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 − 4x) − 3x).
f(x) = √(9x2 − 4x) − 3xx → ∞
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) = x·e−x. Find f′(0.5).
f(x) = x·e−xx = 0.5
f′(0.5) =
4.DerivativesHard
Let f(x) = x3 − 1.5x2 − 6x + 5. Find the x-coordinate x of the local maximum.
f(x) = x3 − 1.5x2 − 6x + 5
x =
5.IntegralsMedium
Evaluate I = ∫28 3/x dx.
f(x) = 3/x[2, 8]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = x.
y = x2y = x
A =
7.Complex numbersMedium
Find the argument θ of z = −5 + j4 in degrees, as the principal value between −180° and 180°.
z = −5 + j4
θ = °
8.Complex numbersHard
Exactly one of the points A–D is a root of z³ = 8∠0°. Click it.
w = 8∠0°n = 3
1A (2∠60°)
2B (2∠0°)
3C (2∠30°)
4D (3.6∠0°)
9.Differential equationsMedium
Solve y″ + 8y′ + 16y = 0 with y(0) = 3 and y′(0) = 1. Find y(2).
y″ + 8y′ + 16y = 0y(0) = 3y′(0) = 1
y(2) =
10.Differential equationsHard
Solve y″ + 3y′ + 27.25y = 0 with y(0) = 1 and y′(0) = 0 (released from rest). Find y(0.25); angles in radians.
y″ + 3y′ + 27.25y = 0y(0) = 1y′(0) = 0
y(0.25) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [−3 5; 6 3]B = [−2 −4; 3 4]
c11 =
12.Matrices & determinantsHard
Solve the system 4x + y = −8 and x + 2y = 5. Find x.
4x + y = −8x + 2y = 5
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (0, −4, −1) and b = (−1, −4, 0), in degrees.
a = (0, −4, −1)b = (−1, −4, 0)
θ = °
14.Vectors: dot & cross productHard
A force F = (−10, 10, 10) N acts at point A(−4, −3, 2). Coordinates are in metres. Find the magnitude |M| of the moment of the force about point O(−3, 1, 3).
F = (−10, 10, 10) NA(−4, −3, 2) mO(−3, 1, 3) m
|M| = N·m
15.ProbabilityMedium
A fair die is rolled 7 times. What is the probability P of getting at least one six?
n = 7
P =
16.ProbabilityHard
X has the distribution shown, and Y = 3X + 5. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 22, 53, 57, 60, 48.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 25. Find the z-score z of a score of 480.
μ = 500σ = 25x = 480
z =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.9707. Find z to two decimals.
A = 0.9707
z =
20.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 90% of them lie (the 90th percentile).
μ = 100σ = 15P = 90%
x = points
3Solutions
1.Limits
Answer: L = −0.6667
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) = −4x/(√(9x2 − 4x) + 3x)
Divide by x
f(x) = −4/(√(9 + (−4)/x) + 3)
Let x → ∞
L = −4/(3 + 3) = −0.6667
Answer
L = −0.6667
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′(0.5) = 0.3033
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′ = e−x·(1 + (−1)·x)
Substitute
f′(0.5) = 0.6065·(1 + (−0.5)) = 0.3033
Answer
f′(0.5) = 0.3033
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 − 3x − 6 = 3·(x2 − x − 2)
Solve f′(x) = 0
x = −1 or x = 2
Which one is which
f″(x) = 6x − 3: negative at a maximum, positive at a minimum.
f″(−1) = −9, f″(2) = 9
x = −1
Answer
x = −1
5.Integrals
Answer: I = 4.159
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 8 − ln 2) = 3·ln 4 = 4.159
Answer
I = 4.159
6.Integrals
Answer: A = 0.1667
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
x2 = x → x = 0, x = 1
Integrate upper minus lower
A = [0.5x2 − 0.3333x3]01 = 0.1667
Answer
A = 0.1667
7.Complex numbers
Answer: θ = 141.3 °
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.
Quadrant
a = −5, b = 4 → II
Reference angle
α = atan(4/5) = 38.66°
Argument
θ = 180° − α = 141.3°
Answer
θ = 141.3°
In polar form z = 6.403∠141.3°.
8.Complex numbers
Answer: B (2∠0°)
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| = 2
θ = 0°, 120°, −120°
The point
B = 2∠0°
Answer
B (2∠0°)
9.Differential equations
Answer: y(2) = 0.009728
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 + 4)² = 0 → r = −4
Constants
A = y(0) = 3
y′(0) = B − 4A → B = 1 + 12 = 13
Evaluate
y(2) = (3 + 13·2)·e−8 = 0.009728
Answer
y(2) = 0.009728
10.Differential equations
Answer: y(0.25) = 0.4124
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.
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
c11 = (−3)·(−2) + 5·3 = 21
Answer
c11 = 21
12.Matrices & determinants
Answer: x = −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.
Determinant of the coefficient matrix
det A = 4·2 − 1·1 = 7
Replace the x column
det Aₓ = (−8)·2 − 1·5 = −21
x = −21/7 = −3
Answer
x = −3
y = 4
13.Vectors: dot & cross product
Answer: θ = 19.75 °
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 = 0·(−1) + (−4)·(−4) + (−1)·0 = 16
Magnitudes
|a| = √(0² + (−4)² + (−1)²) = 4.123
|b| = √((−1)² + (−4)² + 0²) = 4.123
Angle
cos θ = 16/(4.123·4.123) = 0.9412
θ = arccos(0.9412) = 19.75°
Answer
θ = 19.75°
14.Vectors: dot & cross product
Answer: |M| = 61.64 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.
Position vector from O to A
r = A − O = (−4, −3, 2) − (−3, 1, 3) = (−1, −4, −1) m
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.
Complement
P(no six) = (5/6)7 = 0.2791
P = 1 − 0.2791 = 0.7209
Answer
P = 0.7209
16.Probability
Answer: σY = 3.059
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 = 1.6, Var(X) = 1.04, σX = 1.02
Linear change
σY = |3|·1.02 = 3.059
Answer
σY = 3.059
The mean does move: E(Y) = 3·1.6 + 5 = 9.8.
17.Descriptive statistics
Answer: s = 15.22
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.
Mean
x̄ = 240/5 = 48
Squared deviations
Σ(x − x̄)² = (−26)² + 5² + 9² + 12² + 0² = 926
Standard deviation
s = √(926/4) = 15.22
Answer
s = 15.22
18.Descriptive statistics
Answer: z = −0.8
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 = (480 − 500)/25 = −0.8
Answer
z = −0.8
19.Normal distribution
Answer: z = 2.18
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 + 0.9707)/2 = 0.9854
Read the z-table backwards
Find 0.9854 inside the table and read off z.
z = 2.18
Answer
z = 2.18
20.Normal distribution
Answer: x = 119.2 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.