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→∞ (1 − 1/x)x. Give a decimal value.
f(x) = (1 − 1/x)xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e0.5x − 1 − 0.5x)/x².
f(x) = (e0.5x − 1 − 0.5x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(3x), with x in radians. Find f′(1).
f(x) = x·sin(3x)x = 1 rad
f′(1) =
4.DerivativesHard
The tangent line to f(x) = x2 − 2 at x = 2 crosses the y-axis at y = b. Find b.
f(x) = x2 − 2x = 2
b =
5.IntegralsMedium
Evaluate I = ∫26 2/x dx.
f(x) = 2/x[2, 6]
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 9 − x² and the x-axis.
y = 9 − x²
A =
7.Complex numbersMedium
Find the argument θ of z = −7 − j4 in degrees, as the principal value between −180° and 180°.
z = −7 − j4
θ = °
8.Complex numbersHard
Solve z² − 5z + 22.25 = 0. The roots are a ± jb; find b > 0.
p = −5q = 22.25
b =
9.Differential equationsMedium
A can of soda at 5 °C is taken out of the fridge into a room at 27 °C. After 22 min its temperature is 20 °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 = 27 °CT(22) = 20 °CT = 23 °C
t = min
10.Differential equationsHard
A tank holds 1000 L of water in which 20 kg of salt is dissolved. Brine with 0.1 kg of salt per litre flows in at 10 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 90 min?
V = 1000 Lm0 = 20 kgc = 0.1 kg/Lr = 10 L/min
m = kg
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [−5 −6; 2 6]B = [2 1; −4 3]
c21 =
12.Matrices & determinantsHard
Solve the system −x + y = 5 and 3x − 4y = −3. Find x.
−x + y = 53x − 4y = −3
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (6, 3, 1) and b = (1, k, −1) are perpendicular.
a = (6, 3, 1)b = (1, k, −1)
k =
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (3, −5, 4) and b = (5, 3, −5) and has a positive z-component. Find its y-component n̂y.
a = (3, −5, 4)b = (5, 3, −5)n̂z > 0
n̂y =
15.ProbabilityMedium
In how many ways N can a committee of 3 people be chosen from a group of 15? The order does not matter.
n = 15k = 3
N =
16.ProbabilityHard
7 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?
n = 7p = 0.1
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 53, 60, 10, 34, 28, 29.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 40. What proportion P of values is above 550?
μ = 500σ = 40x = 550
P =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.5346. Find z to two decimals.
A = 0.5346
z =
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 = 0.3679
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).
Take the logarithm
ln L = limx→∞ x·ln(1 − 1/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 1·(−1) = −1
Undo the logarithm
L = e−1 = 0.3679
Answer
L = 0.3679
2.Limits
Answer: L = 0.125
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
(0.5·e0.5x − 0.5)/(2x) → 0/0
Second application
0.25·e0.5x/2
Substitute x = 0
L = 0.25·1/2 = 0.125
Answer
L = 0.125
3.Derivatives
Answer: f′(1) = −2.829
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) = 0.1411 + 3·(−0.99) = −2.829
Answer
f′(1) = −2.829
4.Derivatives
Answer: b = −6
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) = 2
f′(x) = 2x → f′(2) = 4
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 2 − 4·2 = −6
Answer
b = −6
5.Integrals
Answer: I = 2.197
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) = 2·ln x
Evaluate at the limits
I = 2·(ln 6 − ln 2) = 2·ln 3 = 2.197
Answer
I = 2.197
6.Integrals
Answer: A = 36
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
9 − x² = 0 → x = ±3
Integrate
A = [9x − x³/3]−33 = 18 − (−18) = 36
Answer
A = 36
7.Complex numbers
Answer: θ = −150.3 °
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.
Quadrant
a = −7, b = −4 → III
Reference angle
α = atan(4/7) = 29.74°
Argument
θ = −(180° − α) = −150.3°
Answer
θ = −150.3°
In polar form z = 8.062∠−150.3°.
8.Complex numbers
Answer: b = 4
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 = 25 − 89 = −64 < 0
Quadratic formula
z = (5 ± j√64)/2 = 2.5 + j4, 2.5 − j4
Answer
b = 4
Answer
b = 4
9.Differential equations
Answer: t = 32.75 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(-22/-7)/22 = 0.05205 min⁻¹
Time to reach the target
t = ln(-22/(−4))/0.05205 = 32.75 min
Answer
t = 32.75 min
10.Differential equations
Answer: m = 67.47 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 = 10·0.1 − (10/1000)·m = 1 − 0.01·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 100 + (−80)·e−t/100
Substitute t
m(90) = 100 + (−80)·0.4066 = 67.47 kg
Answer
m = 67.47 kg
11.Matrices & determinants
Answer: c21 = −20
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 = 2·2 + 6·(−4) = −20
Answer
c21 = −20
12.Matrices & determinants
Answer: x = −17
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 = (−1)·(−4) − 1·3 = 1
Replace the x column
det Aₓ = 5·(−4) − 1·(−3) = −17
x = −17/1 = −17
Answer
x = −17
y = −12
13.Vectors: dot & cross product
Answer: k = −1.667
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 = 6·1 + 3·k + 1·(−1) = 0
5 + 3·k = 0
Solve for k
k = −5/3 = −1.667
Answer
k = −1.667
14.Vectors: dot & cross product
Answer: n̂y = 0.6931
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 = (−5)·(−5) − 4·3 = 13
(a × b)y = azbx − axbz = 4·5 − 3·(−5) = 35
(a × b)z = axby − aybx = 3·3 − (−5)·5 = 34
Normalise
The z-component of a × b is already positive.
|a × b| = √(13² + 35² + 34²) = 50.5
n̂ = (13, 35, 34)/50.5 = (0.2574, 0.6931, 0.6733)
n̂y = 0.6931
Answer
n̂y = 0.6931
15.Probability
Answer: N = 455
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.
Combinations
C(15, 3) = (15·14·13)/(3·2·1) = 455
Answer
N = 455
16.Probability
Answer: P = 0.8503
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) = 0.97 = 0.4783
P(X = 1) = 7·0.1·0.96 = 0.372
Add them
P = 0.4783 + 0.372 = 0.8503
Answer
P = 0.8503
17.Descriptive statistics
Answer: s = 18.21
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 = 1.25
Standard normal table
Φ(1.25) = 0.8944
P = 1 − 0.8944 = 0.1056
Answer
P = 0.1056
19.Normal distribution
Answer: z = 0.73
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 + 0.5346)/2 = 0.7673
Read the z-table backwards
Find 0.7673 inside the table and read off z.
z = 0.73
Answer
z = 0.73
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).