Module 1: Mathematical Foundations · Reference · consolidates Foundations 1–10
Everything from this module in one place — built for scanning, not for reading.
What this module gave you
Ten readings, and underneath them a single argument. Physics describes the world with mathematics, and every description carries a domain outside which it stops being true.
Vectors gave direction to quantities that need it. Coordinate systems gave you a frame, and the warning that the frame is yours rather than nature’s. The two products showed what two vectors can say about each other. Graphs let you see a relationship before computing it, and calculus let you compute what you had learned to see. Approximation, last, admitted what had been true throughout — that every problem you solved was a model, chosen for its usefulness inside a stated range.
What follows is the machinery. The judgment about when to use it is the part that took ten modules.
On this page
The right-hand column of each table gives the Foundation the result came from. If a formula looks unfamiliar, that is where the reasoning is.
1. Vectors
| Quantity | Formula | From |
|---|---|---|
| Magnitude | |A| = √(Ax² + Ay² + Az²) | F2 |
| Unit vector | â = A / |A| | F2 |
| Components | Ax = A cos θ, Ay = A sin θ | F4 |
| Resultant | R = √(A² + B² + 2AB cos θ) | F4 |
| Its direction | tan α = B sin θ / (A + B cos θ) | F4 |
| Equal magnitudes | R = 2A cos(θ/2), along the bisector | F4 |
| Bound on a resultant | |A − B| ≤ R ≤ A + B | F4 |
| Dot product | A·B = AB cos θ = AxBx + AyBy + AzBz | F5 |
| Angle between | cos θ = (A·B) / (AB) | F5 |
| Projection of A on B | scalar: (A·B)/B vector: [(A·B)/B²]B | F5 |
| Cross product | |A×B| = AB sin θ, direction by right hand | F6 |
| In components | (AyBz−AzBy)î − (AxBz−AzBx)ĵ + (AxBy−AyBx)k̂ | F6 |
| Self-products | A·A = A² A×A = 0 | F5, F6 |
| The two together | |A×B|² + (A·B)² = A²B² | F6 |
| Relative velocity | vA/B = vA − vB | F4 |
Unit vector products. î·î = 1 and î·ĵ = 0. For crosses, forward round î → ĵ → k̂ → î gives a plus: î×ĵ = k̂, ĵ×k̂ = î, k̂×î = ĵ. Backwards gives a minus.
2. Coordinates and trigonometry
Cartesian and polar · F3
x = r cos θ y = r sin θ
r = √(x² + y²) tan θ = y/x
Never take θ = tan−1(y/x) straight from a calculator. It returns a value between −90° and +90° and cannot tell the second quadrant from the fourth. Take α = tan−1|y/x|, then: Q1 gives α; Q2 gives 180° − α; Q3 gives 180° + α; Q4 gives 360° − α. Sketch the point first and the quadrant is obvious.
Rotating the frame · F3
A′x = Ax cos θ + Ay sin θ
A′y = −Ax sin θ + Ay cos θ
Trigonometric identities
sin²θ + cos²θ = 1
sin 2θ = 2 sin θ cos θ
cos 2θ = cos²θ − sin²θ = 1 − 2sin²θ = 2cos²θ − 1
sin(A ± B) = sin A cos B ± cos A sin B
3. Calculus
| Rule | Differentiation | Its inverse |
|---|---|---|
| Power | nxn−1 | xn+1/(n+1) |
| Product | u′v + uv′ | ∫u dv = uv − ∫v du |
| Chain | f′(g)·g′ | substitution, u = g(x) |
| Quotient | (u′v − uv′)/v² | — |
| f(x) | df/dx | ∫f dx (+ C) |
|---|---|---|
| c | 0 | cx |
| xn (n ≠ −1) | nxn−1 | xn+1/(n+1) |
| 1/x | −1/x² | ln|x| |
| ekx | kekx | (1/k)ekx |
| sin kx | k cos kx | −(1/k) cos kx |
| cos kx | −k sin kx | (1/k) sin kx |
| ln x | 1/x | x ln x − x |
Definite integrals · F9. ∫ab f dx = F(b) − F(a), no constant needed. Force and energy · F8. F = −dU/dx; equilibria where dU/dx = 0; stable if U″ > 0, unstable if U″ < 0, inconclusive if U″ = 0. SHM · F8. a = −ω²x gives T = 2π/ω.
