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→∞ (√(9x2 − 4x) − 3x).
f(x) = √(9x2 − 4x) − 3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 2x)/(3x²), with x in radians.
f(x) = (1 − cos 2x)/(3x²)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(2x2 + 1). Find f′(3).
f(x) = ln(2x2 + 1)x = 3
f′(3) =
4.DerivativesHard
The tangent line to f(x) = −x2 + x − 5 at x = 3 crosses the y-axis at y = b. Find b.
f(x) = −x2 + x − 5x = 3
b =
5.IntegralsMedium
Evaluate I = ∫02 e−0.5x dx.
f(x) = e−0.5x[0, 2]
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 9 − x² and the x-axis.
y = 9 − x²
A =
7.Complex numbersMedium
Find the argument θ of z = −7 − j in degrees, as the principal value between −180° and 180°.
z = −7 − j
θ = °
8.Complex numbersHard
Exactly one of the points A–D is a root of z⁴ = 16∠120°. Click it.
w = 16∠120°n = 4
1A (3.6∠−15°)
2B (2∠60°)
3C (2∠−60°)
4D (2∠−15°)
9.Differential equationsMedium
Solve y″ + 4y′ + 4y = 0 with y(0) = 2 and y′(0) = 1. Find y(2).
y″ + 4y′ + 4y = 0y(0) = 2y′(0) = 1
y(2) =
10.Differential equationsHard
A tank holds 400 L of pure water. Brine with 0.05 kg of salt per litre flows in at 10 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 15 min?
V = 400 Lm0 = 0 kgc = 0.05 kg/Lr = 10 L/min
m = kg
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [1 −1 0; 0 −2 3; 3 2 3]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [2 −2; −3 1]
λ =
13.Vectors: dot & cross productMedium
A constant force F = (−30, −10, −25) N acts on an object while it moves in a straight line from point A(5, 2, −3) to point B(−4, −3, 4). Coordinates are in metres. Find the work W done by the force.
F = (−30, −10, −25) NA(5, 2, −3) mB(−4, −3, 4) m
W = J
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−3, 0, −5) and b = (4, −3, 0) and has a positive z-component. Find its y-component n̂y.
a = (−3, 0, −5)b = (4, −3, 0)n̂z > 0
n̂y =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4p(x): 0.15, 0.35, 0.15, 0.35
σ =
16.ProbabilityHard
X has the distribution shown, and Y = 5X − 3. Find the standard deviation σY of Y.
x: 0, 2, 4, 6p(x): 0.3, 0.15, 0.25, 0.3Y = 5X − 3
σY =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 49, 34, 52, 52, 39, 52.
n = 6
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 8. Find the z-score z of a score of 166.
μ = 170σ = 8x = 166
z =
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 107.5 points? Give P as a decimal.
μ = 100σ = 15x1 = 85.75x2 = 107.5
P =
20.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 90% is to the left of it. Find the cut-off x.
μ = 65 pointsσ = 10 points90%
x = points
3Solutions
1.Limits
Answer: L = −0.6667
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) = −4x/(√(9x2 − 4x) + 3x)
Divide by x
f(x) = −4/(√(9 + (−4)/x) + 3)
Let x → ∞
L = −4/(3 + 3) = −0.6667
Answer
L = −0.6667
2.Limits
Answer: L = 0.6667
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
2·sin 2x/(6x) → 0/0
Second application
4·cos 2x/6
Substitute x = 0
L = 4/6 = 0.6667
Answer
L = 0.6667
3.Derivatives
Answer: f′(3) = 0.6316
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 for the logarithm
f′(x) = 4x/(2x2 + 1)
Substitute
f′(3) = 12/19 = 0.6316
Answer
f′(3) = 0.6316
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(3) = −11
f′(x) = −2x + 1 → f′(3) = −5
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −11 − (−5)·3 = 4
Answer
b = 4
5.Integrals
Answer: I = 1.264
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.
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
9 − x² = 0 → x = ±3
Integrate
A = [9x − x³/3]−33 = 18 − (−18) = 36
Answer
A = 36
7.Complex numbers
Answer: θ = −171.9 °
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 = −7, b = −1 → III
Reference angle
α = atan(1/7) = 8.13°
Argument
θ = −(180° − α) = −171.9°
Answer
θ = −171.9°
In polar form z = 7.071∠−171.9°.
8.Complex numbers
Answer: C (2∠−60°)
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| = 2
θ = 30°, 120°, −150°, −60°
The point
C = 2∠−60°
Answer
C (2∠−60°)
9.Differential equations
Answer: y(2) = 0.2198
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) = 2
y′(0) = B − 2A → B = 1 + 4 = 5
Evaluate
y(2) = (2 + 5·2)·e−4 = 0.2198
Answer
y(2) = 0.2198
10.Differential equations
Answer: m = 6.254 kg
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.
Set up the equation
dm/dt = 10·0.05 − (10/400)·m = 0.5 − 0.025·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 20 + (−20)·e−t/40
Substitute t
m(15) = 20 + (−20)·0.6873 = 6.254 kg
Answer
m = 6.254 kg
11.Matrices & determinants
Answer: det A = −21
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 = −12, M₁₂ = −9, M₁₃ = 6
Cofactor expansion
det A = 1·(−12) − (−1)·(−9) + 0·6 = −21
Answer
det A = −21
12.Matrices & determinants
Answer: λ = 4
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.
Characteristic equation
λ² − (3)λ + (−4) = 0
Solve
λ = (3 ± √(25))/2 → λ = −1, 4
λ = 4
Answer
λ = 4
13.Vectors: dot & cross product
Answer: W = 145 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 = (−4, −3, 4) − (5, 2, −3) = (−9, −5, 7) m
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.
Mean
μ = 1·0.15 + 2·0.35 + 3·0.15 + 4·0.35 = 2.7
Mean of the squares
E(X²) = 1·0.15 + 4·0.35 + 9·0.15 + 16·0.35 = 8.5
Variance
Var(X) = 8.5 − 2.7² = 1.21
σ = √1.21 = 1.1
Answer
σ = 1.1
16.Probability
Answer: σY = 12.03
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 = 3.1, Var(X) = 5.79, σX = 2.406
Linear change
σY = |5|·2.406 = 12.03
Answer
σY = 12.03
The mean does move: E(Y) = 5·3.1 − 3 = 12.5.
17.Descriptive statistics
Answer: s = 7.866
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 = (166 − 170)/8 = −0.5
Answer
z = −0.5
19.Normal distribution
Answer: P = 0.5204
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 = (107.5 − 100)/15 = 0.5
Area between
P = Φ(0.5) − Φ(−0.95) = 0.6915 − 0.1711 = 0.5204
Answer
P = 0.5204
20.Normal distribution
Answer: x = 77.8 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.