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→∞ (√(9x2 + 6x) − 3x).
f(x) = √(9x2 + 6x) − 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) = (3x − 3)5. Find f′(2).
f(x) = (3x − 3)5x = 2
f′(2) =
4.DerivativesHard
Let f(x) = x3 − 12x + 4. Find the x-coordinate x of the local maximum.
f(x) = x3 − 12x + 4
x =
5.IntegralsMedium
Evaluate I = ∫₁2.5 x·ln x dx.
∫₁2.5 x ln x dx
I =
6.IntegralsHard
Find the area A of the region enclosed by y = 0.5x2 and y = 2x.
y = 0.5x2y = 2x
A =
7.Complex numbersMedium
The number z = 2.5·ej2 has its angle in radians. Find its real part a.
r = 2.5θ = 2 rad
a =
8.Complex numbersHard
Use De Moivre's theorem to find the real part of z⁹ for z = 1 + j.
z = 1 + jn = 9
Re(z⁹) =
9.Differential equationsMedium
A can of soda at 5 °C is taken out of the fridge into a room at 29 °C. After 10 min its temperature is 16 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 22 °C?
T0 = 5 °CTa = 29 °CT(10) = 16 °CT = 22 °C
t = min
10.Differential equationsHard
Solve y′ + y = 4e−3t with y(0) = 0. Find y(2).
y′ + y = 4e−3ty(0) = 0
y(2) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).
A = [3 2; 4 −4]B = [−3 −4; 3 5]
c12 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [1 4; 2 3]
λ =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (−1, 0, 3) and b = (k, 6, 2) are perpendicular.
a = (−1, 0, 3)b = (k, 6, 2)
k =
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(−1, 3, 2), B(4, −1, −3) and C(3, −4, −1). Coordinates are in metres.
A(−1, 3, 2)B(4, −1, −3)C(3, −4, −1)
A = m²
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 10, 20, 30, 40p(x): 0.25, 0.3, 0.35, 0.1
σ =
16.ProbabilityHard
A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 2 correct answers?
n = 5p = 0.3k = 2
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 30, 50, 27, 20, 52, 27, 15.
n = 7
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 8. Find the z-score z of a score of 40.4.
μ = 50σ = 8x = 40.4
z =
19.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.1841. Find z to two decimals.
A = 0.1841
z =
20.Normal distributionHard
Hole diameters D are normal with mean 25.03 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.03 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
3Solutions
1.Limits
Answer: L = 1
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) = 6x/(√(9x2 + 6x) + 3x)
Divide by x
f(x) = 6/(√(9 + 6/x) + 3)
Let x → ∞
L = 6/(3 + 3) = 1
Answer
L = 1
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) = 1215
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·(3x − 3)4·3
Substitute
f′(2) = 5·34·3 = 1215
Answer
f′(2) = 1215
4.Derivatives
Answer: x = −2
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 − 12 = 3·(x2 − 4)
Solve f′(x) = 0
x = −2 or x = 2
Which one is which
f″(x) = 6x: negative at a maximum, positive at a minimum.
f″(−2) = −12, f″(2) = 12
x = −2
Answer
x = −2
5.Integrals
Answer: I = 1.551
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.
Choose u and dv
u = ln x, du = dx/x; dv = x dx, v = x²/2
Antiderivative
∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4
Limits
(3.125·ln 2.5 − 1.5625) − (0 − 0.25) = 1.551
Answer
I = 1.551
6.Integrals
Answer: A = 5.333
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
0.5x2 = 2x → x = 0, x = 4
Integrate upper minus lower
A = [x2 − 0.1667x3]04 = 5.333
Answer
A = 5.333
7.Complex numbers
Answer: a = −1.04
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.
Euler's formula
z = 2.5·(cos 2 + j sin 2)
Components (radians)
a = 2.5·cos 2 = −1.04
b = 2.5·sin 2 = 2.273
a = −1.04
Answer
a = −1.04
8.Complex numbers
Answer: Re(z⁹) = 16
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.
Polar form
z = 1 + j = √2∠45°
De Moivre
z⁹ = (√2)⁹∠405° = 22.63∠45°
Rectangular form
z⁹ = 22.63·(cos 45° + j sin 45°) = 16 + j16
Re(z⁹) = 16
Answer
Re(z⁹) = 16
9.Differential equations
Answer: t = 20.1 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(-24/-13)/10 = 0.06131 min⁻¹
Time to reach the target
t = ln(-24/(−7))/0.06131 = 20.1 min
Answer
t = 20.1 min
10.Differential equations
Answer: y(2) = 0.2657
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.
Integrating factor
μ = et
(ety)′ = 4e−2t
General solution
y = 4/(−2)·e−3t + C·e−t = −2·e−3t + C·e−t
Initial condition
0 = −2 + C → C = 2
Evaluate
y(2) = −2·0.002479 + 2·0.1353 = 0.2657
Answer
y(2) = 0.2657
11.Matrices & determinants
Answer: c12 = −2
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
c12 = 3·(−4) + 2·5 = −2
Answer
c12 = −2
12.Matrices & determinants
Answer: λ = 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.
Characteristic equation
λ² − (4)λ + (−5) = 0
Solve
λ = (4 ± √(36))/2 → λ = −1, 5
λ = 5
Answer
λ = 5
13.Vectors: dot & cross product
Answer: k = 6
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 = (−1)·k + 0·6 + 3·2 = 0
6 + (−1)·k = 0
Solve for k
k = −6/(−1) = 6
Answer
k = 6
14.Vectors: dot & cross product
Answer: A = 15.12 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.
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) = 10·0.32·0.73 = 10·0.09·0.343 = 0.3087
Answer
P = 0.3087
17.Descriptive statistics
Answer: s = 14.2
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 = (40.4 − 50)/8 = −1.2
Answer
z = −1.2
19.Normal distribution
Answer: z = 0.9
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.
Area to the left of z
Φ(z) = 1 − 0.1841 = 0.8159
Read the z-table backwards
Find 0.8159 inside the table and read off z.
z = 0.90
Answer
z = 0.90
20.Normal distribution
Answer: P = 0.9515
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).