Math for ME · Chapter 1 of 19 · Beginner
Algebra, Functions, and Engineering Notation
Manipulate equations confidently before entering real engineering math. Every later course leans on this chapter.
The thread: This is where the journey starts. Before you can model any machine, you need to manipulate the relationships that describe it. Everything ahead is built on this one skill.
Readiness check
School-level skills only. Tick only what you can do closed-notes.
- Add, subtract, multiply, and divide fractions without a calculator.
- Work with negative numbers and inequalities correctly.
- Compute percentages and ratios.
- Read values off a graph with labeled axes.
- Use a scientific calculator's power, root, and log keys.
The core idea
An equation is a model. Rearranging it is engineering, not symbol-pushing.
σ = F/AA = πd²/4Every formula you will ever use can be solved for any variable in it. The skill is doing it without breaking units: whatever you do to one side, you do to the other, and the units must still cancel at the end.
The skills, taught in order
3.1 Isolate any variable
One operation per line, applied to both sides, until the unknown stands alone. To solve P = F/A for A: multiply both sides by A, then divide by P, giving A = F/P. To solve it for F: multiply both sides by A, giving F = PA. The order is mechanical once you commit to writing one step per line.
3.2 Powers, roots, logarithms, exponentials
These four behave as one family. A root is a fractional power: √x = x1/2, and ∛x = x1/3. A logarithm is the inverse of an exponential: if y = aˣ then x = log_a y. The three logarithm rules are the only ones that exist:
log(ab) = log a + log blog(a/b) = log a − log blog(aⁿ) = n log aEngineers use the natural log, ln, almost only to undo eˣ, the way division undoes multiplication. To solve ekt = 5 for t, take ln of both sides: kt = ln 5, so t = (ln 5)/k.
3.3 Engineering and scientific notation
Engineering notation keeps the exponent a multiple of three, so SI prefixes read straight off it.
| Prefix | Symbol | Factor |
|---|---|---|
| giga | G | 10⁹ |
| mega | M | 10⁶ |
| kilo | k | 10³ |
| milli | m | 10⁻³ |
| micro | µ | 10⁻⁶ |
| nano | n | 10⁻⁹ |
So 0.00347 m = 3.47 × 10⁻³ m = 3.47 mm, and 200 MPa = 200 × 10⁶ Pa = 200 N/mm². That last identity, 1 MPa = 1 N/mm², removes power-of-ten errors from every stress problem you will ever do.
3.4 The five function shapes of engineering
Almost every quantity you model follows one of these shapes. Learn the shape and you can predict behaviour before you compute a single value.
| Family | Form | Behaviour | Where it appears |
|---|---|---|---|
| Linear | y = ax + b | constant rate of change | Hooke's law, F = kx |
| Quadratic | y = ax² | one turning point, symmetric | kinetic energy, drag at speed |
| Exponential | y = ekt | growth if k > 0, decay if k < 0 | cooling, capacitor discharge |
| Logarithmic | y = ln x | rises fast then flattens | sound level, entropy |
| Sinusoidal | y = A sin(ωt) | repeating oscillation | vibration, AC, waves |
3.5 Dimensional homogeneity: a free error detector
Every additive term in a correct equation carries identical units. Check any formula before trusting it. For projectile range R = v² sin(2θ)/g, the units are [v²]/[g] = (m²/s²)/(m/s²) = m, which is a length, so the formula is at least dimensionally possible. If a term had come out in seconds, the formula would be wrong before a single number entered.
Engineering connection: every course on this site. Statics Module 1 (units and W = mg) begins exactly where this chapter ends.
Worked example: size a rod from an allowable stress
A steel tie rod must carry a tension of F = 10 kN. The allowable stress is σ = 200 MPa. What rod diameter is required?
- ProblemFind the smallest diameter d for the rod in Figure 1.
- Given / findF = 10 kN = 10 000 N; σ = 200 MPa = 200 N/mm². Find d in mm.
- AssumptionsUniform stress across the section (pure tension); circular cross-section.
- ModelTwo formulas chained: σ = F/A and A = πd²/4. The unknown d is buried inside; algebra digs it out.
- EquationsA = F/σ d = √(4A/π)
- SolveA = 10 000/200 = 50 mm². d = √(4 × 50/π) = √63.66 = 7.98 mm, so specify 8 mm.
- CheckUnits: N divided by N/mm² gives mm²; the root of mm² gives mm. Scale: an 8 mm steel rod holding 1 tonne-force is plausible; 8 m or 0.08 mm would not be.
- ConclusionThe entire calculation was algebra: isolate A, then isolate d, with units checked at each step. This exact chain (load, allowable stress, required size) reappears in Mechanics of Materials as the basic design equation.
Worked example 2: solve an exponential for time
A capacitor in a sensor circuit discharges as V = V₀ e−t/τ, with V₀ = 5.0 V and time constant τ = 20 ms. The logic downstream reads "low" once the voltage falls below 0.5 V. How long after discharge begins does the signal read low?
- Given / findV₀ = 5.0 V, τ = 20 ms, threshold V = 0.5 V. Find the time t.
- ModelOne exponential with the unknown inside the exponent. A logarithm is the tool that frees it.
- Isolate the exponentialV/V₀ = e−t/τ, so 0.5/5.0 = 0.10 = e−t/τ.
