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→∞ (3x3 + 4x − 2)/(5x3 + 5x + 3).
f(x) = (3x3 + 4x − 2)/(5x3 + 5x + 3)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 5x)/(2x²), with x in radians.
f(x) = (1 − cos 5x)/(2x²)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) = 0.5x2 − 4x − 2 at x = 3 crosses the y-axis at y = b. Find b.
f(x) = 0.5x2 − 4x − 2x = 3
b =
5.IntegralsMedium
Evaluate I = ∫14 3/x dx.
f(x) = 3/x[1, 4]
I =
6.IntegralsHard
Evaluate I = ∫₀3 x²·eˣ dx.
∫₀3 x²eˣ dx
I =
7.Complex numbersMedium
Given z₁ = 4∠−35° and z₂ = 1.5∠−50°, find the real part of w = z₁/z₂.
z₁ = 4∠−35°z₂ = 1.5∠−50°
Re(w) =
8.Complex numbersHard
Exactly one of the points A–D is a root of z⁶ = 729∠60°. Click it.
w = 729∠60°n = 6
1A (3∠−20°)
2B (3∠50°)
3C (5.4∠−20°)
4D (3∠−50°)
9.Differential equationsMedium
A forged steel part at 360 °C is left to cool in a workshop at 30 °C. After 59 min its temperature is 47 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 266 °C?
T0 = 360 °CTa = 30 °CT(59) = 47 °CT = 266 °C
t = min
10.Differential equationsHard
A system obeys y″ + 2y′ + 17y = 17 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 2y′ + 17y = 17y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [0 −3 2; −5 −1 −1; −2 1 −5]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [5 −1; −7 −1]
λ =
13.Vectors: dot & cross productMedium
A constant force F = (30, −10, 25) N acts on an object while it moves in a straight line from point A(3, 2, 2) to point B(4, −5, −5). Coordinates are in metres. Find the work W done by the force.
F = (30, −10, 25) NA(3, 2, 2) mB(4, −5, −5) m
W = J
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (−4, −3, −2), b = (−4, 1, 2) and c = (−1, −1, −4).
a = (−4, −3, −2)b = (−4, 1, 2)c = (−1, −1, −4)
V =
15.ProbabilityMedium
A fair die is rolled 8 times. What is the probability P of getting at least one six?
n = 8
P =
16.ProbabilityHard
10 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?
n = 10p = 0.6
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 45, 51, 45, 33, 44, 30.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 60. What proportion P of values is below 524?
μ = 500σ = 60x = 524
P =
19.Normal distributionMedium
The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 5 of them is below 196 g. Round z to two decimals.
μ = 40 gσ = 1.2 gn = 5< 196 g
P =
20.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 5% of them lie (the 5th percentile).
μ = 500σ = 40P = 5%
x = h
3Solutions
1.Limits
Answer: L = 0.6
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
(3 + 4/x2 + (−2)/x3)/(5 + 5/x2 + 3/x3)
Let x → ∞
L = 3/5 = 0.6
Answer
L = 0.6
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 6.25
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
5·sin 5x/(4x) → 0/0
Second application
25·cos 5x/4
Substitute x = 0
L = 25/4 = 6.25
Answer
L = 6.25
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 = −6.5
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) = −9.5
f′(x) = x − 4 → f′(3) = −1
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −9.5 − (−1)·3 = −6.5
Answer
b = −6.5
5.Integrals
Answer: I = 4.159
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) = 3·ln x
Evaluate at the limits
I = 3·(ln 4 − ln 1) = 3·ln 4 = 4.159
Answer
I = 4.159
6.Integrals
Answer: I = 98.43
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
20.09·(5) − 1·2 = 98.43
Answer
I = 98.43
7.Complex numbers
Answer: Re(w) = 2.576
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| = 4 / 1.5 = 2.667
θ = −35° − (−50)° = 15°
Rectangular form
Re(w) = 2.667·cos(15°) = 2.576
Im(w) = 2.667·sin(15°) = 0.6902
Re(w) = 2.576
Answer
Re(w) = 2.576
8.Complex numbers
Answer: D (3∠−50°)
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 = 10°, 70°, 130°, −170°, −110°, −50°
The point
D = 3∠−50°
Answer
D (3∠−50°)
9.Differential equations
Answer: t = 6.669 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(330/17)/59 = 0.05027 min⁻¹
Time to reach the target
t = ln(330/236)/0.05027 = 6.669 min
Answer
t = 6.669 min
10.Differential equations
Answer: ymax = 1.456
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 = 1, r = −1 ± 4j
Constants from the initial values
y(0) = 1 + A = 0 → A = −1
y′(0) = −1A + 4B = 0 → B = −0.25
Peak
t = π/4 = 0.7854
ymax = 1·(1 + e−1·π/4) = 1.456
Answer
ymax = 1.456
11.Matrices & determinants
Answer: det A = 55
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 = 23, M₁₃ = −7
Cofactor expansion
det A = 0·6 − (−3)·23 + 2·(−7) = 55
Answer
det A = 55
12.Matrices & determinants
Answer: λ = 6
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.
Characteristic equation
λ² − (4)λ + (−12) = 0
Solve
λ = (4 ± √(64))/2 → λ = −2, 6
λ = 6
Answer
λ = 6
13.Vectors: dot & cross product
Answer: W = −75 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 = (4, −5, −5) − (3, 2, 2) = (1, −7, −7) m
Work
W = F·d = 30·1 + (−10)·(−7) + 25·(−7) = −75 J
Answer
W = −75 J
Negative work: the force acts mostly against the motion.
14.Vectors: dot & cross product
Answer: V = 52
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 b × c
(b × c)x = bycz − bzcy = 1·(−4) − 2·(−1) = −2
(b × c)y = bzcx − bxcz = 2·(−1) − (−4)·(−4) = −18
(b × c)z = bxcy − bycx = (−4)·(−1) − 1·(−1) = 5
Dot product with a
a·(b × c) = (−4)·(−2) + (−3)·(−18) + (−2)·5 = 52
V = |52| = 52
Answer
V = 52
15.Probability
Answer: P = 0.7674
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.
Complement
P(no six) = (5/6)8 = 0.2326
P = 1 − 0.2326 = 0.7674
Answer
P = 0.7674
16.Probability
Answer: P = 0.001678
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) = 0.410 = 0.0001049
P(X = 1) = 10·0.6·0.49 = 0.001573
Add them
P = 0.0001049 + 0.001573 = 0.001678
Answer
P = 0.001678
17.Descriptive statistics
Answer: s = 8.066
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 = 0.6554
Answer
P = 0.6554
19.Normal distribution
Answer: P = 0.06811
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 = 200 g, σT = 1.2·√5 = 2.683 g
Standardise
z = (196 − 200)/2.683 = −1.49
Probability
P = Φ(−1.49) = 0.06811
Answer
P = 0.06811
20.Normal distribution
Answer: x = 434.2 h
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.