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 + 1/x)3x. Give a decimal value.
f(x) = (1 + 1/x)3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 3x)/(4x²), with x in radians.
f(x) = (1 − cos 3x)/(4x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(x), with x in radians. Find f′(2).
f(x) = x·sin(x)x = 2 rad
f′(2) =
4.DerivativesHard
The tangent line to f(x) = 0.5x2 − x + 6 at x = −1 crosses the y-axis at y = b. Find b.
f(x) = 0.5x2 − x + 6x = −1
b =
5.IntegralsMedium
Evaluate I = ∫24 5/x dx.
f(x) = 5/x[2, 4]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = 2x2 and y = 4x.
y = 2x2y = 4x
A =
7.Complex numbersMedium
Write z = 6∠140° in rectangular form a + jb and give the real part a.
r = 6θ = 140°
a =
8.Complex numbersHard
The roots of zⁿ = w with n = 5 and w = 243∠−90° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, …, 4. Find the real part of z₀.
w = 243∠−90°n = 5k = 0
Re(z₀) =
9.Differential equationsMedium
Solve y″ + 7y′ + 12y = 0 with y(0) = 2 and y′(0) = 1. Find y(1.5).
y″ + 7y′ + 12y = 0y(0) = 2y′(0) = 1
y(1.5) =
10.Differential equationsHard
Solve y″ + 2y′ + 17y = 0 with y(0) = 5 and y′(0) = 0 (released from rest). Find y(1); angles in radians.
y″ + 2y′ + 17y = 0y(0) = 5y′(0) = 0
y(1) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−4 −1 −4; −5 −3 −3; −2 −2 −3]
det A =
12.Matrices & determinantsHard
Solve the system 3x + 5y = −4 and x − 6y = −8. Find x.
3x + 5y = −4x − 6y = −8
x =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (6, 7, −7) and b = (5, 6, 7).
a = (6, 7, −7)b = (5, 6, 7)
a·b =
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (−4, 4, −5), b = (4, 3, 5) and c = (−3, 2, 1).
a = (−4, 4, −5)b = (4, 3, 5)c = (−3, 2, 1)
V =
15.ProbabilityMedium
A fair die is rolled 2 times. What is the probability P of getting at least one six?
n = 2
P =
16.ProbabilityHard
A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 5 correct answers?
n = 6p = 0.1k = 5
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 15, 50, 17, 27, 59.
n = 5
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 56.
μ = 60σ = 8x = 56
z =
19.Normal distributionMedium
X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 200.3 mm. Round z to two decimals.
μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 200.3 mm
P =
20.Normal distributionHard
Hole diameters D are normal with mean 25.04 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.04 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
3Solutions
1.Limits
Answer: L = 20.09
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 + 1/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 3·1 = 3
Undo the logarithm
L = e3 = 20.09
Answer
L = 20.09
2.Limits
Answer: L = 1.125
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
3·sin 3x/(8x) → 0/0
Second application
9·cos 3x/8
Substitute x = 0
L = 9/8 = 1.125
Answer
L = 1.125
3.Derivatives
Answer: f′(2) = 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(x) + x·cos(x)
Substitute
f′(2) = 0.9093 + 2·(−0.4161) = 0.077
Answer
f′(2) = 0.077
4.Derivatives
Answer: b = 5.5
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(−1) = 7.5
f′(x) = x − 1 → f′(−1) = −2
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 7.5 − (−2)·(−1) = 5.5
Answer
b = 5.5
5.Integrals
Answer: I = 3.466
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) = 5·ln x
Evaluate at the limits
I = 5·(ln 4 − ln 2) = 5·ln 2 = 3.466
Answer
I = 3.466
6.Integrals
Answer: A = 2.667
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
2x2 = 4x → x = 0, x = 2
Integrate upper minus lower
A = [2x2 − 0.6667x3]02 = 2.667
Answer
A = 2.667
7.Complex numbers
Answer: a = −4.596
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.
Components
a = 6·cos(140°) = −4.596
b = 6·sin(140°) = 3.857
Rectangular form
z = −4.596 + j3.857
a = −4.596
Answer
a = −4.596
8.Complex numbers
Answer: Re(z₀) = 2.853
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 of the roots
2431/5 = 3
Argument of the root
θ₀ = (−90° + 360°·0)/5 = −18°
Rectangular form
z₀ = 3·(cos −18° + j sin −18°) = 2.853 − j0.9271
Re(z₀) = 2.853
Answer
Re(z₀) = 2.853
The 5 roots lie on the circle of radius 3, 72° apart.
9.Differential equations
Answer: y(1.5) = 0.08263
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 = 2, −3A − 4B = 1
A = 9, B = −7
Evaluate
y(1.5) = 9·e−4.5 + (−7)·e−6 = 0.08263
Answer
y(1.5) = 0.08263
10.Differential equations
Answer: y(1) = −1.55
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.
Complex roots
r = −1 ± 4j
Constants
A = 5, B = 1·5/4 = 1.25
Evaluate
y(1) = e−1(5·cos 4 + 1.25·sin 4) = −1.55
Answer
y(1) = −1.55
11.Matrices & determinants
Answer: det A = −19
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 = 3, M₁₂ = 9, M₁₃ = 4
Cofactor expansion
det A = (−4)·3 − (−1)·9 + (−4)·4 = −19
Answer
det A = −19
12.Matrices & determinants
Answer: x = −2.783
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·(−6) − 5·1 = −23
Replace the x column
det Aₓ = (−4)·(−6) − 5·(−8) = 64
x = 64/(−23) = −2.783
Answer
x = −2.783
y = 0.8696
13.Vectors: dot & cross product
Answer: a·b = 23
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.
Multiply matching components and add
a·b = 6·5 + 7·6 + (−7)·7 = 30 + 42 + (−49) = 23
Answer
a·b = 23
Positive: the angle between the vectors is less than 90°.
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.
Cross product b × c
(b × c)x = bycz − bzcy = 3·1 − 5·2 = −7
(b × c)y = bzcx − bxcz = 5·(−3) − 4·1 = −19
(b × c)z = bxcy − bycx = 4·2 − 3·(−3) = 17
Dot product with a
a·(b × c) = (−4)·(−7) + 4·(−19) + (−5)·17 = −133
V = |−133| = 133
Answer
V = 133
15.Probability
Answer: P = 0.3056
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)2 = 0.6944
P = 1 − 0.6944 = 0.3056
Answer
P = 0.3056
16.Probability
Answer: P = 5.4 × 10⁻⁵
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 = (56 − 60)/8 = −0.5
Answer
z = −0.5
19.Normal distribution
Answer: P = 0.2743
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 sum
μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm
Standardise
z = (200.3 − 200)/0.5 = 0.6
Upper tail
P = 1 − Φ(0.6) = 1 − 0.7257 = 0.2743
Answer
P = 0.2743
20.Normal distribution
Answer: P = 0.9452
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).