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→∞ (√(x2 − 6x) − x).
f(x) = √(x2 − 6x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 5x)/sin(4x), with x in radians.
f(x) = ln(1 + 5x)/sin(4x)x → 0
L =
3.DerivativesMedium
Let f(x) = (3x − 1)4. Find f′(2).
f(x) = (3x − 1)4x = 2
f′(2) =
4.DerivativesHard
Let f(x) = x3 − 9x2 + 15x − 2. Find the x-coordinate x of the local maximum.
f(x) = x3 − 9x2 + 15x − 2
x =
5.IntegralsMedium
Evaluate I = ∫₀π/3 x·cos x dx (x in radians).
∫₀π/3 x·cos x dx
I =
6.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = 4x.
y = x2y = 4x
A =
7.Complex numbersMedium
The number z = 1.5·ej2 has its angle in radians. Find its imaginary part b.
r = 1.5θ = 2 rad
b =
8.Complex numbersHard
The roots of zⁿ = w with n = 3 and w = 8∠−90° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2. Find the real part of z₀.
w = 8∠−90°n = 3k = 0
Re(z₀) =
9.Differential equationsMedium
A can of soda at 5 °C is taken out of the fridge into a room at 22 °C. After 11 min its temperature is 11 °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?
Find the determinant det A of the 3×3 matrix A shown.
A = [3 −4 0; −3 −3 0; −3 2 5]
det A =
12.Matrices & determinantsHard
Solve the system −4x + 4y = −9 and 3x − 6y = −11. Find x.
−4x + 4y = −93x − 6y = −11
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (−2, −5, −2) and b = (−1, k, −1) are perpendicular.
a = (−2, −5, −2)b = (−1, k, −1)
k =
14.Vectors: dot & cross productHard
For a = (−3, 3, 3) and b = (−4, −4, −6), find the x-component (a × b)x of the cross product a × b.
a = (−3, 3, 3)b = (−4, −4, −6)
(a × b)x =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3p(x): 0.3, 0.35, 0.2, 0.15
Var(X) =
16.ProbabilityHard
11 components are tested, and each one fails with probability 0.7, independently. What is the probability P that at most one fails?
n = 11p = 0.7
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 52, 54, 14, 57, 51.
n = 5
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 158.
μ = 170σ = 8x = 158
z =
19.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9761. Find z to two decimals.
A = 0.9761
z =
20.Normal distributionHard
Hole diameters D are normal with mean 25.05 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.05 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
3Solutions
1.Limits
Answer: L = −3
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/(√(x2 − 6x) + x)
Divide by x
f(x) = −6/(√(1 + (−6)/x) + 1)
Let x → ∞
L = −6/(1 + 1) = −3
Answer
L = −3
2.Limits
Answer: L = 1.25
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 [5/(1 + 5x)]/[4·cos 4x]
Substitute x = 0
L = 5/4 = 1.25
Answer
L = 1.25
3.Derivatives
Answer: f′(2) = 1500
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
f′(x) = 4·(3x − 1)3·3
Substitute
f′(2) = 4·53·3 = 1500
Answer
f′(2) = 1500
4.Derivatives
Answer: x = 1
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.
Derivative
f′(x) = 3x2 − 18x + 15 = 3·(x2 − 6x + 5)
Solve f′(x) = 0
x = 1 or x = 5
Which one is which
f″(x) = 6x − 18: negative at a maximum, positive at a minimum.
f″(1) = −12, f″(5) = 12
x = 1
Answer
x = 1
5.Integrals
Answer: I = 0.4069
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 = cos x dx, v = sin x
Antiderivative
∫x cos x dx = x sin x − ∫sin x dx = x sin x + cos x
Limits
(π/3·sin π/3 + cos π/3) − (0 + 1) = 0.4069
Answer
I = 0.4069
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.
Intersection points
x2 = 4x → x = 0, x = 4
Integrate upper minus lower
A = [2x2 − 0.3333x3]04 = 10.67
Answer
A = 10.67
7.Complex numbers
Answer: b = 1.364
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.
Euler's formula
z = 1.5·(cos 2 + j sin 2)
Components (radians)
a = 1.5·cos 2 = −0.6242
b = 1.5·sin 2 = 1.364
b = 1.364
Answer
b = 1.364
8.Complex numbers
Answer: Re(z₀) = 1.732
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 = (−90° + 360°·0)/3 = −30°
Rectangular form
z₀ = 2·(cos −30° + j sin −30°) = 1.732 − j1
Re(z₀) = 1.732
Answer
Re(z₀) = 1.732
The 3 roots lie on the circle of radius 2, 120° apart.
9.Differential equations
Answer: t = 3.163 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(-17/-11)/11 = 0.03957 min⁻¹
Time to reach the target
t = ln(-17/(−15))/0.03957 = 3.163 min
Answer
t = 3.163 min
10.Differential equations
Answer: y(0.5) = 0.8891
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 = e3t
(e3ty)′ = 4e2.5t
General solution
y = 4/2.5·e−0.5t + C·e−3t = 1.6·e−0.5t + C·e−3t
Initial condition
0 = 1.6 + C → C = −1.6
Evaluate
y(0.5) = 1.6·0.7788 + (−1.6)·0.2231 = 0.8891
Answer
y(0.5) = 0.8891
11.Matrices & determinants
Answer: det A = −105
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 = −15, M₁₂ = −15, M₁₃ = −15
Cofactor expansion
det A = 3·(−15) − (−4)·(−15) + 0·(−15) = −105
Answer
det A = −105
12.Matrices & determinants
Answer: x = 8.167
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)·(−6) − 4·3 = 12
Replace the x column
det Aₓ = (−9)·(−6) − 4·(−11) = 98
x = 98/12 = 8.167
Answer
x = 8.167
y = 5.917
13.Vectors: dot & cross product
Answer: k = 0.8
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 = (−2)·(−1) + (−5)·k + (−2)·(−1) = 0
4 + (−5)·k = 0
Solve for k
k = −4/(−5) = 0.8
Answer
k = 0.8
14.Vectors: dot & cross product
Answer: (a × b)x = −6
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 = 3·(−6) − 3·(−4) = −6
(a × b)y = azbx − axbz = 3·(−4) − (−3)·(−6) = −30
(a × b)z = axby − aybx = (−3)·(−4) − 3·(−4) = 24
Requested component
(a × b)x = −6
Answer
(a × b)x = −6
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: Var(X) = 1.06
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 + 1·0.35 + 2·0.2 + 3·0.15 = 1.2
Mean of the squares
E(X²) = 0·0.3 + 1·0.35 + 4·0.2 + 9·0.15 = 2.5
Variance
Var(X) = 2.5 − 1.2² = 1.06
Answer
Var(X) = 1.06
16.Probability
Answer: P = 4.724 × 10⁻⁵
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) = 0.311 = 1.771 × 10⁻⁶
P(X = 1) = 11·0.7·0.310 = 4.547 × 10⁻⁵
Add them
P = 1.771 × 10⁻⁶ + 4.547 × 10⁻⁵ = 4.724 × 10⁻⁵
Answer
P = 4.724 × 10⁻⁵
17.Descriptive statistics
Answer: s = 17.81
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 = (158 − 170)/8 = −1.5
Answer
z = −1.5
19.Normal distribution
Answer: z = −1.98
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.
Area to the left of z
Φ(z) = 1 − 0.9761 = 0.0239
Read the z-table backwards
Find 0.0239 inside the table and read off z.
z = −1.98
Answer
z = −1.98
20.Normal distribution
Answer: P = 0.9909
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).