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→∞ (x2 − x + 7)/(2x2 + x + 1).
f(x) = (x2 − x + 7)/(2x2 + x + 1)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 4x)/(2x²), with x in radians.
f(x) = (1 − cos 4x)/(2x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·ex. Find f′(1).
f(x) = x·exx = 1
f′(1) =
4.DerivativesHard
The tangent line to f(x) = x2 − 2x + 2 at x = 3 crosses the y-axis at y = b. Find b.
f(x) = x2 − 2x + 2x = 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
The region between the line y = 4x and the parabola y = x² is rotated about the x-axis. Find the volume V of the solid.
y = 4xy = x²
V =
7.Complex numbersMedium
The diagram shows z = 1 + j. With w = 0.5∠90°, click the point that represents z/w.
z = 1 + jw = 0.5∠90°
1A (1 − j)
2B (2 − j2)
3C (2 + j2)
4D (−2 + j2)
8.Complex numbersHard
One root of z³ + 2z² − 11z − 52 = 0 is z₁ = −3 + j2. The coefficients are real. Find the real root r.
z₁ = −3 + j2
r =
9.Differential equationsMedium
A can of soda at 3 °C is taken out of the fridge into a room at 27 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.07 min⁻¹, find its temperature T after 5 min.
T0 = 3 °CTa = 27 °Ck = 0.07 min⁻¹t = 5 min
T = °C
10.Differential equationsHard
A tank holds 400 L of pure water. Brine with 0.05 kg of salt per litre flows in at 2 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 15 min?
V = 400 Lm0 = 0 kgc = 0.05 kg/Lr = 2 L/min
m = kg
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [−5 1; 2 −3]B = [−4 2; −3 −4]
c11 =
12.Matrices & determinantsHard
Solve the system 5x − 6y = 7 and −2x + 5y = 12. Find x.
5x − 6y = 7−2x + 5y = 12
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (3, 3, 0) and b = (1, −6, 2), in degrees.
a = (3, 3, 0)b = (1, −6, 2)
θ = °
14.Vectors: dot & cross productHard
For a = (1, 6, 6) and b = (3, −5, 5), find the y-component (a × b)y of the cross product a × b.
a = (1, 6, 6)b = (3, −5, 5)
(a × b)y =
15.ProbabilityMedium
A bag holds 7 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
7 / 5
P =
16.ProbabilityHard
X has the distribution shown, and Y = 3X + 20. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 23, 55, 23, 32, 24, 22, 48.
n = 7
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 75. Find the z-score z of a score of 545.
μ = 500σ = 75x = 545
z =
19.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 57 and 77.5 points? Give P as a decimal.
μ = 65σ = 10x1 = 57x2 = 77.5
P =
20.Normal distributionHard
Hole diameters D are normal with mean 25.05 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.05 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.02 mm
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
(1 + (−1)/x + 7/x2)/(2 + 1/x + 1/x2)
Let x → ∞
L = 1/2 = 0.5
Answer
L = 0.5
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 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).
First application
4·sin 4x/(4x) → 0/0
Second application
16·cos 4x/4
Substitute x = 0
L = 16/4 = 4
Answer
L = 4
3.Derivatives
Answer: f′(1) = 5.437
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·(1 + 1·x)
Substitute
f′(1) = 2.718·(1 + 1) = 5.437
Answer
f′(1) = 5.437
4.Derivatives
Answer: b = −7
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) = 5
f′(x) = 2x − 2 → f′(3) = 4
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 5 − 4·3 = −7
Answer
b = −7
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: V = 428.9
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
4x = 1x² → x = 0, x = 4
Washer method
V = π∫₀4 ((4x)² − (1x²)²) dx
Integrate
V = π(16·4³/3 − 1·4⁵/5) = π·136.5 = 428.9
Answer
V = 428.9
7.Complex numbers
Answer: B (2 − j2)
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.
Effect of w
|z/w| = |z|/0.5, arg(z/w) = arg z − 90°
The point
z/w = 2 − j2 → B
Answer
B (2 − j2)
8.Complex numbers
Answer: r = 4
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 = −6 + 4 = −2 = −(2).
9.Differential equations
Answer: T = 10.09 °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) = 27 + (−24)·e−0.07t
Substitute t
T(5) = 27 + (−24)·0.7047 = 10.09 °C
Answer
T = 10.09 °C
10.Differential equations
Answer: m = 1.445 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 = 2·0.05 − (2/400)·m = 0.1 − 0.005·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 20 + (−20)·e−t/200
Substitute t
m(15) = 20 + (−20)·0.9277 = 1.445 kg
Answer
m = 1.445 kg
11.Matrices & determinants
Answer: c11 = 17
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 = (−5)·(−4) + 1·(−3) = 17
Answer
c11 = 17
12.Matrices & determinants
Answer: x = 8.231
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 = 5·5 − (−6)·(−2) = 13
Replace the x column
det Aₓ = 7·5 − (−6)·12 = 107
x = 107/13 = 8.231
Answer
x = 8.231
y = 5.692
13.Vectors: dot & cross product
Answer: θ = 123.5 °
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 + 3·(−6) + 0·2 = −15
Magnitudes
|a| = √(3² + 3² + 0²) = 4.243
|b| = √(1² + (−6)² + 2²) = 6.403
Angle
cos θ = −15/(4.243·6.403) = −0.5522
θ = arccos(−0.5522) = 123.5°
Answer
θ = 123.5°
14.Vectors: dot & cross product
Answer: (a × b)y = 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, component by component
(a × b)x = aybz − azby = 6·5 − 6·(−5) = 60
(a × b)y = azbx − axbz = 6·3 − 1·5 = 13
(a × b)z = axby − aybx = 1·(−5) − 6·3 = −23
Requested component
(a × b)y = 13
Answer
(a × b)y = 13
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: P = 0.3182
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 = 7/12 · 6/11 = 42/132 = 0.3182
Answer
P = 0.3182
16.Probability
Answer: σY = 28.26
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 = 22.5, Var(X) = 88.75, σX = 9.421
Linear change
σY = |3|·9.421 = 28.26
Answer
σY = 28.26
The mean does move: E(Y) = 3·22.5 + 20 = 87.5.
17.Descriptive statistics
Answer: s = 13.6
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 = (545 − 500)/75 = 0.6
Answer
z = 0.6
19.Normal distribution
Answer: P = 0.6825
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 = (57 − 65)/10 = −0.8
z₂ = (77.5 − 65)/10 = 1.25
Area between
P = Φ(1.25) − Φ(−0.8) = 0.8944 − 0.2119 = 0.6825
Answer
P = 0.6825
20.Normal distribution
Answer: P = 0.9616
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).