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→∞ (4x3 + 3)/(5x3 + 4x + 5).
f(x) = (4x3 + 3)/(5x3 + 4x + 5)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 4x)/(2x²), with x in radians.
f(x) = (1 − cos 4x)/(2x²)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(3x2 + 4). Find f′(2).
f(x) = ln(3x2 + 4)x = 2
f′(2) =
4.DerivativesHard
The tangent line to f(x) = −x2 + x − 1 at x = 3 crosses the y-axis at y = b. Find b.
f(x) = −x2 + x − 1x = 3
b =
5.IntegralsMedium
Evaluate I = ∫02 e2x dx.
f(x) = e2x[0, 2]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = 4x.
y = x2y = 4x
A =
7.Complex numbersMedium
The number z = 1.5·e−j1.8 has its angle in radians. Find its imaginary part b.
r = 1.5θ = −1.8 rad
b =
8.Complex numbersHard
Exactly one of the points A–D is a root of z³ = 27∠0°. Click it.
w = 27∠0°n = 3
1A (3∠−90°)
2B (3∠−60°)
3C (3∠−120°)
4D (1.5∠−120°)
9.Differential equationsMedium
A cup of coffee at 84 °C is left in a room at 23 °C. After 5 min its temperature is 71 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 84 °CTa = 23 °CT(5) = 71 °C
k = min⁻¹
10.Differential equationsHard
Solve y′ + 3y = 6e−0.5t with y(0) = 0. Find y(2).
y′ + 3y = 6e−0.5ty(0) = 0
y(2) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−4 −4 −5; 2 1 −2; −4 −3 0]
det A =
12.Matrices & determinantsHard
Solve the system 3x + y = 4 and 4x + 2y = 7. Find x.
3x + y = 44x + 2y = 7
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (2, −3, −4) and b = (4, 6, 0), in degrees.
a = (2, −3, −4)b = (4, 6, 0)
θ = °
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−1, −3, −1) and b = (−5, 4, 1) and has a positive z-component. Find its y-component n̂y.
a = (−1, −3, −1)b = (−5, 4, 1)n̂z > 0
n̂y =
15.ProbabilityMedium
A bag holds 5 red and 9 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
5 / 9
P =
16.ProbabilityHard
A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 2 correct answers?
n = 11p = 0.4k = 2
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 24, 34, 24, 51, 23, 19, 46.
n = 7
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 60 and standard deviation 8. Find the z-score z of a score of 44.
μ = 60σ = 8x = 44
z =
19.Normal distributionMedium
The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 12 of them is below 149.5 kg. Round z to two decimals.
μ = 12 kgσ = 0.8 kgn = 12< 149.5 kg
P =
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 top 20% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g20%
x = g
3Solutions
1.Limits
Answer: L = 0.8
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).
Divide by the highest power
(4 + 3/x3)/(5 + 4/x2 + 5/x3)
Let x → ∞
L = 4/5 = 0.8
Answer
L = 0.8
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 4
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
4·sin 4x/(4x) → 0/0
Second application
16·cos 4x/4
Substitute x = 0
L = 16/4 = 4
Answer
L = 4
3.Derivatives
Answer: f′(2) = 0.75
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) = 6x/(3x2 + 4)
Substitute
f′(2) = 12/16 = 0.75
Answer
f′(2) = 0.75
4.Derivatives
Answer: b = 8
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) = −7
f′(x) = −2x + 1 → f′(3) = −5
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −7 − (−5)·3 = 8
Answer
b = 8
5.Integrals
Answer: I = 26.8
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) = e2x/2
Evaluate at the limits
I = (e4 − 1)/2 = (54.6 − 1)/2 = 26.8
Answer
I = 26.8
6.Integrals
Answer: A = 10.67
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.
Intersection points
x2 = 4x → x = 0, x = 4
Integrate upper minus lower
A = [2x2 − 0.3333x3]04 = 10.67
Answer
A = 10.67
7.Complex numbers
Answer: b = −1.461
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.
Euler's formula
z = 1.5·(cos −1.8 + j sin −1.8)
Components (radians)
a = 1.5·cos −1.8 = −0.3408
b = 1.5·sin −1.8 = −1.461
b = −1.461
Answer
b = −1.461
8.Complex numbers
Answer: C (3∠−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| = 3
θ = 0°, 120°, −120°
The point
C = 3∠−120°
Answer
C (3∠−120°)
9.Differential equations
Answer: k = 0.04793 min⁻¹
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.
Ratio of temperature differences
e−5k = 48/61
Solve for k
k = ln(61/48)/5 = 0.04793 min⁻¹
Answer
k = 0.04793 min⁻¹
10.Differential equations
Answer: y(2) = 0.877
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 = e3t
(e3ty)′ = 6e2.5t
General solution
y = 6/2.5·e−0.5t + C·e−3t = 2.4·e−0.5t + C·e−3t
Initial condition
0 = 2.4 + C → C = −2.4
Evaluate
y(2) = 2.4·0.3679 + (−2.4)·0.002479 = 0.877
Answer
y(2) = 0.877
11.Matrices & determinants
Answer: det A = 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.
2×2 minors of the first row
M₁₁ = −6, M₁₂ = −8, M₁₃ = −2
Cofactor expansion
det A = (−4)·(−6) − (−4)·(−8) + (−5)·(−2) = 2
Answer
det A = 2
12.Matrices & determinants
Answer: x = 0.5
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.
Determinant of the coefficient matrix
det A = 3·2 − 1·4 = 2
Replace the x column
det Aₓ = 4·2 − 1·7 = 1
x = 1/2 = 0.5
Answer
x = 0.5
y = 2.5
13.Vectors: dot & cross product
Answer: θ = 104.9 °
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 = 2·4 + (−3)·6 + (−4)·0 = −10
Magnitudes
|a| = √(2² + (−3)² + (−4)²) = 5.385
|b| = √(4² + 6² + 0²) = 7.211
Angle
cos θ = −10/(5.385·7.211) = −0.2575
θ = arccos(−0.2575) = 104.9°
Answer
θ = 104.9°
14.Vectors: dot & cross product
Answer: n̂y = −0.3008
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 = (−3)·1 − (−1)·4 = 1
(a × b)y = azbx − axbz = (−1)·(−5) − (−1)·1 = 6
(a × b)z = axby − aybx = (−1)·4 − (−3)·(−5) = −19
Normalise
The z-component of a × b is negative, so use −(a × b).
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 = 5/14 · 4/13 = 20/182 = 0.1099
Answer
P = 0.1099
16.Probability
Answer: P = 0.08868
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 = (44 − 60)/8 = −2
Answer
z = −2
19.Normal distribution
Answer: P = 0.9761
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 = 144 kg, σT = 0.8·√12 = 2.771 kg
Standardise
z = (149.5 − 144)/2.771 = 1.98
Probability
P = Φ(1.98) = 0.9761
Answer
P = 0.9761
20.Normal distribution
Answer: x = 254.2 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.