The most predictable marks on the paper. The syllabus barely moves, the question styles repeat, and none of it depends on remembering a ship.
Rank decides whether a system of equations has a solution at all. Compare the rank of the coefficient matrix with the rank of the augmented matrix.
Eigenvalues come from setting the characteristic determinant to zero, and eigenvectors from substituting each one back.
Two identities make free checks available, and using them habitually is worth more marks than any amount of care:
Useful facts that turn up as one-mark questions: a symmetric real matrix has real eigenvalues and orthogonal eigenvectors; a triangular matrix has its eigenvalues on the diagonal; and the eigenvalues of A⁻¹ are the reciprocals of those of A, with the same eigenvectors.
Three operators, three different kinds of answer.
The directional derivative of f in the direction of a unit vector û is ∇f·û, which is largest when û points along ∇f — that is what "steepest increase" means quantitatively.
The three integral theorems all do the same job: they trade an integral over a boundary for an integral over what the boundary encloses.
Choosing between them is a matter of noticing which dimension you are moving between: line to area, line to surface, or surface to volume.
Two identities worth memorising because they collapse whole questions to a single line: ∇×(∇f) = 0 always, and ∇·(∇×F) = 0 always. A field with zero curl is conservative and has a potential; a field with zero divergence is solenoidal.
First order linear. Put it in standard form and multiply by the integrating factor.
Second order with constant coefficients. The solution is the complementary function plus a particular integral.
For the particular integral, try a form matching the right-hand side: a polynomial for a polynomial, Ae^(kx) for an exponential, A cos + B sin for a sinusoid. The one thing to watch is resonance: if the trial form already appears in the complementary function, multiply it by x.
This is not a coincidence — the complex-root case is the vibration chapter's damped oscillator, and the resonance rule is why a system driven at its own natural frequency grows without limit.
Laplace transforms turn a differential equation into an algebraic one and carry the initial conditions automatically.
A complex function is analytic where it is differentiable, and the test is the Cauchy–Riemann conditions.
An analytic function's real and imaginary parts are both harmonic — they satisfy Laplace's equation — which is exactly why complex analysis appears in potential flow, and therefore in the resistance chapter.
For contour integration, two results carry most of the marks:
The practical routine is: locate the poles, decide which lie inside the contour, compute their residues, sum and multiply by 2πi.
Most marks here come from choosing the right distribution, not from the arithmetic.
Bayes questions are almost always a conditional stated the wrong way round — the probability of a positive test given the disease, when what is wanted is the probability of the disease given a positive test. Drawing the tree first, with actual numbers of cases rather than probabilities, makes the answer obvious and takes thirty seconds.
And two results always worth reaching for: P(at least one) = 1 − P(none), and for independent events P(A∩B) = P(A)·P(B). Mutually exclusive is not the same as independent — mutually exclusive events are maximally dependent, since one occurring guarantees the other did not.
Root finding. Newton–Raphson follows the tangent to the curve.
It converges quadratically — the number of correct digits roughly doubles per step — but it needs a good starting point and fails badly where f′ is near zero. Bisection is slower but cannot fail once a sign change has been bracketed.
Integration. The same rules as the hydrostatics chapter, which is convenient.
Simpson's rule is exact for cubics, and its error falls as h⁴ — halving the interval cuts the error by a factor of sixteen. The trapezoidal rule overestimates for a convex function and underestimates for a concave one, which is a common conceptual question.
Marching an ODE forward. Euler's method uses the slope at the start of the interval and accumulates error quickly. Fourth-order Runge–Kutta samples the slope four times per step and is the practical default.
An eigenvalue problem with the checks that make it self-verifying, a second-order ODE, and the two numerical methods that appear most often.
Find the eigenvalues and eigenvectors of A = [[4, 1], [2, 3]]. Verify the result without repeating the working.
A = | 4 1 | | 2 3 |
Find the eigenvalues and eigenvectors of A = [[4, 1], [2, 3]]
Verify the result without repeating the working
Set up the characteristic equation.
Subtract λ from the leading diagonal and set the determinant to zero.
Solve it.
Check before going further.
The eigenvalues must sum to the trace and multiply to the determinant. Both take five seconds and catch nearly every arithmetic slip.
Eigenvector for λ = 5.
Substitute back and solve (A − λI)x = 0. The two rows will be multiples of each other — if they are not, the eigenvalue is wrong.
Eigenvector for λ = 2.
Verify by multiplication.
One line each, and it removes all doubt.
Note that eigenvectors are only defined up to a scale factor.
[2, 2]ᵀ and [−1, −1]ᵀ are equally correct answers for λ = 5, which matters when matching against multiple-choice options that look different from what you wrote.
