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 − 4x + 8)/(2x2 + x + 5).
f(x) = (−4x2 − 4x + 8)/(2x2 + x + 5)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e−2x − 1 + 2x)/x².
f(x) = (e−2x − 1 + 2x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x2·e0.5x. Find f′(0.5).
f(x) = x2·e0.5xx = 0.5
f′(0.5) =
4.DerivativesHard
Let f(x) = x3 − 4.5x2 − 4. Find the x-coordinate x of the local minimum.
f(x) = x3 − 4.5x2 − 4
x =
5.IntegralsMedium
Evaluate I = ∫03 ex dx.
f(x) = ex[0, 3]
I =
6.IntegralsHard
The region between the line y = 5x and the parabola y = 3x² is rotated about the x-axis. Find the volume V of the solid.
y = 5xy = 3x²
V =
7.Complex numbersMedium
Find the argument θ of z = 3 − j2 in degrees, as the principal value between −180° and 180°.
z = 3 − j2
θ = °
8.Complex numbersHard
The roots of zⁿ = w with n = 4 and w = 16∠−90° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2, 3. Find the real part of z₃.
w = 16∠−90°n = 4k = 3
Re(z₃) =
9.Differential equationsMedium
Solve y″ + 7y′ + 10y = 0 with y(0) = 1 and y′(0) = 6. Find y(0.5).
y″ + 7y′ + 10y = 0y(0) = 1y′(0) = 6
y(0.5) =
10.Differential equationsHard
A tank holds 100 L of water in which 20 kg of salt is dissolved. Brine with 0.05 kg of salt per litre flows in at 5 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 25 min?
V = 100 Lm0 = 20 kgc = 0.05 kg/Lr = 5 L/min
m = kg
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c22 (row 2, column 2).
A = [6 0; −6 0]B = [6 4; −1 5]
c22 =
12.Matrices & determinantsHard
Solve the system −5x − 4y = −10 and −6x + 3y = −12. Find x.
−5x − 4y = −10−6x + 3y = −12
x =
13.Vectors: dot & cross productMedium
A constant force F = (−10, −30, 30) N acts on an object while it moves in a straight line from point A(−3, −1, 0) to point B(2, 4, −3). Coordinates are in metres. Find the work W done by the force.
F = (−10, −30, 30) NA(−3, −1, 0) mB(2, 4, −3) m
W = J
14.Vectors: dot & cross productHard
For a = (0, −5, −5) and b = (4, 4, −6), find the x-component (a × b)x of the cross product a × b.
a = (0, −5, −5)b = (4, 4, −6)
(a × b)x =
15.ProbabilityMedium
A bag holds 7 red and 9 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
7 / 9
P =
16.ProbabilityHard
7 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?
n = 7p = 0.6
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 38, 18, 28, 48, 55, 42.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 5. What proportion P of values is below 97.5?
μ = 100σ = 5x = 97.5
P =
19.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 88 and 118.75 points? Give P as a decimal.
μ = 100σ = 15x1 = 88x2 = 118.75
P =
20.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 75% is to the left of it. Find the cut-off x.
μ = 178 cmσ = 7 cm75%
x = cm
3Solutions
1.Limits
Answer: L = −2
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).
Divide by the highest power
(−4 + (−4)/x + 8/x2)/(2 + 1/x + 5/x2)
Let x → ∞
L = −4/2 = −2
Answer
L = −2
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 2
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·e−2x + 2)/(2x) → 0/0
Second application
4·e−2x/2
Substitute x = 0
L = 4·1/2 = 2
Answer
L = 2
3.Derivatives
Answer: f′(0.5) = 1.445
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′ = e0.5x·(2x + 0.5·x2)
Substitute
f′(0.5) = 1.284·(1 + 0.125) = 1.445
Answer
f′(0.5) = 1.445
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 − 9x = 3·(x2 − 3x)
Solve f′(x) = 0
x = 0 or x = 3
Which one is which
f″(x) = 6x − 9: negative at a maximum, positive at a minimum.
f″(0) = −9, f″(3) = 9
x = 3
Answer
x = 3
5.Integrals
Answer: I = 19.09
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) = ex/1
Evaluate at the limits
I = (e3 − 1)/1 = (20.09 − 1)/1 = 19.09
Answer
I = 19.09
6.Integrals
Answer: V = 48.48
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
5x = 3x² → x = 0, x = 1.667
Washer method
V = π∫₀1.667 ((5x)² − (3x²)²) dx
Integrate
V = π(25·1.667³/3 − 9·1.667⁵/5) = π·15.43 = 48.48
Answer
V = 48.48
7.Complex numbers
Answer: θ = −33.69 °
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 = 3, b = −2 → IV
Reference angle
α = atan(2/3) = 33.69°
Argument
θ = −α = −33.69°
Answer
θ = −33.69°
In polar form z = 3.606∠−33.69°.
8.Complex numbers
Answer: Re(z₃) = −0.7654
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.
The 4 roots lie on the circle of radius 2, 90° apart.
9.Differential equations
Answer: y(0.5) = 1.13
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 + 2)(r + 5) = 0 → r = −2, −5
Constants from the initial values
A + B = 1, −2A − 5B = 6
A = 3.667, B = −2.667
Evaluate
y(0.5) = 3.667·e−1 + (−2.667)·e−2.5 = 1.13
Answer
y(0.5) = 1.13
10.Differential equations
Answer: m = 9.298 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 = 5·0.05 − (5/100)·m = 0.25 − 0.05·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 5 + 15·e−t/20
Substitute t
m(25) = 5 + 15·0.2865 = 9.298 kg
Answer
m = 9.298 kg
11.Matrices & determinants
Answer: c22 = −24
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
c22 = (−6)·4 + 0·5 = −24
Answer
c22 = −24
12.Matrices & determinants
Answer: x = 2
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 = (−5)·3 − (−4)·(−6) = −39
Replace the x column
det Aₓ = (−10)·3 − (−4)·(−12) = −78
x = −78/(−39) = 2
Answer
x = 2
y = 0
13.Vectors: dot & cross product
Answer: W = −290 J
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.
Displacement
d = B − A = (2, 4, −3) − (−3, −1, 0) = (5, 5, −3) m
Work
W = F·d = (−10)·5 + (−30)·5 + 30·(−3) = −290 J
Answer
W = −290 J
Negative work: the force acts mostly against the motion.
14.Vectors: dot & cross product
Answer: (a × b)x = 50
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 = (−5)·(−6) − (−5)·4 = 50
(a × b)y = azbx − axbz = (−5)·4 − 0·(−6) = −20
(a × b)z = axby − aybx = 0·4 − (−5)·4 = 20
Requested component
(a × b)x = 50
Answer
(a × b)x = 50
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: P = 0.175
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.
First and second draw
P = 7/16 · 6/15 = 42/240 = 0.175
Answer
P = 0.175
16.Probability
Answer: P = 0.01884
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) = 0.47 = 0.001638
P(X = 1) = 7·0.6·0.46 = 0.0172
Add them
P = 0.001638 + 0.0172 = 0.01884
Answer
P = 0.01884
17.Descriptive statistics
Answer: s = 13.45
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 = −0.5
Standard normal table
Φ(−0.5) = 0.3085
P = 0.3085
Answer
P = 0.3085
19.Normal distribution
Answer: P = 0.6825
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 = (88 − 100)/15 = −0.8
z₂ = (118.75 − 100)/15 = 1.25
Area between
P = Φ(1.25) − Φ(−0.8) = 0.8944 − 0.2119 = 0.6825
Answer
P = 0.6825
20.Normal distribution
Answer: x = 182.7 cm
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.