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 + 9x) − 2x).
f(x) = √(4x2 + 9x) − 2xx → ∞
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·e0.5x. Find f′(0.5).
f(x) = x2·e0.5xx = 0.5
f′(0.5) =
4.DerivativesHard
The tangent line to f(x) = x2 + x − 5 at x = −2 crosses the y-axis at y = b. Find b.
f(x) = x2 + x − 5x = −2
b =
5.IntegralsMedium
Evaluate I = ∫₁1.5 x·ln x dx.
∫₁1.5 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
Write z = 4∠−40° in rectangular form a + jb and give the real part a.
r = 4θ = −40°
a =
8.Complex numbersHard
Exactly one of the points A–D is a root of z⁴ = 81∠180°. Click it.
w = 81∠180°n = 4
1A (5.4∠90°)
2B (3∠180°)
3C (3∠90°)
4D (3∠45°)
9.Differential equationsMedium
Solve y″ + 4y′ + 4y = 0 with y(0) = 5 and y′(0) = 2. Find y(1.5).
y″ + 4y′ + 4y = 0y(0) = 5y′(0) = 2
y(1.5) =
10.Differential equationsHard
Solve y′ + 0.5y = 5e−t with y(0) = 0. Find y(2).
y′ + 0.5y = 5e−ty(0) = 0
y(2) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [5 −2 −1; −2 −3 5; 3 0 −1]
det A =
12.Matrices & determinantsHard
Solve the system −2x − y = −11 and x − 4y = 0. Find x.
−2x − y = −11x − 4y = 0
x =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (6, −4, 7) onto b = (1, 3, 0), that is, the component of a in the direction of b.
a = (6, −4, 7)b = (1, 3, 0)
compb a =
14.Vectors: dot & cross productHard
For a = (6, 6, −5) and b = (4, 6, 6), find the x-component (a × b)x of the cross product a × b.
a = (6, 6, −5)b = (4, 6, 6)
(a × b)x =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 2, 4, 6p(x): 0.3, 0.2, 0.25, 0.25
Var(X) =
16.ProbabilityHard
X has the distribution shown, and Y = 5X + 5. Find the standard deviation σY of Y.
x: 0, 2, 4, 6p(x): 0.35, 0.3, 0.1, 0.25Y = 5X + 5
σY =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 55, 55, 23, 24, 12.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 12. What proportion P of values is below 188?
μ = 170σ = 12x = 188
P =
19.Normal distributionMedium
X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 114.6 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 114.6 s
P =
20.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 10% of them lie (the 10th percentile).
μ = 100σ = 15P = 10%
x = points
3Solutions
1.Limits
Answer: L = 2.25
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) = 9x/(√(4x2 + 9x) + 2x)
Divide by x
f(x) = 9/(√(4 + 9/x) + 2)
Let x → ∞
L = 9/(2 + 2) = 2.25
Answer
L = 2.25
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′(0.5) = 1.445
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′ = e0.5x·(2x + 0.5·x2)
Substitute
f′(0.5) = 1.284·(1 + 0.125) = 1.445
Answer
f′(0.5) = 1.445
4.Derivatives
Answer: b = −9
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) = −3
f′(x) = 2x + 1 → f′(−2) = −3
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −3 − (−3)·(−2) = −9
Answer
b = −9
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: 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: a = 3.064
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 = 4·cos(−40°) = 3.064
b = 4·sin(−40°) = −2.571
Rectangular form
z = 3.064 − j2.571
a = 3.064
Answer
a = 3.064
8.Complex numbers
Answer: D (3∠45°)
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 and arguments of the roots
|z| = 3
θ = 45°, 135°, −135°, −45°
The point
D = 3∠45°
Answer
D (3∠45°)
9.Differential equations
Answer: y(1.5) = 1.145
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.
Double root
(r + 2)² = 0 → r = −2
Constants
A = y(0) = 5
y′(0) = B − 2A → B = 2 + 10 = 12
Evaluate
y(1.5) = (5 + 12·1.5)·e−3 = 1.145
Answer
y(1.5) = 1.145
10.Differential equations
Answer: y(2) = 2.325
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 = e0.5t
(e0.5ty)′ = 5e−0.5t
General solution
y = 5/(−0.5)·e−t + C·e−0.5t = −10·e−t + C·e−0.5t
Initial condition
0 = −10 + C → C = 10
Evaluate
y(2) = −10·0.1353 + 10·0.3679 = 2.325
Answer
y(2) = 2.325
11.Matrices & determinants
Answer: det A = −20
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 = 3, M₁₂ = −13, M₁₃ = 9
Cofactor expansion
det A = 5·3 − (−2)·(−13) + (−1)·9 = −20
Answer
det A = −20
12.Matrices & determinants
Answer: x = 4.889
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 = (−2)·(−4) − (−1)·1 = 9
Replace the x column
det Aₓ = (−11)·(−4) − (−1)·0 = 44
x = 44/9 = 4.889
Answer
x = 4.889
y = 1.222
13.Vectors: dot & cross product
Answer: compb a = −1.897
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 = 6·1 + (−4)·3 + 7·0 = −6
Length of b
|b| = √(1² + 3² + 0²) = 3.162
Scalar projection
compb a = a·b/|b| = −6/3.162 = −1.897
Answer
compb a = −1.897
Negative: a has a component pointing against b.
14.Vectors: dot & cross product
Answer: (a × b)x = 66
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 = 6·6 − (−5)·6 = 66
(a × b)y = azbx − axbz = (−5)·4 − 6·6 = −56
(a × b)z = axby − aybx = 6·6 − 6·4 = 12
Requested component
(a × b)x = 66
Answer
(a × b)x = 66
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: Var(X) = 5.39
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.
Mean
μ = 0·0.3 + 2·0.2 + 4·0.25 + 6·0.25 = 2.9
Mean of the squares
E(X²) = 0·0.3 + 4·0.2 + 16·0.25 + 36·0.25 = 13.8
Variance
Var(X) = 13.8 − 2.9² = 5.39
Answer
Var(X) = 5.39
16.Probability
Answer: σY = 11.78
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) = 5.55, σX = 2.356
Linear change
σY = |5|·2.356 = 11.78
Answer
σY = 11.78
The mean does move: E(Y) = 5·2.5 + 5 = 17.5.
17.Descriptive statistics
Answer: s = 19.92
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: P = 0.06681
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).
Distribution of the sum
μ = 105 s, σ = √(5² + 4²) = 6.403 s
Standardise
z = (114.6 − 105)/6.403 = 1.5
Upper tail
P = 1 − Φ(1.5) = 1 − 0.9332 = 0.06681
Answer
P = 0.06681
20.Normal distribution
Answer: x = 80.78 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.