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→∞ (√(4x2 − 2x) − 2x).
f(x) = √(4x2 − 2x) − 2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e3x − 1 − 3x)/x².
f(x) = (e3x − 1 − 3x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(3x), with x in radians. Find f′(1).
f(x) = x·sin(3x)x = 1 rad
f′(1) =
4.DerivativesHard
The tangent line to f(x) = 0.5x2 + 2x − 2 at x = 2 crosses the y-axis at y = b. Find b.
f(x) = 0.5x2 + 2x − 2x = 2
b =
5.IntegralsMedium
Evaluate I = ∫₁3 x·ln x dx.
∫₁3 x ln x dx
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 4 − x² and the x-axis.
y = 4 − x²
A =
7.Complex numbersMedium
Find the argument θ of z = −6 + j6 in degrees, as the principal value between −180° and 180°.
z = −6 + j6
θ = °
8.Complex numbersHard
One root of z³ + 8z² + 30z + 36 = 0 is z₁ = −3 + j3. The coefficients are real. Find the real root r.
z₁ = −3 + j3
r =
9.Differential equationsMedium
A can of soda at 4 °C is taken out of the fridge into a room at 22 °C. After 11 min its temperature is 15 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 7 °C?
T0 = 4 °CTa = 22 °CT(11) = 15 °CT = 7 °C
t = min
10.Differential equationsHard
Solve the initial value problem y′ + 3y/x = 5x2 for x > 0, with y(1) = −1. Find y(2).
y′ + 3y/x = 5x2y(1) = −1
y(2) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).
A = [2 3; −4 4]B = [6 −1; −6 −1]
c12 =
12.Matrices & determinantsHard
Solve the system 4x − 5y = 2 and 2x + 3y = −10. Find x.
4x − 5y = 22x + 3y = −10
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (−1, 0, −6) and b = (1, −5, 6), in degrees.
a = (−1, 0, −6)b = (1, −5, 6)
θ = °
14.Vectors: dot & cross productHard
For a = (−1, −1, 0) and b = (−1, 1, 4), find the y-component (a × b)y of the cross product a × b.
a = (−1, −1, 0)b = (−1, 1, 4)
(a × b)y =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4, 5p(x): 0.3, 0.15, 0.25, 0.2, 0.1
Var(X) =
16.ProbabilityHard
A test has 8 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 3 correct answers?
n = 8p = 0.1k = 3
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 28, 55, 54, 42, 60, 33, 35.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 15. What proportion P of values is below 52.5?
μ = 60σ = 15x = 52.5
P =
19.Normal distributionMedium
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 174.5 and 185 cm? Give P as a decimal.
μ = 178σ = 7x1 = 174.5x2 = 185
P =
20.Normal distributionHard
The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
3Solutions
1.Limits
Answer: L = −0.5
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) = −2x/(√(4x2 − 2x) + 2x)
Divide by x
f(x) = −2/(√(4 + (−2)/x) + 2)
Let x → ∞
L = −2/(2 + 2) = −0.5
Answer
L = −0.5
2.Limits
Answer: L = 4.5
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
(3·e3x − 3)/(2x) → 0/0
Second application
9·e3x/2
Substitute x = 0
L = 9·1/2 = 4.5
Answer
L = 4.5
3.Derivatives
Answer: f′(1) = −2.829
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(3x) + 3x·cos(3x)
Substitute
f′(1) = 0.1411 + 3·(−0.99) = −2.829
Answer
f′(1) = −2.829
4.Derivatives
Answer: b = −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.
Point and slope
f(2) = 4
f′(x) = x + 2 → f′(2) = 4
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 4 − 4·2 = −4
Answer
b = −4
5.Integrals
Answer: I = 2.944
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
(4.5·ln 3 − 2.25) − (0 − 0.25) = 2.944
Answer
I = 2.944
6.Integrals
Answer: A = 10.67
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.
Zeros
4 − x² = 0 → x = ±2
Integrate
A = [4x − x³/3]−22 = 5.333 − (−5.333) = 10.67
Answer
A = 10.67
7.Complex numbers
Answer: θ = 135 °
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 = −6, b = 6 → II
Reference angle
α = atan(6/6) = 45°
Argument
θ = 180° − α = 135°
Answer
θ = 135°
In polar form z = 8.485∠135°.
8.Complex numbers
Answer: r = −2
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 = −6 + (−2) = −8 = −(8).
9.Differential equations
Answer: t = 2.123 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(-18/-7)/11 = 0.08586 min⁻¹
Time to reach the target
t = ln(-18/(−15))/0.08586 = 2.123 min
Answer
t = 2.123 min
10.Differential equations
Answer: y(2) = 6.438
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 = e∫3/x dx = x3
Integrate
(x3y)′ = 5x5 → x3y = 5x6/6 + C
Initial condition
−1 = 5/6 + C → C = −1.833
Evaluate
y(2) = (5·26/6 + (−1.833))/23 = 6.438
Answer
y(2) = 6.438
11.Matrices & determinants
Answer: c12 = −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.
Row times column
c12 = 2·(−1) + 3·(−1) = −5
Answer
c12 = −5
12.Matrices & determinants
Answer: x = −2
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 = 4·3 − (−5)·2 = 22
Replace the x column
det Aₓ = 2·3 − (−5)·(−10) = −44
x = −44/22 = −2
Answer
x = −2
y = −2
13.Vectors: dot & cross product
Answer: θ = 140.6 °
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)·1 + 0·(−5) + (−6)·6 = −37
Magnitudes
|a| = √((−1)² + 0² + (−6)²) = 6.083
|b| = √(1² + (−5)² + 6²) = 7.874
Angle
cos θ = −37/(6.083·7.874) = −0.7725
θ = arccos(−0.7725) = 140.6°
Answer
θ = 140.6°
14.Vectors: dot & cross product
Answer: (a × b)y = 4
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)·4 − 0·1 = −4
(a × b)y = azbx − axbz = 0·(−1) − (−1)·4 = 4
(a × b)z = axby − aybx = (−1)·1 − (−1)·(−1) = −2
Requested component
(a × b)y = 4
Answer
(a × b)y = 4
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: Var(X) = 1.827
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.
Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.
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 = −0.5
Standard normal table
Φ(−0.5) = 0.3085
P = 0.3085
Answer
P = 0.3085
19.Normal distribution
Answer: P = 0.5328
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 = (174.5 − 178)/7 = −0.5
z₂ = (185 − 178)/7 = 1
Area between
P = Φ(1) − Φ(−0.5) = 0.8413 − 0.3085 = 0.5328
Answer
P = 0.5328
20.Normal distribution
Answer: P = 0.7389
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).