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→∞ (5x2 − 2x + 9)/(x2 + x + 1).
f(x) = (5x2 − 2x + 9)/(x2 + x + 1)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 5x)/(4x²), with x in radians.
f(x) = (1 − cos 5x)/(4x²)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(x2 + 4). Find f′(2).
f(x) = ln(x2 + 4)x = 2
f′(2) =
4.DerivativesHard
The tangent line to f(x) = −x2 + 5 at x = 3 crosses the y-axis at y = b. Find b.
f(x) = −x2 + 5x = 3
b =
5.IntegralsMedium
The region under y = 2x from x = 0 to x = 1 is rotated about the x-axis. Find the volume V of the solid (a cone).
y = 2x0 ≤ x ≤ 1
V =
6.IntegralsHard
Evaluate I = ∫₀3 x²·eˣ dx.
∫₀3 x²eˣ dx
I =
7.Complex numbersMedium
The number z = 4·ej0.6 has its angle in radians. Find its real part a.
r = 4θ = 0.6 rad
a =
8.Complex numbersHard
Exactly one of the points A–D is a root of z⁴ = 81∠180°. Click it.
w = 81∠180°n = 4
1A (3∠90°)
2B (3∠180°)
3C (3∠45°)
4D (1.5∠45°)
9.Differential equationsMedium
A can of soda at 5 °C is taken out of the fridge into a room at 26 °C. After 20 min its temperature is 21 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).
A = [6 −1; −6 5]B = [−6 −1; 5 −2]
c12 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [4 −1; 7 −4]
λ =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (2, −2, −6) and b = (4, k, −1) are perpendicular.
a = (2, −2, −6)b = (4, k, −1)
k =
14.Vectors: dot & cross productHard
A force F = (−40, 50, −40) N acts at point A(4, 4, 4). Coordinates are in metres. Find the magnitude |M| of the moment of the force about point O(3, 3, −2).
F = (−40, 50, −40) NA(4, 4, 4) mO(3, 3, −2) m
|M| = N·m
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4p(x): 0.35, 0.3, 0.05, 0.3
Var(X) =
16.ProbabilityHard
X has the distribution shown, and Y = 5X + 10. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 13, 16, 27, 47, 46.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 15. What proportion P of values is below 192.5?
μ = 170σ = 15x = 192.5
P =
19.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 65.5 and 80.5 points? Give P as a decimal.
μ = 65σ = 10x1 = 65.5x2 = 80.5
P =
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 top 85% is to the right of it. Find the cut-off x.
μ = 100 pointsσ = 15 points85%
x = points
3Solutions
1.Limits
Answer: L = 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).
Divide by the highest power
(5 + (−2)/x + 9/x2)/(1 + 1/x + 1/x2)
Let x → ∞
L = 5/1 = 5
Answer
L = 5
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 3.125
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
5·sin 5x/(8x) → 0/0
Second application
25·cos 5x/8
Substitute x = 0
L = 25/8 = 3.125
Answer
L = 3.125
3.Derivatives
Answer: f′(2) = 0.5
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′(2) = 4/8 = 0.5
Answer
f′(2) = 0.5
4.Derivatives
Answer: b = 14
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(3) = −4
f′(x) = −2x → f′(3) = −6
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −4 − (−6)·3 = 14
Answer
b = 14
5.Integrals
Answer: V = 4.189
Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.
Disc method
V = π∫₀1 4x² dx
Integrate
V = π·4·1³/3 = 4.189
Answer
V = 4.189
Check: a cone of radius 2 and height 1 has (1/3)πr²h = 4.189.
6.Integrals
Answer: I = 98.43
Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.
First round
∫x²eˣ dx = x²eˣ − ∫2x eˣ dx
Second round
∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)
Limits
20.09·(5) − 1·2 = 98.43
Answer
I = 98.43
7.Complex numbers
Answer: a = 3.301
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 = 4·(cos 0.6 + j sin 0.6)
Components (radians)
a = 4·cos 0.6 = 3.301
b = 4·sin 0.6 = 2.259
a = 3.301
Answer
a = 3.301
8.Complex numbers
Answer: C (3∠45°)
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 and arguments of the roots
|z| = 3
θ = 45°, 135°, −135°, −45°
The point
C = 3∠45°
Answer
C (3∠45°)
9.Differential equations
Answer: k = 0.07175 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−20k = -5/-21
Solve for k
k = ln(-21/-5)/20 = 0.07175 min⁻¹
Answer
k = 0.07175 min⁻¹
10.Differential equations
Answer: y(0.5) = 2.894
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)′ = 8e2.5t
General solution
y = 8/2.5·e−0.5t + C·e−3t = 3.2·e−0.5t + C·e−3t
Initial condition
5 = 3.2 + C → C = 1.8
Evaluate
y(0.5) = 3.2·0.7788 + 1.8·0.2231 = 2.894
Answer
y(0.5) = 2.894
11.Matrices & determinants
Answer: c12 = −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.
Row times column
c12 = 6·(−1) + (−1)·(−2) = −4
Answer
c12 = −4
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 − (0)λ + (−9) = 0
Solve
λ = (0 ± √(36))/2 → λ = −3, 3
λ = 3
Answer
λ = 3
13.Vectors: dot & cross product
Answer: k = 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 equals zero
a·b = 2·4 + (−2)·k + (−6)·(−1) = 0
14 + (−2)·k = 0
Solve for k
k = −14/(−2) = 7
Answer
k = 7
14.Vectors: dot & cross product
Answer: |M| = 404.6 N·m
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.
Mean
μ = 1·0.35 + 2·0.3 + 3·0.05 + 4·0.3 = 2.3
Mean of the squares
E(X²) = 1·0.35 + 4·0.3 + 9·0.05 + 16·0.3 = 6.8
Variance
Var(X) = 6.8 − 2.3² = 1.51
Answer
Var(X) = 1.51
16.Probability
Answer: σY = 4.548
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 = 1.85, Var(X) = 0.8275, σX = 0.9097
Linear change
σY = |5|·0.9097 = 4.548
Answer
σY = 4.548
The mean does move: E(Y) = 5·1.85 + 10 = 19.25.
17.Descriptive statistics
Answer: s = 16.12
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 = 1.5
Standard normal table
Φ(1.5) = 0.9332
P = 0.9332
Answer
P = 0.9332
19.Normal distribution
Answer: P = 0.4195
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 = (65.5 − 65)/10 = 0.05
z₂ = (80.5 − 65)/10 = 1.55
Area between
P = Φ(1.55) − Φ(0.05) = 0.9394 − 0.5199 = 0.4195
Answer
P = 0.4195
20.Normal distribution
Answer: x = 84.4 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.