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→∞ (1 − 2/x)3x. Give a decimal value.
f(x) = (1 − 2/x)3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e3x − 1 − 3x)/x².
f(x) = (e3x − 1 − 3x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = ln(2x2 + 5). Find f′(−2).
f(x) = ln(2x2 + 5)x = −2
f′(−2) =
4.DerivativesHard
Let f(x) = x3 + 6x2 − 4. Find the x-coordinate x of the local maximum.
f(x) = x3 + 6x2 − 4
x =
5.IntegralsMedium
Evaluate I = ∫26 1/x dx.
f(x) = 1/x[2, 6]
I =
6.IntegralsHard
Evaluate I = ∫₀1 x²·eˣ dx.
∫₀1 x²eˣ dx
I =
7.Complex numbersMedium
Write z = 4∠160° in rectangular form a + jb and give the imaginary part b.
r = 4θ = 160°
b =
8.Complex numbersHard
Exactly one of the points A–D is a root of z⁶ = 64∠120°. Click it.
w = 64∠120°n = 6
1A (2∠−10°)
2B (3.6∠−40°)
3C (2∠−40°)
4D (2∠120°)
9.Differential equationsMedium
Solve y″ + 4y′ + 4y = 0 with y(0) = 5 and y′(0) = 4. Find y(0.5).
y″ + 4y′ + 4y = 0y(0) = 5y′(0) = 4
y(0.5) =
10.Differential equationsHard
A system obeys y″ + 4y′ + 13y = 13 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1); angles in radians.
y″ + 4y′ + 13y = 13y(0) = 0y′(0) = 0
y(1) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [3 3 2; 1 −4 3; −1 0 −5]
det A =
12.Matrices & determinantsHard
Solve the system −6x + 6y = −4 and −4x + 6y = 12. Find x.
−6x + 6y = −4−4x + 6y = 12
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (−6, −3, −4) and b = (−3, 5, −1), in degrees.
a = (−6, −3, −4)b = (−3, 5, −1)
θ = °
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(−2, 3, 0), B(−2, −1, 2) and C(−4, 4, 1). Coordinates are in metres.
A(−2, 3, 0)B(−2, −1, 2)C(−4, 4, 1)
A = m²
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 2, 4, 6p(x): 0.3, 0.15, 0.3, 0.25
σ =
16.ProbabilityHard
X has the distribution shown, and Y = 2X − 3. Find the standard deviation σY of Y.
x: 0, 1, 2p(x): 0.35, 0.3, 0.35Y = 2X − 3
σY =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 41, 30, 39, 15, 10.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 5. Find the z-score z of a score of 102.
μ = 100σ = 5x = 102
z =
19.Normal distributionMedium
The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 6 of them exceeds 151.8 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 6> 151.8 kg
P =
20.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 80% of them lie (the 80th percentile).
μ = 65σ = 10P = 80%
x = points
3Solutions
1.Limits
Answer: L = 0.002479
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).
Take the logarithm
ln L = limx→∞ 3·x·ln(1 − 2/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 3·(−2) = −6
Undo the logarithm
L = e−6 = 0.002479
Answer
L = 0.002479
2.Limits
Answer: L = 4.5
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).
First application
(3·e3x − 3)/(2x) → 0/0
Second application
9·e3x/2
Substitute x = 0
L = 9·1/2 = 4.5
Answer
L = 4.5
3.Derivatives
Answer: f′(−2) = −0.6154
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 for the logarithm
f′(x) = 4x/(2x2 + 5)
Substitute
f′(−2) = −8/13 = −0.6154
Answer
f′(−2) = −0.6154
4.Derivatives
Answer: x = −4
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 + 12x = 3·(x2 + 4x)
Solve f′(x) = 0
x = −4 or x = 0
Which one is which
f″(x) = 6x + 12: negative at a maximum, positive at a minimum.
f″(−4) = −12, f″(0) = 12
x = −4
Answer
x = −4
5.Integrals
Answer: I = 1.099
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.
