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 + 5x) − x).
f(x) = √(x2 + 5x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 4x)/(x²), with x in radians.
f(x) = (1 − cos 4x)/(x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·e0.5x. Find f′(0.5).
f(x) = x·e0.5xx = 0.5
f′(0.5) =
4.DerivativesHard
Let f(x) = x3 + 7.5x2 + 12x − 2. Find the x-coordinate x of the local maximum.
f(x) = x3 + 7.5x2 + 12x − 2
x =
5.IntegralsMedium
Evaluate I = ∫01 2x·(x² + 1)3 dx.
f(x) = 2x·(x² + 1)3[0, 1]
I =
6.IntegralsHard
Evaluate I = ∫₀1 x²·eˣ dx.
∫₀1 x²eˣ dx
I =
7.Complex numbersMedium
Write z = 6∠35° in rectangular form a + jb and give the imaginary part b.
r = 6θ = 35°
b =
8.Complex numbersHard
Solve z² − 2z + 7.25 = 0. The roots are a ± jb; find b > 0.
p = −2q = 7.25
b =
9.Differential equationsMedium
A cup of coffee at 89 °C is left in a room at 21 °C. After 22 min its temperature is 39 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 64 °C?
T0 = 89 °CTa = 21 °CT(22) = 39 °CT = 64 °C
t = min
10.Differential equationsHard
Solve y″ + y′ + 25.25y = 0 with y(0) = 5 and y′(0) = 0 (released from rest). Find y(1); angles in radians.
y″ + y′ + 25.25y = 0y(0) = 5y′(0) = 0
y(1) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [3 5 3; 4 −3 −2; 3 2 0]
det A =
12.Matrices & determinantsHard
Solve the system 5x − 5y = 1 and x + 4y = −12. Find x.
5x − 5y = 1x + 4y = −12
x =
13.Vectors: dot & cross productMedium
A constant force F = (30, 30, 0) N acts on an object while it moves in a straight line from point A(−5, −1, −4) to point B(1, −3, −4). Coordinates are in metres. Find the work W done by the force.
F = (30, 30, 0) NA(−5, −1, −4) mB(1, −3, −4) m
W = J
14.Vectors: dot & cross productHard
A force F = (60, 10, −10) N acts at point A(4, −3, −1). Coordinates are in metres. Find the magnitude |M| of the moment of the force about point O(1, 0, −3).
F = (60, 10, −10) NA(4, −3, −1) mO(1, 0, −3) m
|M| = N·m
15.ProbabilityMedium
A bag holds 3 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
3 / 7
P =
16.ProbabilityHard
X has the distribution shown, and Y = 3X − 3. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 29, 39, 57, 56, 20.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 12. What proportion P of values is above 67.2?
μ = 60σ = 12x = 67.2
P =
19.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 452 and 542 h? Give P as a decimal.
μ = 500σ = 40x1 = 452x2 = 542
P =
20.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 30% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g30%
x = g
3Solutions
1.Limits
Answer: L = 2.5
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).
Multiply by the conjugate
f(x) = 5x/(√(x2 + 5x) + x)
Divide by x
f(x) = 5/(√(1 + 5/x) + 1)
Let x → ∞
L = 5/(1 + 1) = 2.5
Answer
L = 2.5
2.Limits
Answer: L = 8
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/(2x) → 0/0
Second application
16·cos 4x/2
Substitute x = 0
L = 16/2 = 8
Answer
L = 8
3.Derivatives
Answer: f′(0.5) = 1.605
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′ = e0.5x·(1 + 0.5·x)
Substitute
f′(0.5) = 1.284·(1 + 0.25) = 1.605
Answer
f′(0.5) = 1.605
4.Derivatives
Answer: x = −4
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 + 15x + 12 = 3·(x2 + 5x + 4)
Solve f′(x) = 0
x = −4 or x = −1
Which one is which
f″(x) = 6x + 15: negative at a maximum, positive at a minimum.
f″(−4) = −9, f″(−1) = 9
x = −4
Answer
x = −4
5.Integrals
Answer: I = 3.75
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.
Substitution
u = x² + 1, du = 2x dx
I = ∫12 u3 du
Integrate and evaluate
I = [u4/4]12 = (16 − 1)/4 = 3.75
Answer
I = 3.75
6.Integrals
Answer: I = 0.7183
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.
First round
∫x²eˣ dx = x²eˣ − ∫2x eˣ dx
Second round
∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)
Limits
2.718·(1) − 1·2 = 0.7183
Answer
I = 0.7183
7.Complex numbers
Answer: b = 3.441
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 = 6·cos(35°) = 4.915
b = 6·sin(35°) = 3.441
Rectangular form
z = 4.915 + j3.441
b = 3.441
Answer
b = 3.441
8.Complex numbers
Answer: b = 2.5
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.
Discriminant
p² − 4q = 4 − 29 = −25 < 0
Quadratic formula
z = (2 ± j√25)/2 = 1 + j2.5, 1 − j2.5
Answer
b = 2.5
Answer
b = 2.5
9.Differential equations
Answer: t = 7.586 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.
Cooling constant
k = ln(68/18)/22 = 0.06042 min⁻¹
Time to reach the target
t = ln(68/43)/0.06042 = 7.586 min
Answer
t = 7.586 min
10.Differential equations
Answer: y(1) = 0.5694
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.
Complex roots
r = −0.5 ± 5j
Constants
A = 5, B = 0.5·5/5 = 0.5
Evaluate
y(1) = e−0.5(5·cos 5 + 0.5·sin 5) = 0.5694
Answer
y(1) = 0.5694
11.Matrices & determinants
Answer: det A = 33
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 = 6, M₁₃ = 17
Cofactor expansion
det A = 3·4 − 5·6 + 3·17 = 33
Answer
det A = 33
12.Matrices & determinants
Answer: x = −2.24
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 = 5·4 − (−5)·1 = 25
Replace the x column
det Aₓ = 1·4 − (−5)·(−12) = −56
x = −56/25 = −2.24
Answer
x = −2.24
y = −2.44
13.Vectors: dot & cross product
Answer: W = 120 J
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.
Displacement
d = B − A = (1, −3, −4) − (−5, −1, −4) = (6, −2, 0) m
Work
W = F·d = 30·6 + 30·(−2) + 0·0 = 120 J
Answer
W = 120 J
14.Vectors: dot & cross product
Answer: |M| = 258.3 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, −1) − (1, 0, −3) = (3, −3, 2) 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.
First and second draw
P = 3/10 · 2/9 = 6/90 = 0.06667
Answer
P = 0.06667
16.Probability
Answer: σY = 4.078
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 = 2.55, Var(X) = 1.848, σX = 1.359
Linear change
σY = |3|·1.359 = 4.078
Answer
σY = 4.078
The mean does move: E(Y) = 3·2.55 − 3 = 4.65.
17.Descriptive statistics
Answer: s = 16.33
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 = 0.6
Standard normal table
Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743
Answer
P = 0.2743
19.Normal distribution
Answer: P = 0.7381
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 = (452 − 500)/40 = −1.2
z₂ = (542 − 500)/40 = 1.05
Area between
P = Φ(1.05) − Φ(−1.2) = 0.8531 − 0.1151 = 0.7381
Answer
P = 0.7381
20.Normal distribution
Answer: x = 252.6 g
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.