Decide which form suits the operation: rectangular a + jb for adding and subtracting, polar r∠θ for multiplying, dividing, powers and roots.
Multiply out with j² = −1; divide by multiplying the top and bottom by the conjugate of the denominator.
Convert with r = √(a² + b²) and θ from a sketch of the point (check the quadrant), and back with a = r cos θ, b = r sin θ.
For powers use De Moivre, (r∠θ)ⁿ = rⁿ∠nθ; for zⁿ = R∠φ list all n roots R1/n∠((φ + 360°k)/n).
Differential equations
Exponential growth/decay
dy/dt = ky → y = y0ekt
Half-life
k = ln 2 / t½
Newton's law of cooling
T(t) = Ta + (T0 − Ta)e−kt
Integrating factor
y′ + p(x)y = q(x): μ = e∫p dx, (μy)′ = μq
Mixing tank
dm/dt = r·c − (r/V)·m
Matrices & determinants
2×2 determinant
det = ad − bc
2×2 inverse
A⁻¹ = (1/det)·[d −b; −c a]
Product
(AB)ij = Σk AikBkj
Eigenvalues
det(A − λI) = 0
Cramer’s rule
xi = det(Ai)/det(A)
Vectors: dot & cross product
Magnitude
|a| = √(ax² + ay² + az²)
Unit vector
û = a/|a|
Components from an angle
Fx = F·cos θ, Fy = F·sin θ
Dot product
a·b = axbx + ayby + azbz = |a||b|cos θ
Scalar projection of a onto b
a·b/|b|
Work
W = F·d
Cross product
a × b = (aybz − azby, azbx − axbz, axby − aybx)
Areas
parallelogram |a × b|, triangle ½|a × b|
Moment about O
MO = r × F, r = A − O
Volume of a parallelepiped
V = |a·(b × c)|
Method
Write every vector in components; for a force given by a magnitude and an angle, use F·cos θ and F·sin θ.
For lengths and directions use the magnitude √(x² + y² + z²) and the unit vector a/|a|.
Use the dot product for angles, projections, perpendicularity and work: a·b = |a||b|cos θ.
Use the cross product for perpendicular vectors, areas and moments: write the three components carefully, keeping the order a × b.
Probability
Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k
Descriptive statistics
Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ
Normal distribution
z-score
z = (x − μ)/σ
Below x
P(X < x) = Φ(z)
Above x
P(X > x) = 1 − Φ(z)
Between
Φ(z2) − Φ(z1)
Percentile
x = μ + z·σ
Method
Convert the value to a z-score: z = (x − μ)/σ.
Look up Φ(z), the area to the left of z.
For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
For a percentile, work backwards: find z for the area, then x = μ + z·σ.
2Practice problems
1.LimitsMedium
Find the limit L = limx→∞ (√(4x2 − 10x) − 2x).
f(x) = √(4x2 − 10x) − 2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 4x)/(2x²), with x in radians.
f(x) = (1 − cos 4x)/(2x²)x → 0
L =
3.DerivativesMedium
Let f(x) = (4x + 1)5. Find f′(−1).
f(x) = (4x + 1)5x = −1
f′(−1) =
4.DerivativesHard
Let f(x) = x3 − 9x2 + 15x − 4. Find the x-coordinate x of the local maximum.
f(x) = x3 − 9x2 + 15x − 4
x =
5.IntegralsMedium
Evaluate I = ∫02 2x·(x² + 1)1 dx.
f(x) = 2x·(x² + 1)1[0, 2]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = 2x2 and y = 3x.
y = 2x2y = 3x
A =
7.Complex numbersMedium
Find the argument θ of z = −2 + j6 in degrees, as the principal value between −180° and 180°.
z = −2 + j6
θ = °
8.Complex numbersHard
The equation z² + 4z + 13 = 0 has two complex roots. Find their modulus |z|.
p = 4q = 13
|z| =
9.Differential equationsMedium
A forged steel part at 440 °C is left to cool in a workshop at 24 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.03 min⁻¹, find its temperature T after 41 min.
T0 = 440 °CTa = 24 °Ck = 0.03 min⁻¹t = 41 min
T = °C
10.Differential equationsHard
A system obeys y″ + 4y′ + 29y = 290 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1); angles in radians.
y″ + 4y′ + 29y = 290y(0) = 0y′(0) = 0
y(1) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [5 1 5; 2 0 −4; 5 3 −1]
det A =
12.Matrices & determinantsHard
Solve the system −2x + y = −12 and −6x − 4y = −1. Find x.
