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 + 10x) − 2x).
f(x) = √(4x2 + 10x) − 2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e1.5x − 1 − 1.5x)/x².
f(x) = (e1.5x − 1 − 1.5x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = (−2x − 1)4. Find f′(1).
f(x) = (−2x − 1)4x = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 − 13.5x2 + 54x + 1. Find the x-coordinate x of the local maximum.
f(x) = x3 − 13.5x2 + 54x + 1
x =
5.IntegralsMedium
Evaluate I = ∫₀2 x·sin x dx (x in radians).
∫₀2 x·sin 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∠140° in rectangular form a + jb and give the imaginary part b.
r = 4θ = 140°
b =
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 (2∠165°)
2B (2∠120°)
3C (1∠120°)
4D (2∠−120°)
9.Differential equationsMedium
A cup of coffee at 84 °C is left in a room at 22 °C. After 30 min its temperature is 28 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 25 °C?
T0 = 84 °CTa = 22 °CT(30) = 28 °CT = 25 °C
t = min
10.Differential equationsHard
Solve the initial value problem y′ + 3y/x = 4x for x > 0, with y(1) = 1. Find y(2).
y′ + 3y/x = 4xy(1) = 1
y(2) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [4 0; −5 −1]B = [−3 −4; −4 −5]
c11 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−3 1; −8 6]
λ =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (2, −3, −6) and b = (5, 6, −3), in degrees.
a = (2, −3, −6)b = (5, 6, −3)
θ = °
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (5, −4, 1), b = (−4, −1, −3) and c = (−5, 3, 2).
a = (5, −4, 1)b = (−4, −1, −3)c = (−5, 3, 2)
V =
15.ProbabilityMedium
A fair die is rolled 5 times. What is the probability P of getting at least one six?
n = 5
P =
16.ProbabilityHard
X has the distribution shown, and Y = −2X + 5. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 26, 21, 57, 26, 27, 31, 14.
n = 7
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 10. Find the z-score z of a score of 35.
μ = 50σ = 10x = 35
z =
19.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 440 and 508 h? Give P as a decimal.
μ = 500σ = 40x1 = 440x2 = 508
P =
20.Normal distributionHard
The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
3Solutions
1.Limits
Answer: L = 2.5
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) = 10x/(√(4x2 + 10x) + 2x)
Divide by x
f(x) = 10/(√(4 + 10/x) + 2)
Let x → ∞
L = 10/(2 + 2) = 2.5
Answer
L = 2.5
2.Limits
Answer: L = 1.125
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
(1.5·e1.5x − 1.5)/(2x) → 0/0
Second application
2.25·e1.5x/2
Substitute x = 0
L = 2.25·1/2 = 1.125
Answer
L = 1.125
3.Derivatives
Answer: f′(1) = 216
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·(−2x − 1)3·(−2)
Substitute
f′(1) = 4·(−3)3·(−2) = 216
Answer
f′(1) = 216
4.Derivatives
Answer: x = 3
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 − 27x + 54 = 3·(x2 − 9x + 18)
Solve f′(x) = 0
x = 3 or x = 6
Which one is which
f″(x) = 6x − 27: negative at a maximum, positive at a minimum.
f″(3) = −9, f″(6) = 9
x = 3
Answer
x = 3
5.Integrals
Answer: I = 1.742
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
(−2·cos 2 + sin 2) − (0 + 0) = 1.742
Answer
I = 1.742
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: b = 2.571
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(140°) = −3.064
b = 4·sin(140°) = 2.571
Rectangular form
z = −3.064 + j2.571
b = 2.571
Answer
b = 2.571
8.Complex numbers
Answer: B (2∠120°)
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
B = 2∠120°
Answer
B (2∠120°)
9.Differential equations
Answer: t = 38.9 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(62/6)/30 = 0.07785 min⁻¹
Time to reach the target
t = ln(62/3)/0.07785 = 38.9 min
Answer
t = 38.9 min
10.Differential equations
Answer: y(2) = 3.225
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 = e∫3/x dx = x3
Integrate
(x3y)′ = 4x4 → x3y = 4x5/5 + C
Initial condition
1 = 4/5 + C → C = 0.2
Evaluate
y(2) = (4·25/5 + 0.2)/23 = 3.225
Answer
y(2) = 3.225
11.Matrices & determinants
Answer: c11 = −12
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 = 4·(−3) + 0·(−4) = −12
Answer
c11 = −12
12.Matrices & determinants
Answer: λ = 5
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 + (−10) = 0
Solve
λ = (3 ± √(49))/2 → λ = −2, 5
λ = 5
Answer
λ = 5
13.Vectors: dot & cross product
Answer: θ = 80.17 °
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 = 2·5 + (−3)·6 + (−6)·(−3) = 10
Magnitudes
|a| = √(2² + (−3)² + (−6)²) = 7
|b| = √(5² + 6² + (−3)²) = 8.367
Angle
cos θ = 10/(7·8.367) = 0.1707
θ = arccos(0.1707) = 80.17°
Answer
θ = 80.17°
14.Vectors: dot & cross product
Answer: V = 74
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 b × c
(b × c)x = bycz − bzcy = (−1)·2 − (−3)·3 = 7
(b × c)y = bzcx − bxcz = (−3)·(−5) − (−4)·2 = 23
(b × c)z = bxcy − bycx = (−4)·3 − (−1)·(−5) = −17
Dot product with a
a·(b × c) = 5·7 + (−4)·23 + 1·(−17) = −74
V = |−74| = 74
Answer
V = 74
15.Probability
Answer: P = 0.5981
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)5 = 0.4019
P = 1 − 0.4019 = 0.5981
Answer
P = 0.5981
16.Probability
Answer: σY = 4.4
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 = 2.6, Var(X) = 4.84, σX = 2.2
Linear change
σY = |−2|·2.2 = 4.4
Answer
σY = 4.4
The mean does move: E(Y) = −2·2.6 + 5 = −0.2.
17.Descriptive statistics
Answer: s = 13.53
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 = (35 − 50)/10 = −1.5
Answer
z = −1.5
19.Normal distribution
Answer: P = 0.5125
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 = (440 − 500)/40 = −1.5
z₂ = (508 − 500)/40 = 0.2
Area between
P = Φ(0.2) − Φ(−1.5) = 0.5793 − 0.06681 = 0.5125
Answer
P = 0.5125
20.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).