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 + 0.5/x)3x. Give a decimal value.
f(x) = (1 + 0.5/x)3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 3x)/sin(4x), with x in radians.
f(x) = ln(1 + 3x)/sin(4x)x → 0
L =
3.DerivativesMedium
Let f(x) = (2x − 1)3. Find f′(1).
f(x) = (2x − 1)3x = 1
f′(1) =
4.DerivativesHard
The tangent line to f(x) = 2x2 − x − 6 at x = 3 crosses the y-axis at y = b. Find b.
f(x) = 2x2 − x − 6x = 3
b =
5.IntegralsMedium
Evaluate I = ∫₀1 x·sin x dx (x in radians).
∫₀1 x·sin x dx
I =
6.IntegralsHard
Evaluate I = ∫₀1 x²·eˣ dx.
∫₀1 x²eˣ dx
I =
7.Complex numbersMedium
Given z₁ = 2.5∠100° and z₂ = 3∠−50°, find the imaginary part of w = z₁·z₂.
z₁ = 2.5∠100°z₂ = 3∠−50°
Im(w) =
8.Complex numbersHard
The roots of zⁿ = w with n = 3 and w = 8∠90° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2. Find the argument of z₂ in degrees, between −180° and 180°.
w = 8∠90°n = 3k = 2
θ₂ = °
9.Differential equationsMedium
A can of soda at 3 °C is taken out of the fridge into a room at 26 °C. After 8 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 8 °C?
T0 = 3 °CTa = 26 °CT(8) = 16 °CT = 8 °C
t = min
10.Differential equationsHard
A tank holds 500 L of pure water. Brine with 0.2 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 30 kg of salt?
V = 500 Lm0 = 0 kgc = 0.2 kg/Lr = 20 L/min
t = min
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [4 4 −1; 5 −2 5; 4 2 2]
det A =
12.Matrices & determinantsHard
Solve the system 5x + 4y = 12 and −4x − y = 1. Find x.
5x + 4y = 12−4x − y = 1
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (0, 4, −1) and b = (3, 2, −5), in degrees.
a = (0, 4, −1)b = (3, 2, −5)
θ = °
14.Vectors: dot & cross productHard
For a = (−5, 5, −3) and b = (−3, −3, −6), find the y-component (a × b)y of the cross product a × b.
a = (−5, 5, −3)b = (−3, −3, −6)
(a × b)y =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2, 3, 4p(x): 0.2, 0.05, 0.3, 0.1, 0.35
σ =
16.ProbabilityHard
A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 5 correct answers?
n = 7p = 0.1k = 5
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 55, 45, 43, 21, 56, 17, 54.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 5. What proportion P of values is below 44?
μ = 50σ = 5x = 44
P =
19.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 62.5 and 67 points? Give P as a decimal.
μ = 65σ = 10x1 = 62.5x2 = 67
P =
20.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 80% is to the left of it. Find the cut-off x.
μ = 65 pointsσ = 10 points80%
x = points
3Solutions
1.Limits
Answer: L = 4.482
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→∞ 3·x·ln(1 + 0.5/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 3·0.5 = 1.5
Undo the logarithm
L = e1.5 = 4.482
Answer
L = 4.482
2.Limits
Answer: L = 0.75
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 [3/(1 + 3x)]/[4·cos 4x]
Substitute x = 0
L = 3/4 = 0.75
Answer
L = 0.75
3.Derivatives
Answer: f′(1) = 6
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·(2x − 1)2·2
Substitute
f′(1) = 3·12·2 = 6
Answer
f′(1) = 6
4.Derivatives
Answer: b = −24
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(3) = 9
f′(x) = 4x − 1 → f′(3) = 11
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 9 − 11·3 = −24
Answer
b = −24
5.Integrals
Answer: I = 0.3012
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
(−1·cos 1 + sin 1) − (0 + 0) = 0.3012
Answer
I = 0.3012
6.Integrals
Answer: I = 0.7183
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
2.718·(1) − 1·2 = 0.7183
Answer
I = 0.7183
7.Complex numbers
Answer: Im(w) = 5.745
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| = 2.5 · 3 = 7.5
θ = 100° + (−50)° = 50°
Rectangular form
Re(w) = 7.5·cos(50°) = 4.821
Im(w) = 7.5·sin(50°) = 5.745
Im(w) = 5.745
Answer
Im(w) = 5.745
8.Complex numbers
Answer: θ₂ = −90 °
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 of the roots
81/3 = 2
Argument of the root
θ₂ = (90° + 360°·2)/3 = 270° → −90°
Argument
θ₂ = −90°
Answer
θ₂ = −90°
The 3 roots lie on the circle of radius 2, 120° apart.
9.Differential equations
Answer: t = 2.354 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(-23/-10)/8 = 0.1041 min⁻¹
Time to reach the target
t = ln(-23/(−18))/0.1041 = 2.354 min
Answer
t = 2.354 min
10.Differential equations
Answer: t = 8.917 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) = 100 + (−100)·e−t/25
Solve for t
e−t/25 = (100 − 30)/(100 − 0) = 0.7
t = 25·ln(1.429) = 8.917 min
Answer
t = 8.917 min
11.Matrices & determinants
Answer: det A = −34
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 = −14, M₁₂ = −10, M₁₃ = 18
Cofactor expansion
det A = 4·(−14) − 4·(−10) + (−1)·18 = −34
Answer
det A = −34
12.Matrices & determinants
Answer: x = −1.455
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 = 5·(−1) − 4·(−4) = 11
Replace the x column
det Aₓ = 12·(−1) − 4·1 = −16
x = −16/11 = −1.455
Answer
x = −1.455
y = 4.818
13.Vectors: dot & cross product
Answer: θ = 59.24 °
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 = 0·3 + 4·2 + (−1)·(−5) = 13
Magnitudes
|a| = √(0² + 4² + (−1)²) = 4.123
|b| = √(3² + 2² + (−5)²) = 6.164
Angle
cos θ = 13/(4.123·6.164) = 0.5115
θ = arccos(0.5115) = 59.24°
Answer
θ = 59.24°
14.Vectors: dot & cross product
Answer: (a × b)y = −21
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.
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: σ = 1.492
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.
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 = −1.2
Standard normal table
Φ(−1.2) = 0.1151
P = 0.1151
Answer
P = 0.1151
19.Normal distribution
Answer: P = 0.178
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.
z-scores
z₁ = (62.5 − 65)/10 = −0.25
z₂ = (67 − 65)/10 = 0.2
Area between
P = Φ(0.2) − Φ(−0.25) = 0.5793 − 0.4013 = 0.178
Answer
P = 0.178
20.Normal distribution
Answer: x = 73.4 points
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.