finally, perform addition and subtraction from left to right in the order they appear.

Example:

Let's simplify the following expression: 10 + 2 × (15 – 5) ÷ 4

Brackets:

15 - 5 = 10

The expression becomes: 10 + 2 × 10 ÷ 4

Multiplication and Division (from left to right):

2 × 10 = 20

The expression becomes: 10 + 20 ÷ 4

20 / 4 = 5

The expression becomes 10 + 5

Addition:

10 + 5 = 15

Therefore, the simplified expression is 15.

Importance:

Equations

Definition of an Equation

Constructing Equations of Letters and Numbers

Examples:

Building Equations:

Rules of Equations (and Transposition)

The fundamental principle governing equation manipulation is that whatever you do to one side of the equation, you must do to the other side to maintain equality. This principle leads to the following rules:

Addition Property of Equality:

If a = b, then a + c = b + c (You can add the same value to both sides).

Subtraction Property of Equality:

If a = b, then a – c = b - c (You can subtract the same value from both sides).

Multiplication Property of Equality:

If a = b, then a × c = b × c (You can multiply both sides by the same value).

Division Property of Equality:

If a = b, then a ÷ c = b ÷ c (You can divide both sides by the same non-zero value).

Transposition

Transposition is a simplified way of applying these rules. It involves moving terms from one side of the equation to the other by changing their signs.

How Transposition Works:

Example:

Solve for x: x + 3 = 7

Using transposition: x = 7 - 3

Therefore, x = 4

Explanation of the steps.

Applying the Rules

Example 1: Solve for x: 2x – 5 = 9

Example 2: Solve for y: y / 3 + 1 = 4

Distinguishing Between Expressions and Equations

Expression:

Examples:

Expressions can be simplified or evaluated, but they don't solve for a specific variable.

Equation:

Examples:

Equations can be solved to find the value of the unknown variable(s).

Key Difference: The presence of an equals sign (=) is the defining characteristic that distinguishes an equation from an expression.

Relating Equations to Formulas

Formula:

Examples:

Relationship:

Transposing Formulas in Motor Vehicle Technology and Science

Transposition involves rearranging a formula to solve for a different variable.

Motor Vehicle Technology:

Formula: Torque (T) = Force (F) × Lever Arm (r)

Formula: Power (P) = Torque (T) × Angular Velocity (ω)

Science (Physics):

Formula: Force (F) = mass (m) × acceleration (a)

Formula: Ohm's Law (Electrical Circuits): Voltage (V) = Current (I) × Resistance (R)

Formula: Kinematic Equation: v = u + at (final velocity = initial velocity + acceleration × time)

General Transposition Steps:

Algebraic expressions involving multiplication and division of polynomials.

1. Multiplication of Polynomials

Examples:

2. Division of Polynomials

Examples:

Polynomial Long Division Example:

3. Combining Multiplication and Division

Example:

Algebraic expressions with different types of indices (exponents)

1. Whole Number Indices

2. Fractional Indices

3. Decimal Indices

Examples of Evaluating Algebraic Expressions:

Key Reminders:

Factorization

1. Factorization by Common Grouping

2. Factorization by Difference of Two Squares

3. Factorizing Quadratic Expressions

4. Determining if a Quadratic Expression is a Perfect Square

Solving Linear Equations with One Unknown

Solving Simultaneous Linear Equations (Two Unknowns)

a) Substitution Method

Equating Coefficients (Elimination Method)

Solving Linear Equations with Three Unknowns

Solving quadratic equations using factoring, completing the square, and the quadratic formula.

1. Solving Quadratic Equations by Factoring

2. Solving Quadratic Equations by Completing the Square

3. Solving Quadratic Equations Using the Quadratic Formula

Again, Evaluating Polynomial Expressions

Example 1:

Example 2:

2. Function Notation

Example:

3. Polynomial Functions

Examples:

Key points:

Math and Algebra Quiz

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    Polynomial Functions

    Example:

    Nesting Method (Horner's Method):

    The "nesting" method, also known as Horner's method or synthetic substitution, is a very efficient way to evaluate polynomial expressions, especially for higher-degree polynomials. It minimizes the number of multiplications needed, which can reduce rounding errors and improve computational speed.

