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→∞ (−3x3 + 2x − 5)/(5x3 + 4x + 4).
f(x) = (−3x3 + 2x − 5)/(5x3 + 4x + 4)x → ∞
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) = (4x − 3)5. Find f′(1).
f(x) = (4x − 3)5x = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 − 7.5x2 + 18x + 5. Find the x-coordinate x of the local maximum.
f(x) = x3 − 7.5x2 + 18x + 5
x =
5.IntegralsMedium
Evaluate I = ∫26 5/x dx.
f(x) = 5/x[2, 6]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = 0.5x2 and y = 3x.
y = 0.5x2y = 3x
A =
7.Complex numbersMedium
The number z = 4·e−j0.5 has its angle in radians. Find its real part a.
r = 4θ = −0.5 rad
a =
8.Complex numbersHard
The roots of zⁿ = w with n = 4 and w = 16∠−60° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2, 3. Find the real part of z₀.
w = 16∠−60°n = 4k = 0
Re(z₀) =
9.Differential equationsMedium
A forged steel part at 320 °C is left to cool in a workshop at 23 °C. After 27 min its temperature is 196 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 258 °C?
T0 = 320 °CTa = 23 °CT(27) = 196 °CT = 258 °C
t = min
10.Differential equationsHard
Solve y″ + 3y′ + 6.25y = 0 with y(0) = 2 and y′(0) = 0 (released from rest). Find y(1.5); angles in radians.
y″ + 3y′ + 6.25y = 0y(0) = 2y′(0) = 0
y(1.5) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [2 −4; −3 −5]B = [5 3; 3 5]
c11 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [3 3; −8 −7]
λ =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (2, 1, −6) onto b = (−2, 2, −3), that is, the component of a in the direction of b.
a = (2, 1, −6)b = (−2, 2, −3)
compb a =
14.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (−2, 2, 4) and b = (4, 5, −1).
a = (−2, 2, 4)b = (4, 5, −1)
A =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4p(x): 0.25, 0.25, 0.25, 0.25
Var(X) =
16.ProbabilityHard
X has the distribution shown, and Y = 10X + 20. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 43, 58, 20, 30, 29, 30, 15.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 20. What proportion P of values is above 60?
μ = 70σ = 20x = 60
P =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.6047. Find z to two decimals.
A = 0.6047
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 = −0.6
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
(−3 + 2/x2 + (−5)/x3)/(5 + 4/x2 + 4/x3)
Let x → ∞
L = −3/5 = −0.6
Answer
L = −0.6
For equal degrees the limit is the ratio of the leading coefficients.
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) = 20
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) = 5·(4x − 3)4·4
Substitute
f′(1) = 5·14·4 = 20
Answer
f′(1) = 20
4.Derivatives
Answer: x = 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.
Derivative
f′(x) = 3x2 − 15x + 18 = 3·(x2 − 5x + 6)
Solve f′(x) = 0
x = 2 or x = 3
Which one is which
f″(x) = 6x − 15: negative at a maximum, positive at a minimum.
f″(2) = −3, f″(3) = 3
x = 2
Answer
x = 2
5.Integrals
Answer: I = 5.493
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) = 5·ln x
Evaluate at the limits
I = 5·(ln 6 − ln 2) = 5·ln 3 = 5.493
Answer
I = 5.493
6.Integrals
Answer: A = 18
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
0.5x2 = 3x → x = 0, x = 6
Integrate upper minus lower
A = [1.5x2 − 0.1667x3]06 = 18
Answer
A = 18
7.Complex numbers
Answer: a = 3.51
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 = 4·(cos −0.5 + j sin −0.5)
Components (radians)
a = 4·cos −0.5 = 3.51
b = 4·sin −0.5 = −1.918
a = 3.51
Answer
a = 3.51
8.Complex numbers
Answer: Re(z₀) = 1.932
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
161/4 = 2
Argument of the root
θ₀ = (−60° + 360°·0)/4 = −15°
Rectangular form
z₀ = 2·(cos −15° + j sin −15°) = 1.932 − j0.5176
Re(z₀) = 1.932
Answer
Re(z₀) = 1.932
The 4 roots lie on the circle of radius 2, 90° apart.
9.Differential equations
Answer: t = 11.7 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(297/173)/27 = 0.02002 min⁻¹
Time to reach the target
t = ln(297/235)/0.02002 = 11.7 min
Answer
t = 11.7 min
10.Differential equations
Answer: y(1.5) = −0.1864
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.
Complex roots
r = −1.5 ± 2j
Constants
A = 2, B = 1.5·2/2 = 1.5
Evaluate
y(1.5) = e−2.25(2·cos 3 + 1.5·sin 3) = −0.1864
Answer
y(1.5) = −0.1864
11.Matrices & determinants
Answer: c11 = −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.
Row times column
c11 = 2·5 + (−4)·3 = −2
Answer
c11 = −2
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 − (−4)λ + (3) = 0
Solve
λ = (−4 ± √(4))/2 → λ = −3, −1
λ = −1
Answer
λ = −1
13.Vectors: dot & cross product
Answer: compb a = 3.881
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·(−2) + 1·2 + (−6)·(−3) = 16
Length of b
|b| = √((−2)² + 2² + (−3)²) = 4.123
Scalar projection
compb a = a·b/|b| = 16/4.123 = 3.881
Answer
compb a = 3.881
14.Vectors: dot & cross product
Answer: A = 31.69
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 = 2·(−1) − 4·5 = −22
(a × b)y = azbx − axbz = 4·4 − (−2)·(−1) = 14
(a × b)z = axby − aybx = (−2)·5 − 2·4 = −18
Area
A = |a × b| = √((−22)² + 14² + (−18)²) = 31.69
Answer
A = 31.69
15.Probability
Answer: Var(X) = 1.25
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 = 1·0.25 + 2·0.25 + 3·0.25 + 4·0.25 = 2.5
Mean of the squares
E(X²) = 1·0.25 + 4·0.25 + 9·0.25 + 16·0.25 = 7.5
Variance
Var(X) = 7.5 − 2.5² = 1.25
Answer
Var(X) = 1.25
16.Probability
Answer: σY = 10.62
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 = 1.35, Var(X) = 1.127, σX = 1.062
Linear change
σY = |10|·1.062 = 10.62
Answer
σY = 10.62
The mean does move: E(Y) = 10·1.35 + 20 = 33.5.
17.Descriptive statistics
Answer: s = 14.42
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 = −0.5
Standard normal table
Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915
Answer
P = 0.6915
19.Normal distribution
Answer: z = 0.85
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 + 0.6047)/2 = 0.8023
Read the z-table backwards
Find 0.8023 inside the table and read off z.
z = 0.85
Answer
z = 0.85
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).