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 − 4x) − x).
f(x) = √(x2 − 4x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 2x)/sin(3x), with x in radians.
f(x) = ln(1 + 2x)/sin(3x)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(x2 + 5). Find f′(1).
f(x) = ln(x2 + 5)x = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 + 10.5x2 + 36x − 6. Find the x-coordinate x of the local minimum.
f(x) = x3 + 10.5x2 + 36x − 6
x =
5.IntegralsMedium
Evaluate I = ∫₁2.5 x·ln x dx.
∫₁2.5 x ln x dx
I =
6.IntegralsHard
Evaluate I = ∫₀3 x²·eˣ dx.
∫₀3 x²eˣ dx
I =
7.Complex numbersMedium
Write z = 4∠110° in rectangular form a + jb and give the imaginary part b.
r = 4θ = 110°
b =
8.Complex numbersHard
The roots of zⁿ = w with n = 3 and w = 64∠90° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2. Find the imaginary part of z₀.
w = 64∠90°n = 3k = 0
Im(z₀) =
9.Differential equationsMedium
A can of soda at 6 °C is taken out of the fridge into a room at 23 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.07 min⁻¹, find its temperature T after 26 min.
T0 = 6 °CTa = 23 °Ck = 0.07 min⁻¹t = 26 min
T = °C
10.Differential equationsHard
Solve the initial value problem y′ + y/x = 5x2 for x > 0, with y(1) = 0. Find y(3).
y′ + y/x = 5x2y(1) = 0
y(3) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [5 −1; 4 −1]B = [0 −2; −2 −6]
c21 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [0 2; −1 3]
λ =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (−3, −3, −6) and b = (−6, −5, k) are perpendicular.
a = (−3, −3, −6)b = (−6, −5, k)
k =
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(−2, 3, 4), B(−3, 3, 1) and C(4, 3, 2). Coordinates are in metres.
A(−2, 3, 4)B(−3, 3, 1)C(4, 3, 2)
A = m²
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4p(x): 0.2, 0.2, 0.3, 0.3
σ =
16.ProbabilityHard
A game costs €5 to play. You win €10 with probability 0.2 and €250 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €5€10: p = 0.2€250: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 39, 51, 27, 12, 20, 28, 31.
n = 7
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 8. Find the z-score z of a score of 93.6.
μ = 100σ = 8x = 93.6
z =
19.Normal distributionMedium
The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 9 of them exceeds 358.2 g. Round z to two decimals.
μ = 40 gσ = 1.2 gn = 9> 358.2 g
P =
20.Normal distributionHard
Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
3Solutions
1.Limits
Answer: L = −2
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) = −4x/(√(x2 − 4x) + x)
Divide by x
f(x) = −4/(√(1 + (−4)/x) + 1)
Let x → ∞
L = −4/(1 + 1) = −2
Answer
L = −2
2.Limits
Answer: L = 0.6667
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)]/[3·cos 3x]
Substitute x = 0
L = 2/3 = 0.6667
Answer
L = 0.6667
3.Derivatives
Answer: f′(1) = 0.3333
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 + 5)
Substitute
f′(1) = 2/6 = 0.3333
Answer
f′(1) = 0.3333
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 + 21x + 36 = 3·(x2 + 7x + 12)
Solve f′(x) = 0
x = −4 or x = −3
Which one is which
f″(x) = 6x + 21: negative at a maximum, positive at a minimum.
f″(−4) = −3, f″(−3) = 3
x = −3
Answer
x = −3
5.Integrals
Answer: I = 1.551
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
(3.125·ln 2.5 − 1.5625) − (0 − 0.25) = 1.551
Answer
I = 1.551
6.Integrals
Answer: I = 98.43
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.
First round
∫x²eˣ dx = x²eˣ − ∫2x eˣ dx
Second round
∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)
Limits
20.09·(5) − 1·2 = 98.43
Answer
I = 98.43
7.Complex numbers
Answer: b = 3.759
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 = 4·cos(110°) = −1.368
b = 4·sin(110°) = 3.759
Rectangular form
z = −1.368 + j3.759
b = 3.759
Answer
b = 3.759
8.Complex numbers
Answer: Im(z₀) = 2
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.
Modulus of the roots
641/3 = 4
Argument of the root
θ₀ = (90° + 360°·0)/3 = 30°
Rectangular form
z₀ = 4·(cos 30° + j sin 30°) = 3.464 + j2
Im(z₀) = 2
Answer
Im(z₀) = 2
The 3 roots lie on the circle of radius 4, 120° apart.
9.Differential equations
Answer: T = 20.25 °C
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.
Solution of the cooling law
T(t) = 23 + (−17)·e−0.07t
Substitute t
T(26) = 23 + (−17)·0.162 = 20.25 °C
Answer
T = 20.25 °C
10.Differential equations
Answer: y(3) = 33.33
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)′ = 5x3 → x1y = 5x4/4 + C
Initial condition
0 = 5/4 + C → C = −1.25
Evaluate
y(3) = (5·34/4 + (−1.25))/31 = 33.33
Answer
y(3) = 33.33
11.Matrices & determinants
Answer: c21 = 2
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
c21 = 4·0 + (−1)·(−2) = 2
Answer
c21 = 2
12.Matrices & determinants
Answer: λ = 2
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.
Characteristic equation
λ² − (3)λ + (2) = 0
Solve
λ = (3 ± √(1))/2 → λ = 1, 2
λ = 2
Answer
λ = 2
13.Vectors: dot & cross product
Answer: k = 5.5
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 = (−3)·(−6) + (−3)·(−5) + (−6)·k = 0
33 + (−6)·k = 0
Solve for k
k = −33/(−6) = 5.5
Answer
k = 5.5
14.Vectors: dot & cross product
Answer: A = 10 m²
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.
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.
Mean
μ = 1·0.2 + 2·0.2 + 3·0.3 + 4·0.3 = 2.7
Mean of the squares
E(X²) = 1·0.2 + 4·0.2 + 9·0.3 + 16·0.3 = 8.5
Variance
Var(X) = 8.5 − 2.7² = 1.21
σ = √1.21 = 1.1
Answer
σ = 1.1
16.Probability
Answer: E(G) = 2 €
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.
Net outcomes
G: 5 (p = 0.2), 245 (p = 0.02), −5 (p = 0.78)
Expected value
E(G) = 5·0.2 + 245·0.02 − 5·0.78 = 2 €
Answer
E(G) = 2 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 12.65
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 = (93.6 − 100)/8 = −0.8
Answer
z = −0.8
19.Normal distribution
Answer: P = 0.6915
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 = 360 g, σT = 1.2·√9 = 3.6 g
Standardise
z = (358.2 − 360)/3.6 = −0.5
Probability
P = 1 − Φ(−0.5) = 0.6915
Answer
P = 0.6915
20.Normal distribution
Answer: P = 0.7881
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).