    How the Nesting Method Works

    Consider a polynomial:

    p(x) = anxn + an-1xn-1 + ... + a1x + a0

    To evaluate p(c) using the nesting method, you rewrite the polynomial as:

    p(c) = a0 + c(a1 + c(a2 + ... + c(an-1 + c(an))...))

    Steps for Using the Nesting Method:

    1. Start with the coefficient of the highest power term (an).
    2. Multiply it by the value you're substituting (c).
    3. Add the next coefficient (an-1)
    4. Multiply the result by c.
    5. Add the next coefficient (an-2).
    6. Continue this process until you add the constant term (a0).

    Examples of Using the Nesting Method

    Example 1: Evaluating a Quadratic Polynomial

    Let's evaluate p(x) = 2x2 - 3x + 5 when x = 4.

    p(4) = 5 + 4(-3 + 4(2))

    Therefore, p(4) = 25.

    Example 2: Evaluating a Cubic Polynomial

    Let's evaluate p(x) = x3 - 4x2 + 6x - 2 when x = 3.

    p(3) = -2 + 3(6 + 3(-4 + 3(1)))

    Therefore, p(3) = 7.

    Example 3: Evaluating a Higher-Degree Polynomial

    Let's evaluate p(x) = 3x4 - 2x3 + 5x2 - x + 7 when x = -2.

    p(-2) = 7 + (-2)(-1 + (-2)(5 + (-2)(-2 + (-2)(3)))))

    Therefore, p(-2) = 93.

    Why the Nesting Method is Useful

    The Remainder Theorem

    Application:

    Example 1:

    Example 2:

    The Factor Theorem

    Application:

    Example 1:

    Example 2:

    Example 3:

    Factorizing Cubic and Quadratic Expressions

    Quadratic Expressions (ax2 + bx + c)

    Example: Factor x2 + 5x + 6

    Cubic Expressions (ax3 + bx2 + cx + d)

    1. Step 1: Find a factor using the Factor Theorem: Try plugging in small integers (like 1, -1, 2, -2) into the polynomial until you get 0. This gives you a linear factor (x − a).
    2. Step 2: Divide: Divide the cubic polynomial by the linear factor you found. This will give you a quadratic expression.
    3. Step 3: Factor the quadratic: Factor the quadratic expression.
    4. Step 4: Write all factors: Write the cubic expression as the product of the linear factor and the two factors of the quadratic.

    Example: Factor x3 - 6x2 + 11x - 6

    Indices (exponents)

    Parts of an Index

    Rules of Indices (Laws of Exponents)

    Here are the fundamental rules of indices:

    Applying Rules of Indices to Simplify Expressions

    Expressing Numbers in Standard Form (Scientific Notation)

    Steps:

    Example 1: Large Number

    Example 2: Small Number

    Example 3:

    Example 4:

    Evaluating Expressions in Standard Form

    When you have expressions involving numbers in standard form, you apply the rules of indices and arithmetic operations.

    Examples:

    Evaluating Expressions in "Preferred" Standard Form

    "Preferred" standard form usually refers to expressing a number in standard form where:

    Why "Preferred" Standard Form Matters:

    Examples:

    Logarithms

    Parts of a Logarithm

    3. Rules of Logarithms (Laws of Logarithms)

    Examples of Applying Rules:

    4. Converting Logarithms to Anti-Logarithms

    Examples:

    Converting Logarithm Bases

    Example:

    Converting Numbers to Natural (Napierian) Logarithms

    Example: Find 𝒍𝒏(𝟔𝟒)

    Applying Logarithm Rules to Solve Indicial Equations

    Steps:

    1. Take the logarithm of both sides of the equation.
    2. Apply the power rule of logarithms to bring the exponent down.
    3. Solve for the variable.

    Example 1: Solve 2ˣ = 16

    Example 2: Solve 32x – 1 = 81

    Example 3: Solve 5ˣ = 20

    Example 4: Solve e2x = 10

    Polynomials, Indices, and Logarithms Quiz

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      Area

      a) Areas of Circles

      b) Annulus (Ring)

      c) Cone

      d) Sphere

      e) Hemisphere

      f) Ellipse

      Calculating the volumes

      1. Cylinder

      2. Cube

      3. Hollow Cylinder (Cylindrical Shell)

      4. Sphere

      5. Cone

      Engine and transmission calculations.

