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→∞ (√(x2 + 2x) − x).
f(x) = √(x2 + 2x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e2x − 1 − 2x)/x².
f(x) = (e2x − 1 − 2x)/x²x → 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 − 3x − 2 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = x2 − 3x − 2x = 1
b =
5.IntegralsMedium
Evaluate I = ∫01 ex dx.
f(x) = ex[0, 1]
I =
6.IntegralsHard
The region between the line y = x and the parabola y = 0.5x² is rotated about the x-axis. Find the volume V of the solid.
y = xy = 0.5x²
V =
7.Complex numbersMedium
Find the argument θ of z = −9 − j6 in degrees, as the principal value between −180° and 180°.
z = −9 − j6
θ = °
8.Complex numbersHard
Use De Moivre's theorem to find the real part of z⁶ for z = √3 + j.
z = √3 + jn = 6
Re(z⁶) =
9.Differential equationsMedium
A can of soda at 6 °C is taken out of the fridge into a room at 22 °C. After 8 min its temperature is 9 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 8 °C?
T0 = 6 °CTa = 22 °CT(8) = 9 °CT = 8 °C
t = min
10.Differential equationsHard
A tank holds 400 L of pure water. Brine with 0.25 kg of salt per litre flows in at 10 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 25 min?
V = 400 Lm0 = 0 kgc = 0.25 kg/Lr = 10 L/min
m = kg
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [0 −2; 6 −2]B = [−6 0; −5 −1]
c21 =
12.Matrices & determinantsHard
Solve the system −x − 5y = 8 and −2x + 5y = −2. Find x.
−x − 5y = 8−2x + 5y = −2
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (−2, −6, −2) and b = (5, −4, 6), in degrees.
a = (−2, −6, −2)b = (5, −4, 6)
θ = °
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(3, 4, −1), B(3, −2, −3) and C(−2, −3, −4). Coordinates are in metres.
A(3, 4, −1)B(3, −2, −3)C(−2, −3, −4)
A = m²
15.ProbabilityMedium
A fair die is rolled 4 times. What is the probability P of getting at least one six?
n = 4
P =
16.ProbabilityHard
X has the distribution shown, and Y = 2X + 5. Find the standard deviation σY of Y.
x: 1, 2, 3, 4p(x): 0.3, 0.2, 0.15, 0.35Y = 2X + 5
σY =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 31, 23, 38, 47, 35.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 10. Find the z-score z of a score of 112.5.
μ = 100σ = 10x = 112.5
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 10 of them is below 251.6 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 10< 251.6 kg
P =
20.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 80% of them lie (the 80th percentile).
μ = 100σ = 15P = 80%
x = points
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) = 2x/(√(x2 + 2x) + x)
Divide by x
f(x) = 2/(√(1 + 2/x) + 1)
Let x → ∞
L = 2/(1 + 1) = 1
Answer
L = 1
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·e2x − 2)/(2x) → 0/0
Second application
4·e2x/2
Substitute x = 0
L = 4·1/2 = 2
Answer
L = 2
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) = 3·32·4 = 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) = −4
f′(x) = 2x − 3 → f′(1) = −1
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −4 − (−1)·1 = −3
Answer
b = −3
5.Integrals
Answer: I = 1.718
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) = ex/1
Evaluate at the limits
I = (e1 − 1)/1 = (2.718 − 1)/1 = 1.718
Answer
I = 1.718
6.Integrals
Answer: V = 3.351
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
1x = 0.5x² → x = 0, x = 2
Washer method
V = π∫₀2 ((1x)² − (0.5x²)²) dx
Integrate
V = π(1·2³/3 − 0.25·2⁵/5) = π·1.067 = 3.351
Answer
V = 3.351
7.Complex numbers
Answer: θ = −146.3 °
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 = −9, b = −6 → III
Reference angle
α = atan(6/9) = 33.69°
Argument
θ = −(180° − α) = −146.3°
Answer
θ = −146.3°
In polar form z = 10.82∠−146.3°.
8.Complex numbers
Answer: Re(z⁶) = −64
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 = √3 + j = 2∠30°
De Moivre
z⁶ = (2)⁶∠180° = 64∠180°
Rectangular form
z⁶ = 64·(cos 180° + j sin 180°) = −64
Re(z⁶) = −64
Answer
Re(z⁶) = −64
9.Differential equations
Answer: t = 5.145 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(-16/-13)/8 = 0.02595 min⁻¹
Time to reach the target
t = ln(-16/(−14))/0.02595 = 5.145 min
Answer
t = 5.145 min
10.Differential equations
Answer: m = 46.47 kg
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.
Set up the equation
dm/dt = 10·0.25 − (10/400)·m = 2.5 − 0.025·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 100 + (−100)·e−t/40
Substitute t
m(25) = 100 + (−100)·0.5353 = 46.47 kg
Answer
m = 46.47 kg
11.Matrices & determinants
Answer: c21 = −26
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
c21 = 6·(−6) + (−2)·(−5) = −26
Answer
c21 = −26
12.Matrices & determinants
Answer: x = −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.
Determinant of the coefficient matrix
det A = (−1)·5 − (−5)·(−2) = −15
Replace the x column
det Aₓ = 8·5 − (−5)·(−2) = 30
x = 30/(−15) = −2
Answer
x = −2
y = −1.2
13.Vectors: dot & cross product
Answer: θ = 88.03 °
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 = (−2)·5 + (−6)·(−4) + (−2)·6 = 2
Magnitudes
|a| = √((−2)² + (−6)² + (−2)²) = 6.633
|b| = √(5² + (−4)² + 6²) = 8.775
Angle
cos θ = 2/(6.633·8.775) = 0.03436
θ = arccos(0.03436) = 88.03°
Answer
θ = 88.03°
14.Vectors: dot & cross product
Answer: A = 15.94 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.
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.
Complement
P(no six) = (5/6)4 = 0.4823
P = 1 − 0.4823 = 0.5177
Answer
P = 0.5177
16.Probability
Answer: σY = 2.488
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 = 2.55, Var(X) = 1.548, σX = 1.244
Linear change
σY = |2|·1.244 = 2.488
Answer
σY = 2.488
The mean does move: E(Y) = 2·2.55 + 5 = 10.1.
17.Descriptive statistics
Answer: s = 8.843
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 = (112.5 − 100)/10 = 1.25
Answer
z = 1.25
19.Normal distribution
Answer: P = 0.8438
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 = 250 kg, σT = 0.5·√10 = 1.581 kg
Standardise
z = (251.6 − 250)/1.581 = 1.01
Probability
P = Φ(1.01) = 0.8438
Answer
P = 0.8438
20.Normal distribution
Answer: x = 112.6 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.