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→∞ (−2x3 − 2x + 7)/(4x3 + 4x + 9).
f(x) = (−2x3 − 2x + 7)/(4x3 + 4x + 9)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e3x − 1 − 3x)/x².
f(x) = (e3x − 1 − 3x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x2·ex. Find f′(1.5).
f(x) = x2·exx = 1.5
f′(1.5) =
4.DerivativesHard
Let f(x) = x3 − 10.5x2 + 30x − 6. Find the x-coordinate x of the local minimum.
f(x) = x3 − 10.5x2 + 30x − 6
x =
5.IntegralsMedium
Evaluate I = ∫₀1 x·sin x dx (x in radians).
∫₀1 x·sin x dx
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 1 − x² and the x-axis.
y = 1 − x²
A =
7.Complex numbersMedium
Write z = 8∠110° in rectangular form a + jb and give the imaginary part b.
r = 8θ = 110°
b =
8.Complex numbersHard
One root of z³ + 9z² + 25z + 25 = 0 is z₁ = −2 + j. The coefficients are real. Find the real root r.
z₁ = −2 + j
r =
9.Differential equationsMedium
Solve y″ + 7y′ + 12y = 0 with y(0) = 3 and y′(0) = -2. Find y(1).
y″ + 7y′ + 12y = 0y(0) = 3y′(0) = -2
y(1) =
10.Differential equationsHard
A tank holds 1000 L of water in which 10 kg of salt is dissolved. Brine with 0.2 kg of salt per litre flows in at 2 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 143 kg of salt?
V = 1000 Lm0 = 10 kgc = 0.2 kg/Lr = 2 L/min
t = min
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−1 −1 4; −3 −5 −4; −5 −4 0]
det A =
12.Matrices & determinantsHard
Solve the system −5x − 3y = 2 and −5x − 2y = −2. Find x.
−5x − 3y = 2−5x − 2y = −2
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (1, 1, 6) and b = (2, −4, k) are perpendicular.
a = (1, 1, 6)b = (2, −4, k)
k =
14.Vectors: dot & cross productHard
A force F = (−50, 50, −40) N acts at point A(0, −4, 1). Coordinates are in metres. Find the x-component Mx of the moment of the force about the origin O(0, 0, 0).
F = (−50, 50, −40) NA(0, −4, 1) mO(0, 0, 0) m
Mx = N·m
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2, 3p(x): 0.35, 0.15, 0.25, 0.25
σ =
16.ProbabilityHard
A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 5 correct answers?
n = 7p = 0.4k = 5
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 21, 45, 53, 37, 30, 35.
n = 6
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 60 and standard deviation 10. Find the z-score z of a score of 70.
μ = 60σ = 10x = 70
z =
19.Normal distributionMedium
The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 12 of them exceeds 298.3 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 12> 298.3 kg
P =
20.Normal distributionHard
Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
3Solutions
1.Limits
Answer: L = −0.5
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
(−2 + (−2)/x2 + 7/x3)/(4 + 4/x2 + 9/x3)
Let x → ∞
L = −2/4 = −0.5
Answer
L = −0.5
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 4.5
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
(3·e3x − 3)/(2x) → 0/0
Second application
9·e3x/2
Substitute x = 0
L = 9·1/2 = 4.5
Answer
L = 4.5
3.Derivatives
Answer: f′(1.5) = 23.53
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′ = ex·(2x + 1·x2)
Substitute
f′(1.5) = 4.482·(3 + 2.25) = 23.53
Answer
f′(1.5) = 23.53
4.Derivatives
Answer: x = 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.
Derivative
f′(x) = 3x2 − 21x + 30 = 3·(x2 − 7x + 10)
Solve f′(x) = 0
x = 2 or x = 5
Which one is which
f″(x) = 6x − 21: negative at a maximum, positive at a minimum.
f″(2) = −9, f″(5) = 9
x = 5
Answer
x = 5
5.Integrals
Answer: I = 0.3012
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.
Choose u and dv
u = x, du = dx; dv = sin x dx, v = −cos x
Antiderivative
∫x sin x dx = −x cos x + ∫cos x dx = −x cos x + sin x
Limits
(−1·cos 1 + sin 1) − (0 + 0) = 0.3012
Answer
I = 0.3012
6.Integrals
Answer: A = 1.333
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.
Zeros
1 − x² = 0 → x = ±1
Integrate
A = [1x − x³/3]−11 = 0.6667 − (−0.6667) = 1.333
Answer
A = 1.333
7.Complex numbers
Answer: b = 7.518
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.
Components
a = 8·cos(110°) = −2.736
b = 8·sin(110°) = 7.518
Rectangular form
z = −2.736 + j7.518
b = 7.518
Answer
b = 7.518
8.Complex numbers
Answer: r = −5
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.
Check with the sum of the roots: z₁ + z₂ + r = −4 + (−5) = −9 = −(9).
9.Differential equations
Answer: y(1) = 0.3697
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.
Characteristic roots
(r + 3)(r + 4) = 0 → r = −3, −4
Constants from the initial values
A + B = 3, −3A − 4B = -2
A = 10, B = −7
Evaluate
y(1) = 10·e−3 + (−7)·e−4 = 0.3697
Answer
y(1) = 0.3697
10.Differential equations
Answer: t = 602 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) = 200 + (−190)·e−t/500
Solve for t
e−t/500 = (200 − 143)/(200 − 10) = 0.3
t = 500·ln(3.333) = 602 min
Answer
t = 602 min
11.Matrices & determinants
Answer: det A = −56
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 = −16, M₁₂ = −20, M₁₃ = −13
Cofactor expansion
det A = (−1)·(−16) − (−1)·(−20) + 4·(−13) = −56
Answer
det A = −56
12.Matrices & determinants
Answer: x = 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.
Determinant of the coefficient matrix
det A = (−5)·(−2) − (−3)·(−5) = −5
Replace the x column
det Aₓ = 2·(−2) − (−3)·(−2) = −10
x = −10/(−5) = 2
Answer
x = 2
y = −4
13.Vectors: dot & cross product
Answer: k = 0.3333
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 = 1·2 + 1·(−4) + 6·k = 0
−2 + 6·k = 0
Solve for k
k = −(−2)/6 = 0.3333
Answer
k = 0.3333
14.Vectors: dot & cross product
Answer: Mx = 110 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 = 0·0.35 + 1·0.15 + 2·0.25 + 3·0.25 = 1.4
Mean of the squares
E(X²) = 0·0.35 + 1·0.15 + 4·0.25 + 9·0.25 = 3.4
Variance
Var(X) = 3.4 − 1.4² = 1.44
σ = √1.44 = 1.2
Answer
σ = 1.2
16.Probability
Answer: P = 0.07741
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.
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 = (70 − 60)/10 = 1
Answer
z = 1
19.Normal distribution
Answer: P = 0.8365
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 = 300 kg, σT = 0.5·√12 = 1.732 kg
Standardise
z = (298.3 − 300)/1.732 = −0.98
Probability
P = 1 − Φ(−0.98) = 0.8365
Answer
P = 0.8365
20.Normal distribution
Answer: P = 0.7881
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).