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 − 2/x)3x. Give a decimal value.
f(x) = (1 − 2/x)3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 4x)/(x²), with x in radians.
f(x) = (1 − cos 4x)/(x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x2·e2x. Find f′(1.5).
f(x) = x2·e2xx = 1.5
f′(1.5) =
4.DerivativesHard
The tangent line to f(x) = x2 + 3x − 5 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = x2 + 3x − 5x = 1
b =
5.IntegralsMedium
Evaluate I = ∫₁1.5 x·ln x dx.
∫₁1.5 x ln x dx
I =
6.IntegralsHard
Evaluate I = ∫₀1 x²·eˣ dx.
∫₀1 x²eˣ dx
I =
7.Complex numbersMedium
The number z = 2·ej2.8 has its angle in radians. Find its real part a.
r = 2θ = 2.8 rad
a =
8.Complex numbersHard
Use De Moivre's theorem to find the imaginary part of z⁶ for z = 1 + j.
z = 1 + jn = 6
Im(z⁶) =
9.Differential equationsMedium
A cup of coffee at 84 °C is left in a room at 22 °C. After 16 min its temperature is 37 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 51 °C?
T0 = 84 °CTa = 22 °CT(16) = 37 °CT = 51 °C
t = min
10.Differential equationsHard
Solve y′ + 2y = 6e−3t with y(0) = 1. Find y(1.5).
y′ + 2y = 6e−3ty(0) = 1
y(1.5) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [1 6; 6 6]B = [−3 3; 1 5]
c11 =
12.Matrices & determinantsHard
Solve the system x + 5y = −3 and 4x + y = 5. Find x.
x + 5y = −34x + y = 5
x =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (−3, 3, −5) and b = (−6, −5, 2).
a = (−3, 3, −5)b = (−6, −5, 2)
a·b =
14.Vectors: dot & cross productHard
For a = (2, −2, −6) and b = (6, 3, −1), find the y-component (a × b)y of the cross product a × b.
a = (2, −2, −6)b = (6, 3, −1)
(a × b)y =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 10, 20, 30, 40p(x): 0.2, 0.35, 0.1, 0.35
Var(X) =
16.ProbabilityHard
X has the distribution shown, and Y = −2X + 1. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 41, 43, 38, 30, 49.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 10. What proportion P of values is below 85?
μ = 100σ = 10x = 85
P =
19.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 65 and 75.5 points? Give P as a decimal.
μ = 65σ = 10x1 = 65x2 = 75.5
P =
20.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 95% of them lie (the 95th percentile).
μ = 65σ = 10P = 95%
x = points
3Solutions
1.Limits
Answer: L = 0.002479
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→∞ 3·x·ln(1 − 2/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 3·(−2) = −6
Undo the logarithm
L = e−6 = 0.002479
Answer
L = 0.002479
2.Limits
Answer: L = 8
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/(2x) → 0/0
Second application
16·cos 4x/2
Substitute x = 0
L = 16/2 = 8
Answer
L = 8
3.Derivatives
Answer: f′(1.5) = 150.6
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′(1.5) = 20.09·(3 + 4.5) = 150.6
Answer
f′(1.5) = 150.6
4.Derivatives
Answer: b = −6
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.
Point and slope
f(1) = −1
f′(x) = 2x + 3 → f′(1) = 5
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −1 − 5·1 = −6
Answer
b = −6
5.Integrals
Answer: I = 0.1436
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
(1.125·ln 1.5 − 0.5625) − (0 − 0.25) = 0.1436
Answer
I = 0.1436
6.Integrals
Answer: I = 0.7183
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.
First round
∫x²eˣ dx = x²eˣ − ∫2x eˣ dx
Second round
∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)
Limits
2.718·(1) − 1·2 = 0.7183
Answer
I = 0.7183
7.Complex numbers
Answer: a = −1.884
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.
Euler's formula
z = 2·(cos 2.8 + j sin 2.8)
Components (radians)
a = 2·cos 2.8 = −1.884
b = 2·sin 2.8 = 0.67
a = −1.884
Answer
a = −1.884
8.Complex numbers
Answer: Im(z⁶) = −8
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.
Polar form
z = 1 + j = √2∠45°
De Moivre
z⁶ = (√2)⁶∠270° = 8∠−90°
Rectangular form
z⁶ = 8·(cos −90° + j sin −90°) = −j8
Im(z⁶) = −8
Answer
Im(z⁶) = −8
9.Differential equations
Answer: t = 8.567 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(62/15)/16 = 0.08869 min⁻¹
Time to reach the target
t = ln(62/29)/0.08869 = 8.567 min
Answer
t = 8.567 min
10.Differential equations
Answer: y(1.5) = 0.2819
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 = e2t
(e2ty)′ = 6e−t
General solution
y = 6/(−1)·e−3t + C·e−2t = −6·e−3t + C·e−2t
Initial condition
1 = −6 + C → C = 7
Evaluate
y(1.5) = −6·0.01111 + 7·0.04979 = 0.2819
Answer
y(1.5) = 0.2819
11.Matrices & determinants
Answer: c11 = 3
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 = 1·(−3) + 6·1 = 3
Answer
c11 = 3
12.Matrices & determinants
Answer: x = 1.474
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 = 1·1 − 5·4 = −19
Replace the x column
det Aₓ = (−3)·1 − 5·5 = −28
x = −28/(−19) = 1.474
Answer
x = 1.474
y = −0.8947
13.Vectors: dot & cross product
Answer: a·b = −7
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.
Negative: the angle between the vectors is more than 90°.
14.Vectors: dot & cross product
Answer: (a × b)y = −34
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 = (−2)·(−1) − (−6)·3 = 20
(a × b)y = azbx − axbz = (−6)·6 − 2·(−1) = −34
(a × b)z = axby − aybx = 2·3 − (−2)·6 = 18
Requested component
(a × b)y = −34
Answer
(a × b)y = −34
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: Var(X) = 134
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 = 2.3, Var(X) = 1.11, σX = 1.054
Linear change
σY = |−2|·1.054 = 2.107
Answer
σY = 2.107
The mean does move: E(Y) = −2·2.3 + 1 = −3.6.
17.Descriptive statistics
Answer: s = 6.979
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 = −1.5
Standard normal table
Φ(−1.5) = 0.06681
P = 0.06681
Answer
P = 0.06681
19.Normal distribution
Answer: P = 0.3531
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 = (65 − 65)/10 = 0
z₂ = (75.5 − 65)/10 = 1.05
Area between
P = Φ(1.05) − Φ(0) = 0.8531 − 0.5 = 0.3531
Answer
P = 0.3531
20.Normal distribution
Answer: x = 81.45 points
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.