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→∞ (√(9x2 + 7x) − 3x).
f(x) = √(9x2 + 7x) − 3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e−0.5x − 1 + 0.5x)/x².
f(x) = (e−0.5x − 1 + 0.5x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x2·e2x. Find f′(1).
f(x) = x2·e2xx = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 − 4.5x2 + 6x − 1. Find the x-coordinate x of the local maximum.
f(x) = x3 − 4.5x2 + 6x − 1
x =
5.IntegralsMedium
Evaluate I = ∫12 1/x dx.
f(x) = 1/x[1, 2]
I =
6.IntegralsHard
The region between the line y = 5x and the parabola y = 3x² is rotated about the x-axis. Find the volume V of the solid.
y = 5xy = 3x²
V =
7.Complex numbersMedium
Find the argument θ of z = −3 − j7 in degrees, as the principal value between −180° and 180°.
z = −3 − j7
θ = °
8.Complex numbersHard
The roots of zⁿ = w with n = 2 and w = 16∠−60° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1. Find the argument of z₁ in degrees, between −180° and 180°.
w = 16∠−60°n = 2k = 1
θ₁ = °
9.Differential equationsMedium
A forged steel part at 420 °C is left to cool in a workshop at 21 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.04 min⁻¹, find its temperature T after 33 min.
T0 = 420 °CTa = 21 °Ck = 0.04 min⁻¹t = 33 min
T = °C
10.Differential equationsHard
A system obeys y″ + 4y′ + 13y = 13 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.5); angles in radians.
y″ + 4y′ + 13y = 13y(0) = 0y′(0) = 0
y(0.5) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [−3 5; 6 5]B = [−1 5; 4 0]
c11 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−1 4; −5 8]
λ =
13.Vectors: dot & cross productMedium
A constant force F = (0, −10, −20) N acts on an object while it moves in a straight line from point A(−5, −1, 2) to point B(2, 2, −4). Coordinates are in metres. Find the work W done by the force.
F = (0, −10, −20) NA(−5, −1, 2) mB(2, 2, −4) m
W = J
14.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (−6, 2, 0) and b = (1, −2, 6).
a = (−6, 2, 0)b = (1, −2, 6)
A =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 2, 4, 6p(x): 0.35, 0.25, 0.05, 0.35
Var(X) =
16.ProbabilityHard
X has the distribution shown, and Y = 0.5X + 5. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 34, 49, 22, 56, 42.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 12. Find the z-score z of a score of 82.
μ = 70σ = 12x = 82
z =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.8098. Find z to two decimals.
A = 0.8098
z =
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 bottom 60% is to the left of it. Find the cut-off x.
μ = 250 gσ = 5 g60%
x = g
3Solutions
1.Limits
Answer: L = 1.167
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) = 7x/(√(9x2 + 7x) + 3x)
Divide by x
f(x) = 7/(√(9 + 7/x) + 3)
Let x → ∞
L = 7/(3 + 3) = 1.167
Answer
L = 1.167
2.Limits
Answer: L = 0.125
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
(−0.5·e−0.5x + 0.5)/(2x) → 0/0
Second application
0.25·e−0.5x/2
Substitute x = 0
L = 0.25·1/2 = 0.125
Answer
L = 0.125
3.Derivatives
Answer: f′(1) = 29.56
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′ = e2x·(2x + 2·x2)
Substitute
f′(1) = 7.389·(2 + 2) = 29.56
Answer
f′(1) = 29.56
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 − 9x + 6 = 3·(x2 − 3x + 2)
Solve f′(x) = 0
x = 1 or x = 2
Which one is which
f″(x) = 6x − 9: negative at a maximum, positive at a minimum.
f″(1) = −3, f″(2) = 3
x = 1
Answer
x = 1
5.Integrals
Answer: I = 0.6931
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) = ln x
Evaluate at the limits
I = (ln 2 − ln 1) = ln 2 = 0.6931
Answer
I = 0.6931
6.Integrals
Answer: V = 48.48
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
5x = 3x² → x = 0, x = 1.667
Washer method
V = π∫₀1.667 ((5x)² − (3x²)²) dx
Integrate
V = π(25·1.667³/3 − 9·1.667⁵/5) = π·15.43 = 48.48
Answer
V = 48.48
7.Complex numbers
Answer: θ = −113.2 °
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 = −3, b = −7 → III
Reference angle
α = atan(7/3) = 66.8°
Argument
θ = −(180° − α) = −113.2°
Answer
θ = −113.2°
In polar form z = 7.616∠−113.2°.
8.Complex numbers
Answer: θ₁ = 150 °
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.
Modulus of the roots
161/2 = 4
Argument of the root
θ₁ = (−60° + 360°·1)/2 = 150°
Argument
θ₁ = 150°
Answer
θ₁ = 150°
The 2 roots lie on the circle of radius 4, 180° apart.
9.Differential equations
Answer: T = 127.6 °C
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.
Solution of the cooling law
T(t) = 21 + 399·e−0.04t
Substitute t
T(33) = 21 + 399·0.2671 = 127.6 °C
Answer
T = 127.6 °C
10.Differential equations
Answer: y(0.5) = 0.7293
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 = (−3)·(−1) + 5·4 = 23
Answer
c11 = 23
12.Matrices & determinants
Answer: λ = 4
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 − (7)λ + (12) = 0
Solve
λ = (7 ± √(1))/2 → λ = 3, 4
λ = 4
Answer
λ = 4
13.Vectors: dot & cross product
Answer: W = 90 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 = (2, 2, −4) − (−5, −1, 2) = (7, 3, −6) m
Work
W = F·d = 0·7 + (−10)·3 + (−20)·(−6) = 90 J
Answer
W = 90 J
14.Vectors: dot & cross product
Answer: A = 39.24
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
(a × b)x = aybz − azby = 2·6 − 0·(−2) = 12
(a × b)y = azbx − axbz = 0·1 − (−6)·6 = 36
(a × b)z = axby − aybx = (−6)·(−2) − 2·1 = 10
Area
A = |a × b| = √(12² + 36² + 10²) = 39.24
Answer
A = 39.24
15.Probability
Answer: Var(X) = 6.56
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 = 25.5, Var(X) = 114.8, σX = 10.71
Linear change
σY = |0.5|·10.71 = 5.356
Answer
σY = 5.356
The mean does move: E(Y) = 0.5·25.5 + 5 = 17.75.
17.Descriptive statistics
Answer: s = 13.22
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 = (82 − 70)/12 = 1
Answer
z = 1
19.Normal distribution
Answer: z = 1.31
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 + 0.8098)/2 = 0.9049
Read the z-table backwards
Find 0.9049 inside the table and read off z.
z = 1.31
Answer
z = 1.31
20.Normal distribution
Answer: x = 251.3 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.