−2x + y = −12−6x − 4y = −1
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (−4, −2, −3) and b = (k, 6, 3) are perpendicular.
a = (−4, −2, −3)b = (k, 6, 3)
k =
14.Vectors: dot & cross productHard
A force F = (0, 20, 30) N acts at point A(−1, 1, 3). Coordinates are in metres. Find the z-component Mz of the moment of the force about point O(−2, −2, 0).
F = (0, 20, 30) NA(−1, 1, 3) mO(−2, −2, 0) m
Mz = N·m
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 2, 4, 6p(x): 0.2, 0.25, 0.2, 0.35
σ =
16.ProbabilityHard
12 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?
n = 12p = 0.25
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 15, 39, 60, 12, 37.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is below 125?
μ = 100σ = 20x = 125
P =
19.Normal distributionMedium
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 242.5 and 256.25 g? Give P as a decimal.
μ = 250σ = 5x1 = 242.5x2 = 256.25
P =
20.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 80% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g80%
x = g
3Solutions
1.Limits
Answer: L = −2.5
Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).
Multiply by the conjugate
f(x) = −10x/(√(4x2 − 10x) + 2x)
Divide by x
f(x) = −10/(√(4 + (−10)/x) + 2)
Let x → ∞
L = −10/(2 + 2) = −2.5
Answer
L = −2.5
2.Limits
Answer: L = 4
Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).
First application
4·sin 4x/(4x) → 0/0
Second application
16·cos 4x/4
Substitute x = 0
L = 16/4 = 4
Answer
L = 4
3.Derivatives
Answer: f′(−1) = 1620
Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.
Chain rule
f′(x) = 5·(4x + 1)4·4
Substitute
f′(−1) = 5·(−3)4·4 = 1620
Answer
f′(−1) = 1620
4.Derivatives
Answer: x = 1
Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.
Derivative
f′(x) = 3x2 − 18x + 15 = 3·(x2 − 6x + 5)
Solve f′(x) = 0
x = 1 or x = 5
Which one is which
f″(x) = 6x − 18: negative at a maximum, positive at a minimum.
f″(1) = −12, f″(5) = 12
x = 1
Answer
x = 1
5.Integrals
Answer: I = 12
Definite integrals: ∫ₐᵇ f(x) dx is the signed area between the graph and the x-axis from a to b. Find an antiderivative F (with F′ = f) and evaluate F(b) − F(a). Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1). The area between two curves is ∫(upper − lower) dx.
Substitution
u = x² + 1, du = 2x dx
I = ∫15 u1 du
Integrate and evaluate
I = [u2/2]15 = (25 − 1)/2 = 12
Answer
I = 12
6.Integrals
Answer: A = 1.125
Definite integrals: ∫ₐᵇ f(x) dx is the signed area between the graph and the x-axis from a to b. Find an antiderivative F (with F′ = f) and evaluate F(b) − F(a). Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1). The area between two curves is ∫(upper − lower) dx.
Intersection points
2x2 = 3x → x = 0, x = 1.5
Integrate upper minus lower
A = [1.5x2 − 0.6667x3]01.5 = 1.125
Answer
A = 1.125
7.Complex numbers
Answer: θ = 108.4 °
Polar form: z = r∠θ = r(cos θ + j sin θ) = r·ejθ, with r = |z| and θ = arg z. Back to rectangular form: a = r cos θ, b = r sin θ. The argument needs the quadrant: θ = atan(b/a) only when a > 0; for a < 0 add or subtract 180°. In polar form multiplication is easy: multiply the moduli and add the arguments; for division divide the moduli and subtract the arguments.
Quadrant
a = −2, b = 6 → II
Reference angle
α = atan(6/2) = 71.57°
Argument
θ = 180° − α = 108.4°
Answer
θ = 108.4°
In polar form z = 6.325∠108.4°.
8.Complex numbers
Answer: |z| = 3.606
De Moivre and roots: (r∠θ)ⁿ = rⁿ∠nθ. The equation zⁿ = R∠φ has exactly n roots, all with modulus R1/n, at the arguments (φ + 360°·k)/n for k = 0, 1, …, n − 1: they lie evenly spaced on a circle, 360°/n apart. A quadratic z² + pz + q = 0 with p² < 4q has the complex roots z = −p/2 ± j√(q − p²/4); a polynomial with real coefficients always has its complex roots in conjugate pairs a ± jb.