Polar derivatives · F8. dr̂/dt = θ̇θ̂ and dθ̂/dt = −θ̇r̂, giving v = ṙr̂ + rθ̇θ̂ and a = (r̈ − rθ̇²)r̂ + (rθ̈ + 2ṙθ̇)θ̂. For circular motion r is constant, leaving centripetal rω² inward and tangential rθ̈ around.
4. Approximations
Valid only for small values · F10
(1 ± x)n ≈ 1 ± nx any n, for |x| « 1
1/(1 ± x) ≈ 1 ∓ x √(1 ± x) ≈ 1 ± x/2
sin θ ≈ θ tan θ ≈ θ cos θ ≈ 1 − θ²/2 θ in radians
f(x) ≈ f(a) + f′(a)(x − a) near x = a
Error in sin θ ≈ θ: about 0.5% at 10°, 2% at 20°, 4.5% at 30°.
5. Kinematic linkage
Down the ladder · F8
v = dx/dt a = dv/dt = d²x/dt²
Up the ladder · F9
v = v0 + ∫a dt x = x0 + ∫v dt
The one worth memorising
a = v dv/dx so a dx = v dv
From the chain rule: dv/dt = (dv/dx)(dx/dt) = v dv/dx. Use it whenever acceleration is given as a function of position rather than time — it removes t from the problem entirely and lets you integrate a dx against v dv directly. This single identity solves a large share of JEE kinematics questions that look intractable at first sight.
Accumulated quantities · F9
Work: W = ∫F dx Impulse: J = ∫F dt
Potential energy: ΔU = −∫F dr Gravitational: U = −GMm/r
6. Traps that cost marks
Every item below appeared as a misconception somewhere in this module. Read this section before an exam even if you skip everything else.
- Calculator in degrees. Every derivative and approximation involving angles assumes radians. Check the mode before you start.
- Arctangent without a quadrant check. The single most common error in vector questions. Sketch the point.
- The middle sign of a cross product. The ĵ term carries a minus from the determinant expansion. Verify by dotting your answer with both inputs — both must give zero.
- Confusing the two self-products. A·A = A², but A×A = 0. And a zero dot product means perpendicular, not zero.
- Order in a cross product. r×F, never F×r. The wrong order sends the body the wrong way round.
- Dropping the + C. It is the initial condition. Omitting it silently assumes the body started from rest at the origin.
- Displacement mistaken for distance. ∫v dt is displacement. For distance, integrate |v| and split wherever the velocity changes sign.
- Taking the modulus of negative work. The sign says energy left the object. Removing it destroys the physics.
- Assuming cosine for the x-component. Cosine belongs to the axis adjacent to the given angle. If the angle is measured from the y-axis, the roles swap.
- Reading a graph without reading the axes. The same parabola is a trajectory on y–x axes and a history on y–t axes.
- Using an approximation past its range. sin θ ≈ θ is fine at 5° and poor at 30°. Know roughly how large the error is.
- Sign collisions in potential energy. The minus in the definition ΔU = −∫F dr and the minus inside an attractive force are two different things. Keep them apart.
Two free checks on any answer. Do the units come out right? And is the size plausible — compared not against nothing, but against the quantity it changed? Both take seconds and catch most errors.
Where this goes next
Module 1 is complete. Everything here exists to be used, not admired — and from Module 2 onward it will be, on units and errors, then kinematics, then Newton’s laws and the rest of mechanics. If a formula on this page still feels unfamiliar, the Foundation it came from is named beside it.
Consolidates: Foundations 1–10 · Type: Reference · Use: Before problem sets and before exams