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 − 4x) − 2x).
f(x) = √(4x2 − 4x) − 2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (ex − 1 − x)/x².
f(x) = (ex − 1 − x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(x), with x in radians. Find f′(1).
f(x) = x·sin(x)x = 1 rad
f′(1) =
4.DerivativesHard
The tangent line to f(x) = 2x2 − 3x + 6 at x = 3 crosses the y-axis at y = b. Find b.
f(x) = 2x2 − 3x + 6x = 3
b =
5.IntegralsMedium
Evaluate I = ∫03 e−x dx.
f(x) = e−x[0, 3]
I =
6.IntegralsHard
Evaluate I = ∫₀2 x²·eˣ dx.
∫₀2 x²eˣ dx
I =
7.Complex numbersMedium
Given z₁ = 3∠70° and z₂ = 2∠65°, find the imaginary part of w = z₁/z₂.
z₁ = 3∠70°z₂ = 2∠65°
Im(w) =
8.Complex numbersHard
One root of z³ − z² − z − 15 = 0 is z₁ = −1 + j2. The coefficients are real. Find the real root r.
z₁ = −1 + j2
r =
9.Differential equationsMedium
Solve y″ + 5y′ + 6y = 0 with y(0) = 2 and y′(0) = -2. Find y(0.5).
y″ + 5y′ + 6y = 0y(0) = 2y′(0) = -2
y(0.5) =
10.Differential equationsHard
A tank holds 1000 L of water in which 5 kg of salt is dissolved. Brine with 0.1 kg of salt per litre flows in at 5 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 70 min?
V = 1000 Lm0 = 5 kgc = 0.1 kg/Lr = 5 L/min
m = kg
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−2 −1 −3; 0 1 3; 4 −1 4]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [3 2; 6 2]
λ =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (0, 1, 4) and b = (2, 5, −2).
a = (0, 1, 4)b = (2, 5, −2)
a·b =
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (2, 3, 2) and b = (3, 1, 0) and has a positive z-component. Find its x-component n̂x.
a = (2, 3, 2)b = (3, 1, 0)n̂z > 0
n̂x =
15.ProbabilityMedium
A bag holds 3 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
3 / 4
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: 60, 18, 29, 18, 17.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 40. What proportion P of values is above 420?
μ = 500σ = 40x = 420
P =
19.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 57.5 and 67.5 points? Give P as a decimal.
μ = 65σ = 10x1 = 57.5x2 = 67.5
P =
20.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 85% is to the left of it. Find the cut-off x.
μ = 500 hσ = 40 h85%
x = h
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).
Multiply by the conjugate
f(x) = −4x/(√(4x2 − 4x) + 2x)
Divide by x
f(x) = −4/(√(4 + (−4)/x) + 2)
Let x → ∞
L = −4/(2 + 2) = −1
Answer
L = −1
2.Limits
Answer: L = 0.5
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).
First application
(1·ex − 1)/(2x) → 0/0
Second application
1·ex/2
Substitute x = 0
L = 1·1/2 = 0.5
Answer
L = 0.5
3.Derivatives
Answer: f′(1) = 1.382
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′(1) = 0.8415 + 1·0.5403 = 1.382
Answer
f′(1) = 1.382
4.Derivatives
Answer: b = −12
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) = 15
f′(x) = 4x − 3 → f′(3) = 9
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 15 − 9·3 = −12
Answer
b = −12
5.Integrals
Answer: I = 0.9502
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) = e−x/(−1)
Evaluate at the limits
I = (e−3 − 1)/(−1) = (0.04979 − 1)/(−1) = 0.9502
Answer
I = 0.9502
6.Integrals
Answer: I = 12.78
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
7.389·(2) − 1·2 = 12.78
Answer
I = 12.78
7.Complex numbers
Answer: Im(w) = 0.1307
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| = 3 / 2 = 1.5
θ = 70° − 65° = 5°
Rectangular form
Re(w) = 1.5·cos(5°) = 1.494
Im(w) = 1.5·sin(5°) = 0.1307
Im(w) = 0.1307
Answer
Im(w) = 0.1307
8.Complex numbers
Answer: r = 3
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.
Check with the sum of the roots: z₁ + z₂ + r = −2 + 3 = 1 = −(−1).
9.Differential equations
Answer: y(0.5) = 1.025
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 + 3) = 0 → r = −2, −3
Constants from the initial values
A + B = 2, −2A − 3B = -2
A = 4, B = −2
Evaluate
y(0.5) = 4·e−1 + (−2)·e−1.5 = 1.025
Answer
y(0.5) = 1.025
10.Differential equations
Answer: m = 33.05 kg
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.
Set up the equation
dm/dt = 5·0.1 − (5/1000)·m = 0.5 − 0.005·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 100 + (−95)·e−t/200
Substitute t
m(70) = 100 + (−95)·0.7047 = 33.05 kg
Answer
m = 33.05 kg
11.Matrices & determinants
Answer: det A = −14
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 = 7, M₁₂ = −12, M₁₃ = −4
Cofactor expansion
det A = (−2)·7 − (−1)·(−12) + (−3)·(−4) = −14
Answer
det A = −14
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 − (5)λ + (−6) = 0
Solve
λ = (5 ± √(49))/2 → λ = −1, 6
λ = 6
Answer
λ = 6
13.Vectors: dot & cross product
Answer: a·b = −3
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 = 0·2 + 1·5 + 4·(−2) = 0 + 5 + (−8) = −3
Answer
a·b = −3
Negative: the angle between the vectors is more than 90°.
14.Vectors: dot & cross product
Answer: n̂x = 0.212
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 = 3·0 − 2·1 = −2
(a × b)y = azbx − axbz = 2·3 − 2·0 = 6
(a × b)z = axby − aybx = 2·1 − 3·3 = −7
Normalise
The z-component of a × b is negative, so use −(a × b).
|a × b| = √((−2)² + 6² + (−7)²) = 9.434
n̂ = −(−2, 6, −7)/9.434 = (0.212, −0.636, 0.742)
n̂x = 0.212
Answer
n̂x = 0.212
15.Probability
Answer: P = 0.1429
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.
First and second draw
P = 3/7 · 2/6 = 6/42 = 0.1429
Answer
P = 0.1429
16.Probability
Answer: σY = 2.644
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 = 1.95, Var(X) = 1.748, σX = 1.322
Linear change
σY = |2|·1.322 = 2.644
Answer
σY = 2.644
The mean does move: E(Y) = 2·1.95 + 4 = 7.9.
17.Descriptive statistics
Answer: s = 18.34
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) = 0.02275
P = 1 − 0.02275 = 0.9772
Answer
P = 0.9772
19.Normal distribution
Answer: P = 0.3721
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 = (57.5 − 65)/10 = −0.75
z₂ = (67.5 − 65)/10 = 0.25
Area between
P = Φ(0.25) − Φ(−0.75) = 0.5987 − 0.2266 = 0.3721
Answer
P = 0.3721
20.Normal distribution
Answer: x = 542 h
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.