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→∞ (3x2 − x + 2)/(4x2 + 2x + 8).
f(x) = (3x2 − x + 2)/(4x2 + 2x + 8)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + x)/sin(4x), with x in radians.
f(x) = ln(1 + x)/sin(4x)x → 0
L =
3.DerivativesMedium
Let f(x) = x3·e2x. Find f′(0.5).
f(x) = x3·e2xx = 0.5
f′(0.5) =
4.DerivativesHard
The tangent line to f(x) = x2 + 2x + 3 at x = −1 crosses the y-axis at y = b. Find b.
f(x) = x2 + 2x + 3x = −1
b =
5.IntegralsMedium
Evaluate I = ∫28 1/x dx.
f(x) = 1/x[2, 8]
I =
6.IntegralsHard
Evaluate I = ∫₀2 x²·eˣ dx.
∫₀2 x²eˣ dx
I =
7.Complex numbersMedium
Write z = 8∠−130° in rectangular form a + jb and give the real part a.
r = 8θ = −130°
a =
8.Complex numbersHard
The equation z² + 2z + 10 = 0 has two complex roots. Find their modulus |z|.
p = 2q = 10
|z| =
9.Differential equationsMedium
Solve y″ + 8y′ + 15y = 0 with y(0) = 3 and y′(0) = 4. Find y(0.5).
y″ + 8y′ + 15y = 0y(0) = 3y′(0) = 4
y(0.5) =
10.Differential equationsHard
A tank holds 100 L of pure water. Brine with 0.1 kg of salt per litre flows in at 20 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 5 min?
V = 100 Lm0 = 0 kgc = 0.1 kg/Lr = 20 L/min
m = kg
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [0 −6; −5 −4]B = [−6 −5; −3 −5]
c11 =
12.Matrices & determinantsHard
Solve the system 2x − 6y = −11 and 6x − y = −5. Find x.
2x − 6y = −116x − y = −5
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (0, 5, 1) and b = (5, k, 1) are perpendicular.
a = (0, 5, 1)b = (5, k, 1)
k =
14.Vectors: dot & cross productHard
For a = (−1, 5, 5) and b = (−1, −3, 2), find the z-component (a × b)z of the cross product a × b.
a = (−1, 5, 5)b = (−1, −3, 2)
(a × b)z =
15.ProbabilityMedium
A bag holds 7 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
7 / 6
P =
16.ProbabilityHard
X has the distribution shown, and Y = 5X + 4. Find the standard deviation σY of Y.
x: 0, 1, 2, 3p(x): 0.35, 0.2, 0.15, 0.3Y = 5X + 4
σY =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 12, 44, 22, 11, 25, 25, 15.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is below 74.8?
μ = 70σ = 8x = 74.8
P =
19.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 49 and 75.5 points? Give P as a decimal.
μ = 65σ = 10x1 = 49x2 = 75.5
P =
20.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 95% is to the left of it. Find the cut-off x.
μ = 100 pointsσ = 15 points95%
x = points
3Solutions
1.Limits
Answer: L = 0.75
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).
Divide by the highest power
(3 + (−1)/x + 2/x2)/(4 + 2/x + 8/x2)
Let x → ∞
L = 3/4 = 0.75
Answer
L = 0.75
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 0.25
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).
L'Hôpital's rule
L = limx→0 [1/(1 + x)]/[4·cos 4x]
Substitute x = 0
L = 1/4 = 0.25
Answer
L = 0.25
3.Derivatives
Answer: f′(0.5) = 2.718
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·(3x2 + 2·x3)
Substitute
f′(0.5) = 2.718·(0.75 + 0.25) = 2.718
Answer
f′(0.5) = 2.718
4.Derivatives
Answer: b = 2
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) = 2
f′(x) = 2x + 2 → f′(−1) = 0
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 2 − 0·(−1) = 2
Answer
b = 2
5.Integrals
Answer: I = 1.386
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) = ln x
Evaluate at the limits
I = (ln 8 − ln 2) = ln 4 = 1.386
Answer
I = 1.386
6.Integrals
Answer: I = 12.78
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
7.389·(2) − 1·2 = 12.78
Answer
I = 12.78
7.Complex numbers
Answer: a = −5.142
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 = 8·cos(−130°) = −5.142
b = 8·sin(−130°) = −6.128
Rectangular form
z = −5.142 − j6.128
a = −5.142
Answer
a = −5.142
8.Complex numbers
Answer: |z| = 3.162
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.
Discriminant
p² − 4q = 4 − 40 = −36 < 0
Quadratic formula
z = (−2 ± j√36)/2 = −1 + j3, −1 − j3
Answer
|z| = √((−1)² + 3²) = √10 = 3.162
Answer
|z| = 3.162
9.Differential equations
Answer: y(0.5) = 1.586
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.
Characteristic roots
(r + 3)(r + 5) = 0 → r = −3, −5
Constants from the initial values
A + B = 3, −3A − 5B = 4
A = 9.5, B = −6.5
Evaluate
y(0.5) = 9.5·e−1.5 + (−6.5)·e−2.5 = 1.586
Answer
y(0.5) = 1.586
10.Differential equations
Answer: m = 6.321 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 = 20·0.1 − (20/100)·m = 2 − 0.2·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 10 + (−10)·e−t/5
Substitute t
m(5) = 10 + (−10)·0.3679 = 6.321 kg
Answer
m = 6.321 kg
11.Matrices & determinants
Answer: c11 = 18
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 = 0·(−6) + (−6)·(−3) = 18
Answer
c11 = 18
12.Matrices & determinants
Answer: x = −0.5588
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·(−1) − (−6)·6 = 34
Replace the x column
det Aₓ = (−11)·(−1) − (−6)·(−5) = −19
x = −19/34 = −0.5588
Answer
x = −0.5588
y = 1.647
13.Vectors: dot & cross product
Answer: k = −0.2
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 equals zero
a·b = 0·5 + 5·k + 1·1 = 0
1 + 5·k = 0
Solve for k
k = −1/5 = −0.2
Answer
k = −0.2
14.Vectors: dot & cross product
Answer: (a × b)z = 8
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 = 5·2 − 5·(−3) = 25
(a × b)y = azbx − axbz = 5·(−1) − (−1)·2 = −3
(a × b)z = axby − aybx = (−1)·(−3) − 5·(−1) = 8
Requested component
(a × b)z = 8
Answer
(a × b)z = 8
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: P = 0.2692
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.
First and second draw
P = 7/13 · 6/12 = 42/156 = 0.2692
Answer
P = 0.2692
16.Probability
Answer: σY = 6.205
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 = 1.4, Var(X) = 1.54, σX = 1.241
Linear change
σY = |5|·1.241 = 6.205
Answer
σY = 6.205
The mean does move: E(Y) = 5·1.4 + 4 = 11.
17.Descriptive statistics
Answer: s = 11.34
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 = 0.6
Standard normal table
Φ(0.6) = 0.7257
P = 0.7257
Answer
P = 0.7257
19.Normal distribution
Answer: P = 0.7983
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 = (49 − 65)/10 = −1.6
z₂ = (75.5 − 65)/10 = 1.05
Area between
P = Φ(1.05) − Φ(−1.6) = 0.8531 − 0.0548 = 0.7983
Answer
P = 0.7983
20.Normal distribution
Answer: x = 124.6 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.