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 + x − 7)/(3x2 + 2x + 8).
f(x) = (−4x2 + x − 7)/(3x2 + 2x + 8)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 2x)/sin(5x), with x in radians.
f(x) = ln(1 + 2x)/sin(5x)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(x2 + 1). Find f′(−1).
f(x) = ln(x2 + 1)x = −1
f′(−1) =
4.DerivativesHard
The tangent line to f(x) = x2 − 3x + 6 at x = −1 crosses the y-axis at y = b. Find b.
f(x) = x2 − 3x + 6x = −1
b =
5.IntegralsMedium
Evaluate I = ∫01 2x·(x² + 1)1 dx.
f(x) = 2x·(x² + 1)1[0, 1]
I =
6.IntegralsHard
The region between the line y = 2x and the parabola y = 2x² is rotated about the x-axis. Find the volume V of the solid.
y = 2xy = 2x²
V =
7.Complex numbersMedium
The number z = 4·ej2.8 has its angle in radians. Find its imaginary part b.
r = 4θ = 2.8 rad
b =
8.Complex numbersHard
One root of z³ + 3z² + 9z − 13 = 0 is z₁ = −2 + j3. The coefficients are real. Find the real root r.
z₁ = −2 + j3
r =
9.Differential equationsMedium
Solve y″ + 7y′ + 12y = 0 with y(0) = 4 and y′(0) = 0. Find y(1).
y″ + 7y′ + 12y = 0y(0) = 4y′(0) = 0
y(1) =
10.Differential equationsHard
A system obeys y″ + y′ + 16.25y = 162.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + y′ + 16.25y = 162.5y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).
A = [3 0; 2 6]B = [3 −1; −1 4]
c12 =
12.Matrices & determinantsHard
Solve the system 6x − 2y = 7 and −3x + 5y = −10. Find x.
6x − 2y = 7−3x + 5y = −10
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (2, −4, 3) and b = (4, k, −1) are perpendicular.
a = (2, −4, 3)b = (4, k, −1)
k =
14.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (3, 5, 0) and b = (−2, −6, 3).
a = (3, 5, 0)b = (−2, −6, 3)
A =
15.ProbabilityMedium
A bag holds 6 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
6 / 7
P =
16.ProbabilityHard
6 components are tested, and each one fails with probability 0.4, independently. What is the probability P that at most one fails?
n = 6p = 0.4
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 33, 47, 29, 56, 45, 56, 49.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 15. What proportion P of values is above 109?
μ = 100σ = 15x = 109
P =
19.Normal distributionMedium
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 164 and 188.5 cm? Give P as a decimal.
μ = 178σ = 7x1 = 164x2 = 188.5
P =
20.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 20% of them lie (the 20th percentile).
μ = 65σ = 10P = 20%
x = points
3Solutions
1.Limits
Answer: L = −1.333
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
(−4 + 1/x + (−7)/x2)/(3 + 2/x + 8/x2)
Let x → ∞
L = −4/3 = −1.333
Answer
L = −1.333
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 0.4
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 [2/(1 + 2x)]/[5·cos 5x]
Substitute x = 0
L = 2/5 = 0.4
Answer
L = 0.4
3.Derivatives
Answer: f′(−1) = −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.
Chain rule for the logarithm
f′(x) = 2x/(x2 + 1)
Substitute
f′(−1) = −2/2 = −1
Answer
f′(−1) = −1
4.Derivatives
Answer: b = 5
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) = 10
f′(x) = 2x − 3 → f′(−1) = −5
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 10 − (−5)·(−1) = 5
Answer
b = 5
5.Integrals
Answer: I = 1.5
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.
Substitution
u = x² + 1, du = 2x dx
I = ∫12 u1 du
Integrate and evaluate
I = [u2/2]12 = (4 − 1)/2 = 1.5
Answer
I = 1.5
6.Integrals
Answer: V = 1.676
Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.
Intersection points
2x = 2x² → x = 0, x = 1
Washer method
V = π∫₀1 ((2x)² − (2x²)²) dx
Integrate
V = π(4·1³/3 − 4·1⁵/5) = π·0.5333 = 1.676
Answer
V = 1.676
7.Complex numbers
Answer: b = 1.34
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.
Euler's formula
z = 4·(cos 2.8 + j sin 2.8)
Components (radians)
a = 4·cos 2.8 = −3.769
b = 4·sin 2.8 = 1.34
b = 1.34
Answer
b = 1.34
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 = −4 + 1 = −3 = −(3).
9.Differential equations
Answer: y(1) = 0.5768
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 = 4, −3A − 4B = 0
A = 16, B = −12
Evaluate
y(1) = 16·e−3 + (−12)·e−4 = 0.5768
Answer
y(1) = 0.5768
10.Differential equations
Answer: ymax = 16.75
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.
Steady value and roots
yss = 10, r = −0.5 ± 4j
Constants from the initial values
y(0) = 10 + A = 0 → A = −10
y′(0) = −0.5A + 4B = 0 → B = −1.25
Peak
t = π/4 = 0.7854
ymax = 10·(1 + e−0.5·π/4) = 16.75
Answer
ymax = 16.75
11.Matrices & determinants
Answer: c12 = −3
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
c12 = 3·(−1) + 0·4 = −3
Answer
c12 = −3
12.Matrices & determinants
Answer: x = 0.625
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 = 6·5 − (−2)·(−3) = 24
Replace the x column
det Aₓ = 7·5 − (−2)·(−10) = 15
x = 15/24 = 0.625
Answer
x = 0.625
y = −1.625
13.Vectors: dot & cross product
Answer: k = 1.25
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 = 2·4 + (−4)·k + 3·(−1) = 0
5 + (−4)·k = 0
Solve for k
k = −5/(−4) = 1.25
Answer
k = 1.25
14.Vectors: dot & cross product
Answer: A = 19.24
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 = 5·3 − 0·(−6) = 15
(a × b)y = azbx − axbz = 0·(−2) − 3·3 = −9
(a × b)z = axby − aybx = 3·(−6) − 5·(−2) = −8
Area
A = |a × b| = √(15² + (−9)² + (−8)²) = 19.24
Answer
A = 19.24
15.Probability
Answer: P = 0.1923
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 = 6/13 · 5/12 = 30/156 = 0.1923
Answer
P = 0.1923
16.Probability
Answer: P = 0.2333
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) = 0.66 = 0.04666
P(X = 1) = 6·0.4·0.65 = 0.1866
Add them
P = 0.04666 + 0.1866 = 0.2333
Answer
P = 0.2333
17.Descriptive statistics
Answer: s = 10.5
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 = 0.6
Standard normal table
Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743
Answer
P = 0.2743
19.Normal distribution
Answer: P = 0.9104
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 = (164 − 178)/7 = −2
z₂ = (188.5 − 178)/7 = 1.5
Area between
P = Φ(1.5) − Φ(−2) = 0.9332 − 0.02275 = 0.9104
Answer
P = 0.9104
20.Normal distribution
Answer: x = 56.58 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.