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 + 6x) − 2x).
f(x) = √(4x2 + 6x) − 2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 4x)/(2x²), with x in radians.
f(x) = (1 − cos 4x)/(2x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(3x), with x in radians. Find f′(1.5).
f(x) = x·sin(3x)x = 1.5 rad
f′(1.5) =
4.DerivativesHard
The tangent line to f(x) = 2x2 − x − 2 at x = 2 crosses the y-axis at y = b. Find b.
f(x) = 2x2 − x − 2x = 2
b =
5.IntegralsMedium
Evaluate I = ∫₀π x·sin x dx (x in radians).
∫₀π x·sin x dx
I =
6.IntegralsHard
Find the area A of the region enclosed by y = 0.5x2 and y = x.
y = 0.5x2y = x
A =
7.Complex numbersMedium
Write z = 12∠35° in rectangular form a + jb and give the imaginary part b.
r = 12θ = 35°
b =
8.Complex numbersHard
The roots of zⁿ = w with n = 3 and w = 8∠45° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2. Find the real part of z₀.
w = 8∠45°n = 3k = 0
Re(z₀) =
9.Differential equationsMedium
Solve y″ + 7y′ + 12y = 0 with y(0) = 5 and y′(0) = 4. Find y(1.5).
y″ + 7y′ + 12y = 0y(0) = 5y′(0) = 4
y(1.5) =
10.Differential equationsHard
Solve y′ + y = 8e−2t with y(0) = 3. Find y(1).
y′ + y = 8e−2ty(0) = 3
y(1) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [0 −3 5; −5 3 −5; 5 4 −5]
det A =
12.Matrices & determinantsHard
Solve the system 3x + 4y = −6 and −5x − 3y = 12. Find x.
3x + 4y = −6−5x − 3y = 12
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (3, 2, 5) and b = (−1, −5, k) are perpendicular.
a = (3, 2, 5)b = (−1, −5, k)
k =
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−2, 4, 2) and b = (1, −1, 3) and has a positive z-component. Find its x-component n̂x.
a = (−2, 4, 2)b = (1, −1, 3)n̂z > 0
n̂x =
15.ProbabilityMedium
A fair die is rolled 2 times. What is the probability P of getting at least one six?
n = 2
P =
16.ProbabilityHard
5 components are tested, and each one fails with probability 0.4, independently. What is the probability P that at most one fails?
n = 5p = 0.4
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 21, 31, 16, 17, 11.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 20. What proportion P of values is above 110?
μ = 70σ = 20x = 110
P =
19.Normal distributionMedium
X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 740.2 g. Round z to two decimals.
μX = 250, σX = 4μY = 500, σY = 9X + Y > 740.2 g
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 30% is to the left of it. Find the cut-off x.
μ = 100 pointsσ = 15 points30%
x = points
3Solutions
1.Limits
Answer: L = 1.5
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) = 6x/(√(4x2 + 6x) + 2x)
Divide by x
f(x) = 6/(√(4 + 6/x) + 2)
Let x → ∞
L = 6/(2 + 2) = 1.5
Answer
L = 1.5
2.Limits
Answer: L = 4
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/(4x) → 0/0
Second application
16·cos 4x/4
Substitute x = 0
L = 16/4 = 4
Answer
L = 4
3.Derivatives
Answer: f′(1.5) = −1.926
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′(1.5) = −0.9775 + 4.5·(−0.2108) = −1.926
Answer
f′(1.5) = −1.926
4.Derivatives
Answer: b = −10
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) = 4x − 1 → f′(2) = 7
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 4 − 7·2 = −10
Answer
b = −10
5.Integrals
Answer: I = 3.142
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 = x, du = dx; dv = sin x dx, v = −cos x
Antiderivative
∫x sin x dx = −x cos x + ∫cos x dx = −x cos x + sin x
Limits
(−π·cos π + sin π) − (0 + 0) = 3.142
Answer
I = 3.142
6.Integrals
Answer: A = 0.6667
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.
Intersection points
0.5x2 = x → x = 0, x = 2
Integrate upper minus lower
A = [0.5x2 − 0.1667x3]02 = 0.6667
Answer
A = 0.6667
7.Complex numbers
Answer: b = 6.883
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 = 12·cos(35°) = 9.83
b = 12·sin(35°) = 6.883
Rectangular form
z = 9.83 + j6.883
b = 6.883
Answer
b = 6.883
8.Complex numbers
Answer: Re(z₀) = 1.932
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
81/3 = 2
Argument of the root
θ₀ = (45° + 360°·0)/3 = 15°
Rectangular form
z₀ = 2·(cos 15° + j sin 15°) = 1.932 + j0.5176
Re(z₀) = 1.932
Answer
Re(z₀) = 1.932
The 3 roots lie on the circle of radius 2, 120° apart.
9.Differential equations
Answer: y(1.5) = 0.2195
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 + 4) = 0 → r = −3, −4
Constants from the initial values
A + B = 5, −3A − 4B = 4
A = 24, B = −19
Evaluate
y(1.5) = 24·e−4.5 + (−19)·e−6 = 0.2195
Answer
y(1.5) = 0.2195
10.Differential equations
Answer: y(1) = 2.964
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 = et
(ety)′ = 8e−t
General solution
y = 8/(−1)·e−2t + C·e−t = −8·e−2t + C·e−t
Initial condition
3 = −8 + C → C = 11
Evaluate
y(1) = −8·0.1353 + 11·0.3679 = 2.964
Answer
y(1) = 2.964
11.Matrices & determinants
Answer: det A = −25
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 = 5, M₁₂ = 50, M₁₃ = −35
Cofactor expansion
det A = 0·5 − (−3)·50 + 5·(−35) = −25
Answer
det A = −25
12.Matrices & determinants
Answer: x = −2.727
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 = 3·(−3) − 4·(−5) = 11
Replace the x column
det Aₓ = (−6)·(−3) − 4·12 = −30
x = −30/11 = −2.727
Answer
x = −2.727
y = 0.5455
13.Vectors: dot & cross product
Answer: k = 2.6
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 equals zero
a·b = 3·(−1) + 2·(−5) + 5·k = 0
−13 + 5·k = 0
Solve for k
k = −(−13)/5 = 2.6
Answer
k = 2.6
14.Vectors: dot & cross product
Answer: n̂x = −0.8616
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
(a × b)x = aybz − azby = 4·3 − 2·(−1) = 14
(a × b)y = azbx − axbz = 2·1 − (−2)·3 = 8
(a × b)z = axby − aybx = (−2)·(−1) − 4·1 = −2
Normalise
The z-component of a × b is negative, so use −(a × b).
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.
Complement
P(no six) = (5/6)2 = 0.6944
P = 1 − 0.6944 = 0.3056
Answer
P = 0.3056
16.Probability
Answer: P = 0.337
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.
Zero and one failure
P(X = 0) = 0.65 = 0.07776
P(X = 1) = 5·0.4·0.64 = 0.2592
Add them
P = 0.07776 + 0.2592 = 0.337
Answer
P = 0.337
17.Descriptive statistics
Answer: s = 7.497
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 = 2
Standard normal table
Φ(2) = 0.9772
P = 1 − 0.9772 = 0.02275
Answer
P = 0.02275
19.Normal distribution
Answer: P = 0.8413
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 = 750 g, σ = √(4² + 9²) = 9.849 g
Standardise
z = (740.2 − 750)/9.849 = −1
Upper tail
P = 1 − Φ(−1) = 1 − 0.1587 = 0.8413
Answer
P = 0.8413
20.Normal distribution
Answer: x = 92.2 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.