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 + 6)/(4x3 + x + 1).
f(x) = (−3x3 − 4x + 6)/(4x3 + x + 1)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 3x)/sin(2x), with x in radians.
f(x) = ln(1 + 3x)/sin(2x)x → 0
L =
3.DerivativesMedium
Let f(x) = (2x + 1)4. Find f′(−1).
f(x) = (2x + 1)4x = −1
f′(−1) =
4.DerivativesHard
The tangent line to f(x) = 3x2 − x + 5 at x = 3 crosses the y-axis at y = b. Find b.
f(x) = 3x2 − x + 5x = 3
b =
5.IntegralsMedium
Evaluate I = ∫28 1/x dx.
f(x) = 1/x[2, 8]
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 4 − x² and the x-axis.
y = 4 − x²
A =
7.Complex numbersMedium
Given z₁ = 1.5∠125° and z₂ = 3∠−50°, find the real part of w = z₁·z₂.
z₁ = 1.5∠125°z₂ = 3∠−50°
Re(w) =
8.Complex numbersHard
Use De Moivre's theorem to find the real part of z⁴ for z = −1 + j.
z = −1 + jn = 4
Re(z⁴) =
9.Differential equationsMedium
A forged steel part at 460 °C is left to cool in a workshop at 29 °C. After 31 min its temperature is 345 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 380 °C?
T0 = 460 °CTa = 29 °CT(31) = 345 °CT = 380 °C
t = min
10.Differential equationsHard
A system obeys y″ + 3y′ + 18.25y = 182.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.25); angles in radians.
y″ + 3y′ + 18.25y = 182.5y(0) = 0y′(0) = 0
y(0.25) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [−5 −6; −4 1]B = [−1 5; −2 −4]
c21 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [0 −2; −6 1]
λ =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (3, −5, −4) and b = (−5, −2, 5), in degrees.
a = (3, −5, −4)b = (−5, −2, 5)
θ = °
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (0, 1, −4) and b = (−5, 3, 4) and has a positive z-component. Find its y-component n̂y.
a = (0, 1, −4)b = (−5, 3, 4)n̂z > 0
n̂y =
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
X has the distribution shown, and Y = −2X + 4. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 35, 33, 46, 15, 28.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 60. What proportion P of values is below 620?
μ = 500σ = 60x = 620
P =
19.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 65 and 75 points? Give P as a decimal.
μ = 65σ = 10x1 = 65x2 = 75
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 10% of them lie (the 10th percentile).
μ = 500σ = 40P = 10%
x = h
3Solutions
1.Limits
Answer: L = −0.75
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 + 6/x3)/(4 + 1/x2 + 1/x3)
Let x → ∞
L = −3/4 = −0.75
Answer
L = −0.75
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 1.5
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)]/[2·cos 2x]
Substitute x = 0
L = 3/2 = 1.5
Answer
L = 1.5
3.Derivatives
Answer: f′(−1) = −8
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) = 4·(2x + 1)3·2
Substitute
f′(−1) = 4·(−1)3·2 = −8
Answer
f′(−1) = −8
4.Derivatives
Answer: b = −22
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) = 29
f′(x) = 6x − 1 → f′(3) = 17
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 29 − 17·3 = −22
Answer
b = −22
5.Integrals
Answer: I = 1.386
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) = ln x
Evaluate at the limits
I = (ln 8 − ln 2) = ln 4 = 1.386
Answer
I = 1.386
6.Integrals
Answer: A = 10.67
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
4 − x² = 0 → x = ±2
Integrate
A = [4x − x³/3]−22 = 5.333 − (−5.333) = 10.67
Answer
A = 10.67
7.Complex numbers
Answer: Re(w) = 1.165
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 · 3 = 4.5
θ = 125° + (−50)° = 75°
Rectangular form
Re(w) = 4.5·cos(75°) = 1.165
Im(w) = 4.5·sin(75°) = 4.347
Re(w) = 1.165
Answer
Re(w) = 1.165
8.Complex numbers
Answer: Re(z⁴) = −4
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.
Polar form
z = −1 + j = √2∠135°
De Moivre
z⁴ = (√2)⁴∠540° = 4∠180°
Rectangular form
z⁴ = 4·(cos 180° + j sin 180°) = −4
Re(z⁴) = −4
Answer
Re(z⁴) = −4
9.Differential equations
Answer: t = 20.51 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(431/316)/31 = 0.01001 min⁻¹
Time to reach the target
t = ln(431/351)/0.01001 = 20.51 min
Answer
t = 20.51 min
10.Differential equations
Answer: y(0.25) = 4.118
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.
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.
Row times column
c21 = (−4)·(−1) + 1·(−2) = 2
Answer
c21 = 2
12.Matrices & determinants
Answer: λ = 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.
Characteristic equation
λ² − (1)λ + (−12) = 0
Solve
λ = (1 ± √(49))/2 → λ = −3, 4
λ = 4
Answer
λ = 4
13.Vectors: dot & cross product
Answer: θ = 118.8 °
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 = 3·(−5) + (−5)·(−2) + (−4)·5 = −25
Magnitudes
|a| = √(3² + (−5)² + (−4)²) = 7.071
|b| = √((−5)² + (−2)² + 5²) = 7.348
Angle
cos θ = −25/(7.071·7.348) = −0.4811
θ = arccos(−0.4811) = 118.8°
Answer
θ = 118.8°
14.Vectors: dot & cross product
Answer: n̂y = 0.7664
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
(a × b)x = aybz − azby = 1·4 − (−4)·3 = 16
(a × b)y = azbx − axbz = (−4)·(−5) − 0·4 = 20
(a × b)z = axby − aybx = 0·3 − 1·(−5) = 5
Normalise
The z-component of a × b is already positive.
|a × b| = √(16² + 20² + 5²) = 26.1
n̂ = (16, 20, 5)/26.1 = (0.6131, 0.7664, 0.1916)
n̂y = 0.7664
Answer
n̂y = 0.7664
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: σY = 2.39
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 = 1.35, Var(X) = 1.428, σX = 1.195
Linear change
σY = |−2|·1.195 = 2.39
Answer
σY = 2.39
The mean does move: E(Y) = −2·1.35 + 4 = 1.3.
17.Descriptive statistics
Answer: s = 11.28
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 = 2
Standard normal table
Φ(2) = 0.9772
P = 0.9772
Answer
P = 0.9772
19.Normal distribution
Answer: P = 0.3413
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 = (65 − 65)/10 = 0
z₂ = (75 − 65)/10 = 1
Area between
P = Φ(1) − Φ(0) = 0.8413 − 0.5 = 0.3413
Answer
P = 0.3413
20.Normal distribution
Answer: x = 448.7 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.