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→∞ (x3 − 6x − 5)/(5x3 + 3x + 7).
f(x) = (x3 − 6x − 5)/(5x3 + 3x + 7)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 5x)/(3x²), with x in radians.
f(x) = (1 − cos 5x)/(3x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(2x), with x in radians. Find f′(0.5).
f(x) = x·sin(2x)x = 0.5 rad
f′(0.5) =
4.DerivativesHard
Let f(x) = x3 − 12x + 5. Find the x-coordinate x of the local maximum.
f(x) = x3 − 12x + 5
x =
5.IntegralsMedium
Evaluate I = ∫01 2x·(x² + 1)2 dx.
f(x) = 2x·(x² + 1)2[0, 1]
I =
6.IntegralsHard
The region between the line y = x and the parabola y = 0.5x² is rotated about the x-axis. Find the volume V of the solid.
y = xy = 0.5x²
V =
7.Complex numbersMedium
Given z₁ = 1.5∠70° and z₂ = 4∠−50°, find the argument θ of w = z₁·z₂ in degrees, between −180° and 180°.
z₁ = 1.5∠70°z₂ = 4∠−50°
θ = °
8.Complex numbersHard
One root of z³ − 5z² + 24z − 20 = 0 is z₁ = 2 + j4. The coefficients are real. Find the real root r.
z₁ = 2 + j4
r =
9.Differential equationsMedium
Solve y″ + 7y′ + 12y = 0 with y(0) = 5 and y′(0) = 4. Find y(1).
y″ + 7y′ + 12y = 0y(0) = 5y′(0) = 4
y(1) =
10.Differential equationsHard
Solve y′ + y = 7e−3t with y(0) = 2. Find y(1).
y′ + y = 7e−3ty(0) = 2
y(1) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [−4 −6; 2 6]B = [−4 −5; 0 −2]
c11 =
12.Matrices & determinantsHard
Solve the system −5x − y = 11 and 3x − 6y = 1. Find x.
−5x − y = 113x − 6y = 1
x =
13.Vectors: dot & cross productMedium
A constant force F = (−30, 5, 30) N acts on an object while it moves in a straight line from point A(2, 3, −1) to point B(5, 3, −5). Coordinates are in metres. Find the work W done by the force.
F = (−30, 5, 30) NA(2, 3, −1) mB(5, 3, −5) m
W = J
14.Vectors: dot & cross productHard
For a = (−1, −4, 3) and b = (−4, 4, −2), find the z-component (a × b)z of the cross product a × b.
a = (−1, −4, 3)b = (−4, 4, −2)
(a × b)z =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4, 5p(x): 0.1, 0.35, 0.3, 0.1, 0.15
σ =
16.ProbabilityHard
X has the distribution shown, and Y = 0.5X − 3. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 10, 37, 57, 17, 56, 18.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 15. What proportion P of values is above 55?
μ = 70σ = 15x = 55
P =
19.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 85.75 and 130.75 points? Give P as a decimal.
μ = 100σ = 15x1 = 85.75x2 = 130.75
P =
20.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 30% is to the left of it. Find the cut-off x.
μ = 100 pointsσ = 15 points30%
x = points
3Solutions
1.Limits
Answer: L = 0.2
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).
Divide by the highest power
(1 + (−6)/x2 + (−5)/x3)/(5 + 3/x2 + 7/x3)
Let x → ∞
L = 1/5 = 0.2
Answer
L = 0.2
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 4.167
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).
First application
5·sin 5x/(6x) → 0/0
Second application
25·cos 5x/6
Substitute x = 0
L = 25/6 = 4.167
Answer
L = 4.167
3.Derivatives
Answer: f′(0.5) = 1.382
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 and chain rule
f′(x) = sin(2x) + 2x·cos(2x)
Substitute
f′(0.5) = 0.8415 + 1·0.5403 = 1.382
Answer
f′(0.5) = 1.382
4.Derivatives
Answer: x = −2
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 − 12 = 3·(x2 − 4)
Solve f′(x) = 0
x = −2 or x = 2
Which one is which
f″(x) = 6x: negative at a maximum, positive at a minimum.
f″(−2) = −12, f″(2) = 12
x = −2
Answer
x = −2
5.Integrals
Answer: I = 2.333
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.
