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 − 5x) − 3x).
f(x) = √(9x2 − 5x) − 3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 2x)/(x²), with x in radians.
f(x) = (1 − cos 2x)/(x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(3x), with x in radians. Find f′(1.5).
f(x) = x·sin(3x)x = 1.5 rad
f′(1.5) =
4.DerivativesHard
The tangent line to f(x) = −x2 − 5x + 6 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = −x2 − 5x + 6x = 1
b =
5.IntegralsMedium
Evaluate I = ∫₀π x·sin x dx (x in radians).
∫₀π x·sin x dx
I =
6.IntegralsHard
The region between the line y = 3x and the parabola y = x² is rotated about the x-axis. Find the volume V of the solid.
y = 3xy = x²
V =
7.Complex numbersMedium
Given z₁ = 5∠55° and z₂ = 1.5∠20°, find the argument θ of w = z₁·z₂ in degrees, between −180° and 180°.
z₁ = 5∠55°z₂ = 1.5∠20°
θ = °
8.Complex numbersHard
Exactly one of the points A–D is a root of z³ = 8∠0°. Click it.
w = 8∠0°n = 3
1A (3.6∠−120°)
2B (2∠−120°)
3C (2∠−90°)
4D (2∠−60°)
9.Differential equationsMedium
A forged steel part at 340 °C is left to cool in a workshop at 20 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.01 min⁻¹, find its temperature T after 58 min.
T0 = 340 °CTa = 20 °Ck = 0.01 min⁻¹t = 58 min
T = °C
10.Differential equationsHard
Solve the initial value problem y′ + y/x = 5x2 for x > 0, with y(1) = 1. Find y(2).
y′ + y/x = 5x2y(1) = 1
y(2) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−2 5 0; −4 −5 −1; 3 −5 0]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−3 2; 1 −2]
λ =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (5, −7, 2) onto b = (−2, 5, −4), that is, the component of a in the direction of b.
a = (5, −7, 2)b = (−2, 5, −4)
compb a =
14.Vectors: dot & cross productHard
For a = (5, −3, 2) and b = (−1, −3, 5), find the x-component (a × b)x of the cross product a × b.
a = (5, −3, 2)b = (−1, −3, 5)
(a × b)x =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2, 3, 4p(x): 0.1, 0.15, 0.3, 0.25, 0.2
σ =
16.ProbabilityHard
10 components are tested, and each one fails with probability 0.4, independently. What is the probability P that at most one fails?
n = 10p = 0.4
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 28, 12, 43, 40, 35.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is below 66?
μ = 70σ = 8x = 66
P =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.8690. Find z to two decimals.
A = 0.8690
z =
20.Normal distributionHard
Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
3Solutions
1.Limits
Answer: L = −0.8333
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) = −5x/(√(9x2 − 5x) + 3x)
Divide by x
f(x) = −5/(√(9 + (−5)/x) + 3)
Let x → ∞
L = −5/(3 + 3) = −0.8333
Answer
L = −0.8333
2.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).
First application
2·sin 2x/(2x) → 0/0
Second application
4·cos 2x/2
Substitute x = 0
L = 4/2 = 2
Answer
L = 2
3.Derivatives
Answer: f′(1.5) = −1.926
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.
Product and chain rule
f′(x) = sin(3x) + 3x·cos(3x)
Substitute
f′(1.5) = −0.9775 + 4.5·(−0.2108) = −1.926
Answer
f′(1.5) = −1.926
4.Derivatives
Answer: b = 7
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) = 0
f′(x) = −2x − 5 → f′(1) = −7
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 0 − (−7)·1 = 7
Answer
b = 7
5.Integrals
Answer: I = 3.142
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 = x, du = dx; dv = sin x dx, v = −cos x
Antiderivative
∫x sin x dx = −x cos x + ∫cos x dx = −x cos x + sin x
Limits
(−π·cos π + sin π) − (0 + 0) = 3.142
Answer
I = 3.142
6.Integrals
Answer: V = 101.8
Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.
Intersection points
3x = 1x² → x = 0, x = 3
Washer method
V = π∫₀3 ((3x)² − (1x²)²) dx
Integrate
V = π(9·3³/3 − 1·3⁵/5) = π·32.4 = 101.8
Answer
V = 101.8
7.Complex numbers
Answer: θ = 75 °
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 and argument of w
|w| = 5 · 1.5 = 7.5
θ = 55° + 20° = 75°
Argument
θ = 75°
Answer
θ = 75°
8.Complex numbers
Answer: B (2∠−120°)
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 = 0°, 120°, −120°
The point
B = 2∠−120°
Answer
B (2∠−120°)
9.Differential equations
Answer: T = 199.2 °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) = 20 + 320·e−0.01t
Substitute t
T(58) = 20 + 320·0.5599 = 199.2 °C
Answer
T = 199.2 °C
10.Differential equations
Answer: y(2) = 9.875
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 = e∫1/x dx = x1
Integrate
(x1y)′ = 5x3 → x1y = 5x4/4 + C
Initial condition
1 = 5/4 + C → C = −0.25
Evaluate
y(2) = (5·24/4 + (−0.25))/21 = 9.875
Answer
y(2) = 9.875
11.Matrices & determinants
Answer: det A = −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.
2×2 minors of the first row
M₁₁ = −5, M₁₂ = 3, M₁₃ = 35
Cofactor expansion
det A = (−2)·(−5) − 5·3 + 0·35 = −5
Answer
det A = −5
12.Matrices & determinants
Answer: λ = −1
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 − (−5)λ + (4) = 0
Solve
λ = (−5 ± √(9))/2 → λ = −4, −1
λ = −1
Answer
λ = −1
13.Vectors: dot & cross product
Answer: compb a = −7.901
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 = 5·(−2) + (−7)·5 + 2·(−4) = −53
Length of b
|b| = √((−2)² + 5² + (−4)²) = 6.708
Scalar projection
compb a = a·b/|b| = −53/6.708 = −7.901
Answer
compb a = −7.901
Negative: a has a component pointing against b.
14.Vectors: dot & cross product
Answer: (a × b)x = −9
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.
Cross product, component by component
(a × b)x = aybz − azby = (−3)·5 − 2·(−3) = −9
(a × b)y = azbx − axbz = 2·(−1) − 5·5 = −27
(a × b)z = axby − aybx = 5·(−3) − (−3)·(−1) = −18
Requested component
(a × b)x = −9
Answer
(a × b)x = −9
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: σ = 1.229
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.
Zero and one failure
P(X = 0) = 0.610 = 0.006047
P(X = 1) = 10·0.4·0.69 = 0.04031
Add them
P = 0.006047 + 0.04031 = 0.04636
Answer
P = 0.04636
17.Descriptive statistics
Answer: s = 12.34
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 = −0.5
Standard normal table
Φ(−0.5) = 0.3085
P = 0.3085
Answer
P = 0.3085
19.Normal distribution
Answer: z = 1.51
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 + 0.8690)/2 = 0.9345
Read the z-table backwards
Find 0.9345 inside the table and read off z.
z = 1.51
Answer
z = 1.51
20.Normal distribution
Answer: P = 0.7881
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).