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→∞ (−6x3 + x + 8)/(3x3 + 5x + 9).
f(x) = (−6x3 + x + 8)/(3x3 + 5x + 9)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 2x)/sin(3x), with x in radians.
f(x) = ln(1 + 2x)/sin(3x)x → 0
L =
3.DerivativesMedium
Let f(x) = (4x − 3)3. Find f′(0).
f(x) = (4x − 3)3x = 0
f′(0) =
4.DerivativesHard
The tangent line to f(x) = x2 + 5x + 4 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = x2 + 5x + 4x = 1
b =
5.IntegralsMedium
Evaluate I = ∫02 2x·(x² + 1)2 dx.
f(x) = 2x·(x² + 1)2[0, 2]
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 9 − x² and the x-axis.
y = 9 − x²
A =
7.Complex numbersMedium
Write z = −6 − j3 in exponential form r·ejθ. Give θ in radians, between −π and π.
z = −6 − j3
θ = rad
8.Complex numbersHard
The roots of zⁿ = w with n = 2 and w = 4∠45° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1. Find the imaginary part of z₀.
w = 4∠45°n = 2k = 0
Im(z₀) =
9.Differential equationsMedium
A cup of coffee at 90 °C is left in a room at 18 °C. After 23 min its temperature is 41 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 27 °C?
T0 = 90 °CTa = 18 °CT(23) = 41 °CT = 27 °C
t = min
10.Differential equationsHard
A system obeys y″ + y′ + 16.25y = 16.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.25); angles in radians.
y″ + y′ + 16.25y = 16.25y(0) = 0y′(0) = 0
y(0.25) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c22 (row 2, column 2).
A = [−1 6; 4 3]B = [−6 −1; −1 5]
c22 =
12.Matrices & determinantsHard
Solve the system −5x − 4y = 4 and 3x + 6y = 4. Find x.
−5x − 4y = 43x + 6y = 4
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (3, −1, 3) and b = (k, −1, −6) are perpendicular.
a = (3, −1, 3)b = (k, −1, −6)
k =
14.Vectors: dot & cross productHard
A force F = (30, 30, −40) N acts at point A(−2, 1, −4). Coordinates are in metres. Find the magnitude |M| of the moment of the force about point O(2, 0, −2).
F = (30, 30, −40) NA(−2, 1, −4) mO(2, 0, −2) m
|M| = N·m
15.ProbabilityMedium
A bag holds 3 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
3 / 6
P =
16.ProbabilityHard
A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.7, independently. What is the probability P of exactly 2 correct answers?
n = 5p = 0.7k = 2
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 47, 16, 40, 44, 15, 25, 41.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 12. What proportion P of values is above 177.2?
μ = 170σ = 12x = 177.2
P =
19.Normal distributionMedium
The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 4 of them is below 50.4 kg. Round z to two decimals.
μ = 12 kgσ = 0.8 kgn = 4< 50.4 kg
P =
20.Normal distributionHard
Hole diameters D are normal with mean 25.02 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.02 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
3Solutions
1.Limits
Answer: L = −2
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).
Divide by the highest power
(−6 + 1/x2 + 8/x3)/(3 + 5/x2 + 9/x3)
Let x → ∞
L = −6/3 = −2
Answer
L = −2
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 0.6667
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).
L'Hôpital's rule
L = limx→0 [2/(1 + 2x)]/[3·cos 3x]
Substitute x = 0
L = 2/3 = 0.6667
Answer
L = 0.6667
3.Derivatives
Answer: f′(0) = 108
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) = 3·(4x − 3)2·4
Substitute
f′(0) = 3·(−3)2·4 = 108
Answer
f′(0) = 108
4.Derivatives
Answer: b = 3
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.
Point and slope
f(1) = 10
f′(x) = 2x + 5 → f′(1) = 7
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 10 − 7·1 = 3
Answer
b = 3
5.Integrals
Answer: I = 41.33
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 u2 du
Integrate and evaluate
I = [u3/3]15 = (125 − 1)/3 = 41.33
Answer
I = 41.33
6.Integrals
Answer: A = 36
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.
Zeros
9 − x² = 0 → x = ±3
Integrate
A = [9x − x³/3]−33 = 18 − (−18) = 36
Answer
A = 36
7.Complex numbers
Answer: θ = −2.678 rad
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.
Modulus
r = √(36 + 9) = 6.708
Argument in radians
θ = −153.4° = −153.4·π/180 = −2.678 rad
Exponential form
z = 6.708·e−j2.678
Answer
θ = −2.678 rad
8.Complex numbers
Answer: Im(z₀) = 0.7654
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.
The 2 roots lie on the circle of radius 2, 180° apart.
9.Differential equations
Answer: t = 41.91 min
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.
Cooling constant
k = ln(72/23)/23 = 0.04962 min⁻¹
Time to reach the target
t = ln(72/9)/0.04962 = 41.91 min
Answer
t = 41.91 min
10.Differential equations
Answer: y(0.25) = 0.4304
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.
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.
Row times column
c22 = 4·(−1) + 3·5 = 11
Answer
c22 = 11
12.Matrices & determinants
Answer: x = −2.222
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 = (−5)·6 − (−4)·3 = −18
Replace the x column
det Aₓ = 4·6 − (−4)·4 = 40
x = 40/(−18) = −2.222
Answer
x = −2.222
y = 1.778
13.Vectors: dot & cross product
Answer: k = 5.667
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 = 3·k + (−1)·(−1) + 3·(−6) = 0
−17 + 3·k = 0
Solve for k
k = −(−17)/3 = 5.667
Answer
k = 5.667
14.Vectors: dot & cross product
Answer: |M| = 267 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 = (−2, 1, −4) − (2, 0, −2) = (−4, 1, −2) m
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.
First and second draw
P = 3/9 · 2/8 = 6/72 = 0.08333
Answer
P = 0.08333
16.Probability
Answer: P = 0.1323
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.
Binomial formula
P(X = 2) = 10·0.72·0.33 = 10·0.49·0.027 = 0.1323
Answer
P = 0.1323
17.Descriptive statistics
Answer: s = 13.58
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 = 0.6
Standard normal table
Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743
Answer
P = 0.2743
19.Normal distribution
Answer: P = 0.9332
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the total
μT = 48 kg, σT = 0.8·√4 = 1.6 kg
Standardise
z = (50.4 − 48)/1.6 = 1.5
Probability
P = Φ(1.5) = 0.9332
Answer
P = 0.9332
20.Normal distribution
Answer: P = 0.8133
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).