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→∞ (1 + 3/x)2x. Give a decimal value.
f(x) = (1 + 3/x)2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + x)/sin(5x), with x in radians.
f(x) = ln(1 + x)/sin(5x)x → 0
L =
3.DerivativesMedium
Let f(x) = x3·e−0.5x. Find f′(1.5).
f(x) = x3·e−0.5xx = 1.5
f′(1.5) =
4.DerivativesHard
The tangent line to f(x) = 3x2 + 5x at x = −1 crosses the y-axis at y = b. Find b.
f(x) = 3x2 + 5xx = −1
b =
5.IntegralsMedium
The region under y = 3x from x = 0 to x = 4 is rotated about the x-axis. Find the volume V of the solid (a cone).
y = 3x0 ≤ x ≤ 4
V =
6.IntegralsHard
The region between the line y = 2x and the parabola y = 2x² is rotated about the x-axis. Find the volume V of the solid.
y = 2xy = 2x²
V =
7.Complex numbersMedium
Find the argument θ of z = −6 − j2 in degrees, as the principal value between −180° and 180°.
z = −6 − j2
θ = °
8.Complex numbersHard
Exactly one of the points A–D is a root of z⁴ = 16∠90°. Click it.
w = 16∠90°n = 4
1A (1∠−157.5°)
2B (2∠157.5°)
3C (2∠−157.5°)
4D (2∠−112.5°)
9.Differential equationsMedium
A forged steel part at 420 °C is left to cool in a workshop at 22 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.05 min⁻¹, find its temperature T after 18 min.
T0 = 420 °CTa = 22 °Ck = 0.05 min⁻¹t = 18 min
T = °C
10.Differential equationsHard
Solve the initial value problem y′ + 3y/x = 6x for x > 0, with y(1) = 5. Find y(2).
y′ + 3y/x = 6xy(1) = 5
y(2) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [2 1 4; −2 −1 4; 1 0 −4]
det A =
12.Matrices & determinantsHard
Solve the system 5x + 2y = −10 and 5x − y = 2. Find x.
5x + 2y = −105x − y = 2
x =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (2, 6, 3) onto b = (0, 3, 5), that is, the component of a in the direction of b.
a = (2, 6, 3)b = (0, 3, 5)
compb a =
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (5, −1, −2), b = (−3, 1, 1) and c = (0, −4, 3).
a = (5, −1, −2)b = (−3, 1, 1)c = (0, −4, 3)
V =
15.ProbabilityMedium
A bag holds 8 red and 3 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
8 / 3
P =
16.ProbabilityHard
5 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?
n = 5p = 0.1
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 14, 39, 31, 28, 31.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 20. What proportion P of values is above 46?
μ = 70σ = 20x = 46
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 10 of them exceeds 407.6 g. Round z to two decimals.
μ = 40 gσ = 1.2 gn = 10> 407.6 g
P =
20.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 90% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points90%
x = points
3Solutions
1.Limits
Answer: L = 403.4
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).
Take the logarithm
ln L = limx→∞ 2·x·ln(1 + 3/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 2·3 = 6
Undo the logarithm
L = e6 = 403.4
Answer
L = 403.4
2.Limits
Answer: L = 0.2
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).
L'Hôpital's rule
L = limx→0 [1/(1 + x)]/[5·cos 5x]
Substitute x = 0
L = 1/5 = 0.2
Answer
L = 0.2
3.Derivatives
Answer: f′(1.5) = 2.391
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−0.5x·(3x2 + (−0.5)·x3)
Substitute
f′(1.5) = 0.4724·(6.75 + (−1.688)) = 2.391
Answer
f′(1.5) = 2.391
4.Derivatives
Answer: b = −3
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(−1) = −2
f′(x) = 6x + 5 → f′(−1) = −1
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −2 − (−1)·(−1) = −3
Answer
b = −3
5.Integrals
Answer: V = 603.2
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∫₀4 9x² dx
Integrate
V = π·9·4³/3 = 603.2
Answer
V = 603.2
Check: a cone of radius 12 and height 4 has (1/3)πr²h = 603.2.
6.Integrals
Answer: V = 1.676
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.
Intersection points
2x = 2x² → x = 0, x = 1
Washer method
V = π∫₀1 ((2x)² − (2x²)²) dx
Integrate
V = π(4·1³/3 − 4·1⁵/5) = π·0.5333 = 1.676
Answer
V = 1.676
7.Complex numbers
Answer: θ = −161.6 °
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 = −6, b = −2 → III
Reference angle
α = atan(2/6) = 18.43°
Argument
θ = −(180° − α) = −161.6°
Answer
θ = −161.6°
In polar form z = 6.325∠−161.6°.
8.Complex numbers
Answer: C (2∠−157.5°)
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.
Modulus and arguments of the roots
|z| = 2
θ = 22.5°, 112.5°, −157.5°, −67.5°
The point
C = 2∠−157.5°
Answer
C (2∠−157.5°)
9.Differential equations
Answer: T = 183.8 °C
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.
Solution of the cooling law
T(t) = 22 + 398·e−0.05t
Substitute t
T(18) = 22 + 398·0.4066 = 183.8 °C
Answer
T = 183.8 °C
10.Differential equations
Answer: y(2) = 5.275
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 = e∫3/x dx = x3
Integrate
(x3y)′ = 6x4 → x3y = 6x5/5 + C
Initial condition
5 = 6/5 + C → C = 3.8
Evaluate
y(2) = (6·25/5 + 3.8)/23 = 5.275
Answer
y(2) = 5.275
11.Matrices & determinants
Answer: det A = 8
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 = 4, M₁₂ = 4, M₁₃ = 1
Cofactor expansion
det A = 2·4 − 1·4 + 4·1 = 8
Answer
det A = 8
12.Matrices & determinants
Answer: x = −0.4
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.
Determinant of the coefficient matrix
det A = 5·(−1) − 2·5 = −15
Replace the x column
det Aₓ = (−10)·(−1) − 2·2 = 6
x = 6/(−15) = −0.4
Answer
x = −0.4
y = −4
13.Vectors: dot & cross product
Answer: compb a = 5.659
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 = 2·0 + 6·3 + 3·5 = 33
Length of b
|b| = √(0² + 3² + 5²) = 5.831
Scalar projection
compb a = a·b/|b| = 33/5.831 = 5.659
Answer
compb a = 5.659
14.Vectors: dot & cross product
Answer: V = 2
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 = 1·3 − 1·(−4) = 7
(b × c)y = bzcx − bxcz = 1·0 − (−3)·3 = 9
(b × c)z = bxcy − bycx = (−3)·(−4) − 1·0 = 12
Dot product with a
a·(b × c) = 5·7 + (−1)·9 + (−2)·12 = 2
V = |2| = 2
Answer
V = 2
15.Probability
Answer: P = 0.5091
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.
First and second draw
P = 8/11 · 7/10 = 56/110 = 0.5091
Answer
P = 0.5091
16.Probability
Answer: P = 0.9185
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) = 0.95 = 0.5905
P(X = 1) = 5·0.1·0.94 = 0.3281
Add them
P = 0.5905 + 0.3281 = 0.9185
Answer
P = 0.9185
17.Descriptive statistics
Answer: s = 9.127
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 = 0.02275
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 = 400 g, σT = 1.2·√10 = 3.795 g
Standardise
z = (407.6 − 400)/3.795 = 2
Probability
P = 1 − Φ(2) = 0.02275
Answer
P = 0.02275
20.Normal distribution
Answer: x = 52.2 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.