Substitution
u = x² + 1, du = 2x dx
I = ∫12 u2 du
Integrate and evaluate
I = [u3/3]12 = (8 − 1)/3 = 2.333
Answer
I = 2.333
6.Integrals
Answer: V = 3.351
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
1x = 0.5x² → x = 0, x = 2
Washer method
V = π∫₀2 ((1x)² − (0.5x²)²) dx
Integrate
V = π(1·2³/3 − 0.25·2⁵/5) = π·1.067 = 3.351
Answer
V = 3.351
7.Complex numbers
Answer: θ = 20 °
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| = 1.5 · 4 = 6
θ = 70° + (−50)° = 20°
Argument
θ = 20°
Answer
θ = 20°
8.Complex numbers
Answer: r = 1
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 = 4 + 1 = 5 = −(−5).
9.Differential equations
Answer: y(1) = 0.8469
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.
Characteristic roots
(r + 3)(r + 4) = 0 → r = −3, −4
Constants from the initial values
A + B = 5, −3A − 4B = 4
A = 24, B = −19
Evaluate
y(1) = 24·e−3 + (−19)·e−4 = 0.8469
Answer
y(1) = 0.8469
10.Differential equations
Answer: y(1) = 1.849
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 = et
(ety)′ = 7e−2t
General solution
y = 7/(−2)·e−3t + C·e−t = −3.5·e−3t + C·e−t
Initial condition
2 = −3.5 + C → C = 5.5
Evaluate
y(1) = −3.5·0.04979 + 5.5·0.3679 = 1.849
Answer
y(1) = 1.849
11.Matrices & determinants
Answer: c11 = 16
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 = (−4)·(−4) + (−6)·0 = 16
Answer
c11 = 16
12.Matrices & determinants
Answer: x = −1.97
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.
Determinant of the coefficient matrix
det A = (−5)·(−6) − (−1)·3 = 33
Replace the x column
det Aₓ = 11·(−6) − (−1)·1 = −65
x = −65/33 = −1.97
Answer
x = −1.97
y = −1.152
13.Vectors: dot & cross product
Answer: W = −210 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, 3, −5) − (2, 3, −1) = (3, 0, −4) m
Work
W = F·d = (−30)·3 + 5·0 + 30·(−4) = −210 J
Answer
W = −210 J
Negative work: the force acts mostly against the motion.
14.Vectors: dot & cross product
Answer: (a × b)z = −20
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 = (−4)·(−2) − 3·4 = −4
(a × b)y = azbx − axbz = 3·(−4) − (−1)·(−2) = −14
(a × b)z = axby − aybx = (−1)·4 − (−4)·(−4) = −20
Requested component
(a × b)z = −20
Answer
(a × b)z = −20
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: σ = 1.195
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 = 3.35, Var(X) = 1.628, σX = 1.276
Linear change
σY = |0.5|·1.276 = 0.6379
Answer
σY = 0.6379
The mean does move: E(Y) = 0.5·3.35 − 3 = −1.325.
17.Descriptive statistics
Answer: s = 20.64
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) = 0.1587
P = 1 − 0.1587 = 0.8413
Answer
P = 0.8413
19.Normal distribution
Answer: P = 0.8088
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 = (85.75 − 100)/15 = −0.95
z₂ = (130.75 − 100)/15 = 2.05
Area between
P = Φ(2.05) − Φ(−0.95) = 0.9798 − 0.1711 = 0.8088
Answer
P = 0.8088
20.Normal distribution
Answer: x = 92.2 points
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.