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→∞ (x2 − 2x − 7)/(x2 + x + 8).
f(x) = (x2 − 2x − 7)/(x2 + x + 8)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 4x)/(4x²), with x in radians.
f(x) = (1 − cos 4x)/(4x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(2x), with x in radians. Find f′(1.5).
f(x) = x·sin(2x)x = 1.5 rad
f′(1.5) =
4.DerivativesHard
The tangent line to f(x) = x2 + 3x at x = −1 crosses the y-axis at y = b. Find b.
f(x) = x2 + 3xx = −1
b =
5.IntegralsMedium
The region under y = √(2x) from x = 0 to x = 3 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).
y = √(2x)0 ≤ x ≤ 3
V =
6.IntegralsHard
The region between the line y = 5x and the parabola y = 2x² is rotated about the x-axis. Find the volume V of the solid.
y = 5xy = 2x²
V =
7.Complex numbersMedium
Find the argument θ of z = −9 − j5 in degrees, as the principal value between −180° and 180°.
z = −9 − j5
θ = °
8.Complex numbersHard
Solve z² − 3z + 27.25 = 0. The roots are a ± jb; find b > 0.
p = −3q = 27.25
b =
9.Differential equationsMedium
A forged steel part at 500 °C is left to cool in a workshop at 26 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.03 min⁻¹, find its temperature T after 52 min.
T0 = 500 °CTa = 26 °Ck = 0.03 min⁻¹t = 52 min
T = °C
10.Differential equationsHard
A system obeys y″ + 3y′ + 6.25y = 6.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1); angles in radians.
y″ + 3y′ + 6.25y = 6.25y(0) = 0y′(0) = 0
y(1) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [5 1 −4; −3 2 0; −2 4 3]
det A =
12.Matrices & determinantsHard
Solve the system 3x + 3y = 4 and −5x + 5y = 0. Find x.
3x + 3y = 4−5x + 5y = 0
x =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (5, 0, −4) onto b = (−1, 1, 6), that is, the component of a in the direction of b.
a = (5, 0, −4)b = (−1, 1, 6)
compb a =
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (4, −5, 0) and b = (0, 5, 4) and has a positive z-component. Find its y-component n̂y.
a = (4, −5, 0)b = (0, 5, 4)n̂z > 0
n̂y =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 2, 4, 6p(x): 0.35, 0.15, 0.2, 0.3
Var(X) =
16.ProbabilityHard
X has the distribution shown, and Y = 2X + 4. Find the standard deviation σY of Y.
x: 1, 2, 3, 4p(x): 0.35, 0.3, 0.05, 0.3Y = 2X + 4
σY =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 59, 10, 55, 15, 19, 58, 20.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 75. What proportion P of values is above 612.5?
μ = 500σ = 75x = 612.5
P =
19.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.3594. Find z to two decimals.
A = 0.3594
z =
20.Normal distributionHard
Hole diameters D are normal with mean 25.03 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 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 = 0.015 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
3Solutions
1.Limits
Answer: L = 1
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
(1 + (−2)/x + (−7)/x2)/(1 + 1/x + 8/x2)
Let x → ∞
L = 1/1 = 1
Answer
L = 1
For equal degrees the limit is the ratio of the leading coefficients.
2.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).
First application
4·sin 4x/(8x) → 0/0
Second application
16·cos 4x/8
Substitute x = 0
L = 16/8 = 2
Answer
L = 2
3.Derivatives
Answer: f′(1.5) = −2.829
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′(1.5) = 0.1411 + 3·(−0.99) = −2.829
Answer
f′(1.5) = −2.829
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(−1) = −2
f′(x) = 2x + 3 → f′(−1) = 1
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −2 − 1·(−1) = −1
Answer
b = −1
5.Integrals
Answer: V = 28.27
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∫₀3 2x dx
Integrate
V = π·2·3²/2 = 28.27
Answer
V = 28.27
6.Integrals
Answer: V = 163.6
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.
Intersection points
5x = 2x² → x = 0, x = 2.5
Washer method
V = π∫₀2.5 ((5x)² − (2x²)²) dx
Integrate
V = π(25·2.5³/3 − 4·2.5⁵/5) = π·52.08 = 163.6
Answer
V = 163.6
7.Complex numbers
Answer: θ = −150.9 °
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.
Quadrant
a = −9, b = −5 → III
Reference angle
α = atan(5/9) = 29.05°
Argument
θ = −(180° − α) = −150.9°
Answer
θ = −150.9°
In polar form z = 10.3∠−150.9°.
8.Complex numbers
Answer: b = 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.
Discriminant
p² − 4q = 9 − 109 = −100 < 0
Quadratic formula
z = (3 ± j√100)/2 = 1.5 + j5, 1.5 − j5
Answer
b = 5
Answer
b = 5
9.Differential equations
Answer: T = 125.6 °C
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.
Solution of the cooling law
T(t) = 26 + 474·e−0.03t
Substitute t
T(52) = 26 + 474·0.2101 = 125.6 °C
Answer
T = 125.6 °C
10.Differential equations
Answer: y(1) = 0.9407
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 = −1.5 ± 2j
Constants from the initial values
y(0) = 1 + A = 0 → A = −1
y′(0) = −1.5A + 2B = 0 → B = −0.75
Evaluate
y(1) = 1·(1 − e−1.5(cos 2 + 0.75·sin 2)) = 0.9407
Answer
y(1) = 0.9407
11.Matrices & determinants
Answer: det A = 71
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 = −9, M₁₃ = −8
Cofactor expansion
det A = 5·6 − 1·(−9) + (−4)·(−8) = 71
Answer
det A = 71
12.Matrices & determinants
Answer: x = 0.6667
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 = 3·5 − 3·(−5) = 30
Replace the x column
det Aₓ = 4·5 − 3·0 = 20
x = 20/30 = 0.6667
Answer
x = 0.6667
y = 0.6667
13.Vectors: dot & cross product
Answer: compb a = −4.704
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 = 5·(−1) + 0·1 + (−4)·6 = −29
Length of b
|b| = √((−1)² + 1² + 6²) = 6.164
Scalar projection
compb a = a·b/|b| = −29/6.164 = −4.704
Answer
compb a = −4.704
Negative: a has a component pointing against b.
14.Vectors: dot & cross product
Answer: n̂y = −0.4924
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 + 2·0.15 + 4·0.2 + 6·0.3 = 2.9
Mean of the squares
E(X²) = 0·0.35 + 4·0.15 + 16·0.2 + 36·0.3 = 14.6
Variance
Var(X) = 14.6 − 2.9² = 6.19
Answer
Var(X) = 6.19
16.Probability
Answer: σY = 2.458
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 = 2.3, Var(X) = 1.51, σX = 1.229
Linear change
σY = |2|·1.229 = 2.458
Answer
σY = 2.458
The mean does move: E(Y) = 2·2.3 + 4 = 8.6.
17.Descriptive statistics
Answer: s = 22.36
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 = 1.5
Standard normal table
Φ(1.5) = 0.9332
P = 1 − 0.9332 = 0.06681
Answer
P = 0.06681
19.Normal distribution
Answer: z = 0.36
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 − 0.3594 = 0.6406
Read the z-table backwards
Find 0.6406 inside the table and read off z.
z = 0.36
Answer
z = 0.36
20.Normal distribution
Answer: P = 0.9207
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).