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)2x. Give a decimal value.
f(x) = (1 + 2/x)2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + x)/sin(3x), with x in radians.
f(x) = ln(1 + x)/sin(3x)x → 0
L =
3.DerivativesMedium
Let f(x) = (−2x + 3)4. Find f′(2).
f(x) = (−2x + 3)4x = 2
f′(2) =
4.DerivativesHard
The tangent line to f(x) = 2x2 − 5x − 5 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = 2x2 − 5x − 5x = 1
b =
5.IntegralsMedium
Evaluate I = ∫26 3/x dx.
f(x) = 3/x[2, 6]
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
Write z = 5∠160° in rectangular form a + jb and give the real part a.
r = 5θ = 160°
a =
8.Complex numbersHard
The roots of zⁿ = w with n = 2 and w = 36∠−60° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1. Find the argument of z₀ in degrees, between −180° and 180°.
w = 36∠−60°n = 2k = 0
θ₀ = °
9.Differential equationsMedium
A forged steel part at 500 °C is left to cool in a workshop at 20 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.02 min⁻¹, find its temperature T after 26 min.
T0 = 500 °CTa = 20 °Ck = 0.02 min⁻¹t = 26 min
T = °C
10.Differential equationsHard
Solve y′ + 3y = 8e−t with y(0) = 3. Find y(2).
y′ + 3y = 8e−ty(0) = 3
y(2) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−3 −3 −4; 1 2 −1; 0 −4 2]
det A =
12.Matrices & determinantsHard
Solve the system −x − 3y = −6 and −x − 5y = 8. Find x.
−x − 3y = −6−x − 5y = 8
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (−4, −4, −2) and b = (−2, −1, 0), in degrees.
a = (−4, −4, −2)b = (−2, −1, 0)
θ = °
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−3, 1, −3) and b = (−2, −4, 4) and has a positive z-component. Find its x-component n̂x.
a = (−3, 1, −3)b = (−2, −4, 4)n̂z > 0
n̂x =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 2, 4, 6p(x): 0.25, 0.05, 0.35, 0.35
σ =
16.ProbabilityHard
X has the distribution shown, and Y = 3X − 3. Find the standard deviation σY of Y.
x: 0, 2, 4, 6p(x): 0.3, 0.3, 0.25, 0.15Y = 3X − 3
σY =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 42, 44, 18, 38, 13, 20, 31.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is below 130?
μ = 100σ = 20x = 130
P =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.9146. Find z to two decimals.
A = 0.9146
z =
20.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 10% of them lie (the 10th percentile).
μ = 250σ = 5P = 10%
x = g
3Solutions
1.Limits
Answer: L = 54.6
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→∞ 2·x·ln(1 + 2/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 2·2 = 4
Undo the logarithm
L = e4 = 54.6
Answer
L = 54.6
2.Limits
Answer: L = 0.3333
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 [1/(1 + x)]/[3·cos 3x]
Substitute x = 0
L = 1/3 = 0.3333
Answer
L = 0.3333
3.Derivatives
Answer: f′(2) = 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 + 3)3·(−2)
Substitute
f′(2) = 4·(−1)3·(−2) = 8
Answer
f′(2) = 8
4.Derivatives
Answer: b = −7
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) = −8
f′(x) = 4x − 5 → f′(1) = −1
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −8 − (−1)·1 = −7
Answer
b = −7
5.Integrals
Answer: I = 3.296
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) = 3·ln x
Evaluate at the limits
I = 3·(ln 6 − ln 2) = 3·ln 3 = 3.296
Answer
I = 3.296
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: a = −4.698
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.
Components
a = 5·cos(160°) = −4.698
b = 5·sin(160°) = 1.71
Rectangular form
z = −4.698 + j1.71
a = −4.698
Answer
a = −4.698
8.Complex numbers
Answer: θ₀ = −30 °
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
361/2 = 6
Argument of the root
θ₀ = (−60° + 360°·0)/2 = −30°
Argument
θ₀ = −30°
Answer
θ₀ = −30°
The 2 roots lie on the circle of radius 6, 180° apart.
9.Differential equations
Answer: T = 305.4 °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) = 20 + 480·e−0.02t
Substitute t
T(26) = 20 + 480·0.5945 = 305.4 °C
Answer
T = 305.4 °C
10.Differential equations
Answer: y(2) = 0.5389
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 = e3t
(e3ty)′ = 8e2t
General solution
y = 8/2·e−t + C·e−3t = 4·e−t + C·e−3t
Initial condition
3 = 4 + C → C = −1
Evaluate
y(2) = 4·0.1353 + (−1)·0.002479 = 0.5389
Answer
y(2) = 0.5389
11.Matrices & determinants
Answer: det A = 22
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.
2×2 minors of the first row
M₁₁ = 0, M₁₂ = 2, M₁₃ = −4
Cofactor expansion
det A = (−3)·0 − (−3)·2 + (−4)·(−4) = 22
Answer
det A = 22
12.Matrices & determinants
Answer: x = 27
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)·(−5) − (−3)·(−1) = 2
Replace the x column
det Aₓ = (−6)·(−5) − (−3)·8 = 54
x = 54/2 = 27
Answer
x = 27
y = −7
13.Vectors: dot & cross product
Answer: θ = 26.57 °
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 = (−4)·(−2) + (−4)·(−1) + (−2)·0 = 12
Magnitudes
|a| = √((−4)² + (−4)² + (−2)²) = 6
|b| = √((−2)² + (−1)² + 0²) = 2.236
Angle
cos θ = 12/(6·2.236) = 0.8944
θ = arccos(0.8944) = 26.57°
Answer
θ = 26.57°
14.Vectors: dot & cross product
Answer: n̂x = −0.331
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.
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 = 2.5, Var(X) = 4.35, σX = 2.086
Linear change
σY = |3|·2.086 = 6.257
Answer
σY = 6.257
The mean does move: E(Y) = 3·2.5 − 3 = 4.5.
17.Descriptive statistics
Answer: s = 12.49
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) = 0.9332
P = 0.9332
Answer
P = 0.9332
19.Normal distribution
Answer: z = 1.72
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 + 0.9146)/2 = 0.9573
Read the z-table backwards
Find 0.9573 inside the table and read off z.
z = 1.72
Answer
z = 1.72
20.Normal distribution
Answer: x = 243.6 g
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.