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→∞ (√(x2 + 7x) − x).
f(x) = √(x2 + 7x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 2x)/(4x²), with x in radians.
f(x) = (1 − cos 2x)/(4x²)x → 0
L =
3.DerivativesMedium
Let f(x) = (4x + 1)4. Find f′(1).
f(x) = (4x + 1)4x = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 + 3x2 − 9x + 1. Find the x-coordinate x of the local maximum.
f(x) = x3 + 3x2 − 9x + 1
x =
5.IntegralsMedium
The region under y = √(2x) from x = 0 to x = 2 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).
y = √(2x)0 ≤ x ≤ 2
V =
6.IntegralsHard
The region between the line y = 2x and the parabola y = 0.5x² is rotated about the x-axis. Find the volume V of the solid.
y = 2xy = 0.5x²
V =
7.Complex numbersMedium
Find the argument θ of z = −7 − j9 in degrees, as the principal value between −180° and 180°.
z = −7 − j9
θ = °
8.Complex numbersHard
One root of z³ + 5z² + 24z + 20 = 0 is z₁ = −2 + j4. The coefficients are real. Find the real root r.
z₁ = −2 + j4
r =
9.Differential equationsMedium
A can of soda at 6 °C is taken out of the fridge into a room at 26 °C. After 13 min its temperature is 20 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 23 °C?
T0 = 6 °CTa = 26 °CT(13) = 20 °CT = 23 °C
t = min
10.Differential equationsHard
A tank holds 100 L of water in which 5 kg of salt is dissolved. Brine with 0.25 kg of salt per litre flows in at 2 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 11 kg of salt?
V = 100 Lm0 = 5 kgc = 0.25 kg/Lr = 2 L/min
t = min
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [4 −3 −5; 1 2 −2; 5 −4 2]
det A =
12.Matrices & determinantsHard
Solve the system 6x + 6y = −4 and −4x − 6y = 7. Find x.
6x + 6y = −4−4x − 6y = 7
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (4, −4, 3) and b = (−5, 1, k) are perpendicular.
a = (4, −4, 3)b = (−5, 1, k)
k =
14.Vectors: dot & cross productHard
For a = (0, 1, 6) and b = (1, −1, 6), find the x-component (a × b)x of the cross product a × b.
a = (0, 1, 6)b = (1, −1, 6)
(a × b)x =
15.ProbabilityMedium
A fair die is rolled 4 times. What is the probability P of getting at least one six?
n = 4
P =
16.ProbabilityHard
X has the distribution shown, and Y = 10X + 5. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 22, 36, 17, 43, 57, 40.
n = 6
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 12. Find the z-score z of a score of 94.
μ = 100σ = 12x = 94
z =
19.Normal distributionMedium
The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 9 of them exceeds 111.6 kg. Round z to two decimals.
μ = 12 kgσ = 0.8 kgn = 9> 111.6 kg
P =
20.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 25% of them lie (the 25th percentile).
μ = 32σ = 6P = 25%
x = min
3Solutions
1.Limits
Answer: L = 3.5
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) = 7x/(√(x2 + 7x) + x)
Divide by x
f(x) = 7/(√(1 + 7/x) + 1)
Let x → ∞
L = 7/(1 + 1) = 3.5
Answer
L = 3.5
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
2·sin 2x/(8x) → 0/0
Second application
4·cos 2x/8
Substitute x = 0
L = 4/8 = 0.5
Answer
L = 0.5
3.Derivatives
Answer: f′(1) = 2000
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) = 4·(4x + 1)3·4
Substitute
f′(1) = 4·53·4 = 2000
Answer
f′(1) = 2000
4.Derivatives
Answer: x = −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.
Derivative
f′(x) = 3x2 + 6x − 9 = 3·(x2 + 2x − 3)
Solve f′(x) = 0
x = −3 or x = 1
Which one is which
f″(x) = 6x + 6: negative at a maximum, positive at a minimum.
f″(−3) = −12, f″(1) = 12
x = −3
Answer
x = −3
5.Integrals
Answer: V = 12.57
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.
Disc method
V = π∫₀2 2x dx
Integrate
V = π·2·2²/2 = 12.57
Answer
V = 12.57
6.Integrals
Answer: V = 107.2
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 = 0.5x² → x = 0, x = 4
Washer method
V = π∫₀4 ((2x)² − (0.5x²)²) dx
Integrate
V = π(4·4³/3 − 0.25·4⁵/5) = π·34.13 = 107.2
Answer
V = 107.2
7.Complex numbers
Answer: θ = −127.9 °
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.
Quadrant
a = −7, b = −9 → III
Reference angle
α = atan(9/7) = 52.13°
Argument
θ = −(180° − α) = −127.9°
Answer
θ = −127.9°
In polar form z = 11.4∠−127.9°.
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) = −5 = −(5).
9.Differential equations
Answer: t = 20.48 min
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.
Cooling constant
k = ln(-20/-6)/13 = 0.09261 min⁻¹
Time to reach the target
t = ln(-20/(−3))/0.09261 = 20.48 min
Answer
t = 20.48 min
10.Differential equations
Answer: t = 17.83 min
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.
Result
m(t) = 25 + (−20)·e−t/50
Solve for t
e−t/50 = (25 − 11)/(25 − 5) = 0.7
t = 50·ln(1.429) = 17.83 min
Answer
t = 17.83 min
11.Matrices & determinants
Answer: det A = 90
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 = −4, M₁₂ = 12, M₁₃ = −14
Cofactor expansion
det A = 4·(−4) − (−3)·12 + (−5)·(−14) = 90
Answer
det A = 90
12.Matrices & determinants
Answer: x = 1.5
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·(−6) − 6·(−4) = −12
Replace the x column
det Aₓ = (−4)·(−6) − 6·7 = −18
x = −18/(−12) = 1.5
Answer
x = 1.5
y = −2.167
13.Vectors: dot & cross product
Answer: k = 8
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 = 4·(−5) + (−4)·1 + 3·k = 0
−24 + 3·k = 0
Solve for k
k = −(−24)/3 = 8
Answer
k = 8
14.Vectors: dot & cross product
Answer: (a × b)x = 12
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 = 1·6 − 6·(−1) = 12
(a × b)y = azbx − axbz = 6·1 − 0·6 = 6
(a × b)z = axby − aybx = 0·(−1) − 1·1 = −1
Requested component
(a × b)x = 12
Answer
(a × b)x = 12
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: P = 0.5177
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.
Complement
P(no six) = (5/6)4 = 0.4823
P = 1 − 0.4823 = 0.5177
Answer
P = 0.5177
16.Probability
Answer: σY = 10.62
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 = 2.15, Var(X) = 1.128, σX = 1.062
Linear change
σY = |10|·1.062 = 10.62
Answer
σY = 10.62
The mean does move: E(Y) = 10·2.15 + 5 = 26.5.
17.Descriptive statistics
Answer: s = 14.58
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 = (94 − 100)/12 = −0.5
Answer
z = −0.5
19.Normal distribution
Answer: P = 0.06681
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).
Distribution of the total
μT = 108 kg, σT = 0.8·√9 = 2.4 kg
Standardise
z = (111.6 − 108)/2.4 = 1.5
Probability
P = 1 − Φ(1.5) = 0.06681
Answer
P = 0.06681
20.Normal distribution
Answer: x = 27.95 min
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.