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→∞ (√(4x2 + 10x) − 2x).
f(x) = √(4x2 + 10x) − 2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 3x)/sin(2x), with x in radians.
f(x) = ln(1 + 3x)/sin(2x)x → 0
L =
3.DerivativesMedium
Let f(x) = (2x − 1)4. Find f′(1).
f(x) = (2x − 1)4x = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 + 1.5x2 − 6x − 6. Find the x-coordinate x of the local minimum.
f(x) = x3 + 1.5x2 − 6x − 6
x =
5.IntegralsMedium
Evaluate I = ∫03 e0.5x dx.
f(x) = e0.5x[0, 3]
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
Given z₁ = 4∠−35° and z₂ = 4∠20°, find the real part of w = z₁/z₂.
z₁ = 4∠−35°z₂ = 4∠20°
Re(w) =
8.Complex numbersHard
Solve z² − 6z + 13 = 0. The roots are a ± jb; find b > 0.
p = −6q = 13
b =
9.Differential equationsMedium
A cup of coffee at 89 °C is left in a room at 20 °C. After 23 min its temperature is 47 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 89 °CTa = 20 °CT(23) = 47 °C
k = min⁻¹
10.Differential equationsHard
Solve y″ + y′ + 25.25y = 0 with y(0) = 10 and y′(0) = 0 (released from rest). Find y(0.25); angles in radians.
y″ + y′ + 25.25y = 0y(0) = 10y′(0) = 0
y(0.25) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [−5 −1; −2 −1]B = [−6 6; 2 4]
c11 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−1 2; −3 6]
λ =
13.Vectors: dot & cross productMedium
A constant force F = (−10, 15, −30) N acts on an object while it moves in a straight line from point A(−1, −5, 3) to point B(−5, 0, −3). Coordinates are in metres. Find the work W done by the force.
F = (−10, 15, −30) NA(−1, −5, 3) mB(−5, 0, −3) m
W = J
14.Vectors: dot & cross productHard
For a = (−6, 0, −5) and b = (−3, −1, 3), find the x-component (a × b)x of the cross product a × b.
a = (−6, 0, −5)b = (−3, −1, 3)
(a × b)x =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 10, 20, 30, 40p(x): 0.1, 0.35, 0.35, 0.2
σ =
16.ProbabilityHard
X has the distribution shown, and Y = 10X + 1. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 53, 58, 24, 38, 34.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 40. What proportion P of values is above 540?
μ = 500σ = 40x = 540
P =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.3328. Find z to two decimals.
A = 0.3328
z =
20.Normal distributionHard
Hole diameters D are normal with mean 25.04 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 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 = 0.01 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
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) = 10x/(√(4x2 + 10x) + 2x)
Divide by x
f(x) = 10/(√(4 + 10/x) + 2)
Let x → ∞
L = 10/(2 + 2) = 2.5
Answer
L = 2.5
2.Limits
Answer: L = 1.5
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).
L'Hôpital's rule
L = limx→0 [3/(1 + 3x)]/[2·cos 2x]
Substitute x = 0
L = 3/2 = 1.5
Answer
L = 1.5
3.Derivatives
Answer: f′(1) = 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
f′(x) = 4·(2x − 1)3·2
Substitute
f′(1) = 4·13·2 = 8
Answer
f′(1) = 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 + 3x − 6 = 3·(x2 + x − 2)
Solve f′(x) = 0
x = −2 or x = 1
Which one is which
f″(x) = 6x + 3: negative at a maximum, positive at a minimum.
f″(−2) = −9, f″(1) = 9
x = 1
Answer
x = 1
5.Integrals
Answer: I = 6.963
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.
Antiderivative
F(x) = e0.5x/0.5
Evaluate at the limits
I = (e1.5 − 1)/0.5 = (4.482 − 1)/0.5 = 6.963
Answer
I = 6.963
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: Re(w) = 0.5736
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.
Modulus and argument of w
|w| = 4 / 4 = 1
θ = −35° − 20° = −55°
Rectangular form
Re(w) = 1·cos(−55°) = 0.5736
Im(w) = 1·sin(−55°) = −0.8192
Re(w) = 0.5736
Answer
Re(w) = 0.5736
8.Complex numbers
Answer: b = 2
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 = 36 − 52 = −16 < 0
Quadratic formula
z = (6 ± j√16)/2 = 3 + j2, 3 − j2
Answer
b = 2
Answer
b = 2
9.Differential equations
Answer: k = 0.04079 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−23k = 27/69
Solve for k
k = ln(69/27)/23 = 0.04079 min⁻¹
Answer
k = 0.04079 min⁻¹
10.Differential equations
Answer: y(0.25) = 3.62
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.
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)·(−6) + (−1)·2 = 28
Answer
c11 = 28
12.Matrices & determinants
Answer: λ = 5
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 − (5)λ + (0) = 0
Solve
λ = (5 ± √(25))/2 → λ = 0, 5
λ = 5
Answer
λ = 5
13.Vectors: dot & cross product
Answer: W = 295 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 = (−5, 0, −3) − (−1, −5, 3) = (−4, 5, −6) m
Work
W = F·d = (−10)·(−4) + 15·5 + (−30)·(−6) = 295 J
Answer
W = 295 J
14.Vectors: dot & cross product
Answer: (a × b)x = −5
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 = 0·3 − (−5)·(−1) = −5
(a × b)y = azbx − axbz = (−5)·(−3) − (−6)·3 = 33
(a × b)z = axby − aybx = (−6)·(−1) − 0·(−3) = 6
Requested component
(a × b)x = −5
Answer
(a × b)x = −5
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: σ = 9.097
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.
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 = 2.75, Var(X) = 0.9875, σX = 0.9937
Linear change
σY = |10|·0.9937 = 9.937
Answer
σY = 9.937
The mean does move: E(Y) = 10·2.75 + 1 = 28.5.
17.Descriptive statistics
Answer: s = 13.96
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
Standard normal table
Φ(1) = 0.8413
P = 1 − 0.8413 = 0.1587
Answer
P = 0.1587
19.Normal distribution
Answer: z = 0.43
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.
Area to the left of z
Φ(z) = (1 + 0.3328)/2 = 0.6664
Read the z-table backwards
Find 0.6664 inside the table and read off z.
z = 0.43
Answer
z = 0.43
20.Normal distribution
Answer: P = 0.9868
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).