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→∞ (4x2 − 2x + 1)/(3x2 + 2x + 4).
f(x) = (4x2 − 2x + 1)/(3x2 + 2x + 4)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 3x)/sin(5x), with x in radians.
f(x) = ln(1 + 3x)/sin(5x)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(x), with x in radians. Find f′(0.5).
f(x) = x·sin(x)x = 0.5 rad
f′(0.5) =
4.DerivativesHard
The tangent line to f(x) = x2 − x − 1 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = x2 − x − 1x = 1
b =
5.IntegralsMedium
The region under y = 0.5x from x = 0 to x = 1 is rotated about the x-axis. Find the volume V of the solid (a cone).
y = 0.5x0 ≤ x ≤ 1
V =
6.IntegralsHard
Find the area A of the region between the parabola y = 9 − x² and the x-axis.
y = 9 − x²
A =
7.Complex numbersMedium
Write z = 6∠160° in rectangular form a + jb and give the imaginary part b.
r = 6θ = 160°
b =
8.Complex numbersHard
The roots of zⁿ = w with n = 5 and w = 243∠180° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, …, 4. Find the real part of z₁.
w = 243∠180°n = 5k = 1
Re(z₁) =
9.Differential equationsMedium
Solve y″ + 7y′ + 10y = 0 with y(0) = 5 and y′(0) = 2. Find y(1.5).
y″ + 7y′ + 10y = 0y(0) = 5y′(0) = 2
y(1.5) =
10.Differential equationsHard
Solve y″ + 4y′ + 8y = 0 with y(0) = 2 and y′(0) = 0 (released from rest). Find y(1.5); angles in radians.
y″ + 4y′ + 8y = 0y(0) = 2y′(0) = 0
y(1.5) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [2 5 −2; 2 −1 4; 5 −5 −2]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−1 3; −7 9]
λ =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (−1, −6, 0) and b = (k, 3, −6) are perpendicular.
a = (−1, −6, 0)b = (k, 3, −6)
k =
14.Vectors: dot & cross productHard
For a = (−5, 0, 2) and b = (−4, 6, 3), find the x-component (a × b)x of the cross product a × b.
a = (−5, 0, 2)b = (−4, 6, 3)
(a × b)x =
15.ProbabilityMedium
In how many ways N can a committee of 3 people be chosen from a group of 6? The order does not matter.
n = 6k = 3
N =
16.ProbabilityHard
X has the distribution shown, and Y = 2X + 1. Find the standard deviation σY of Y.
x: 1, 2, 3, 4p(x): 0.25, 0.25, 0.2, 0.3Y = 2X + 1
σY =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 48, 31, 29, 23, 12, 33.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 20. What proportion P of values is above 68?
μ = 60σ = 20x = 68
P =
19.Normal distributionMedium
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 172.4 and 181.5 cm? Give P as a decimal.
μ = 178σ = 7x1 = 172.4x2 = 181.5
P =
20.Normal distributionHard
Hole diameters D are normal with mean 25.03 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.03 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
3Solutions
1.Limits
Answer: L = 1.333
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
(4 + (−2)/x + 1/x2)/(3 + 2/x + 4/x2)
Let x → ∞
L = 4/3 = 1.333
Answer
L = 1.333
For equal degrees the limit is the ratio of the leading coefficients.
2.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).
L'Hôpital's rule
L = limx→0 [3/(1 + 3x)]/[5·cos 5x]
Substitute x = 0
L = 3/5 = 0.6
Answer
L = 0.6
3.Derivatives
Answer: f′(0.5) = 0.9182
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′(0.5) = 0.4794 + 0.5·0.8776 = 0.9182
Answer
f′(0.5) = 0.9182
4.Derivatives
Answer: b = −2
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) = −1
f′(x) = 2x − 1 → f′(1) = 1
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −1 − 1·1 = −2
Answer
b = −2
5.Integrals
Answer: V = 0.2618
Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.
Disc method
V = π∫₀1 0.25x² dx
Integrate
V = π·0.25·1³/3 = 0.2618
Answer
V = 0.2618
Check: a cone of radius 0.5 and height 1 has (1/3)πr²h = 0.2618.
6.Integrals
Answer: A = 36
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.
Zeros
9 − x² = 0 → x = ±3
Integrate
A = [9x − x³/3]−33 = 18 − (−18) = 36
Answer
A = 36
7.Complex numbers
Answer: b = 2.052
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(160°) = −5.638
b = 6·sin(160°) = 2.052
Rectangular form
z = −5.638 + j2.052
b = 2.052
Answer
b = 2.052
8.Complex numbers
Answer: Re(z₁) = −0.9271
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
θ₁ = (180° + 360°·1)/5 = 108°
Rectangular form
z₁ = 3·(cos 108° + j sin 108°) = −0.9271 + j2.853
Re(z₁) = −0.9271
Answer
Re(z₁) = −0.9271
The 5 roots lie on the circle of radius 3, 72° apart.
9.Differential equations
Answer: y(1.5) = 0.4459
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 + 2)(r + 5) = 0 → r = −2, −5
Constants from the initial values
A + B = 5, −2A − 5B = 2
A = 9, B = −4
Evaluate
y(1.5) = 9·e−3 + (−4)·e−7.5 = 0.4459
Answer
y(1.5) = 0.4459
10.Differential equations
Answer: y(1.5) = −0.08453
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 = −2 ± 2j
Constants
A = 2, B = 2·2/2 = 2
Evaluate
y(1.5) = e−3(2·cos 3 + 2·sin 3) = −0.08453
Answer
y(1.5) = −0.08453
11.Matrices & determinants
Answer: det A = 174
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 = 22, M₁₂ = −24, M₁₃ = −5
Cofactor expansion
det A = 2·22 − 5·(−24) + (−2)·(−5) = 174
Answer
det A = 174
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 − (8)λ + (12) = 0
Solve
λ = (8 ± √(16))/2 → λ = 2, 6
λ = 6
Answer
λ = 6
13.Vectors: dot & cross product
Answer: k = −18
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 equals zero
a·b = (−1)·k + (−6)·3 + 0·(−6) = 0
−18 + (−1)·k = 0
Solve for k
k = −(−18)/(−1) = −18
Answer
k = −18
14.Vectors: dot & cross product
Answer: (a × b)x = −12
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, component by component
(a × b)x = aybz − azby = 0·3 − 2·6 = −12
(a × b)y = azbx − axbz = 2·(−4) − (−5)·3 = 7
(a × b)z = axby − aybx = (−5)·6 − 0·(−4) = −30
Requested component
(a × b)x = −12
Answer
(a × b)x = −12
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: N = 20
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.
Combinations
C(6, 3) = (6·5·4)/(3·2·1) = 20
Answer
N = 20
16.Probability
Answer: σY = 2.322
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.
Spread of X
μ = 2.55, Var(X) = 1.348, σX = 1.161
Linear change
σY = |2|·1.161 = 2.322
Answer
σY = 2.322
The mean does move: E(Y) = 2·2.55 + 1 = 6.1.
17.Descriptive statistics
Answer: s = 11.88
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 = 0.4796
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.
z-scores
z₁ = (172.4 − 178)/7 = −0.8
z₂ = (181.5 − 178)/7 = 0.5
Area between
P = Φ(0.5) − Φ(−0.8) = 0.6915 − 0.2119 = 0.4796
Answer
P = 0.4796
20.Normal distribution
Answer: P = 0.9099
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).