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 − 1/x)2x. Give a decimal value.
f(x) = (1 − 1/x)2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 5x)/sin(3x), with x in radians.
f(x) = ln(1 + 5x)/sin(3x)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(x2 + 1). Find f′(1).
f(x) = ln(x2 + 1)x = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 − 9x2 + 15x + 6. Find the x-coordinate x of the local minimum.
f(x) = x3 − 9x2 + 15x + 6
x =
5.IntegralsMedium
Evaluate I = ∫₁4 x·ln x dx.
∫₁4 x ln x dx
I =
6.IntegralsHard
Evaluate I = ∫₀2 x²·eˣ dx.
∫₀2 x²eˣ dx
I =
7.Complex numbersMedium
Write z = 12∠160° in rectangular form a + jb and give the imaginary part b.
r = 12θ = 160°
b =
8.Complex numbersHard
One root of z³ − 13z² + 60z − 100 = 0 is z₁ = 4 + j2. The coefficients are real. Find the real root r.
z₁ = 4 + j2
r =
9.Differential equationsMedium
A cup of coffee at 93 °C is left in a room at 24 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.12 min⁻¹, find its temperature T after 17 min.
T0 = 93 °CTa = 24 °Ck = 0.12 min⁻¹t = 17 min
T = °C
10.Differential equationsHard
A tank holds 200 L of pure water. Brine with 0.05 kg of salt per litre flows in at 20 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 5 kg of salt?
V = 200 Lm0 = 0 kgc = 0.05 kg/Lr = 20 L/min
t = min
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [6 −1; −6 5]B = [3 −5; −4 0]
c11 =
12.Matrices & determinantsHard
Solve the system 6x − 4y = −7 and 6x + 5y = −6. Find x.
6x − 4y = −76x + 5y = −6
x =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (3, 2, 3) and b = (4, −5, −7).
a = (3, 2, 3)b = (4, −5, −7)
a·b =
14.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (0, −4, 6) and b = (0, −2, −2).
a = (0, −4, 6)b = (0, −2, −2)
A =
15.ProbabilityMedium
A bag holds 8 red and 9 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
8 / 9
P =
16.ProbabilityHard
X has the distribution shown, and Y = 0.5X + 20. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 11, 48, 34, 24, 17, 50, 46.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 10. What proportion P of values is above 40?
μ = 50σ = 10x = 40
P =
19.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0132. Find z to two decimals.
A = 0.0132
z =
20.Normal distributionHard
Hole diameters D are normal with mean 25.03 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 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 = 0.02 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
3Solutions
1.Limits
Answer: L = 0.1353
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 − 1/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 2·(−1) = −2
Undo the logarithm
L = e−2 = 0.1353
Answer
L = 0.1353
2.Limits
Answer: L = 1.667
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).
L'Hôpital's rule
L = limx→0 [5/(1 + 5x)]/[3·cos 3x]
Substitute x = 0
L = 5/3 = 1.667
Answer
L = 1.667
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) = 2x/(x2 + 1)
Substitute
f′(1) = 2/2 = 1
Answer
f′(1) = 1
4.Derivatives
Answer: x = 5
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 − 18x + 15 = 3·(x2 − 6x + 5)
Solve f′(x) = 0
x = 1 or x = 5
Which one is which
f″(x) = 6x − 18: negative at a maximum, positive at a minimum.
f″(1) = −12, f″(5) = 12
x = 5
Answer
x = 5
5.Integrals
Answer: I = 7.34
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
(8·ln 4 − 4) − (0 − 0.25) = 7.34
Answer
I = 7.34
6.Integrals
Answer: I = 12.78
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
7.389·(2) − 1·2 = 12.78
Answer
I = 12.78
7.Complex numbers
Answer: b = 4.104
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 = 12·cos(160°) = −11.28
b = 12·sin(160°) = 4.104
Rectangular form
z = −11.28 + j4.104
b = 4.104
Answer
b = 4.104
8.Complex numbers
Answer: r = 5
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.
Check with the sum of the roots: z₁ + z₂ + r = 8 + 5 = 13 = −(−13).
9.Differential equations
Answer: T = 32.97 °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) = 24 + 69·e−0.12t
Substitute t
T(17) = 24 + 69·0.13 = 32.97 °C
Answer
T = 32.97 °C
10.Differential equations
Answer: t = 6.931 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.
Result
m(t) = 10 + (−10)·e−t/10
Solve for t
e−t/10 = (10 − 5)/(10 − 0) = 0.5
t = 10·ln(2) = 6.931 min
Answer
t = 6.931 min
11.Matrices & determinants
Answer: c11 = 22
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
c11 = 6·3 + (−1)·(−4) = 22
Answer
c11 = 22
12.Matrices & determinants
Answer: x = −1.093
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 = 6·5 − (−4)·6 = 54
Replace the x column
det Aₓ = (−7)·5 − (−4)·(−6) = −59
x = −59/54 = −1.093
Answer
x = −1.093
y = 0.1111
13.Vectors: dot & cross product
Answer: a·b = −19
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.
Negative: the angle between the vectors is more than 90°.
14.Vectors: dot & cross product
Answer: A = 20
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
(a × b)x = aybz − azby = (−4)·(−2) − 6·(−2) = 20
(a × b)y = azbx − axbz = 6·0 − 0·(−2) = 0
(a × b)z = axby − aybx = 0·(−2) − (−4)·0 = 0
Area
A = |a × b| = √(20² + 0² + 0²) = 20
Answer
A = 20
15.Probability
Answer: P = 0.2059
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 = 8/17 · 7/16 = 56/272 = 0.2059
Answer
P = 0.2059
16.Probability
Answer: σY = 5.59
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 = 25, Var(X) = 125, σX = 11.18
Linear change
σY = |0.5|·11.18 = 5.59
Answer
σY = 5.59
The mean does move: E(Y) = 0.5·25 + 20 = 32.5.
17.Descriptive statistics
Answer: s = 15.84
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 = −1
Standard normal table
Φ(−1) = 0.1587
P = 1 − 0.1587 = 0.8413
Answer
P = 0.8413
19.Normal distribution
Answer: z = 2.22
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 − 0.0132 = 0.9868
Read the z-table backwards
Find 0.9868 inside the table and read off z.
z = 2.22
Answer
z = 2.22
20.Normal distribution
Answer: P = 0.8849
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).