Antiderivative
F(x) = ln x
Evaluate at the limits
I = (ln 6 − ln 2) = ln 3 = 1.099
Answer
I = 1.099
6.Integrals
Answer: I = 0.7183
Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.
First round
∫x²eˣ dx = x²eˣ − ∫2x eˣ dx
Second round
∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)
Limits
2.718·(1) − 1·2 = 0.7183
Answer
I = 0.7183
7.Complex numbers
Answer: b = 1.368
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.
Components
a = 4·cos(160°) = −3.759
b = 4·sin(160°) = 1.368
Rectangular form
z = −3.759 + j1.368
b = 1.368
Answer
b = 1.368
8.Complex numbers
Answer: C (2∠−40°)
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.
Modulus and arguments of the roots
|z| = 2
θ = 20°, 80°, 140°, −160°, −100°, −40°
The point
C = 2∠−40°
Answer
C (2∠−40°)
9.Differential equations
Answer: y(0.5) = 4.415
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.
Double root
(r + 2)² = 0 → r = −2
Constants
A = y(0) = 5
y′(0) = B − 2A → B = 4 + 10 = 14
Evaluate
y(0.5) = (5 + 14·0.5)·e−1 = 4.415
Answer
y(0.5) = 4.415
10.Differential equations
Answer: y(1) = 1.121
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 = 1, r = −2 ± 3j
Constants from the initial values
y(0) = 1 + A = 0 → A = −1
y′(0) = −2A + 3B = 0 → B = −0.6667
Evaluate
y(1) = 1·(1 − e−2(cos 3 + 0.6667·sin 3)) = 1.121
Answer
y(1) = 1.121
11.Matrices & determinants
Answer: det A = 58
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 = 20, M₁₂ = −2, M₁₃ = −4
Cofactor expansion
det A = 3·20 − 3·(−2) + 2·(−4) = 58
Answer
det A = 58
12.Matrices & determinants
Answer: x = 8
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 = (−6)·6 − 6·(−4) = −12
Replace the x column
det Aₓ = (−4)·6 − 6·12 = −96
x = −96/(−12) = 8
Answer
x = 8
y = 7.333
13.Vectors: dot & cross product
Answer: θ = 81.29 °
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
a·b = (−6)·(−3) + (−3)·5 + (−4)·(−1) = 7
Magnitudes
|a| = √((−6)² + (−3)² + (−4)²) = 7.81
|b| = √((−3)² + 5² + (−1)²) = 5.916
Angle
cos θ = 7/(7.81·5.916) = 0.1515
θ = arccos(0.1515) = 81.29°
Answer
θ = 81.29°
14.Vectors: dot & cross product
Answer: A = 5.385 m²
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.
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 = 0·0.3 + 2·0.15 + 4·0.3 + 6·0.25 = 3
Mean of the squares
E(X²) = 0·0.3 + 4·0.15 + 16·0.3 + 36·0.25 = 14.4
Variance
Var(X) = 14.4 − 3² = 5.4
σ = √5.4 = 2.324
Answer
σ = 2.324
16.Probability
Answer: σY = 1.673
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.
Spread of X
μ = 1, Var(X) = 0.7, σX = 0.8367
Linear change
σY = |2|·0.8367 = 1.673
Answer
σY = 1.673
The mean does move: E(Y) = 2·1 − 3 = −1.
17.Descriptive statistics
Answer: s = 13.98
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 = (102 − 100)/5 = 0.4
Answer
z = 0.4
19.Normal distribution
Answer: P = 0.07078
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 = 150 kg, σT = 0.5·√6 = 1.225 kg
Standardise
z = (151.8 − 150)/1.225 = 1.47
Probability
P = 1 − Φ(1.47) = 0.07078
Answer
P = 0.07078
20.Normal distribution
Answer: x = 73.42 points
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.