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 − 7x) − 2x).
f(x) = √(4x2 − 7x) − 2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 2x)/(4x²), with x in radians.
f(x) = (1 − cos 2x)/(4x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x3·e2x. Find f′(1.5).
f(x) = x3·e2xx = 1.5
f′(1.5) =
4.DerivativesHard
The tangent line to f(x) = 0.5x2 + 5x at x = 1 crosses the y-axis at y = b. Find b.
f(x) = 0.5x2 + 5xx = 1
b =
5.IntegralsMedium
Evaluate I = ∫02 2x·(x² + 1)2 dx.
f(x) = 2x·(x² + 1)2[0, 2]
I =
6.IntegralsHard
The region between the line y = 4x and the parabola y = 3x² is rotated about the x-axis. Find the volume V of the solid.
y = 4xy = 3x²
V =
7.Complex numbersMedium
The diagram shows z = −4 − j4. With w = 2∠−90°, click the point that represents z·w.
z = −4 − j4w = 2∠−90°
1A (−8 + j8)
2B (−8 − j8)
3C (−2 + j2)
4D (−4 + j4)
8.Complex numbersHard
Find the real part of (1 + j2)³.
z = 1 + j2n = 3
Re(z³) =
9.Differential equationsMedium
A can of soda at 5 °C is taken out of the fridge into a room at 26 °C. After 15 min its temperature is 16 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 23 °C?
T0 = 5 °CTa = 26 °CT(15) = 16 °CT = 23 °C
t = min
10.Differential equationsHard
A tank holds 100 L of pure water. Brine with 0.1 kg of salt per litre flows in at 2 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 45 min?
V = 100 Lm0 = 0 kgc = 0.1 kg/Lr = 2 L/min
m = kg
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [−6 −4; 2 6]B = [3 4; 2 −2]
c21 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−2 2; 1 −3]
λ =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (−1, −7, −5) onto b = (3, −6, 1), that is, the component of a in the direction of b.
a = (−1, −7, −5)b = (3, −6, 1)
compb a =
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (−5, −3, −1), b = (2, −4, 1) and c = (5, −1, 5).
a = (−5, −3, −1)b = (2, −4, 1)c = (5, −1, 5)
V =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 2, 4, 6p(x): 0.3, 0.2, 0.25, 0.25
Var(X) =
16.ProbabilityHard
X has the distribution shown, and Y = 3X + 1. Find the standard deviation σY of Y.
x: 0, 1, 2p(x): 0.3, 0.35, 0.35Y = 3X + 1
σY =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 25, 36, 60, 19, 52.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 100. Find the z-score z of a score of 380.
μ = 500σ = 100x = 380
z =
19.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.7291. Find z to two decimals.
A = 0.7291
z =
20.Normal distributionHard
Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
3Solutions
1.Limits
Answer: L = −1.75
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).
Multiply by the conjugate
f(x) = −7x/(√(4x2 − 7x) + 2x)
Divide by x
f(x) = −7/(√(4 + (−7)/x) + 2)
Let x → ∞
L = −7/(2 + 2) = −1.75
Answer
L = −1.75
2.Limits
Answer: L = 0.5
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).
First application
2·sin 2x/(8x) → 0/0
Second application
4·cos 2x/8
Substitute x = 0
L = 4/8 = 0.5
Answer
L = 0.5
3.Derivatives
Answer: f′(1.5) = 271.2
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.
Product rule
f′(x) = u′v + uv′ = e2x·(3x2 + 2·x3)
Substitute
f′(1.5) = 20.09·(6.75 + 6.75) = 271.2
Answer
f′(1.5) = 271.2
4.Derivatives
Answer: b = −0.5
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) = 5.5
f′(x) = x + 5 → f′(1) = 6
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 5.5 − 6·1 = −0.5
Answer
b = −0.5
5.Integrals
Answer: I = 41.33
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.
Substitution
u = x² + 1, du = 2x dx
I = ∫15 u2 du
Integrate and evaluate
I = [u3/3]15 = (125 − 1)/3 = 41.33
Answer
I = 41.33
6.Integrals
Answer: V = 15.89
Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.
Intersection points
4x = 3x² → x = 0, x = 1.333
Washer method
V = π∫₀1.333 ((4x)² − (3x²)²) dx
Integrate
V = π(16·1.333³/3 − 9·1.333⁵/5) = π·5.057 = 15.89
Answer
V = 15.89
7.Complex numbers
Answer: A (−8 + j8)
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.
Effect of w
|z·w| = 2·|z|, arg(z·w) = arg z + (−90)°
The point
z·w = −8 + j8 → A
Answer
A (−8 + j8)
8.Complex numbers
Answer: Re(z³) = −11
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.
Powers step by step
z² = (1 + j2)(1 + j2) = −3 + j4
z³ = (−3 + j4)(1 + j2) = −11 − j2
Check with De Moivre
z = 2.236∠63.43° → z³ = 11.18∠−169.7°
Re(z³) = −11
Answer
Re(z³) = −11
9.Differential equations
Answer: t = 39.34 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(-21/-10)/15 = 0.04946 min⁻¹
Time to reach the target
t = ln(-21/(−3))/0.04946 = 39.34 min
Answer
t = 39.34 min
10.Differential equations
Answer: m = 5.934 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 = 2·0.1 − (2/100)·m = 0.2 − 0.02·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 10 + (−10)·e−t/50
Substitute t
m(45) = 10 + (−10)·0.4066 = 5.934 kg
Answer
m = 5.934 kg
11.Matrices & determinants
Answer: c21 = 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
c21 = 2·3 + 6·2 = 18
Answer
c21 = 18
12.Matrices & determinants
Answer: λ = −1
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 − (−5)λ + (4) = 0
Solve
λ = (−5 ± √(9))/2 → λ = −4, −1
λ = −1
Answer
λ = −1
13.Vectors: dot & cross product
Answer: compb a = 5.013
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 = (−1)·3 + (−7)·(−6) + (−5)·1 = 34
Length of b
|b| = √(3² + (−6)² + 1²) = 6.782
Scalar projection
compb a = a·b/|b| = 34/6.782 = 5.013
Answer
compb a = 5.013
14.Vectors: dot & cross product
Answer: V = 92
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 = (−4)·5 − 1·(−1) = −19
(b × c)y = bzcx − bxcz = 1·5 − 2·5 = −5
(b × c)z = bxcy − bycx = 2·(−1) − (−4)·5 = 18
Dot product with a
a·(b × c) = (−5)·(−19) + (−3)·(−5) + (−1)·18 = 92
V = |92| = 92
Answer
V = 92
15.Probability
Answer: Var(X) = 5.39
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 + 2·0.2 + 4·0.25 + 6·0.25 = 2.9
Mean of the squares
E(X²) = 0·0.3 + 4·0.2 + 16·0.25 + 36·0.25 = 13.8
Variance
Var(X) = 13.8 − 2.9² = 5.39
Answer
Var(X) = 5.39
16.Probability
Answer: σY = 2.414
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 = 1.05, Var(X) = 0.6475, σX = 0.8047
Linear change
σY = |3|·0.8047 = 2.414
Answer
σY = 2.414
The mean does move: E(Y) = 3·1.05 + 1 = 4.15.
17.Descriptive statistics
Answer: s = 17.42
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 = (380 − 500)/100 = −1.2
Answer
z = −1.2
19.Normal distribution
Answer: z = −0.61
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 − 0.7291 = 0.2709
Read the z-table backwards
Find 0.2709 inside the table and read off z.
z = −0.61
Answer
z = −0.61
20.Normal distribution
Answer: P = 0.7881
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).