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 − 2/x)x. Give a decimal value.
f(x) = (1 − 2/x)xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e0.5x − 1 − 0.5x)/x².
f(x) = (e0.5x − 1 − 0.5x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = (4x − 3)3. Find f′(0).
f(x) = (4x − 3)3x = 0
f′(0) =
4.DerivativesHard
Let f(x) = x3 − 7.5x2 + 12x − 2. Find the x-coordinate x of the local maximum.
f(x) = x3 − 7.5x2 + 12x − 2
x =
5.IntegralsMedium
Evaluate I = ∫01 e2x dx.
f(x) = e2x[0, 1]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = x.
y = x2y = x
A =
7.Complex numbersMedium
The diagram shows z = −4 − j4. With w = 2∠180°, click the point that represents z·w.
z = −4 − j4w = 2∠180°
1A (8 + j8)
2B (2 + j2)
3C (−8 − j8)
4D (4 + j4)
8.Complex numbersHard
Use De Moivre's theorem to find the imaginary part of z⁹ for z = −2 − j2.
z = −2 − j2n = 9
Im(z⁹) =
9.Differential equationsMedium
Solve y″ + 7y′ + 10y = 0 with y(0) = 3 and y′(0) = 0. Find y(0.5).
y″ + 7y′ + 10y = 0y(0) = 3y′(0) = 0
y(0.5) =
10.Differential equationsHard
Solve the initial value problem y′ + 2y/x = 3x for x > 0, with y(1) = 3. Find y(3).
y′ + 2y/x = 3xy(1) = 3
y(3) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−3 1 −4; −4 −1 4; 1 5 1]
det A =
12.Matrices & determinantsHard
Solve the system x − 5y = 10 and −3x + y = 9. Find x.
x − 5y = 10−3x + y = 9
x =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (−6, 0, −2) and b = (−4, −4, 2).
a = (−6, 0, −2)b = (−4, −4, 2)
a·b =
14.Vectors: dot & cross productHard
A force F = (−40, −50, 50) N acts at point A(4, 2, −3). Coordinates are in metres. Find the x-component Mx of the moment of the force about the origin O(0, 0, 0).
F = (−40, −50, 50) NA(4, 2, −3) mO(0, 0, 0) m
Mx = N·m
15.ProbabilityMedium
In how many ways N can a committee of 5 people be chosen from a group of 11? The order does not matter.
n = 11k = 5
N =
16.ProbabilityHard
X has the distribution shown, and Y = 2X + 10. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 36, 48, 47, 11, 26, 48.
n = 6
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 60 and standard deviation 15. Find the z-score z of a score of 37.5.
μ = 60σ = 15x = 37.5
z =
19.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0329. Find z to two decimals.
A = 0.0329
z =
20.Normal distributionHard
Hole diameters D are normal with mean 25.05 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 = 25.05 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
3Solutions
1.Limits
Answer: L = 0.1353
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 − 2/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 1·(−2) = −2
Undo the logarithm
L = e−2 = 0.1353
Answer
L = 0.1353
2.Limits
Answer: L = 0.125
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
(0.5·e0.5x − 0.5)/(2x) → 0/0
Second application
0.25·e0.5x/2
Substitute x = 0
L = 0.25·1/2 = 0.125
Answer
L = 0.125
3.Derivatives
Answer: f′(0) = 108
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) = 3·(4x − 3)2·4
Substitute
f′(0) = 3·(−3)2·4 = 108
Answer
f′(0) = 108
4.Derivatives
Answer: x = 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.
Derivative
f′(x) = 3x2 − 15x + 12 = 3·(x2 − 5x + 4)
Solve f′(x) = 0
x = 1 or x = 4
Which one is which
f″(x) = 6x − 15: negative at a maximum, positive at a minimum.
f″(1) = −9, f″(4) = 9
x = 1
Answer
x = 1
5.Integrals
Answer: I = 3.195
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) = e2x/2
Evaluate at the limits
I = (e2 − 1)/2 = (7.389 − 1)/2 = 3.195
Answer
I = 3.195
6.Integrals
Answer: A = 0.1667
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.
Intersection points
x2 = x → x = 0, x = 1
Integrate upper minus lower
A = [0.5x2 − 0.3333x3]01 = 0.1667
Answer
A = 0.1667
7.Complex numbers
Answer: A (8 + j8)
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| = 2·|z|, arg(z·w) = arg z + 180°
The point
z·w = 8 + j8 → A
Answer
A (8 + j8)
8.Complex numbers
Answer: Im(z⁹) = −8192
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.
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 = 3, −2A − 5B = 0
A = 5, B = −2
Evaluate
y(0.5) = 5·e−1 + (−2)·e−2.5 = 1.675
Answer
y(0.5) = 1.675
10.Differential equations
Answer: y(3) = 7
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∫2/x dx = x2
Integrate
(x2y)′ = 3x3 → x2y = 3x4/4 + C
Initial condition
3 = 3/4 + C → C = 2.25
Evaluate
y(3) = (3·34/4 + 2.25)/32 = 7
Answer
y(3) = 7
11.Matrices & determinants
Answer: det A = 147
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 = −21, M₁₂ = −8, M₁₃ = −19
Cofactor expansion
det A = (−3)·(−21) − 1·(−8) + (−4)·(−19) = 147
Answer
det A = 147
12.Matrices & determinants
Answer: x = −3.929
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 = 1·1 − (−5)·(−3) = −14
Replace the x column
det Aₓ = 10·1 − (−5)·9 = 55
x = 55/(−14) = −3.929
Answer
x = −3.929
y = −2.786
13.Vectors: dot & cross product
Answer: a·b = 20
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.
Positive: the angle between the vectors is less than 90°.
14.Vectors: dot & cross product
Answer: Mx = −50 N·m
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.
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(11, 5) = (11·10·9·8·7)/(5·4·3·2·1) = 462
Answer
N = 462
16.Probability
Answer: σY = 2.027
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 = 1.15, Var(X) = 1.028, σX = 1.014
Linear change
σY = |2|·1.014 = 2.027
Answer
σY = 2.027
The mean does move: E(Y) = 2·1.15 + 10 = 12.3.
17.Descriptive statistics
Answer: s = 15.06
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 = (37.5 − 60)/15 = −1.5
Answer
z = −1.5
19.Normal distribution
Answer: z = 1.84
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 − 0.0329 = 0.9671
Read the z-table backwards
Find 0.9671 inside the table and read off z.
z = 1.84
Answer
z = 1.84
20.Normal distribution
Answer: P = 0.9909
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).