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)2x. Give a decimal value.
f(x) = (1 + 2/x)2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e−x − 1 + x)/x².
f(x) = (e−x − 1 + x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = ln(2x2 + 2). Find f′(−1).
f(x) = ln(2x2 + 2)x = −1
f′(−1) =
4.DerivativesHard
Let f(x) = x3 − 3x2 − 1. Find the x-coordinate x of the local minimum.
f(x) = x3 − 3x2 − 1
x =
5.IntegralsMedium
Evaluate I = ∫03 ex dx.
f(x) = ex[0, 3]
I =
6.IntegralsHard
The region between the line y = 2x and the parabola y = 0.5x² is rotated about the x-axis. Find the volume V of the solid.
y = 2xy = 0.5x²
V =
7.Complex numbersMedium
Write z = −2 − j4 in exponential form r·ejθ. Give θ in radians, between −π and π.
z = −2 − j4
θ = rad
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 forged steel part at 320 °C is left to cool in a workshop at 29 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.03 min⁻¹, find its temperature T after 46 min.
T0 = 320 °CTa = 29 °Ck = 0.03 min⁻¹t = 46 min
T = °C
10.Differential equationsHard
A tank holds 500 L of pure water. Brine with 0.25 kg of salt per litre flows in at 20 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 30 min?
V = 500 Lm0 = 0 kgc = 0.25 kg/Lr = 20 L/min
m = kg
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c22 (row 2, column 2).
A = [6 5; 2 5]B = [5 6; −2 5]
c22 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−2 2; 7 3]
λ =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (−2, 2, 5) and b = (−5, k, −4) are perpendicular.
a = (−2, 2, 5)b = (−5, k, −4)
k =
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (−5, 3, 4), b = (1, −3, 1) and c = (−3, −5, 5).
a = (−5, 3, 4)b = (1, −3, 1)c = (−3, −5, 5)
V =
15.ProbabilityMedium
A bag holds 7 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
7 / 4
P =
16.ProbabilityHard
A test has 8 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 5 correct answers?
n = 8p = 0.4k = 5
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 21, 59, 29, 22, 24, 51, 59.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 12. What proportion P of values is below 64.8?
μ = 60σ = 12x = 64.8
P =
19.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.8962. Find z to two decimals.
A = 0.8962
z =
20.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 30% is to the left of it. Find the cut-off x.
μ = 250 gσ = 5 g30%
x = g
3Solutions
1.Limits
Answer: L = 54.6
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→∞ 2·x·ln(1 + 2/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 2·2 = 4
Undo the logarithm
L = e4 = 54.6
Answer
L = 54.6
2.Limits
Answer: L = 0.5
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
(−1·e−x + 1)/(2x) → 0/0
Second application
1·e−x/2
Substitute x = 0
L = 1·1/2 = 0.5
Answer
L = 0.5
3.Derivatives
Answer: f′(−1) = −1
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 + 2)
Substitute
f′(−1) = −4/4 = −1
Answer
f′(−1) = −1
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 − 6x = 3·(x2 − 2x)
Solve f′(x) = 0
x = 0 or x = 2
Which one is which
f″(x) = 6x − 6: negative at a maximum, positive at a minimum.
f″(0) = −6, f″(2) = 6
x = 2
Answer
x = 2
5.Integrals
Answer: I = 19.09
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 = (e3 − 1)/1 = (20.09 − 1)/1 = 19.09
Answer
I = 19.09
6.Integrals
Answer: V = 107.2
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
2x = 0.5x² → x = 0, x = 4
Washer method
V = π∫₀4 ((2x)² − (0.5x²)²) dx
Integrate
V = π(4·4³/3 − 0.25·4⁵/5) = π·34.13 = 107.2
Answer
V = 107.2
7.Complex numbers
Answer: θ = −2.034 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 = √(4 + 16) = 4.472
Argument in radians
θ = −116.6° = −116.6·π/180 = −2.034 rad
Exponential form
z = 4.472·e−j2.034
Answer
θ = −2.034 rad
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 = 102.2 °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) = 29 + 291·e−0.03t
Substitute t
T(46) = 29 + 291·0.2516 = 102.2 °C
Answer
T = 102.2 °C
10.Differential equations
Answer: m = 87.35 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 = 20·0.25 − (20/500)·m = 5 − 0.04·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 125 + (−125)·e−t/25
Substitute t
m(30) = 125 + (−125)·0.3012 = 87.35 kg
Answer
m = 87.35 kg
11.Matrices & determinants
Answer: c22 = 37
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 = 2·6 + 5·5 = 37
Answer
c22 = 37
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 − (1)λ + (−20) = 0
Solve
λ = (1 ± √(81))/2 → λ = −4, 5
λ = 5
Answer
λ = 5
13.Vectors: dot & cross product
Answer: k = 5
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 = (−2)·(−5) + 2·k + 5·(−4) = 0
−10 + 2·k = 0
Solve for k
k = −(−10)/2 = 5
Answer
k = 5
14.Vectors: dot & cross product
Answer: V = 30
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 b × c
(b × c)x = bycz − bzcy = (−3)·5 − 1·(−5) = −10
(b × c)y = bzcx − bxcz = 1·(−3) − 1·5 = −8
(b × c)z = bxcy − bycx = 1·(−5) − (−3)·(−3) = −14
Dot product with a
a·(b × c) = (−5)·(−10) + 3·(−8) + 4·(−14) = −30
V = |−30| = 30
Answer
V = 30
15.Probability
Answer: P = 0.3818
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 = 7/11 · 6/10 = 42/110 = 0.3818
Answer
P = 0.3818
16.Probability
Answer: P = 0.1239
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.
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 = 0.4
Standard normal table
Φ(0.4) = 0.6554
P = 0.6554
Answer
P = 0.6554
19.Normal distribution
Answer: z = −1.26
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 − 0.8962 = 0.1038
Read the z-table backwards
Find 0.1038 inside the table and read off z.
z = −1.26
Answer
z = −1.26
20.Normal distribution
Answer: x = 247.4 g
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.