- Take ln of both sidesln(0.10) = −t/τ, so −2.303 = −t/τ.
- Solvet = 2.303 × τ = 2.303 × 20 ms = 46 ms.
- CheckUnits: τ is in ms and ln is dimensionless, so t comes out in ms. Scale: after one τ the signal sits at 37%, and reaching 10% takes a little over 2 time constants, so 46 ms ≈ 2.3 × 20 ms is consistent.
- ConclusionA drop by one decade, 5 V to 0.5 V, always takes ln(10) ≈ 2.30 time constants, whatever τ happens to be. Recognising that pattern is worth more than the single number, and it reappears in RC circuits, cooling, and first-order systems throughout the course.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Operating on one side only | Equation "magically" changes meaning mid-solution | "Did I apply that operation to the entire other side too?" | Write each step on a new line; one operation per line, both sides. |
| Square-rooting term by term | √(a² + b²) becomes a + b | "Is the root of a sum the sum of the roots?" | No. Test with numbers: √(9 + 16) = 5, not 7. Numbers expose fake rules instantly. |
| Logarithm rules invented on the spot | log(a + b) treated as log a + log b | "Which rule am I using, exactly?" | Only three: log(ab) = log a + log b; log(a/b) = log a − log b; log(aⁿ) = n log a. |
| Dropping units during algebra | Answer has no units, or absurd magnitude | "Can I write the units of every line I just wrote?" | Carry units like algebra symbols; they cancel or they expose the error. |
Practice ladder
Solve v² = u² + 2as for s, then evaluate with v = 30 m/s, u = 10 m/s, a = 4 m/s².
Show answer
s = (v² − u²)/(2a) = (900 − 100)/8 = 100 m. Units: (m²/s²)/(m/s²) = m.
Then solve A = πd²/4 for d, and find d when A = 78.5 mm².
Show answer
d = √(4A/π) = √(4 × 78.5/π) = √99.95 = 10.0 mm. The square root undoes the square; the factor 4/π is simply carried along.
The belt friction formula is T₁ = T₂ eμβ. Solve it for μ, then evaluate with T₁ = 600 N, T₂ = 200 N, β = π rad.
Show answer
μ = ln(T₁/T₂)/β = ln(3)/π = 1.0986/3.1416 = 0.350. The logarithm undoes the exponential; that is the only reason engineers care about ln.
A pressure is given as p = 2.5 × 10⁶ Pa. Express it in MPa and in N/mm², then confirm the units of p = F/A agree with it.
Show answer
2.5 × 10⁶ Pa = 2.5 MPa = 2.5 N/mm², using 1 MPa = 1 N/mm². And [F]/[A] = N/mm² = Pa, so the units are consistent.
The period of a pendulum is T = 2π√(L/g). A clock pendulum must have T = 2.00 s. Find the required length L, and state what happens to T if L is accidentally made 10% too long.
Show answer
L = g(T/2π)² = 9.81 × (2/6.2832)² = 9.81 × 0.1013 = 0.994 m. Since T scales with √L, a 10% length error gives roughly a 5% period error (√1.1 ≈ 1.049): the clock runs about 5% slow.
Take any formula from a datasheet or textbook (gear ratio, pipe flow, battery capacity). Solve it for each variable in turn, state the units of every form, and write one sentence on which rearranged form a designer would actually use.
What good work looks like
All rearranged forms algebraically correct, units stated and consistent for each, and a sensible note like "the designer knows the load and the allowable stress, so the d = ... form is the design equation."
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Build a one-page "Formula Fluency Sheet": pick five real engineering formulas, show each solved for every variable, with units verified per form. Add the five function-shape sketches (linear, quadratic, exponential, logarithmic, sinusoidal) drawn by hand with one engineering example each.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. What does it mean for an equation to be dimensionally homogeneous?
Every additive term has the same units. If terms disagree, the equation is wrong before any numbers enter.
2. Write 0.00347 A in engineering notation and with an SI prefix.
3.47 × 10⁻³ A = 3.47 mA. Engineering notation uses exponents in multiples of 3 so prefixes read directly.
3. State the three logarithm rules.
log(ab) = log a + log b; log(a/b) = log a − log b; log(aⁿ) = n log a.
4. Why is √(a² + b²) not equal to a + b?
Squaring and adding do not commute with the root. Test: a = 3, b = 4 gives 5, not 7. This sum is the magnitude formula used for vectors all through Statics.
5. Which function shape describes free cooling, and which describes resonance growth?
Cooling: decaying exponential e−kt. Resonant or unstable growth: rising exponential. Recognizing shapes predicts system behavior before solving.
Textbook mapping
| Item | Mapping |
|---|---|
| Main sources | Stewart, Redlin and Watson, Precalculus: Mathematics for Calculus (algebra, exponentials, logarithms, functions); Stewart, Calculus: Early Transcendentals (review appendix) |
| Core topics | 1.1 Algebraic manipulation · 1.2 Fractions, powers, roots, logs, exponentials · 1.3 Engineering notation · 1.4 Functions and graphs · 1.5 The five function families · 1.6 Units and dimensional consistency · 1.7 Rearranging engineering formulas |
| Engineering connection | Every course. Most directly: Statics Module 1 (units and W = mg) starts exactly where this chapter ends. |
| Skip on first pass | Nothing. Weakness here taxes every later chapter. |
| Read next | Trigonometry and Geometry. |