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→∞ (−2x3 − 6x + 2)/(x3 + 5x + 1).
f(x) = (−2x3 − 6x + 2)/(x3 + 5x + 1)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 2x)/(3x²), with x in radians.
f(x) = (1 − cos 2x)/(3x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(2x), with x in radians. Find f′(0.5).
f(x) = x·sin(2x)x = 0.5 rad
f′(0.5) =
4.DerivativesHard
The tangent line to f(x) = 0.5x2 + 3x + 1 at x = −2 crosses the y-axis at y = b. Find b.
f(x) = 0.5x2 + 3x + 1x = −2
b =
5.IntegralsMedium
Evaluate I = ∫₁2 x·ln x dx.
∫₁2 x ln x dx
I =
6.IntegralsHard
Evaluate I = ∫₀1.5 x²·eˣ dx.
∫₀1.5 x²eˣ dx
I =
7.Complex numbersMedium
Given z₁ = 1.5∠−60° and z₂ = 2∠−140°, find the real part of w = z₁/z₂.
z₁ = 1.5∠−60°z₂ = 2∠−140°
Re(w) =
8.Complex numbersHard
Exactly one of the points A–D is a root of z⁶ = 729∠180°. Click it.
w = 729∠180°n = 6
1A (3∠60°)
2B (3∠30°)
3C (5.4∠60°)
4D (3∠180°)
9.Differential equationsMedium
A cup of coffee at 80 °C is left in a room at 19 °C. After 18 min its temperature is 33 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 27 °C?
T0 = 80 °CTa = 19 °CT(18) = 33 °CT = 27 °C
t = min
10.Differential equationsHard
A system obeys y″ + y′ + 16.25y = 162.5 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 = 162.5y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [3 3 −2; 3 −3 −2; −1 −2 4]
det A =
12.Matrices & determinantsHard
Solve the system x − 3y = −9 and 4x − 2y = 12. Find x.
x − 3y = −94x − 2y = 12
x =
13.Vectors: dot & cross productMedium
A constant force F = (0, 15, 30) N acts on an object while it moves in a straight line from point A(0, 0, 4) to point B(0, 3, −2). Coordinates are in metres. Find the work W done by the force.
F = (0, 15, 30) NA(0, 0, 4) mB(0, 3, −2) m
W = J
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (5, −3, 1), b = (−2, 1, −4) and c = (−4, −5, −1).
a = (5, −3, 1)b = (−2, 1, −4)c = (−4, −5, −1)
V =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 2, 4, 6p(x): 0.35, 0.2, 0.3, 0.15
σ =
16.ProbabilityHard
5 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?
n = 5p = 0.6
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 53, 41, 33, 12, 48.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 10. Find the z-score z of a score of 58.
μ = 70σ = 10x = 58
z =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.6629. Find z to two decimals.
A = 0.6629
z =
20.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 10% of them lie (the 10th percentile).
μ = 178σ = 7P = 10%
x = cm
3Solutions
1.Limits
Answer: L = −2
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
(−2 + (−6)/x2 + 2/x3)/(1 + 5/x2 + 1/x3)
Let x → ∞
L = −2/1 = −2
Answer
L = −2
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 0.6667
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
2·sin 2x/(6x) → 0/0
Second application
4·cos 2x/6
Substitute x = 0
L = 4/6 = 0.6667
Answer
L = 0.6667
3.Derivatives
Answer: f′(0.5) = 1.382
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′(0.5) = 0.8415 + 1·0.5403 = 1.382
Answer
f′(0.5) = 1.382
4.Derivatives
Answer: b = −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.
Point and slope
f(−2) = −3
f′(x) = x + 3 → f′(−2) = 1
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −3 − 1·(−2) = −1
Answer
b = −1
5.Integrals
Answer: I = 0.6363
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
(2·ln 2 − 1) − (0 − 0.25) = 0.6363
Answer
I = 0.6363
6.Integrals
Answer: I = 3.602
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
4.482·(1.25) − 1·2 = 3.602
Answer
I = 3.602
7.Complex numbers
Answer: Re(w) = 0.1302
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| = 1.5 / 2 = 0.75
θ = −60° − (−140)° = 80°
Rectangular form
Re(w) = 0.75·cos(80°) = 0.1302
Im(w) = 0.75·sin(80°) = 0.7386
Re(w) = 0.1302
Answer
Re(w) = 0.1302
8.Complex numbers
Answer: B (3∠30°)
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 = 30°, 90°, 150°, −150°, −90°, −30°
The point
B = 3∠30°
Answer
B (3∠30°)
9.Differential equations
Answer: t = 24.84 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(61/14)/18 = 0.08177 min⁻¹
Time to reach the target
t = ln(61/8)/0.08177 = 24.84 min
Answer
t = 24.84 min
10.Differential equations
Answer: ymax = 16.75
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 = 10, r = −0.5 ± 4j
Constants from the initial values
y(0) = 10 + A = 0 → A = −10
y′(0) = −0.5A + 4B = 0 → B = −1.25
Peak
t = π/4 = 0.7854
ymax = 10·(1 + e−0.5·π/4) = 16.75
Answer
ymax = 16.75
11.Matrices & determinants
Answer: det A = −60
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 = −16, M₁₂ = 10, M₁₃ = −9
Cofactor expansion
det A = 3·(−16) − 3·10 + (−2)·(−9) = −60
Answer
det A = −60
12.Matrices & determinants
Answer: x = 5.4
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 = 1·(−2) − (−3)·4 = 10
Replace the x column
det Aₓ = (−9)·(−2) − (−3)·12 = 54
x = 54/10 = 5.4
Answer
x = 5.4
y = 4.8
13.Vectors: dot & cross product
Answer: W = −135 J
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.
Displacement
d = B − A = (0, 3, −2) − (0, 0, 4) = (0, 3, −6) m
Work
W = F·d = 0·0 + 15·3 + 30·(−6) = −135 J
Answer
W = −135 J
Negative work: the force acts mostly against the motion.
14.Vectors: dot & cross product
Answer: V = 133
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.
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 + 2·0.2 + 4·0.3 + 6·0.15 = 2.5
Mean of the squares
E(X²) = 0·0.35 + 4·0.2 + 16·0.3 + 36·0.15 = 11
Variance
Var(X) = 11 − 2.5² = 4.75
σ = √4.75 = 2.179
Answer
σ = 2.179
16.Probability
Answer: P = 0.08704
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.
Zero and one failure
P(X = 0) = 0.45 = 0.01024
P(X = 1) = 5·0.6·0.44 = 0.0768
Add them
P = 0.01024 + 0.0768 = 0.08704
Answer
P = 0.08704
17.Descriptive statistics
Answer: s = 16.07
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 = (58 − 70)/10 = −1.2
Answer
z = −1.2
19.Normal distribution
Answer: z = 0.96
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 + 0.6629)/2 = 0.8315
Read the z-table backwards
Find 0.8315 inside the table and read off z.
z = 0.96
Answer
z = 0.96
20.Normal distribution
Answer: x = 169 cm
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.