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 (1 − cos 4x)/(2x²), with x in radians.
f(x) = (1 − cos 4x)/(2x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·e2x. Find f′(1.5).
f(x) = x·e2xx = 1.5
f′(1.5) =
4.DerivativesHard
Let f(x) = x3 − 10.5x2 + 36x + 6. Find the x-coordinate x of the local minimum.
f(x) = x3 − 10.5x2 + 36x + 6
x =
5.IntegralsMedium
Evaluate I = ∫13 4/x dx.
f(x) = 4/x[1, 3]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = 0.5x2 and y = 2x.
y = 0.5x2y = 2x
A =
7.Complex numbersMedium
Given z₁ = 2.5∠40° and z₂ = 2∠45°, find the argument θ of w = z₁/z₂ in degrees, between −180° and 180°.
z₁ = 2.5∠40°z₂ = 2∠45°
θ = °
8.Complex numbersHard
Exactly one of the points A–D is a root of z⁴ = 81∠−90°. Click it.
w = 81∠−90°n = 4
1A (3∠22.5°)
2B (3∠−22.5°)
3C (5.4∠−22.5°)
4D (3∠−90°)
9.Differential equationsMedium
A cup of coffee at 80 °C is left in a room at 19 °C. After 20 min its temperature is 29 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 23 °C?
T0 = 80 °CTa = 19 °CT(20) = 29 °CT = 23 °C
t = min
10.Differential equationsHard
A system obeys y″ + y′ + 16.25y = 32.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + y′ + 16.25y = 32.5y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [−5 3; −4 5]B = [6 −4; −2 0]
c21 =
12.Matrices & determinantsHard
Solve the system −4x − y = −4 and 5x − 2y = −9. Find x.
−4x − y = −45x − 2y = −9
x =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (−1, 2, −2) onto b = (6, 3, 5), that is, the component of a in the direction of b.
a = (−1, 2, −2)b = (6, 3, 5)
compb a =
14.Vectors: dot & cross productHard
A force F = (40, 0, 30) N acts at point A(0, 2, −3). Coordinates are in metres. Find the y-component My of the moment of the force about point O(2, 3, −1).
F = (40, 0, 30) NA(0, 2, −3) mO(2, 3, −1) m
My = N·m
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3p(x): 0.35, 0.1, 0.25, 0.3
Var(X) =
16.ProbabilityHard
A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 4 correct answers?
n = 9p = 0.25k = 4
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 30, 13, 20, 29, 30, 56.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 5. What proportion P of values is above 62?
μ = 60σ = 5x = 62
P =
19.Normal distributionMedium
The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 6 of them is below 495.3 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 6< 495.3 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 60% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g60%
x = g
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 = 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′(1.5) = 80.34
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 rule
f′(x) = u′v + uv′ = e2x·(1 + 2·x)
Substitute
f′(1.5) = 20.09·(1 + 3) = 80.34
Answer
f′(1.5) = 80.34
4.Derivatives
Answer: x = 4
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 − 21x + 36 = 3·(x2 − 7x + 12)
Solve f′(x) = 0
x = 3 or x = 4
Which one is which
f″(x) = 6x − 21: negative at a maximum, positive at a minimum.
f″(3) = −3, f″(4) = 3
x = 4
Answer
x = 4
5.Integrals
Answer: I = 4.394
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) = 4·ln x
Evaluate at the limits
I = 4·(ln 3 − ln 1) = 4·ln 3 = 4.394
Answer
I = 4.394
6.Integrals
Answer: A = 5.333
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
0.5x2 = 2x → x = 0, x = 4
Integrate upper minus lower
A = [x2 − 0.1667x3]04 = 5.333
Answer
A = 5.333
7.Complex numbers
Answer: θ = −5 °
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 / 2 = 1.25
θ = 40° − 45° = −5°
Argument
θ = −5°
Answer
θ = −5°
8.Complex numbers
Answer: B (3∠−22.5°)
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.
Modulus and arguments of the roots
|z| = 3
θ = −22.5°, 67.5°, 157.5°, −112.5°
The point
B = 3∠−22.5°
Answer
B (3∠−22.5°)
9.Differential equations
Answer: t = 30.13 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(61/10)/20 = 0.09041 min⁻¹
Time to reach the target
t = ln(61/4)/0.09041 = 30.13 min
Answer
t = 30.13 min
10.Differential equations
Answer: ymax = 3.35
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.
Steady value and roots
yss = 2, r = −0.5 ± 4j
Constants from the initial values
y(0) = 2 + A = 0 → A = −2
y′(0) = −0.5A + 4B = 0 → B = −0.25
Peak
t = π/4 = 0.7854
ymax = 2·(1 + e−0.5·π/4) = 3.35
Answer
ymax = 3.35
11.Matrices & determinants
Answer: c21 = −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.
Row times column
c21 = (−4)·6 + 5·(−2) = −34
Answer
c21 = −34
12.Matrices & determinants
Answer: x = −0.07692
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 = (−4)·(−2) − (−1)·5 = 13
Replace the x column
det Aₓ = (−4)·(−2) − (−1)·(−9) = −1
x = −1/13 = −0.07692
Answer
x = −0.07692
y = 4.308
13.Vectors: dot & cross product
Answer: compb a = −1.195
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 = (−1)·6 + 2·3 + (−2)·5 = −10
Length of b
|b| = √(6² + 3² + 5²) = 8.367
Scalar projection
compb a = a·b/|b| = −10/8.367 = −1.195
Answer
compb a = −1.195
Negative: a has a component pointing against b.
14.Vectors: dot & cross product
Answer: My = −20 N·m
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.
Position vector from O to A
r = A − O = (0, 2, −3) − (2, 3, −1) = (−2, −1, −2) m
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.
Mean
μ = 0·0.35 + 1·0.1 + 2·0.25 + 3·0.3 = 1.5
Mean of the squares
E(X²) = 0·0.35 + 1·0.1 + 4·0.25 + 9·0.3 = 3.8
Variance
Var(X) = 3.8 − 1.5² = 1.55
Answer
Var(X) = 1.55
16.Probability
Answer: P = 0.1168
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 = 1 − 0.6554 = 0.3446
Answer
P = 0.3446
19.Normal distribution
Answer: P = 0.3085
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 = 510 kg, σT = 12·√6 = 29.39 kg
Standardise
z = (495.3 − 510)/29.39 = −0.5
Probability
P = Φ(−0.5) = 0.3085
Answer
P = 0.3085
20.Normal distribution
Answer: x = 248.8 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.