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 − 2x) − x).
f(x) = √(x2 − 2x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e2x − 1 − 2x)/x².
f(x) = (e2x − 1 − 2x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(x), with x in radians. Find f′(1.5).
f(x) = x·sin(x)x = 1.5 rad
f′(1.5) =
4.DerivativesHard
Let f(x) = x3 − 6x2 + 1. Find the x-coordinate x of the local minimum.
f(x) = x3 − 6x2 + 1
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 = 0.5x2 and y = 2x.
y = 0.5x2y = 2x
A =
7.Complex numbersMedium
Given z₁ = 1.5∠125° and z₂ = 4∠45°, find the real part of w = z₁/z₂.
z₁ = 1.5∠125°z₂ = 4∠45°
Re(w) =
8.Complex numbersHard
The roots of zⁿ = w with n = 4 and w = 16∠60° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2, 3. Find the imaginary part of z₃.
w = 16∠60°n = 4k = 3
Im(z₃) =
9.Differential equationsMedium
A can of soda at 5 °C is taken out of the fridge into a room at 27 °C. After 17 min its temperature is 19 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 5 °CTa = 27 °CT(17) = 19 °C
k = min⁻¹
10.Differential equationsHard
A system obeys y″ + 4y′ + 29y = 58 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 4y′ + 29y = 58y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−3 5 5; 4 −4 −1; 4 −1 0]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [0 −3; 2 7]
λ =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (2, 5, −4) onto b = (−6, 3, 0), that is, the component of a in the direction of b.
a = (2, 5, −4)b = (−6, 3, 0)
compb a =
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (4, −3, −1), b = (−1, 2, 2) and c = (−1, −2, 1).
a = (4, −3, −1)b = (−1, 2, 2)c = (−1, −2, 1)
V =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2p(x): 0.35, 0.3, 0.35
σ =
16.ProbabilityHard
X has the distribution shown, and Y = −2X + 10. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 32, 50, 38, 40, 53.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 60 and standard deviation 8. Find the z-score z of a score of 68.
μ = 60σ = 8x = 68
z =
19.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.2033. Find z to two decimals.
A = 0.2033
z =
20.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 60% is to the left of it. Find the cut-off x.
μ = 100 pointsσ = 15 points60%
x = points
3Solutions
1.Limits
Answer: L = −1
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/(√(x2 − 2x) + x)
Divide by x
f(x) = −2/(√(1 + (−2)/x) + 1)
Let x → ∞
L = −2/(1 + 1) = −1
Answer
L = −1
2.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).
First application
(2·e2x − 2)/(2x) → 0/0
Second application
4·e2x/2
Substitute x = 0
L = 4·1/2 = 2
Answer
L = 2
3.Derivatives
Answer: f′(1.5) = 1.104
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 and chain rule
f′(x) = sin(x) + x·cos(x)
Substitute
f′(1.5) = 0.9975 + 1.5·0.07074 = 1.104
Answer
f′(1.5) = 1.104
4.Derivatives
Answer: x = 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.
Derivative
f′(x) = 3x2 − 12x = 3·(x2 − 4x)
Solve f′(x) = 0
x = 0 or x = 4
Which one is which
f″(x) = 6x − 12: negative at a maximum, positive at a minimum.
f″(0) = −12, f″(4) = 12
x = 4
Answer
x = 4
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 = 5.333
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.
Intersection points
0.5x2 = 2x → x = 0, x = 4
Integrate upper minus lower
A = [x2 − 0.1667x3]04 = 5.333
Answer
A = 5.333
7.Complex numbers
Answer: Re(w) = 0.06512
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| = 1.5 / 4 = 0.375
θ = 125° − 45° = 80°
Rectangular form
Re(w) = 0.375·cos(80°) = 0.06512
Im(w) = 0.375·sin(80°) = 0.3693
Re(w) = 0.06512
Answer
Re(w) = 0.06512
8.Complex numbers
Answer: Im(z₃) = −1.932
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
θ₃ = (60° + 360°·3)/4 = 285° → −75°
Rectangular form
z₃ = 2·(cos −75° + j sin −75°) = 0.5176 − j1.932
Im(z₃) = −1.932
Answer
Im(z₃) = −1.932
The 4 roots lie on the circle of radius 2, 90° apart.
9.Differential equations
Answer: k = 0.05951 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−17k = -8/-22
Solve for k
k = ln(-22/-8)/17 = 0.05951 min⁻¹
Answer
k = 0.05951 min⁻¹
10.Differential equations
Answer: ymax = 2.569
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.
Steady value and roots
yss = 2, r = −2 ± 5j
Constants from the initial values
y(0) = 2 + A = 0 → A = −2
y′(0) = −2A + 5B = 0 → B = −0.8
Peak
t = π/5 = 0.6283
ymax = 2·(1 + e−2·π/5) = 2.569
Answer
ymax = 2.569
11.Matrices & determinants
Answer: det A = 43
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 = −1, M₁₂ = 4, M₁₃ = 12
Cofactor expansion
det A = (−3)·(−1) − 5·4 + 5·12 = 43
Answer
det A = 43
12.Matrices & determinants
Answer: λ = 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.
Characteristic equation
λ² − (7)λ + (6) = 0
Solve
λ = (7 ± √(25))/2 → λ = 1, 6
λ = 6
Answer
λ = 6
13.Vectors: dot & cross product
Answer: compb a = 0.4472
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·(−6) + 5·3 + (−4)·0 = 3
Length of b
|b| = √((−6)² + 3² + 0²) = 6.708
Scalar projection
compb a = a·b/|b| = 3/6.708 = 0.4472
Answer
compb a = 0.4472
14.Vectors: dot & cross product
Answer: V = 23
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 = 2·1 − 2·(−2) = 6
(b × c)y = bzcx − bxcz = 2·(−1) − (−1)·1 = −1
(b × c)z = bxcy − bycx = (−1)·(−2) − 2·(−1) = 4
Dot product with a
a·(b × c) = 4·6 + (−3)·(−1) + (−1)·4 = 23
V = |23| = 23
Answer
V = 23
15.Probability
Answer: σ = 0.8367
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 = 0·0.35 + 1·0.3 + 2·0.35 = 1
Mean of the squares
E(X²) = 0·0.35 + 1·0.3 + 4·0.35 = 1.7
Variance
Var(X) = 1.7 − 1² = 0.7
σ = √0.7 = 0.8367
Answer
σ = 0.8367
16.Probability
Answer: σY = 2.522
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 = 1.9, Var(X) = 1.59, σX = 1.261
Linear change
σY = |−2|·1.261 = 2.522
Answer
σY = 2.522
The mean does move: E(Y) = −2·1.9 + 10 = 6.2.
17.Descriptive statistics
Answer: s = 8.706
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 = (68 − 60)/8 = 1
Answer
z = 1
19.Normal distribution
Answer: z = 0.83
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.
Area to the left of z
Φ(z) = 1 − 0.2033 = 0.7967
Read the z-table backwards
Find 0.7967 inside the table and read off z.
z = 0.83
Answer
z = 0.83
20.Normal distribution
Answer: x = 103.8 points
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.