      1. Swept Volume

      2. Total Volume

      3. Clearance Volume

      4. Compression Ratio

      5. Camshaft Drives

      6. Air-Fuel Ratio (AFR)

      7. Gear Ratios (Manual Gearbox and Final Drive)

      8. Belt Drive Ratios in Relation to Gear Ratios

      Types of angles and triangles

      1. Types of Angles

      2. Describing and Identifying Different Angles

      Definition of a Triangle

      Types of Triangles

      Ignition Timing (Spark Ignition Engines)

      Piston Travel

      Angle of Obliquity

      Crank Angle Using Sine and Cosine Formulas

      Mechanical Engineering Quiz

      1 of 15

        Coordinate Geometry and Calculus Concepts

        1. Calculating the Midpoint of a Straight Line

        2. Calculating the Slope or Gradient of a Straight Line

        3. Finding the Equation of a Tangent of a Curve at a Point

        Sketching Straight Line Graphs

        Sketching Quadratic Graphs (Use of Maxima and Minima)

        Sketching Straight Line Graphs

        Sketching Quadratic Graphs (Use of Maxima and Minima)

        Explanation of Illustrations:

        Differentiation

        1. Determining the Slope of a Curve Algebraically
          • Concept:
            • Differentiation allows us to find the instantaneous rate of change of a function, which represents the slope of the tangent line to the curve at any given point.
            • The derivative of a function f(x), denoted as f'(x) or dy/dx, gives the slope of the curve at any x-value.
          • Process:
            • Find the derivative: Use differentiation rules to find the derivative of the function.
            • Evaluate at a point: Substitute the x-coordinate of the point where you want to find the slope into the derivative.
          • Example:
            • Find the slope of the curve y = x² at the point x = 3.
            • Find the derivative: dy/dx = 2x
            • Evaluate: At x = 3, dy/dx = 2(3) = 6.
            • Therefore, the slope of the curve at x = 3 is 6.
        2. Differentiating Coefficients of Powers of x and Polynomials Only
          • Power Rule:
            • If y = xⁿ, then dy/dx = nxⁿ⁻¹
            • If y = axⁿ, then dy/dx = an xⁿ⁻¹ (where 'a' is a constant)
          • Constant Rule:
            • If y = c (a constant), then dy/dx = 0
          • Sum/Difference Rule:
            • If y = u(x) ± v(x), then dy/dx = u'(x) ± v'(x)
          • Polynomials:
            • A polynomial is a function of the form f(x) = a0 + a1x + a2x² + … + anxⁿ
            • To differentiate a polynomial, differentiate each term separately using the power and constant rules.
          • Examples:
            • y = x³: dy/dx = 3x²
            • y = 5x⁴: dy/dx = 20x³
            • y = 7: dy/dx = 0
            • y = 2x³ + 4x² - 6x + 1: dy/dx = 6x² + 8x - 6
        3. Determining Velocity and Acceleration Through Differentiation
          • Position, Velocity, and Acceleration:
            • If s(t) represents the position of an object at time t, then:
              • Velocity (v(t)) is the rate of change of position with respect to time: v(t) = ds/dt.
              • Acceleration (a(t)) is the rate of change of velocity with respect to time: a(t) = dv/dt = d²s/dt².
          • Steps:
            • Find the velocity: Differentiate the position function s(t) to find the velocity function v(t).
            • Find the acceleration: Differentiate the velocity function v(t) to find the acceleration function a(t).
            • Evaluate at a time: Substitute the given time (t) into the velocity and acceleration functions to find the velocity and acceleration at that time.
          • Example:
            • The position of an object is given by s(t) = t³ - 6t² + 9t. Find the velocity and acceleration at t = 2.
            • Velocity: v(t) = ds/dt = 3t² - 12t + 9
            • Acceleration: a(t) = dv/dt = 6t - 12
            • Evaluate:
              • At t = 2, v(2) = 3(2)² - 12(2) + 9 = -3
              • At t = 2, a(2) = 6(2) - 12 = 0
            • Thus, at t=2, the velocity is -3, and the acceleration is 0.