Answerλ = 5 with [1, 1]ᵀ; λ = 2 with [1, −2]ᵀ
The trap: in a multiple-choice question, rejecting the right option because it is scaled differently from your answer. Check whether it is a multiple of yours before moving on.
Solve y″ − 5y′ + 6y = e^x with y(0) = 1 and y′(0) = 0.
y″ − 5y′ + 6y = eˣ y(0) = 1, y′(0) = 0
Solve y″ − 5y′ + 6y = e^x with y(0) = 1 and y′(0) = 0
A linear ODE with constant coefficients splits into two parts.
The complementary function, which solves the equation set to zero, and a particular integral, which handles the forcing term.
Complementary function.
Replace y″ by m², y′ by m and y by 1, and solve the auxiliary equation.
Particular integral.
The forcing term is e^x, so try y = Ae^x. Note that 1 is not a root of the auxiliary equation, so no extra factor of x is needed — if the forcing exponent had matched a root, it would have been.
General solution.
Apply the initial conditions.
Differentiate first, then substitute x = 0 in both.
Solve the pair.
Write the answer, then check it against both conditions.
Answery = 2e^(2x) − 1.5e^(3x) + 0.5e^x
The trap: forgetting to check whether the forcing exponent is already a root of the auxiliary equation. Here it is not, but if the right-hand side had been e^(2x), the trial solution would need to be Axe^(2x) — and Ae^(2x) would have produced 0 = 1.
(a) Evaluate ∫₀¹ dx/(1 + x²) using Simpson's first rule with four intervals, and compare with the exact value. (b) Find the root of x³ − 2x − 5 = 0 near x = 2 by Newton–Raphson, to four decimal places.
(a) f(x) = 1/(1 + x²) on [0, 1], n = 4 (b) f(x) = x³ − 2x − 5, x₀ = 2
(a) Evaluate ∫₀¹ dx/(1 + x²) using Simpson's first rule with four intervals, and compare with the exact value
(b) Find the root of x³ − 2x − 5 = 0 near x = 2 by Newton–Raphson, to four decimal places
Part (a).
| x | f(x) = 1/(1+x²) | SM | Product |
|---|---|---|---|
| 0.00 | 1.000000 | 1 | 1.000000 |
| 0.25 | 0.941176 | 4 | 3.764706 |
| 0.50 | 0.800000 | 2 | 1.600000 |
| 0.75 | 0.640000 | 4 | 2.560000 |
| 1.00 | 0.500000 | 1 | 0.500000 |
| Σ | 9.424706 |
Interval width and function values.
Apply Simpson's first rule.
Multipliers 1, 4, 2, 4, 1 for four intervals.
Compare with the exact value, which is arctan(1).
That accuracy from five function evaluations is why Simpson's rule is the default.
It is exact for cubics, and the error falls as h⁴ — halving the interval cuts the error by sixteen.
Part (b).
Newton–Raphson iterates along the tangent to the curve.
First iteration from x₀ = 2.
Second iteration.
Third iteration — and note how the correction collapses.
| n | xₙ | f(xₙ) | f′(xₙ) | Correction | xₙ₊₁ |
|---|---|---|---|---|---|
| 0 | 2.000000 | −1.000000 | 10.000 | 0.100000 | 2.100000 |
| 1 | 2.100000 | 0.061000 | 11.230 | 0.005432 | 2.094568 |
| 2 | 2.094568 | 0.000186 | 11.162 | 0.0000167 | 2.094551 |
Newton–Raphson converges quadratically: the number of correct digits roughly doubles each step.
Answer(a) 0.785392 against an exact π/4 = 0.785398. (b) root = 2.0946
The trap: in part (a), using Simpson's rule with an odd number of intervals. And in part (b), starting from a point where f′ is near zero — the tangent is nearly horizontal and the next iterate is thrown far away, which is how Newton–Raphson diverges.
|A − λI| = 0Σλ = trace, Πλ = det — free arithmetic checksrank(A) vs rank(A|b)Decides unique / infinite / no solution∇f, ∇·F, ∇×F∇×(∇f) = 0 and ∇·(∇×F) = 0 always∮F·dr = ∬(∇×F)·dSStokes; Gauss for the volume formy′ + Py = Q ⇒ IF = e^∫P dxFirst-order linearam² + bm + c = 0CF from the roots; multiply PI by x if it duplicates the CFL{f′} = sF(s) − f(0)Laplace carries the initial conditionsu_x = v_y, u_y = −v_xCauchy–Riemann; ∮f dz = 2πi·Σ residuesP(A|B) = P(B|A)P(A)/P(B)Bayes; draw the tree in case counts firstx_{n+1} = x_n − f/f′Newton–Raphson; quadratic convergence