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→∞ (√(9x2 + 2x) − 3x).
f(x) = √(9x2 + 2x) − 3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e0.5x − 1 − 0.5x)/x².
f(x) = (e0.5x − 1 − 0.5x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x2·e2x. Find f′(1).
f(x) = x2·e2xx = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 − 3x − 1. Find the x-coordinate x of the local minimum.
f(x) = x3 − 3x − 1
x =
5.IntegralsMedium
Evaluate I = ∫01 2x·(x² + 1)3 dx.
f(x) = 2x·(x² + 1)3[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
Given z₁ = 5∠100° and z₂ = 1.5∠−25°, find the argument θ of w = z₁/z₂ in degrees, between −180° and 180°.
z₁ = 5∠100°z₂ = 1.5∠−25°
θ = °
8.Complex numbersHard
One root of z³ − 13z² + 65z − 125 = 0 is z₁ = 4 + j3. The coefficients are real. Find the real root r.
z₁ = 4 + j3
r =
9.Differential equationsMedium
A cup of coffee at 92 °C is left in a room at 22 °C. After 6 min its temperature is 71 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 92 °CTa = 22 °CT(6) = 71 °C
k = min⁻¹
10.Differential equationsHard
Solve y′ + 3y = 2e−0.5t with y(0) = 2. Find y(2).
y′ + 3y = 2e−0.5ty(0) = 2
y(2) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−3 −1 −5; −2 −4 2; −5 −2 −2]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−2 −1; −1 −2]
λ =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (0, 2, 6) and b = (−4, k, 2) are perpendicular.
a = (0, 2, 6)b = (−4, k, 2)
k =
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (−4, −4, −4), b = (3, 4, 5) and c = (−4, 3, 1).
a = (−4, −4, −4)b = (3, 4, 5)c = (−4, 3, 1)
V =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3, 4p(x): 0.25, 0.1, 0.25, 0.15, 0.25
Var(X) =
16.ProbabilityHard
X has the distribution shown, and Y = 0.5X + 1. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 31, 41, 51, 47, 30, 25.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 10. What proportion P of values is below 62.5?
μ = 50σ = 10x = 62.5
P =
19.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 62 and 70 points? Give P as a decimal.
μ = 65σ = 10x1 = 62x2 = 70
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 10% of them lie (the 10th percentile).
μ = 32σ = 6P = 10%
x = min
3Solutions
1.Limits
Answer: L = 0.3333
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) = 2x/(√(9x2 + 2x) + 3x)
Divide by x
f(x) = 2/(√(9 + 2/x) + 3)
Let x → ∞
L = 2/(3 + 3) = 0.3333
Answer
L = 0.3333
2.Limits
Answer: L = 0.125
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
(0.5·e0.5x − 0.5)/(2x) → 0/0
Second application
0.25·e0.5x/2
Substitute x = 0
L = 0.25·1/2 = 0.125
Answer
L = 0.125
3.Derivatives
Answer: f′(1) = 29.56
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′ = e2x·(2x + 2·x2)
Substitute
f′(1) = 7.389·(2 + 2) = 29.56
Answer
f′(1) = 29.56
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 − 3 = 3·(x2 − 1)
Solve f′(x) = 0
x = −1 or x = 1
Which one is which
f″(x) = 6x: negative at a maximum, positive at a minimum.
f″(−1) = −6, f″(1) = 6
x = 1
Answer
x = 1
5.Integrals
Answer: I = 3.75
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 u3 du
Integrate and evaluate
I = [u4/4]12 = (16 − 1)/4 = 3.75
Answer
I = 3.75
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: θ = 125 °
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.
Modulus and argument of w
|w| = 5 / 1.5 = 3.333
θ = 100° − (−25)° = 125°
Argument
θ = 125°
Answer
θ = 125°
8.Complex numbers
Answer: r = 5
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 = 8 + 5 = 13 = −(−13).
9.Differential equations
Answer: k = 0.05945 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−6k = 49/70
Solve for k
k = ln(70/49)/6 = 0.05945 min⁻¹
Answer
k = 0.05945 min⁻¹
10.Differential equations
Answer: y(2) = 0.2973
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 = e3t
(e3ty)′ = 2e2.5t
General solution
y = 2/2.5·e−0.5t + C·e−3t = 0.8·e−0.5t + C·e−3t
Initial condition
2 = 0.8 + C → C = 1.2
Evaluate
y(2) = 0.8·0.3679 + 1.2·0.002479 = 0.2973
Answer
y(2) = 0.2973
11.Matrices & determinants
Answer: det A = 58
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 = 12, M₁₂ = 14, M₁₃ = −16
Cofactor expansion
det A = (−3)·12 − (−1)·14 + (−5)·(−16) = 58
Answer
det A = 58
12.Matrices & determinants
Answer: λ = −1
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 − (−4)λ + (3) = 0
Solve
λ = (−4 ± √(4))/2 → λ = −3, −1
λ = −1
Answer
λ = −1
13.Vectors: dot & cross product
Answer: k = −6
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 = 0·(−4) + 2·k + 6·2 = 0
12 + 2·k = 0
Solve for k
k = −12/2 = −6
Answer
k = −6
14.Vectors: dot & cross product
Answer: V = 36
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.
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 = 2.45, Var(X) = 1.247, σX = 1.117
Linear change
σY = |0.5|·1.117 = 0.5585
Answer
σY = 0.5585
The mean does move: E(Y) = 0.5·2.45 + 1 = 2.225.
17.Descriptive statistics
Answer: s = 10.39
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 = 1.25
Standard normal table
Φ(1.25) = 0.8944
P = 0.8944
Answer
P = 0.8944
19.Normal distribution
Answer: P = 0.3094
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 = (62 − 65)/10 = −0.3
z₂ = (70 − 65)/10 = 0.5
Area between
P = Φ(0.5) − Φ(−0.3) = 0.6915 − 0.3821 = 0.3094
Answer
P = 0.3094
20.Normal distribution
Answer: x = 24.31 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.