        Integration

        1. Determining Indefinite Integrals of Polynomials
          • Concept:
            • Integration is the reverse process of differentiation.
            • An indefinite integral represents a family of functions whose derivative is the given function.
            • For a polynomial, we use the power rule of integration.
          • Power Rule of Integration:
            • If f(x) = xⁿ, then ∫f(x)dx = (xⁿ⁺¹)/(n + 1) + C (where n ≠ -1 and C is the constant of integration).
            • If f(x) = axⁿ, then ∫f(x)dx = a(xⁿ⁺¹)/(n + 1) + C
          • Constant Rule of Integration:
            • If f(x) = c (a constant), then ∫f(x)dx = cx + C.
          • Sum/Difference Rule of Integration:
            • ∫[u(x)± v(x)]dx = ∫u(x)dx ± ∫v(x)dx
          • Polynomials:
            • Integrate each term of the polynomial separately using the power and constant rules.
          • Examples:
            • ∫x²dx = (x³)/3 + C
            • ∫4x³dx = 4(x⁴)/4 + C = x⁴ + C
            • ∫5dx = 5x + C
            • ∫(3x² + 2x - 1)dx = (3x³)/3 + (2x²)/2 - x + C
        2. Determining the Constant (C) of Integration
          • Concept:
            • When finding an indefinite integral, we always add a constant of integration (C) because the derivative of a constant is zero.
            • To determine the specific value of C, we need additional information, usually a given point on the function (an initial condition).
          • Steps:
            • Find the indefinite integral: Integrate the given function.
            • Use the given point: Substitute the x and y coordinates of the given point into the indefinite integral.
            • Solve for C: Solve the resulting equation for C.
          • Example:
            • Find the integral of f(x) = 2x, given that the function passes through the point (1, 5).
            • Find Indefinite integral: ∫2xdx = x² + C
            • Use the given point: Substitute x = 1 and y = 5 into the integral: 5 = 1² + C
            • Solve for C: 5 = 1 + C → C = 4
            • Therefore, the specific integral is y = x² + 4.
          • Another Example:
            • Find the integral of f(x) = 3x² + 2x given that the function passes through the point (2,10).
            • Find Indefinite integral: ∫(3x² + 2x)dx = x³ + x² + C
            • Use the given point: 10 = 2³ + 2² + C
            • Solve for C: 10 = 8 + 4 + C → C = -2
            • Therefore, the specific integral is y = x³ + x² - 2.

        Matrices

        A matrix is a rectangular array of numbers, symbols, or expressions arranged in rows and columns. 1

        Types of Matrices

        Matrix Notation

        Matrix Addition and Subtraction

        Examples:

        Example 1: Addition

        Example 2: Subtraction

        Example 3: Addition with 3x3 Matrices

        Matrix Multiplication

        Scalar Multiplication

        Multiplication of Two Matrices

        Scalar Multiplication Examples

        Example 1:

        Example 2:

        Example 3:

        Matrix Multiplication Examples

        Example 1:

        Example 2:

        Example 3:

        Solving Simultaneous Equations Using Matrices

        Simultaneous Equations

        These are sets of equations where you need to find values for the unknowns that satisfy all equations at the same time.

        Matrices as Organizers

        Matrices help organize the coefficients and constants of these equations, making the solving process more systematic.

        Representing Equations as Matrices

        Solving with the Inverse Matrix

        2x2 Example

        3x3 Example (Conceptual)

        Vectors

        Scalars

        Differentiating Between Scalar and Vector Quantities

        Feature Scalar Quantity Vector Quantity
        Magnitude Yes Yes
        Direction No Yes
        Examples Temperature, mass, time, speed, distance Velocity, force, displacement, acceleration
        Mathematical Operations Normal arithmetic Vector algebra (vector addition, dot product, cross product)

        Denoting a Vector by Notation

        Types of Vectors

        Vector Addition (Without Proofs)

        Component Method Examples

        Example 1: 2D Vectors

        Example 2: 3D Vectors

        Example 3: Adding Multiple Vectors

        Graphical Method Examples (Conceptual)

        Example 1: Two Vectors in 2D

        Example 2: Vectors with Opposite Directions

        Example 3: Vectors at Right Angles

        Next Chapter →

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