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→∞ (√(9x2 + 3x) − 3x).
f(x) = √(9x2 + 3x) − 3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 5x)/sin(3x), with x in radians.
f(x) = ln(1 + 5x)/sin(3x)x → 0
L =
3.DerivativesMedium
Let f(x) = x2·e−0.5x. Find f′(2).
f(x) = x2·e−0.5xx = 2
f′(2) =
4.DerivativesHard
The tangent line to f(x) = x2 − 4x − 2 at x = −1 crosses the y-axis at y = b. Find b.
f(x) = x2 − 4x − 2x = −1
b =
5.IntegralsMedium
Evaluate I = ∫26 3/x dx.
f(x) = 3/x[2, 6]
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
Write z = 5∠160° in rectangular form a + jb and give the real part a.
r = 5θ = 160°
a =
8.Complex numbersHard
One root of z³ + 7z² + 24z − 32 = 0 is z₁ = −4 + j4. The coefficients are real. Find the real root r.
z₁ = −4 + j4
r =
9.Differential equationsMedium
Solve y″ + 6y′ + 8y = 0 with y(0) = 3 and y′(0) = 1. Find y(0.5).
y″ + 6y′ + 8y = 0y(0) = 3y′(0) = 1
y(0.5) =
10.Differential equationsHard
A system obeys y″ + y′ + 25.25y = 252.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.5); angles in radians.
y″ + y′ + 25.25y = 252.5y(0) = 0y′(0) = 0
y(0.5) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [0 0 2; −4 1 −3; 5 2 5]
det A =
12.Matrices & determinantsHard
Solve the system 3x + 2y = 0 and 5x − 6y = 4. Find x.
3x + 2y = 05x − 6y = 4
x =
13.Vectors: dot & cross productMedium
A constant force F = (−10, 10, −25) N acts on an object while it moves in a straight line from point A(−3, 2, 2) to point B(3, 0, 5). Coordinates are in metres. Find the work W done by the force.
F = (−10, 10, −25) NA(−3, 2, 2) mB(3, 0, 5) m
W = J
14.Vectors: dot & cross productHard
For a = (−6, 6, −3) and b = (2, 1, 2), find the y-component (a × b)y of the cross product a × b.
a = (−6, 6, −3)b = (2, 1, 2)
(a × b)y =
15.ProbabilityMedium
In how many ways N can a committee of 2 people be chosen from a group of 6? The order does not matter.
n = 6k = 2
N =
16.ProbabilityHard
X has the distribution shown, and Y = 0.5X + 4. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 34, 45, 52, 19, 50, 51, 28.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 15. What proportion P of values is above 38?
μ = 50σ = 15x = 38
P =
19.Normal distributionMedium
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 29 and 33.5 min? Give P as a decimal.
μ = 32σ = 6x1 = 29x2 = 33.5
P =
20.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 10% of them lie (the 10th percentile).
μ = 100σ = 15P = 10%
x = points
3Solutions
1.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).
Multiply by the conjugate
f(x) = 3x/(√(9x2 + 3x) + 3x)
Divide by x
f(x) = 3/(√(9 + 3/x) + 3)
Let x → ∞
L = 3/(3 + 3) = 0.5
Answer
L = 0.5
2.Limits
Answer: L = 1.667
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)]/[3·cos 3x]
Substitute x = 0
L = 5/3 = 1.667
Answer
L = 1.667
3.Derivatives
Answer: f′(2) = 0.7358
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′ = e−0.5x·(2x + (−0.5)·x2)
Substitute
f′(2) = 0.3679·(4 + (−2)) = 0.7358
Answer
f′(2) = 0.7358
4.Derivatives
Answer: b = −3
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) = 3
f′(x) = 2x − 4 → f′(−1) = −6
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 3 − (−6)·(−1) = −3
Answer
b = −3
5.Integrals
Answer: I = 3.296
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) = 3·ln x
Evaluate at the limits
I = 3·(ln 6 − ln 2) = 3·ln 3 = 3.296
Answer
I = 3.296
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: a = −4.698
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.
Components
a = 5·cos(160°) = −4.698
b = 5·sin(160°) = 1.71
Rectangular form
z = −4.698 + j1.71
a = −4.698
Answer
a = −4.698
8.Complex numbers
Answer: r = 1
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 = −8 + 1 = −7 = −(7).
9.Differential equations
Answer: y(0.5) = 1.918
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 + 4) = 0 → r = −2, −4
Constants from the initial values
A + B = 3, −2A − 4B = 1
A = 6.5, B = −3.5
Evaluate
y(0.5) = 6.5·e−1 + (−3.5)·e−2 = 1.918
Answer
y(0.5) = 1.918
10.Differential equations
Answer: y(0.5) = 15.77
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.
2×2 minors of the first row
M₁₁ = 11, M₁₂ = −5, M₁₃ = −13
Cofactor expansion
det A = 0·11 − 0·(−5) + 2·(−13) = −26
Answer
det A = −26
12.Matrices & determinants
Answer: x = 0.2857
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·(−6) − 2·5 = −28
Replace the x column
det Aₓ = 0·(−6) − 2·4 = −8
x = −8/(−28) = 0.2857
Answer
x = 0.2857
y = −0.4286
13.Vectors: dot & cross product
Answer: W = −155 J
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.
Displacement
d = B − A = (3, 0, 5) − (−3, 2, 2) = (6, −2, 3) m
Work
W = F·d = (−10)·6 + 10·(−2) + (−25)·3 = −155 J
Answer
W = −155 J
Negative work: the force acts mostly against the motion.
14.Vectors: dot & cross product
Answer: (a × b)y = 6
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 = 6·2 − (−3)·1 = 15
(a × b)y = azbx − axbz = (−3)·2 − (−6)·2 = 6
(a × b)z = axby − aybx = (−6)·1 − 6·2 = −18
Requested component
(a × b)y = 6
Answer
(a × b)y = 6
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: N = 15
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(6, 2) = (6·5)/(2·1) = 15
Answer
N = 15
16.Probability
Answer: σY = 0.7416
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, Var(X) = 2.2, σX = 1.483
Linear change
σY = |0.5|·1.483 = 0.7416
Answer
σY = 0.7416
The mean does move: E(Y) = 0.5·2 + 4 = 5.
17.Descriptive statistics
Answer: s = 12.98
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 = −0.8
Standard normal table
Φ(−0.8) = 0.2119
P = 1 − 0.2119 = 0.7881
Answer
P = 0.7881
19.Normal distribution
Answer: P = 0.2902
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 = (29 − 32)/6 = −0.5
z₂ = (33.5 − 32)/6 = 0.25
Area between
P = Φ(0.25) − Φ(−0.5) = 0.5987 − 0.3085 = 0.2902
Answer
P = 0.2902
20.Normal distribution
Answer: x = 80.78 points
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.