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 − 1/x)x. Give a decimal value.
f(x) = (1 − 1/x)xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e2x − 1 − 2x)/x².
f(x) = (e2x − 1 − 2x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = (2x + 1)3. Find f′(1).
f(x) = (2x + 1)3x = 1
f′(1) =
4.DerivativesHard
The tangent line to f(x) = 2x2 − 3x − 5 at x = −1 crosses the y-axis at y = b. Find b.
f(x) = 2x2 − 3x − 5x = −1
b =
5.IntegralsMedium
Evaluate I = ∫₁2.5 x·ln x dx.
∫₁2.5 x ln x dx
I =
6.IntegralsHard
Find the area A of the region enclosed by y = 2x2 and y = 4x.
y = 2x2y = 4x
A =
7.Complex numbersMedium
The diagram shows z = 2 + j3. With w = 0.5∠90°, click the point that represents z/w.
z = 2 + j3w = 0.5∠90°
1A (4 + j6)
2B (−6 + j4)
3C (3 − j2)
4D (6 − j4)
8.Complex numbersHard
One root of z³ + 2z² − 6z − 72 = 0 is z₁ = −3 + j3. The coefficients are real. Find the real root r.
z₁ = −3 + j3
r =
9.Differential equationsMedium
A cup of coffee at 87 °C is left in a room at 22 °C. After 5 min its temperature is 58 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 87 °CTa = 22 °CT(5) = 58 °C
k = min⁻¹
10.Differential equationsHard
Solve y′ + 2y = 7e−3t with y(0) = 3. Find y(2).
y′ + 2y = 7e−3ty(0) = 3
y(2) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [1 −5 −1; 0 5 −1; 4 −2 −5]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [0 4; −6 10]
λ =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (5, 1, 4) onto b = (3, 1, −6), that is, the component of a in the direction of b.
a = (5, 1, 4)b = (3, 1, −6)
compb a =
14.Vectors: dot & cross productHard
A force F = (−60, −40, −10) N acts at point A(2, −1, 2). Coordinates are in metres. Find the z-component Mz of the moment of the force about point O(3, 2, 3).
F = (−60, −40, −10) NA(2, −1, 2) mO(3, 2, 3) m
Mz = N·m
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3p(x): 0.25, 0.3, 0.25, 0.2
Var(X) =
16.ProbabilityHard
11 components are tested, and each one fails with probability 0.5, independently. What is the probability P that at most one fails?
n = 11p = 0.5
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 48, 51, 27, 36, 34, 38.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 8. What proportion P of values is below 186?
μ = 170σ = 8x = 186
P =
19.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 438 and 508 h? Give P as a decimal.
μ = 500σ = 40x1 = 438x2 = 508
P =
20.Normal distributionHard
Hole diameters D are normal with mean 25.04 mm and standard deviation 0.015 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 = 25.04 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
3Solutions
1.Limits
Answer: L = 0.3679
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→∞ x·ln(1 − 1/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 1·(−1) = −1
Undo the logarithm
L = e−1 = 0.3679
Answer
L = 0.3679
2.Limits
Answer: L = 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).
First application
(2·e2x − 2)/(2x) → 0/0
Second application
4·e2x/2
Substitute x = 0
L = 4·1/2 = 2
Answer
L = 2
3.Derivatives
Answer: f′(1) = 54
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·(2x + 1)2·2
Substitute
f′(1) = 3·32·2 = 54
Answer
f′(1) = 54
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(−1) = 0
f′(x) = 4x − 3 → f′(−1) = −7
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 0 − (−7)·(−1) = −7
Answer
b = −7
5.Integrals
Answer: I = 1.551
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 = ln x, du = dx/x; dv = x dx, v = x²/2
Antiderivative
∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4
Limits
(3.125·ln 2.5 − 1.5625) − (0 − 0.25) = 1.551
Answer
I = 1.551
6.Integrals
Answer: A = 2.667
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.
Intersection points
2x2 = 4x → x = 0, x = 2
Integrate upper minus lower
A = [2x2 − 0.6667x3]02 = 2.667
Answer
A = 2.667
7.Complex numbers
Answer: D (6 − j4)
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 = 6 − j4 → D
Answer
D (6 − j4)
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: k = 0.1182 min⁻¹
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.
Ratio of temperature differences
e−5k = 36/65
Solve for k
k = ln(65/36)/5 = 0.1182 min⁻¹
Answer
k = 0.1182 min⁻¹
10.Differential equations
Answer: y(2) = 0.1658
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 = e2t
(e2ty)′ = 7e−t
General solution
y = 7/(−1)·e−3t + C·e−2t = −7·e−3t + C·e−2t
Initial condition
3 = −7 + C → C = 10
Evaluate
y(2) = −7·0.002479 + 10·0.01832 = 0.1658
Answer
y(2) = 0.1658
11.Matrices & determinants
Answer: det A = 13
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 = −27, M₁₂ = 4, M₁₃ = −20
Cofactor expansion
det A = 1·(−27) − (−5)·4 + (−1)·(−20) = 13
Answer
det A = 13
12.Matrices & determinants
Answer: λ = 6
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 − (10)λ + (24) = 0
Solve
λ = (10 ± √(4))/2 → λ = 4, 6
λ = 6
Answer
λ = 6
13.Vectors: dot & cross product
Answer: compb a = −1.18
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 = 5·3 + 1·1 + 4·(−6) = −8
Length of b
|b| = √(3² + 1² + (−6)²) = 6.782
Scalar projection
compb a = a·b/|b| = −8/6.782 = −1.18
Answer
compb a = −1.18
Negative: a has a component pointing against b.
14.Vectors: dot & cross product
Answer: Mz = −140 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 = (2, −1, 2) − (3, 2, 3) = (−1, −3, −1) m
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 = 0·0.25 + 1·0.3 + 2·0.25 + 3·0.2 = 1.4
Mean of the squares
E(X²) = 0·0.25 + 1·0.3 + 4·0.25 + 9·0.2 = 3.1
Variance
Var(X) = 3.1 − 1.4² = 1.14
Answer
Var(X) = 1.14
16.Probability
Answer: P = 0.005859
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) = 0.511 = 0.0004883
P(X = 1) = 11·0.5·0.510 = 0.005371
Add them
P = 0.0004883 + 0.005371 = 0.005859
Answer
P = 0.005859
17.Descriptive statistics
Answer: s = 8.989
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 = 2
Standard normal table
Φ(2) = 0.9772
P = 0.9772
Answer
P = 0.9772
19.Normal distribution
Answer: P = 0.5187
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 = (438 − 500)/40 = −1.55
z₂ = (508 − 500)/40 = 0.2
Area between
P = Φ(0.2) − Φ(−1.55) = 0.5793 − 0.06057 = 0.5187
Answer
P = 0.5187
20.Normal distribution
Answer: P = 0.9452
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).