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 + 3/x)x. Give a decimal value.
f(x) = (1 + 3/x)xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 5x)/sin(x), with x in radians.
f(x) = ln(1 + 5x)/sin(x)x → 0
L =
3.DerivativesMedium
Let f(x) = x2·ex. Find f′(2).
f(x) = x2·exx = 2
f′(2) =
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 = ∫₁4 x·ln x dx.
∫₁4 x ln x dx
I =
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
The diagram shows z = −1 − j4. With w = 0.5∠−90°, click the point that represents z/w.
z = −1 − j4w = 0.5∠−90°
1A (−8 + j2)
2B (−2 − j8)
3C (4 − j)
4D (8 − j2)
8.Complex numbersHard
Find the real part of (−3 − j)⁵.
z = −3 − jn = 5
Re(z⁵) =
9.Differential equationsMedium
Solve y″ + 7y′ + 12y = 0 with y(0) = 1 and y′(0) = 0. Find y(1.5).
y″ + 7y′ + 12y = 0y(0) = 1y′(0) = 0
y(1.5) =
10.Differential equationsHard
Solve the initial value problem y′ + y/x = 6x for x > 0, with y(1) = 3. Find y(3).
y′ + y/x = 6xy(1) = 3
y(3) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [4 3 4; 1 −3 −5; 2 1 −3]
det A =
12.Matrices & determinantsHard
Solve the system −2x + 5y = 7 and 6x − 3y = 9. Find x.
−2x + 5y = 76x − 3y = 9
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (2, 5, −5) and b = (6, k, −1) are perpendicular.
a = (2, 5, −5)b = (6, k, −1)
k =
14.Vectors: dot & cross productHard
For a = (3, −3, 4) and b = (−2, 5, −1), find the x-component (a × b)x of the cross product a × b.
a = (3, −3, 4)b = (−2, 5, −1)
(a × b)x =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 10, 20, 30, 40p(x): 0.35, 0.15, 0.25, 0.25
Var(X) =
16.ProbabilityHard
9 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?
n = 9p = 0.1
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 57, 51, 53, 21, 57.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 10. What proportion P of values is below 62.5?
μ = 50σ = 10x = 62.5
P =
19.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 100 and 130.75 points? Give P as a decimal.
μ = 100σ = 15x1 = 100x2 = 130.75
P =
20.Normal distributionHard
Hole diameters D are normal with mean 25.03 mm and standard deviation 0.01 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 = 25.03 mm, σ = 0.01 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→∞ x·ln(1 + 3/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 1·3 = 3
Undo the logarithm
L = e3 = 20.09
Answer
L = 20.09
2.Limits
Answer: L = 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 [5/(1 + 5x)]/[cos x]
Substitute x = 0
L = 5/1 = 5
Answer
L = 5
3.Derivatives
Answer: f′(2) = 59.11
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 rule
f′(x) = u′v + uv′ = ex·(2x + 1·x2)
Substitute
f′(2) = 7.389·(4 + 4) = 59.11
Answer
f′(2) = 59.11
4.Derivatives
Answer: b = −24
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) = 21
f′(x) = 4x + 3 → f′(3) = 15
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 21 − 15·3 = −24
Answer
b = −24
5.Integrals
Answer: I = 7.34
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.
Choose u and dv
u = ln x, du = dx/x; dv = x dx, v = x²/2
Antiderivative
∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4
Limits
(8·ln 4 − 4) − (0 − 0.25) = 7.34
Answer
I = 7.34
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: D (8 − j2)
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.
Effect of w
|z/w| = |z|/0.5, arg(z/w) = arg z − (−90)°
The point
z/w = 8 − j2 → D
Answer
D (8 − j2)
8.Complex numbers
Answer: Re(z⁵) = 12
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.
Powers step by step
z² = (−3 − j)(−3 − j) = 8 + j6
z³ = (8 + j6)(−3 − j) = −18 − j26
z⁴ = (−18 − j26)(−3 − j) = 28 + j96
z⁵ = (28 + j96)(−3 − j) = 12 − j316
Check with De Moivre
z = 3.162∠−161.6° → z⁵ = 316.2∠−87.83°
Re(z⁵) = 12
Answer
Re(z⁵) = 12
9.Differential equations
Answer: y(1.5) = 0.037
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 = 1, −3A − 4B = 0
A = 4, B = −3
Evaluate
y(1.5) = 4·e−4.5 + (−3)·e−6 = 0.037
Answer
y(1.5) = 0.037
10.Differential equations
Answer: y(3) = 18.33
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.
Integrating factor
μ = e∫1/x dx = x1
Integrate
(x1y)′ = 6x2 → x1y = 6x3/3 + C
Initial condition
3 = 6/3 + C → C = 1
Evaluate
y(3) = (6·33/3 + 1)/31 = 18.33
Answer
y(3) = 18.33
11.Matrices & determinants
Answer: det A = 63
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 = 14, M₁₂ = 7, M₁₃ = 7
Cofactor expansion
det A = 4·14 − 3·7 + 4·7 = 63
Answer
det A = 63
12.Matrices & determinants
Answer: x = 2.75
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 = (−2)·(−3) − 5·6 = −24
Replace the x column
det Aₓ = 7·(−3) − 5·9 = −66
x = −66/(−24) = 2.75
Answer
x = 2.75
y = 2.5
13.Vectors: dot & cross product
Answer: k = −3.4
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 = 2·6 + 5·k + (−5)·(−1) = 0
17 + 5·k = 0
Solve for k
k = −17/5 = −3.4
Answer
k = −3.4
14.Vectors: dot & cross product
Answer: (a × b)x = −17
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 = (−3)·(−1) − 4·5 = −17
(a × b)y = azbx − axbz = 4·(−2) − 3·(−1) = −5
(a × b)z = axby − aybx = 3·5 − (−3)·(−2) = 9
Requested component
(a × b)x = −17
Answer
(a × b)x = −17
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: Var(X) = 144
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.
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.
Zero and one failure
P(X = 0) = 0.99 = 0.3874
P(X = 1) = 9·0.1·0.98 = 0.3874
Add them
P = 0.3874 + 0.3874 = 0.7748
Answer
P = 0.7748
17.Descriptive statistics
Answer: s = 15.21
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 = 1.25
Standard normal table
Φ(1.25) = 0.8944
P = 0.8944
Answer
P = 0.8944
19.Normal distribution
Answer: P = 0.4798
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 = (100 − 100)/15 = 0
z₂ = (130.75 − 100)/15 = 2.05
Area between
P = Φ(2.05) − Φ(0) = 0.9798 − 0.5 = 0.4798
Answer
P = 0.4798
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).