Part I · Mathematical Language, Functions, and Algebraic Models

Chapter 1 — Mathematics as an Engineering Language

Why This Matters

An electronic product is specified by quantities: supply voltage, current, resistance, frequency, power, gain, timing error, and noise level. Numbers alone are not enough. Engineering requires relationships between quantities, together with units and assumptions. Mathematics gives those relationships a form that can be checked and used.

1.1 Quantities, numbers, and units

A physical quantity is a measurable property together with a unit. For example, $$5$$ is only a number; $5\ \mathrm{V}$ is a voltage. The same care is needed when writing equations: both sides of a physically meaningful equation must have the same units.

Real numbers are the main number system of this course. They include natural numbers such as $0,1,2,\ldots$ , integers such as $$-2,-1,0,1,2$$ , rational numbers that can be written as $$p/q$$ with $q\ne0$ , and irrational numbers such as $\sqrt2$ and $\pi$ . Visualise the real numbers as points on a number line: values increase as we move to the right, decrease as we move to the left, and there is no largest or smallest real number.

Scientific notation keeps very large and very small quantities readable:

4700\ \Omega = 4.7\times10^3\ \Omega, \qquad 0.000,002\ \mathrm{F}=2\times10^{-6}\ \mathrm{F}.

The prefixes kilo-, milli-, micro-, and nano- are shorthand for powers of ten. Always estimate the order of magnitude before trusting a numerical output. A calculated capacitor voltage of $3.3\times10^9\ \mathrm{V}$ in a $5\ \mathrm{V}$ laboratory circuit is usually a model or unit error, not an exciting discovery.

From first principles. A unit travels through an equation in the same way as a numerical factor. This makes dimensional checking a calculation: multiplying volts by amperes gives watts, so a claimed power value must have the unit V A, which is the watt.

1\,\mathrm{W}=1\,\mathrm{V}\times1\,\mathrm{A}

A local unit chain from milliamperes to amperes and from volts per ampere to ohms.

1.2 Variables and equations

A variable is a symbol representing a quantity that may vary or whose value is currently unknown. An equation asserts that two expressions have the same value. In the resistor model

\[V=IR,\]

$$V$$ is voltage, $$I$$ is current, and $$R$$ is resistance. If $$R$$ is constant and $$V$$ changes, then $$I$$ changes as well. The equation makes that relationship testable.

A disciplined modelling workflow is:

State the situation and the assumptions.
Name each relevant quantity and include units.
Write the basic relationships.
Solve only for quantities that the model actually determines.
Check the result using units, signs, magnitude, and limiting cases.

Example — Turning words into equations

Two components cost ₹12 and ₹20 each. A purchase contains $$x$$ of the first component and $$y$$ of the second, has $$8$$ items in total, and costs ₹136.

The count and cost statements become

x+y=8, \qquad 12x+20y=136.

The mathematics has not yet solved the problem, but it has separated two pieces of information that can now be combined systematically in Chapter 4.

Keep the statement balanced. An equation says that two expressions name the same quantity. Applying the same reversible operation to both sides preserves that statement; applying it to only one side changes the problem. This is the reason an unknown may be isolated without changing the physical relation it represents.

Start with the relation containing the unknown.
Undo additions or subtractions on both sides.
Undo multiplication or division by a non-zero quantity on both sides, then check the unit of the result.

1.3 Expressions and algebraic equivalence

An expression is a meaningful combination of numbers, variables, and operations, such as $$3x^2-4x+1$$ . Two expressions are equivalent if they have the same value for every input where both are defined. For example,

\frac{x^2-9}{x-3}=x+3

is true only for $x\ne3$ . The left-hand expression is undefined at $$x=3$$ , while the right-hand expression is defined there. Algebraic cancellation simplifies an expression; it does not silently repair a forbidden input.

Example — Unit-aware rearrangement

From $$V=IR$$ , solving for resistance gives

R=\frac{V}{I}, \qquad I\ne0.

If $V=9\ \mathrm{V}$ and $I=3\ \mathrm{mA}=3\times10^{-3}\ \mathrm A$ , then

R=\frac{9}{3\times10^{-3}}\ \Omega=3\times10^3\ \Omega=3\ \mathrm{k}\Omega.

The unit calculation confirms that volt per ampere is an ohm.

Equivalence has a condition. Two expressions may produce the same value wherever both are defined, yet one may conceal a restriction of the other. Factoring and cancelling are useful because they expose structure, but the original denominator or square-root condition still controls the permissible inputs.

Record the original domain before simplifying.
Apply a reversible algebraic transformation and state its condition.
Use the simplified form for calculation without silently enlarging the original domain.

1.4 Geometry as a model-building tool

Equations often encode geometry. A rectangle of base $$b$$ and perpendicular height $$h$$ has area $$A=bh$$ . A parallelogram with the same base and perpendicular height has the same area: cut a triangular piece from one side and move it to the other side to form a rectangle. This is why its area is also $$bh$$ .

The word perpendicular matters. A slanted side of a parallelogram is not automatically its height. The area of a triangle with base $$b$$ and perpendicular height $$h$$ is half the matching parallelogram area:

A=\frac12 bh.

This kind of rearrangement argument is a useful habit: replace a difficult shape or equation by an equivalent one whose meaning is easier to see.

Area is controlled by perpendicular height. A triangle can be paired with a matching copy to form a parallelogram, and a sloping side can be cut and shifted to make a rectangle. These rearrangements preserve area because they preserve the same base and perpendicular height; they do not turn an arbitrary side length into a height.

A_{\triangle}=\frac12 bh,\qquad A_{\text{parallelogram}}=bh

Choose a base and draw its perpendicular height.
Rearrange the matching pieces into a rectangle or parallelogram with that base and height.
Apply the half-area relation only after identifying the perpendicular height.

Common Mistakes

Treating a unitless number as a completed engineering answer.
Cancelling a factor without recording the value at which that factor is zero.
Calling a slanted length a height when it is not perpendicular to the base.
Using an equation outside the assumptions that produced it, such as applying the ideal-resistor law to a component whose resistance changes strongly with temperature without saying so.

Chapter Summary

Engineering mathematics starts with quantities, units, variables, relationships, and checks. Real numbers provide the main numerical setting. Equations express relationships; algebra changes their form without changing their permitted meaning. Units and domain restrictions are part of the mathematics, not an afterthought.

Key Formulas

V=IR,\qquad A_{\text{rectangle}}=bh,\qquad A_{\text{triangle}}=\frac12bh.

Exercises

Conceptual Questions

Explain why $\frac{x^2-1}{x-1}=x+1$ is not an equality of functions on all real numbers.
State the unit of $$V/I$$ and the restriction needed to compute it.
A parallelogram has side lengths $$8$$ cm and $$5$$ cm. Why is its area not necessarily $40\ \mathrm{cm}^2$ ?

Basic Problems

Write $0.000,047\ \mathrm F$ in scientific notation and convenient engineering-prefix form.
A $12\ \mathrm V$ source drives $4\ \mathrm{mA}$ through a resistor. Find its resistance.
A triangular sensor region has base $$18$$ mm and perpendicular height $$7$$ mm. Find its area.