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→∞ (1 + 3/x)x. Give a decimal value.
f(x) = (1 + 3/x)xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 3x)/sin(5x), with x in radians.
f(x) = ln(1 + 3x)/sin(5x)x → 0
L =
3.DerivativesMedium
Let f(x) = x2·ex. Find f′(1.5).
f(x) = x2·exx = 1.5
f′(1.5) =
4.DerivativesHard
Let f(x) = x3 − 12x2 + 45x + 4. Find the x-coordinate x of the local maximum.
f(x) = x3 − 12x2 + 45x + 4
x =
5.IntegralsMedium
Evaluate I = ∫01 2x·(x² + 1)2 dx.
f(x) = 2x·(x² + 1)2[0, 1]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = 2x2 and y = 4x.
y = 2x2y = 4x
A =
7.Complex numbersMedium
The number z = 5·ej0.6 has its angle in radians. Find its real part a.
r = 5θ = 0.6 rad
a =
8.Complex numbersHard
The roots of zⁿ = w with n = 4 and w = 81∠135° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2, 3. Find the argument of z₂ in degrees, between −180° and 180°.
w = 81∠135°n = 4k = 2
θ₂ = °
9.Differential equationsMedium
A forged steel part at 400 °C is left to cool in a workshop at 29 °C. After 21 min its temperature is 189 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 400 °CTa = 29 °CT(21) = 189 °C
k = min⁻¹
10.Differential equationsHard
Solve the initial value problem y′ + y/x = 2x for x > 0, with y(1) = −1. Find y(3).
y′ + y/x = 2xy(1) = −1
y(3) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c22 (row 2, column 2).
A = [0 −1; 6 6]B = [2 5; 0 −2]
c22 =
12.Matrices & determinantsHard
Solve the system 5x + 2y = 12 and −5x − 5y = 12. Find x.
5x + 2y = 12−5x − 5y = 12
x =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (1, 3, 5) onto b = (−2, −2, −5), that is, the component of a in the direction of b.
a = (1, 3, 5)b = (−2, −2, −5)
compb a =
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−4, 5, 3) and b = (−2, 3, 0) and has a positive z-component. Find its y-component n̂y.
a = (−4, 5, 3)b = (−2, 3, 0)n̂z > 0
n̂y =
15.ProbabilityMedium
A bag holds 8 red and 3 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
8 / 3
P =
16.ProbabilityHard
X has the distribution shown, and Y = 0.5X − 3. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 11, 38, 50, 13, 60, 26.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 15. What proportion P of values is below 69?
μ = 60σ = 15x = 69
P =
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 8 of them exceeds 92.6 kg. Round z to two decimals.
μ = 12 kgσ = 0.8 kgn = 8> 92.6 kg
P =
20.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 95% of them lie (the 95th percentile).
μ = 500σ = 40P = 95%
x = h
3Solutions
1.Limits
Answer: L = 20.09
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).
Take the logarithm
ln L = limx→∞ x·ln(1 + 3/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 1·3 = 3
Undo the logarithm
L = e3 = 20.09
Answer
L = 20.09
2.Limits
Answer: L = 0.6
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)]/[5·cos 5x]
Substitute x = 0
L = 3/5 = 0.6
Answer
L = 0.6
3.Derivatives
Answer: f′(1.5) = 23.53
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′ = ex·(2x + 1·x2)
Substitute
f′(1.5) = 4.482·(3 + 2.25) = 23.53
Answer
f′(1.5) = 23.53
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 − 24x + 45 = 3·(x2 − 8x + 15)
Solve f′(x) = 0
x = 3 or x = 5
Which one is which
f″(x) = 6x − 24: negative at a maximum, positive at a minimum.
f″(3) = −6, f″(5) = 6
x = 3
Answer
x = 3
5.Integrals
Answer: I = 2.333
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 u2 du
Integrate and evaluate
I = [u3/3]12 = (8 − 1)/3 = 2.333
Answer
I = 2.333
6.Integrals
Answer: A = 2.667
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
2x2 = 4x → x = 0, x = 2
Integrate upper minus lower
A = [2x2 − 0.6667x3]02 = 2.667
Answer
A = 2.667
7.Complex numbers
Answer: a = 4.127
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 0.6 + j sin 0.6)
Components (radians)
a = 5·cos 0.6 = 4.127
b = 5·sin 0.6 = 2.823
a = 4.127
Answer
a = 4.127
8.Complex numbers
Answer: θ₂ = −146.3 °
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.
Modulus of the roots
811/4 = 3
Argument of the root
θ₂ = (135° + 360°·2)/4 = 213.8° → −146.3°
Argument
θ₂ = −146.3°
Answer
θ₂ = −146.3°
The 4 roots lie on the circle of radius 3, 90° apart.
9.Differential equations
Answer: k = 0.04005 min⁻¹
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.
Ratio of temperature differences
e−21k = 160/371
Solve for k
k = ln(371/160)/21 = 0.04005 min⁻¹
Answer
k = 0.04005 min⁻¹
10.Differential equations
Answer: y(3) = 5.444
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)′ = 2x2 → x1y = 2x3/3 + C
Initial condition
−1 = 2/3 + C → C = −1.667
Evaluate
y(3) = (2·33/3 + (−1.667))/31 = 5.444
Answer
y(3) = 5.444
11.Matrices & determinants
Answer: c22 = 18
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
c22 = 6·5 + 6·(−2) = 18
Answer
c22 = 18
12.Matrices & determinants
Answer: x = 5.6
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 = 5·(−5) − 2·(−5) = −15
Replace the x column
det Aₓ = 12·(−5) − 2·12 = −84
x = −84/(−15) = 5.6
Answer
x = 5.6
y = −8
13.Vectors: dot & cross product
Answer: compb a = −5.745
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 = 1·(−2) + 3·(−2) + 5·(−5) = −33
Length of b
|b| = √((−2)² + (−2)² + (−5)²) = 5.745
Scalar projection
compb a = a·b/|b| = −33/5.745 = −5.745
Answer
compb a = −5.745
Negative: a has a component pointing against b.
14.Vectors: dot & cross product
Answer: n̂y = 0.5455
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 = 5·0 − 3·3 = −9
(a × b)y = azbx − axbz = 3·(−2) − (−4)·0 = −6
(a × b)z = axby − aybx = (−4)·3 − 5·(−2) = −2
Normalise
The z-component of a × b is negative, so use −(a × b).
|a × b| = √((−9)² + (−6)² + (−2)²) = 11
n̂ = −(−9, −6, −2)/11 = (0.8182, 0.5455, 0.1818)
n̂y = 0.5455
Answer
n̂y = 0.5455
15.Probability
Answer: P = 0.5091
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 = 8/11 · 7/10 = 56/110 = 0.5091
Answer
P = 0.5091
16.Probability
Answer: σY = 0.7517
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 = 1.8, Var(X) = 2.26, σX = 1.503
Linear change
σY = |0.5|·1.503 = 0.7517
Answer
σY = 0.7517
The mean does move: E(Y) = 0.5·1.8 − 3 = −2.1.
17.Descriptive statistics
Answer: s = 19.88
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 = 0.7257
Answer
P = 0.7257
19.Normal distribution
Answer: P = 0.9332
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 = 96 kg, σT = 0.8·√8 = 2.263 kg
Standardise
z = (92.6 − 96)/2.263 = −1.5
Probability
P = 1 − Φ(−1.5) = 0.9332
Answer
P = 0.9332
20.Normal distribution
Answer: x = 565.8 h
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.