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 + 0.5/x)2x. Give a decimal value.
f(x) = (1 + 0.5/x)2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 5x)/(x²), with x in radians.
f(x) = (1 − cos 5x)/(x²)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(x2 + 4). Find f′(−1).
f(x) = ln(x2 + 4)x = −1
f′(−1) =
4.DerivativesHard
The tangent line to f(x) = −x2 − x + 2 at x = −2 crosses the y-axis at y = b. Find b.
f(x) = −x2 − x + 2x = −2
b =
5.IntegralsMedium
Evaluate I = ∫24 3/x dx.
f(x) = 3/x[2, 4]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = 2x2 and y = x.
y = 2x2y = x
A =
7.Complex numbersMedium
Write z = −5 + j7 in exponential form r·ejθ. Give θ in radians, between −π and π.
z = −5 + j7
θ = rad
8.Complex numbersHard
The roots of zⁿ = w with n = 4 and w = 16∠120° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2, 3. Find the imaginary part of z₀.
w = 16∠120°n = 4k = 0
Im(z₀) =
9.Differential equationsMedium
Solve y″ + 2y′ + y = 0 with y(0) = 2 and y′(0) = 4. Find y(1).
y″ + 2y′ + y = 0y(0) = 2y′(0) = 4
y(1) =
10.Differential equationsHard
A tank holds 400 L of pure water. Brine with 0.2 kg of salt per litre flows in at 20 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 64 kg of salt?
V = 400 Lm0 = 0 kgc = 0.2 kg/Lr = 20 L/min
t = min
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).
A = [3 −5; 4 4]B = [0 −4; 5 −5]
c12 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−3 −4; 7 8]
λ =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (2, −2, 2) and b = (1, −5, 0), in degrees.
a = (2, −2, 2)b = (1, −5, 0)
θ = °
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (−4, 0, 3), b = (1, −2, 4) and c = (5, 5, 4).
a = (−4, 0, 3)b = (1, −2, 4)c = (5, 5, 4)
V =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3p(x): 0.25, 0.1, 0.3, 0.35
Var(X) =
16.ProbabilityHard
X has the distribution shown, and Y = 2X + 4. Find the standard deviation σY of Y.
x: 0, 1, 2, 3p(x): 0.1, 0.2, 0.35, 0.35Y = 2X + 4
σY =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 56, 51, 36, 58, 31, 56.
n = 6
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 5. Find the z-score z of a score of 102.
μ = 100σ = 5x = 102
z =
19.Normal distributionMedium
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 246 and 254 g? Give P as a decimal.
μ = 250σ = 5x1 = 246x2 = 254
P =
20.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 80% of them lie (the 80th percentile).
μ = 65σ = 10P = 80%
x = points
3Solutions
1.Limits
Answer: L = 2.718
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→∞ 2·x·ln(1 + 0.5/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 2·0.5 = 1
Undo the logarithm
L = e1 = 2.718
Answer
L = 2.718
2.Limits
Answer: L = 12.5
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
5·sin 5x/(2x) → 0/0
Second application
25·cos 5x/2
Substitute x = 0
L = 25/2 = 12.5
Answer
L = 12.5
3.Derivatives
Answer: f′(−1) = −0.4
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 + 4)
Substitute
f′(−1) = −2/5 = −0.4
Answer
f′(−1) = −0.4
4.Derivatives
Answer: b = 6
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.
Point and slope
f(−2) = 0
f′(x) = −2x − 1 → f′(−2) = 3
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 0 − 3·(−2) = 6
Answer
b = 6
5.Integrals
Answer: I = 2.079
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.
Antiderivative
F(x) = 3·ln x
Evaluate at the limits
I = 3·(ln 4 − ln 2) = 3·ln 2 = 2.079
Answer
I = 2.079
6.Integrals
Answer: A = 0.04167
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 = x → x = 0, x = 0.5
Integrate upper minus lower
A = [0.5x2 − 0.6667x3]00.5 = 0.04167
Answer
A = 0.04167
7.Complex numbers
Answer: θ = 2.191 rad
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
r = √(25 + 49) = 8.602
Argument in radians
θ = 125.5° = 125.5·π/180 = 2.191 rad
Exponential form
z = 8.602·ej2.191
Answer
θ = 2.191 rad
8.Complex numbers
Answer: Im(z₀) = 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.
Modulus of the roots
161/4 = 2
Argument of the root
θ₀ = (120° + 360°·0)/4 = 30°
Rectangular form
z₀ = 2·(cos 30° + j sin 30°) = 1.732 + j1
Im(z₀) = 1
Answer
Im(z₀) = 1
The 4 roots lie on the circle of radius 2, 90° apart.
9.Differential equations
Answer: y(1) = 2.943
Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.
Double root
(r + 1)² = 0 → r = −1
Constants
A = y(0) = 2
y′(0) = B − 1A → B = 4 + 2 = 6
Evaluate
y(1) = (2 + 6·1)·e−1 = 2.943
Answer
y(1) = 2.943
10.Differential equations
Answer: t = 32.19 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) = 80 + (−80)·e−t/20
Solve for t
e−t/20 = (80 − 64)/(80 − 0) = 0.2
t = 20·ln(5) = 32.19 min
Answer
t = 32.19 min
11.Matrices & determinants
Answer: c12 = 13
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
c12 = 3·(−4) + (−5)·(−5) = 13
Answer
c12 = 13
12.Matrices & determinants
Answer: λ = 4
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 − (5)λ + (4) = 0
Solve
λ = (5 ± √(9))/2 → λ = 1, 4
λ = 4
Answer
λ = 4
13.Vectors: dot & cross product
Answer: θ = 47.21 °
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 = 2·1 + (−2)·(−5) + 2·0 = 12
Magnitudes
|a| = √(2² + (−2)² + 2²) = 3.464
|b| = √(1² + (−5)² + 0²) = 5.099
Angle
cos θ = 12/(3.464·5.099) = 0.6794
θ = arccos(0.6794) = 47.21°
Answer
θ = 47.21°
14.Vectors: dot & cross product
Answer: V = 157
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 b × c
(b × c)x = bycz − bzcy = (−2)·4 − 4·5 = −28
(b × c)y = bzcx − bxcz = 4·5 − 1·4 = 16
(b × c)z = bxcy − bycx = 1·5 − (−2)·5 = 15
Dot product with a
a·(b × c) = (−4)·(−28) + 0·16 + 3·15 = 157
V = |157| = 157
Answer
V = 157
15.Probability
Answer: Var(X) = 1.388
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 = 0·0.25 + 1·0.1 + 2·0.3 + 3·0.35 = 1.75
Mean of the squares
E(X²) = 0·0.25 + 1·0.1 + 4·0.3 + 9·0.35 = 4.45
Variance
Var(X) = 4.45 − 1.75² = 1.388
Answer
Var(X) = 1.388
16.Probability
Answer: σY = 1.947
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 = 1.95, Var(X) = 0.9475, σX = 0.9734
Linear change
σY = |2|·0.9734 = 1.947
Answer
σY = 1.947
The mean does move: E(Y) = 2·1.95 + 4 = 7.9.
17.Descriptive statistics
Answer: s = 11.58
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 = (102 − 100)/5 = 0.4
Answer
z = 0.4
19.Normal distribution
Answer: P = 0.5763
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 = (246 − 250)/5 = −0.8
z₂ = (254 − 250)/5 = 0.8
Area between
P = Φ(0.8) − Φ(−0.8) = 0.7881 − 0.2119 = 0.5763
Answer
P = 0.5763
20.Normal distribution
Answer: x = 73.42 points
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.