Discriminant
p² − 4q = 16 − 52 = −36 < 0
Quadratic formula
z = (−4 ± j√36)/2 = −2 + j3, −2 − j3
Answer
|z| = √((−2)² + 3²) = √13 = 3.606
Answer
|z| = 3.606
9.Differential equations
Answer: T = 145.6 °C
First-order differential equations: dy/dt = k·y has the solution y = y₀·ekt (growth for k > 0, decay for k < 0); a half-life T gives k = ln 2/T. Newton's law of cooling, dT/dt = −k·(T − Tₐ), gives T(t) = Tₐ + (T₀ − Tₐ)·e−kt. A linear equation y′ + p(x)·y = q(x) is solved with the integrating factor μ = e∫p dx: (μy)′ = μq, so μy = ∫μq dx + C.
Solution of the cooling law
T(t) = 24 + 416·e−0.03t
Substitute t
T(41) = 24 + 416·0.2923 = 145.6 °C
Answer
T = 145.6 °C
10.Differential equations
Answer: y(1) = 10.14
Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.
Steady value and roots
yss = 10, r = −2 ± 5j
Constants from the initial values
y(0) = 10 + A = 0 → A = −10
y′(0) = −2A + 5B = 0 → B = −4
Evaluate
y(1) = 10·(1 − e−2(cos 5 + 0.4·sin 5)) = 10.14
Answer
y(1) = 10.14
11.Matrices & determinants
Answer: det A = 72
Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.
2×2 minors of the first row
M₁₁ = 12, M₁₂ = 18, M₁₃ = 6
Cofactor expansion
det A = 5·12 − 1·18 + 5·6 = 72
Answer
det A = 72
12.Matrices & determinants
Answer: x = 3.5
Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.
Determinant of the coefficient matrix
det A = (−2)·(−4) − 1·(−6) = 14
Replace the x column
det Aₓ = (−12)·(−4) − 1·(−1) = 49
x = 49/14 = 3.5
Answer
x = 3.5
y = −5
13.Vectors: dot & cross product
Answer: k = −5.25
Dot product:a·b = axbx + ayby + azbz = |a|·|b|·cos θ, a scalar. It gives the angle between two vectors, cos θ = a·b/(|a|·|b|), and it is zero when they are perpendicular. The scalar projection of a onto b is a·b/|b|, and a constant force F moving through a displacement d does work W = F·d.
Dot product equals zero
a·b = (−4)·k + (−2)·6 + (−3)·3 = 0
−21 + (−4)·k = 0
Solve for k
k = −(−21)/(−4) = −5.25
Answer
k = −5.25
14.Vectors: dot & cross product
Answer: Mz = 20 N·m
Cross product:a × b = (aybz − azby, azbx − axbz, axby − aybx) is perpendicular to both vectors, and |a × b| = |a|·|b|·sin θ is the area of the parallelogram they span (the triangle has half of it). Order matters: b × a = −(a × b). The moment of a force F applied at point A about point O is MO = r × F, with r = A − O.
Position vector from O to A
r = A − O = (−1, 1, 3) − (−2, −2, 0) = (1, 3, 3) m
Moment M = r × F
(r × F)x = ryFz − rzFy = 3·30 − 3·20 = 30 N·m
(r × F)y = rzFx − rxFz = 3·0 − 1·30 = −30 N·m
(r × F)z = rxFy − ryFx = 1·20 − 3·0 = 20 N·m
Requested component
Mz = 20 N·m
Answer
Mz = 20 N·m
15.Probability
Answer: σ = 2.289
Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.
Mean
μ = 0·0.2 + 2·0.25 + 4·0.2 + 6·0.35 = 3.4
Mean of the squares
E(X²) = 0·0.2 + 4·0.25 + 16·0.2 + 36·0.35 = 16.8
Variance
Var(X) = 16.8 − 3.4² = 5.24
σ = √5.24 = 2.289
Answer
σ = 2.289
16.Probability
Answer: P = 0.1584
Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.
Zero and one failure
P(X = 0) = 0.7512 = 0.03168
P(X = 1) = 12·0.25·0.7511 = 0.1267
Add them
P = 0.03168 + 0.1267 = 0.1584
Answer
P = 0.1584
17.Descriptive statistics
Answer: s = 19.65
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Standardise
z = 1.25
Standard normal table
Φ(1.25) = 0.8944
P = 0.8944
Answer
P = 0.8944
19.Normal distribution
Answer: P = 0.8275
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
z-scores
z₁ = (242.5 − 250)/5 = −1.5
z₂ = (256.25 − 250)/5 = 1.25
Area between
P = Φ(1.25) − Φ(−1.5) = 0.8944 − 0.06681 = 0.8275
Answer
P = 0.8275
20.Normal distribution
Answer: x = 245.8 g
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.