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 + 8x) − 3x).
f(x) = √(9x2 + 8x) − 3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 5x)/sin(x), with x in radians.
f(x) = ln(1 + 5x)/sin(x)x → 0
L =
3.DerivativesMedium
Let f(x) = (3x + 3)3. Find f′(1).
f(x) = (3x + 3)3x = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 − 4.5x2 + 6x + 2. Find the x-coordinate x of the local maximum.
f(x) = x3 − 4.5x2 + 6x + 2
x =
5.IntegralsMedium
Evaluate I = ∫03 ex dx.
f(x) = ex[0, 3]
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 4 − x² and the x-axis.
y = 4 − x²
A =
7.Complex numbersMedium
Find the argument θ of z = −5 + j5 in degrees, as the principal value between −180° and 180°.
z = −5 + j5
θ = °
8.Complex numbersHard
Exactly one of the points A–D is a root of z³ = 27∠180°. Click it.
w = 27∠180°n = 3
1A (3∠−30°)
2B (5.4∠0°)
3C (3∠−90°)
4D (3∠−60°)
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
A tank holds 400 L of pure water. Brine with 0.25 kg of salt per litre flows in at 5 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 35 min?
V = 400 Lm0 = 0 kgc = 0.25 kg/Lr = 5 L/min
m = kg
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [6 6; −2 3]B = [4 −1; −6 −6]
c21 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [5 −3; 7 −5]
λ =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (3, 0, 3) onto b = (1, 4, 4), that is, the component of a in the direction of b.
a = (3, 0, 3)b = (1, 4, 4)
compb a =
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (3, 4, 1), b = (1, 2, 2) and c = (2, −1, −2).
a = (3, 4, 1)b = (1, 2, 2)c = (2, −1, −2)
V =
15.ProbabilityMedium
In how many ways N can a committee of 3 people be chosen from a group of 14? The order does not matter.
n = 14k = 3
N =
16.ProbabilityHard
12 components are tested, and each one fails with probability 0.5, independently. What is the probability P that at most one fails?
n = 12p = 0.5
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 24, 32, 24, 25, 45.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 12. What proportion P of values is above 35.6?
μ = 50σ = 12x = 35.6
P =
19.Normal distributionMedium
The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 9 of them is below 358.2 g. Round z to two decimals.
μ = 40 gσ = 1.2 gn = 9< 358.2 g
P =
20.Normal distributionHard
The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
3Solutions
1.Limits
Answer: L = 1.333
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) = 8x/(√(9x2 + 8x) + 3x)
Divide by x
f(x) = 8/(√(9 + 8/x) + 3)
Let x → ∞
L = 8/(3 + 3) = 1.333
Answer
L = 1.333
2.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).
L'Hôpital's rule
L = limx→0 [5/(1 + 5x)]/[cos x]
Substitute x = 0
L = 5/1 = 5
Answer
L = 5
3.Derivatives
Answer: f′(1) = 324
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
f′(x) = 3·(3x + 3)2·3
Substitute
f′(1) = 3·62·3 = 324
Answer
f′(1) = 324
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 − 9x + 6 = 3·(x2 − 3x + 2)
Solve f′(x) = 0
x = 1 or x = 2
Which one is which
f″(x) = 6x − 9: negative at a maximum, positive at a minimum.
f″(1) = −3, f″(2) = 3
x = 1
Answer
x = 1
5.Integrals
Answer: I = 19.09
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) = ex/1
Evaluate at the limits
I = (e3 − 1)/1 = (20.09 − 1)/1 = 19.09
Answer
I = 19.09
6.Integrals
Answer: A = 10.67
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.
Zeros
4 − x² = 0 → x = ±2
Integrate
A = [4x − x³/3]−22 = 5.333 − (−5.333) = 10.67
Answer
A = 10.67
7.Complex numbers
Answer: θ = 135 °
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 = 5 → II
Reference angle
α = atan(5/5) = 45°
Argument
θ = 180° − α = 135°
Answer
θ = 135°
In polar form z = 7.071∠135°.
8.Complex numbers
Answer: D (3∠−60°)
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 = 60°, 180°, −60°
The point
D = 3∠−60°
Answer
D (3∠−60°)
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: m = 35.44 kg
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.
Set up the equation
dm/dt = 5·0.25 − (5/400)·m = 1.25 − 0.0125·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 100 + (−100)·e−t/80
Substitute t
m(35) = 100 + (−100)·0.6456 = 35.44 kg
Answer
m = 35.44 kg
11.Matrices & determinants
Answer: c21 = −26
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
c21 = (−2)·4 + 3·(−6) = −26
Answer
c21 = −26
12.Matrices & determinants
Answer: λ = 2
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 + (−4) = 0
Solve
λ = (0 ± √(16))/2 → λ = −2, 2
λ = 2
Answer
λ = 2
13.Vectors: dot & cross product
Answer: compb a = 2.611
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 = 3·1 + 0·4 + 3·4 = 15
Length of b
|b| = √(1² + 4² + 4²) = 5.745
Scalar projection
compb a = a·b/|b| = 15/5.745 = 2.611
Answer
compb a = 2.611
14.Vectors: dot & cross product
Answer: V = 13
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·(−2) − 2·(−1) = −2
(b × c)y = bzcx − bxcz = 2·2 − 1·(−2) = 6
(b × c)z = bxcy − bycx = 1·(−1) − 2·2 = −5
Dot product with a
a·(b × c) = 3·(−2) + 4·6 + 1·(−5) = 13
V = |13| = 13
Answer
V = 13
15.Probability
Answer: N = 364
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.
Combinations
C(14, 3) = (14·13·12)/(3·2·1) = 364
Answer
N = 364
16.Probability
Answer: P = 0.003174
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.
Zero and one failure
P(X = 0) = 0.512 = 0.0002441
P(X = 1) = 12·0.5·0.511 = 0.00293
Add them
P = 0.0002441 + 0.00293 = 0.003174
Answer
P = 0.003174
17.Descriptive statistics
Answer: s = 9.028
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 = −1.2
Standard normal table
Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849
Answer
P = 0.8849
19.Normal distribution
Answer: P = 0.3085
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the total
μT = 360 g, σT = 1.2·√9 = 3.6 g
Standardise
z = (358.2 − 360)/3.6 = −0.5
Probability
P = Φ(−0.5) = 0.3085
Answer
P = 0.3085
20.Normal distribution
Answer: P = 0.7389
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).