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 ln(1 + 2x)/sin(x), with x in radians.
f(x) = ln(1 + 2x)/sin(x)x → 0
L =
3.DerivativesMedium
Let f(x) = (4x + 5)4. Find f′(0).
f(x) = (4x + 5)4x = 0
f′(0) =
4.DerivativesHard
Let f(x) = x3 + 6x2 + 4. Find the x-coordinate x of the local maximum.
f(x) = x3 + 6x2 + 4
x =
5.IntegralsMedium
Evaluate I = ∫01 2x·(x² + 1)1 dx.
f(x) = 2x·(x² + 1)1[0, 1]
I =
6.IntegralsHard
The region between the line y = 4x and the parabola y = 3x² is rotated about the x-axis. Find the volume V of the solid.
y = 4xy = 3x²
V =
7.Complex numbersMedium
Find the argument θ of z = 9 + j2 in degrees, as the principal value between −180° and 180°.
z = 9 + j2
θ = °
8.Complex numbersHard
Solve z² + 3z + 3.25 = 0. The roots are a ± jb; find the real part a.
p = 3q = 3.25
a =
9.Differential equationsMedium
A cup of coffee at 81 °C is left in a room at 23 °C. After 13 min its temperature is 41 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 81 °CTa = 23 °CT(13) = 41 °C
k = min⁻¹
10.Differential equationsHard
A system obeys y″ + 3y′ + 6.25y = 6.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 3y′ + 6.25y = 6.25y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [−5 3; −5 −1]B = [2 4; 3 −5]
c11 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [2 −3; −3 2]
λ =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (−1, 5, −6) and b = (3, 0, −5), in degrees.
a = (−1, 5, −6)b = (3, 0, −5)
θ = °
14.Vectors: dot & cross productHard
For a = (2, 1, 4) and b = (0, 5, −6), find the z-component (a × b)z of the cross product a × b.
a = (2, 1, 4)b = (0, 5, −6)
(a × b)z =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2p(x): 0.35, 0.3, 0.35
σ =
16.ProbabilityHard
X has the distribution shown, and Y = 3X + 4. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 46, 19, 46, 34, 41.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 75. What proportion P of values is above 530?
μ = 500σ = 75x = 530
P =
19.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9738. Find z to two decimals.
A = 0.9738
z =
20.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 95% is to the right of it. Find the cut-off x.
μ = 500 hσ = 40 h95%
x = h
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 = 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).
L'Hôpital's rule
L = limx→0 [2/(1 + 2x)]/[cos x]
Substitute x = 0
L = 2/1 = 2
Answer
L = 2
3.Derivatives
Answer: f′(0) = 2000
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·(4x + 5)3·4
Substitute
f′(0) = 4·53·4 = 2000
Answer
f′(0) = 2000
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 + 12x = 3·(x2 + 4x)
Solve f′(x) = 0
x = −4 or x = 0
Which one is which
f″(x) = 6x + 12: negative at a maximum, positive at a minimum.
f″(−4) = −12, f″(0) = 12
x = −4
Answer
x = −4
5.Integrals
Answer: I = 1.5
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 u1 du
Integrate and evaluate
I = [u2/2]12 = (4 − 1)/2 = 1.5
Answer
I = 1.5
6.Integrals
Answer: V = 15.89
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 = 3x² → x = 0, x = 1.333
Washer method
V = π∫₀1.333 ((4x)² − (3x²)²) dx
Integrate
V = π(16·1.333³/3 − 9·1.333⁵/5) = π·5.057 = 15.89
Answer
V = 15.89
7.Complex numbers
Answer: θ = 12.53 °
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.
Quadrant
a = 9, b = 2 → I
Reference angle
α = atan(2/9) = 12.53°
Argument
θ = α = 12.53°
Answer
θ = 12.53°
In polar form z = 9.22∠12.53°.
8.Complex numbers
Answer: a = −1.5
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 = 9 − 13 = −4 < 0
Quadratic formula
z = (−3 ± j√4)/2 = −1.5 + j, −1.5 − j
Answer
a = −1.5
Answer
a = −1.5
9.Differential equations
Answer: k = 0.09001 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−13k = 18/58
Solve for k
k = ln(58/18)/13 = 0.09001 min⁻¹
Answer
k = 0.09001 min⁻¹
10.Differential equations
Answer: ymax = 1.095
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.
Steady value and roots
yss = 1, r = −1.5 ± 2j
Constants from the initial values
y(0) = 1 + A = 0 → A = −1
y′(0) = −1.5A + 2B = 0 → B = −0.75
Peak
t = π/2 = 1.571
ymax = 1·(1 + e−1.5·π/2) = 1.095
Answer
ymax = 1.095
11.Matrices & determinants
Answer: c11 = −1
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)·2 + 3·3 = −1
Answer
c11 = −1
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 − (4)λ + (−5) = 0
Solve
λ = (4 ± √(36))/2 → λ = −1, 5
λ = 5
Answer
λ = 5
13.Vectors: dot & cross product
Answer: θ = 53.98 °
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 = (−1)·3 + 5·0 + (−6)·(−5) = 27
Magnitudes
|a| = √((−1)² + 5² + (−6)²) = 7.874
|b| = √(3² + 0² + (−5)²) = 5.831
Angle
cos θ = 27/(7.874·5.831) = 0.5881
θ = arccos(0.5881) = 53.98°
Answer
θ = 53.98°
14.Vectors: dot & cross product
Answer: (a × b)z = 10
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 = 1·(−6) − 4·5 = −26
(a × b)y = azbx − axbz = 4·0 − 2·(−6) = 12
(a × b)z = axby − aybx = 2·5 − 1·0 = 10
Requested component
(a × b)z = 10
Answer
(a × b)z = 10
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: σ = 0.8367
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 = 0·0.35 + 1·0.3 + 2·0.35 = 1
Mean of the squares
E(X²) = 0·0.35 + 1·0.3 + 4·0.35 = 1.7
Variance
Var(X) = 1.7 − 1² = 0.7
σ = √0.7 = 0.8367
Answer
σ = 0.8367
16.Probability
Answer: σY = 3.186
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 = 2.15, Var(X) = 1.128, σX = 1.062
Linear change
σY = |3|·1.062 = 3.186
Answer
σY = 3.186
The mean does move: E(Y) = 3·2.15 + 4 = 10.45.
17.Descriptive statistics
Answer: s = 11.3
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 = 0.4
Standard normal table
Φ(0.4) = 0.6554
P = 1 − 0.6554 = 0.3446
Answer
P = 0.3446
19.Normal distribution
Answer: z = −1.94
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 − 0.9738 = 0.0262
Read the z-table backwards
Find 0.0262 inside the table and read off z.
z = −1.94
Answer
z = −1.94
20.Normal distribution
Answer: x = 434 h
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.