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→∞ (6x2 + 4x − 6)/(3x2 + 6x + 7).
f(x) = (6x2 + 4x − 6)/(3x2 + 6x + 7)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 3x)/sin(4x), with x in radians.
f(x) = ln(1 + 3x)/sin(4x)x → 0
L =
3.DerivativesMedium
Let f(x) = (−2x + 5)5. Find f′(−1).
f(x) = (−2x + 5)5x = −1
f′(−1) =
4.DerivativesHard
Let f(x) = x3 − 1.5x2 − 6x − 3. Find the x-coordinate x of the local maximum.
f(x) = x3 − 1.5x2 − 6x − 3
x =
5.IntegralsMedium
Evaluate I = ∫02 2x·(x² + 1)3 dx.
f(x) = 2x·(x² + 1)3[0, 2]
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 number z = 5·e−j0.5 has its angle in radians. Find its real part a.
r = 5θ = −0.5 rad
a =
8.Complex numbersHard
One root of z³ − 9z² + 43z − 75 = 0 is z₁ = 3 + j4. The coefficients are real. Find the real root r.
z₁ = 3 + j4
r =
9.Differential equationsMedium
A cup of coffee at 84 °C is left in a room at 23 °C. After 25 min its temperature is 37 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 31 °C?
T0 = 84 °CTa = 23 °CT(25) = 37 °CT = 31 °C
t = min
10.Differential equationsHard
Solve y′ + 2y = 7e−t with y(0) = 5. Find y(2).
y′ + 2y = 7e−ty(0) = 5
y(2) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−2 1 0; −2 5 −3; 4 1 4]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [6 6; −3 −3]
λ =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (2, 3, 6) and b = (−2, 5, −5), in degrees.
a = (2, 3, 6)b = (−2, 5, −5)
θ = °
14.Vectors: dot & cross productHard
For a = (−5, −1, −2) and b = (−1, 1, 4), find the z-component (a × b)z of the cross product a × b.
a = (−5, −1, −2)b = (−1, 1, 4)
(a × b)z =
15.ProbabilityMedium
A fair die is rolled 5 times. What is the probability P of getting at least one six?
n = 5
P =
16.ProbabilityHard
A game costs €3 to play. You win €15 with probability 0.2 and €150 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€15: p = 0.2€150: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 38, 48, 34, 10, 12, 35.
n = 6
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 15. Find the z-score z of a score of 92.5.
μ = 100σ = 15x = 92.5
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 6 of them exceeds 241.5 g. Round z to two decimals.
μ = 40 gσ = 1.2 gn = 6> 241.5 g
P =
20.Normal distributionHard
The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
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).
Divide by the highest power
(6 + 4/x + (−6)/x2)/(3 + 6/x + 7/x2)
Let x → ∞
L = 6/3 = 2
Answer
L = 2
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 0.75
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 [3/(1 + 3x)]/[4·cos 4x]
Substitute x = 0
L = 3/4 = 0.75
Answer
L = 0.75
3.Derivatives
Answer: f′(−1) = −24010
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) = 5·(−2x + 5)4·(−2)
Substitute
f′(−1) = 5·74·(−2) = −24010
Answer
f′(−1) = −24010
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 − 3x − 6 = 3·(x2 − x − 2)
Solve f′(x) = 0
x = −1 or x = 2
Which one is which
f″(x) = 6x − 3: negative at a maximum, positive at a minimum.
f″(−1) = −9, f″(2) = 9
x = −1
Answer
x = −1
5.Integrals
Answer: I = 156
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 = ∫15 u3 du
Integrate and evaluate
I = [u4/4]15 = (625 − 1)/4 = 156
Answer
I = 156
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 = 4.388
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 = 5·(cos −0.5 + j sin −0.5)
Components (radians)
a = 5·cos −0.5 = 4.388
b = 5·sin −0.5 = −2.397
a = 4.388
Answer
a = 4.388
8.Complex numbers
Answer: r = 3
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 = 6 + 3 = 9 = −(−9).
9.Differential equations
Answer: t = 34.51 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(61/14)/25 = 0.05887 min⁻¹
Time to reach the target
t = ln(61/8)/0.05887 = 34.51 min
Answer
t = 34.51 min
10.Differential equations
Answer: y(2) = 0.9107
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 = e2t
(e2ty)′ = 7et
General solution
y = 7/1·e−t + C·e−2t = 7·e−t + C·e−2t
Initial condition
5 = 7 + C → C = −2
Evaluate
y(2) = 7·0.1353 + (−2)·0.01832 = 0.9107
Answer
y(2) = 0.9107
11.Matrices & determinants
Answer: det A = −50
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 = 23, M₁₂ = 4, M₁₃ = −22
Cofactor expansion
det A = (−2)·23 − 1·4 + 0·(−22) = −50
Answer
det A = −50
12.Matrices & determinants
Answer: λ = 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.
Characteristic equation
λ² − (3)λ + (0) = 0
Solve
λ = (3 ± √(9))/2 → λ = 0, 3
λ = 3
Answer
λ = 3
13.Vectors: dot & cross product
Answer: θ = 111.7 °
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
a·b = 2·(−2) + 3·5 + 6·(−5) = −19
Magnitudes
|a| = √(2² + 3² + 6²) = 7
|b| = √((−2)² + 5² + (−5)²) = 7.348
Angle
cos θ = −19/(7·7.348) = −0.3694
θ = arccos(−0.3694) = 111.7°
Answer
θ = 111.7°
14.Vectors: dot & cross product
Answer: (a × b)z = −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 = (−1)·4 − (−2)·1 = −2
(a × b)y = azbx − axbz = (−2)·(−1) − (−5)·4 = 22
(a × b)z = axby − aybx = (−5)·1 − (−1)·(−1) = −6
Requested component
(a × b)z = −6
Answer
(a × b)z = −6
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: P = 0.5981
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)5 = 0.4019
P = 1 − 0.4019 = 0.5981
Answer
P = 0.5981
16.Probability
Answer: E(G) = 3 €
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: 12 (p = 0.2), 147 (p = 0.02), −3 (p = 0.78)
Expected value
E(G) = 12·0.2 + 147·0.02 − 3·0.78 = 3 €
Answer
E(G) = 3 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 15.18
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 = (92.5 − 100)/15 = −0.5
Answer
z = −0.5
19.Normal distribution
Answer: P = 0.305
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 = 240 g, σT = 1.2·√6 = 2.939 g
Standardise
z = (241.5 − 240)/2.939 = 0.51
Probability
P = 1 − Φ(0.51) = 0.305
Answer
P = 0.305
20.Normal distribution
Answer: P = 0.8413
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).