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 + 0.5/x)x. Give a decimal value.
f(x) = (1 + 0.5/x)xx → ∞
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) = ln(2x2 + 2). Find f′(−2).
f(x) = ln(2x2 + 2)x = −2
f′(−2) =
4.DerivativesHard
Let f(x) = x3 + 6x2 + 9x + 2. Find the x-coordinate x of the local minimum.
f(x) = x3 + 6x2 + 9x + 2
x =
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 between the parabola y = 1 − x² and the x-axis.
y = 1 − x²
A =
7.Complex numbersMedium
The diagram shows z = 2 + j2. With w = 2∠90°, click the point that represents z·w.
z = 2 + j2w = 2∠90°
1A (−4 + j4)
2B (4 + j4)
3C (4 − j4)
4D (−1 + j)
8.Complex numbersHard
Find the real part of (−2 + j2)³.
z = −2 + j2n = 3
Re(z³) =
9.Differential equationsMedium
A can of soda at 4 °C is taken out of the fridge into a room at 27 °C. After 30 min its temperature is 22 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 25 °C?
T0 = 4 °CTa = 27 °CT(30) = 22 °CT = 25 °C
t = min
10.Differential equationsHard
Solve y″ + 3y′ + 18.25y = 0 with y(0) = 10 and y′(0) = 0 (released from rest). Find y(0.75); angles in radians.
y″ + 3y′ + 18.25y = 0y(0) = 10y′(0) = 0
y(0.75) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−5 0 0; 0 −4 3; 0 −1 0]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [5 −3; 8 −6]
λ =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (3, −5, −4) and b = (1, k, 3) are perpendicular.
a = (3, −5, −4)b = (1, k, 3)
k =
14.Vectors: dot & cross productHard
A force F = (−40, −40, 60) N acts at point A(1, 4, −3). Coordinates are in metres. Find the y-component My of the moment of the force about point O(−2, 1, −3).
F = (−40, −40, 60) NA(1, 4, −3) mO(−2, 1, −3) m
My = N·m
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4p(x): 0.3, 0.1, 0.35, 0.25
σ =
16.ProbabilityHard
X has the distribution shown, and Y = −2X + 20. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 32, 48, 31, 14, 41.
n = 5
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 410.
μ = 500σ = 75x = 410
z =
19.Normal distributionMedium
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 248.5 and 252.5 g? Give P as a decimal.
μ = 250σ = 5x1 = 248.5x2 = 252.5
P =
20.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 95% of them lie (the 95th percentile).
μ = 178σ = 7P = 95%
x = cm
3Solutions
1.Limits
Answer: L = 1.649
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 + 0.5/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 1·0.5 = 0.5
Undo the logarithm
L = e0.5 = 1.649
Answer
L = 1.649
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′(−2) = −0.8
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 for the logarithm
f′(x) = 4x/(2x2 + 2)
Substitute
f′(−2) = −8/10 = −0.8
Answer
f′(−2) = −0.8
4.Derivatives
Answer: x = −1
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 + 12x + 9 = 3·(x2 + 4x + 3)
Solve f′(x) = 0
x = −3 or x = −1
Which one is which
f″(x) = 6x + 12: negative at a maximum, positive at a minimum.
f″(−3) = −6, f″(−1) = 6
x = −1
Answer
x = −1
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 = 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: A (−4 + 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| = 2·|z|, arg(z·w) = arg z + 90°
The point
z·w = −4 + j4 → A
Answer
A (−4 + j4)
8.Complex numbers
Answer: Re(z³) = 16
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.
Powers step by step
z² = (−2 + j2)(−2 + j2) = −j8
z³ = (−j8)(−2 + j2) = 16 + j16
Check with De Moivre
z = 2.828∠135° → z³ = 22.63∠45°
Re(z³) = 16
Answer
Re(z³) = 16
9.Differential equations
Answer: t = 48.01 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(-23/-5)/30 = 0.05087 min⁻¹
Time to reach the target
t = ln(-23/(−2))/0.05087 = 48.01 min
Answer
t = 48.01 min
10.Differential equations
Answer: y(0.75) = −3.042
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 = −1.5 ± 4j
Constants
A = 10, B = 1.5·10/4 = 3.75
Evaluate
y(0.75) = e−1.125(10·cos 3 + 3.75·sin 3) = −3.042
Answer
y(0.75) = −3.042
11.Matrices & determinants
Answer: det A = −15
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 = 3, M₁₂ = 0, M₁₃ = 0
Cofactor expansion
det A = (−5)·3 − 0·0 + 0·0 = −15
Answer
det A = −15
12.Matrices & determinants
Answer: λ = 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.
Characteristic equation
λ² − (−1)λ + (−6) = 0
Solve
λ = (−1 ± √(25))/2 → λ = −3, 2
λ = 2
Answer
λ = 2
13.Vectors: dot & cross product
Answer: k = −1.8
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 = 3·1 + (−5)·k + (−4)·3 = 0
−9 + (−5)·k = 0
Solve for k
k = −(−9)/(−5) = −1.8
Answer
k = −1.8
14.Vectors: dot & cross product
Answer: My = −180 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 = (1, 4, −3) − (−2, 1, −3) = (3, 3, 0) 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 = 1·0.3 + 2·0.1 + 3·0.35 + 4·0.25 = 2.55
Mean of the squares
E(X²) = 1·0.3 + 4·0.1 + 9·0.35 + 16·0.25 = 7.85
Variance
Var(X) = 7.85 − 2.55² = 1.348
σ = √1.348 = 1.161
Answer
σ = 1.161
16.Probability
Answer: σY = 4.02
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 = 2.6, Var(X) = 4.04, σX = 2.01
Linear change
σY = |−2|·2.01 = 4.02
Answer
σY = 4.02
The mean does move: E(Y) = −2·2.6 + 20 = 14.8.
17.Descriptive statistics
Answer: s = 12.79
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 = (410 − 500)/75 = −1.2
Answer
z = −1.2
19.Normal distribution
Answer: P = 0.3094
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 = (248.5 − 250)/5 = −0.3
z₂ = (252.5 − 250)/5 = 0.5
Area between
P = Φ(0.5) − Φ(−0.3) = 0.6915 − 0.3821 = 0.3094
Answer
P = 0.3094
20.Normal distribution
Answer: x = 189.5 cm
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.