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→∞ (−5x3 + 2x − 9)/(3x3 + 5x + 4).
f(x) = (−5x3 + 2x − 9)/(3x3 + 5x + 4)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 2x)/sin(5x), with x in radians.
f(x) = ln(1 + 2x)/sin(5x)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(2x), with x in radians. Find f′(1).
f(x) = x·sin(2x)x = 1 rad
f′(1) =
4.DerivativesHard
The tangent line to f(x) = 2x2 + 3x − 2 at x = −2 crosses the y-axis at y = b. Find b.
f(x) = 2x2 + 3x − 2x = −2
b =
5.IntegralsMedium
Evaluate I = ∫01 e−0.5x dx.
f(x) = e−0.5x[0, 1]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = 4x.
y = x2y = 4x
A =
7.Complex numbersMedium
Find the argument θ of z = −9 − j5 in degrees, as the principal value between −180° and 180°.
z = −9 − j5
θ = °
8.Complex numbersHard
The equation z² − 8z + 20 = 0 has two complex roots. Find their modulus |z|.
p = −8q = 20
|z| =
9.Differential equationsMedium
Solve y″ + 7y′ + 12y = 0 with y(0) = 5 and y′(0) = 4. Find y(1).
y″ + 7y′ + 12y = 0y(0) = 5y′(0) = 4
y(1) =
10.Differential equationsHard
A tank holds 500 L of water in which 10 kg of salt is dissolved. Brine with 0.05 kg of salt per litre flows in at 2 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 19 kg of salt?
V = 500 Lm0 = 10 kgc = 0.05 kg/Lr = 2 L/min
t = min
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [−4 −5; 2 2]B = [−3 −4; −3 4]
c11 =
12.Matrices & determinantsHard
Solve the system 3x − 4y = −8 and 3x − 5y = 12. Find x.
3x − 4y = −83x − 5y = 12
x =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (4, 1, 7) onto b = (−2, −4, 0), that is, the component of a in the direction of b.
a = (4, 1, 7)b = (−2, −4, 0)
compb a =
14.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (5, −5, 5) and b = (−4, −1, −1).
a = (5, −5, 5)b = (−4, −1, −1)
A =
15.ProbabilityMedium
A bag holds 6 red and 9 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
6 / 9
P =
16.ProbabilityHard
A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 1 correct answers?
n = 12p = 0.25k = 1
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 46, 10, 44, 24, 57, 41, 37.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is above 62?
μ = 50σ = 8x = 62
P =
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 6 of them is below 74.9 kg. Round z to two decimals.
μ = 12 kgσ = 0.8 kgn = 6< 74.9 kg
P =
20.Normal distributionHard
Hole diameters D are normal with mean 25.05 mm and standard deviation 0.015 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 = 25.05 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
3Solutions
1.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).
Divide by the highest power
(−5 + 2/x2 + (−9)/x3)/(3 + 5/x2 + 4/x3)
Let x → ∞
L = −5/3 = −1.667
Answer
L = −1.667
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 0.4
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).
L'Hôpital's rule
L = limx→0 [2/(1 + 2x)]/[5·cos 5x]
Substitute x = 0
L = 2/5 = 0.4
Answer
L = 0.4
3.Derivatives
Answer: f′(1) = 0.077
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.
Product and chain rule
f′(x) = sin(2x) + 2x·cos(2x)
Substitute
f′(1) = 0.9093 + 2·(−0.4161) = 0.077
Answer
f′(1) = 0.077
4.Derivatives
Answer: b = −10
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(−2) = 0
f′(x) = 4x + 3 → f′(−2) = −5
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 0 − (−5)·(−2) = −10
Answer
b = −10
5.Integrals
Answer: I = 0.7869
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.
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: θ = −150.9 °
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 = −5 → III
Reference angle
α = atan(5/9) = 29.05°
Argument
θ = −(180° − α) = −150.9°
Answer
θ = −150.9°
In polar form z = 10.3∠−150.9°.
8.Complex numbers
Answer: |z| = 4.472
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.
Discriminant
p² − 4q = 64 − 80 = −16 < 0
Quadratic formula
z = (8 ± j√16)/2 = 4 + j2, 4 − j2
Answer
|z| = √(4² + 2²) = √20 = 4.472
Answer
|z| = 4.472
9.Differential equations
Answer: y(1) = 0.8469
Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.
Characteristic roots
(r + 3)(r + 4) = 0 → r = −3, −4
Constants from the initial values
A + B = 5, −3A − 4B = 4
A = 24, B = −19
Evaluate
y(1) = 24·e−3 + (−19)·e−4 = 0.8469
Answer
y(1) = 0.8469
10.Differential equations
Answer: t = 229.1 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) = 25 + (−15)·e−t/250
Solve for t
e−t/250 = (25 − 19)/(25 − 10) = 0.4
t = 250·ln(2.5) = 229.1 min
Answer
t = 229.1 min
11.Matrices & determinants
Answer: c11 = 27
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 = (−4)·(−3) + (−5)·(−3) = 27
Answer
c11 = 27
12.Matrices & determinants
Answer: x = −29.33
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·(−5) − (−4)·3 = −3
Replace the x column
det Aₓ = (−8)·(−5) − (−4)·12 = 88
x = 88/(−3) = −29.33
Answer
x = −29.33
y = −20
13.Vectors: dot & cross product
Answer: compb a = −2.683
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 = 4·(−2) + 1·(−4) + 7·0 = −12
Length of b
|b| = √((−2)² + (−4)² + 0²) = 4.472
Scalar projection
compb a = a·b/|b| = −12/4.472 = −2.683
Answer
compb a = −2.683
Negative: a has a component pointing against b.
14.Vectors: dot & cross product
Answer: A = 30.82
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 = (−5)·(−1) − 5·(−1) = 10
(a × b)y = azbx − axbz = 5·(−4) − 5·(−1) = −15
(a × b)z = axby − aybx = 5·(−1) − (−5)·(−4) = −25
Area
A = |a × b| = √(10² + (−15)² + (−25)²) = 30.82
Answer
A = 30.82
15.Probability
Answer: P = 0.1429
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 = 6/15 · 5/14 = 30/210 = 0.1429
Answer
P = 0.1429
16.Probability
Answer: P = 0.1267
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 = 1.5
Standard normal table
Φ(1.5) = 0.9332
P = 1 − 0.9332 = 0.06681
Answer
P = 0.06681
19.Normal distribution
Answer: P = 0.9306
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 = 72 kg, σT = 0.8·√6 = 1.96 kg
Standardise
z = (74.9 − 72)/1.96 = 1.48
Probability
P = Φ(1.48) = 0.9306
Answer
P = 0.9306
20.Normal distribution
Answer: P = 0.9909
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).