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→∞ (−x3 + 6x − 7)/(3x3 + x + 3).
f(x) = (−x3 + 6x − 7)/(3x3 + x + 3)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + x)/sin(3x), with x in radians.
f(x) = ln(1 + x)/sin(3x)x → 0
L =
3.DerivativesMedium
Let f(x) = (2x − 3)4. Find f′(0).
f(x) = (2x − 3)4x = 0
f′(0) =
4.DerivativesHard
The tangent line to f(x) = x2 + x + 6 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = x2 + x + 6x = 1
b =
5.IntegralsMedium
Evaluate I = ∫02 e−0.5x dx.
f(x) = e−0.5x[0, 2]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = 3x.
y = x2y = 3x
A =
7.Complex numbersMedium
The diagram shows z = −1 − j. With w = 2∠90°, click the point that represents z·w.
z = −1 − jw = 2∠90°
1A (−2 − j2)
2B (1 − j)
3C (−2 + j2)
4D (2 − j2)
8.Complex numbersHard
Solve z² + 2z + 5 = 0. The roots are a ± jb; find the real part a.
p = 2q = 5
a =
9.Differential equationsMedium
A can of soda at 3 °C is taken out of the fridge into a room at 29 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.03 min⁻¹, find its temperature T after 29 min.
T0 = 3 °CTa = 29 °Ck = 0.03 min⁻¹t = 29 min
T = °C
10.Differential equationsHard
Solve y″ + 4y′ + 20y = 0 with y(0) = 2 and y′(0) = 0 (released from rest). Find y(0.25); angles in radians.
y″ + 4y′ + 20y = 0y(0) = 2y′(0) = 0
y(0.25) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [−1 2; −2 2]B = [−2 3; 4 −6]
c21 =
12.Matrices & determinantsHard
Solve the system 5x − y = 0 and 4x + 6y = −2. Find x.
5x − y = 04x + 6y = −2
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (0, 6, −5) and b = (5, −2, −3), in degrees.
a = (0, 6, −5)b = (5, −2, −3)
θ = °
14.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (1, −6, 3) and b = (4, −2, 6).
a = (1, −6, 3)b = (4, −2, 6)
A =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4, 5p(x): 0.05, 0.35, 0.2, 0.35, 0.05
σ =
16.ProbabilityHard
A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.7, independently. What is the probability P of exactly 4 correct answers?
n = 5p = 0.7k = 4
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 13, 44, 52, 35, 50, 52, 29.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 12. What proportion P of values is above 182?
μ = 170σ = 12x = 182
P =
19.Normal distributionMedium
X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 35.1 min. Round z to two decimals.
μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 35.1 min
P =
20.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 99% of them lie (the 99th percentile).
μ = 500σ = 40P = 99%
x = h
3Solutions
1.Limits
Answer: L = −0.3333
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
(−1 + 6/x2 + (−7)/x3)/(3 + 1/x2 + 3/x3)
Let x → ∞
L = −1/3 = −0.3333
Answer
L = −0.3333
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 0.3333
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).
L'Hôpital's rule
L = limx→0 [1/(1 + x)]/[3·cos 3x]
Substitute x = 0
L = 1/3 = 0.3333
Answer
L = 0.3333
3.Derivatives
Answer: f′(0) = −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 − 3)3·2
Substitute
f′(0) = 4·(−3)3·2 = −216
Answer
f′(0) = −216
4.Derivatives
Answer: b = 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) = 8
f′(x) = 2x + 1 → f′(1) = 3
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 8 − 3·1 = 5
Answer
b = 5
5.Integrals
Answer: I = 1.264
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.
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
x2 = 3x → x = 0, x = 3
Integrate upper minus lower
A = [1.5x2 − 0.3333x3]03 = 4.5
Answer
A = 4.5
7.Complex numbers
Answer: D (2 − j2)
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 = 2 − j2 → D
Answer
D (2 − j2)
8.Complex numbers
Answer: a = −1
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 = 4 − 20 = −16 < 0
Quadratic formula
z = (−2 ± j√16)/2 = −1 + j2, −1 − j2
Answer
a = −1
Answer
a = −1
9.Differential equations
Answer: T = 18.11 °C
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.
Solution of the cooling law
T(t) = 29 + (−26)·e−0.03t
Substitute t
T(29) = 29 + (−26)·0.419 = 18.11 °C
Answer
T = 18.11 °C
10.Differential equations
Answer: y(0.25) = 1.166
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 = −2 ± 4j
Constants
A = 2, B = 2·2/4 = 1
Evaluate
y(0.25) = e−0.5(2·cos 1 + 1·sin 1) = 1.166
Answer
y(0.25) = 1.166
11.Matrices & determinants
Answer: c21 = 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
c21 = (−2)·(−2) + 2·4 = 12
Answer
c21 = 12
12.Matrices & determinants
Answer: x = −0.05882
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 = 5·6 − (−1)·4 = 34
Replace the x column
det Aₓ = 0·6 − (−1)·(−2) = −2
x = −2/34 = −0.05882
Answer
x = −0.05882
y = −0.2941
13.Vectors: dot & cross product
Answer: θ = 86.43 °
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 = 0·5 + 6·(−2) + (−5)·(−3) = 3
Magnitudes
|a| = √(0² + 6² + (−5)²) = 7.81
|b| = √(5² + (−2)² + (−3)²) = 6.164
Angle
cos θ = 3/(7.81·6.164) = 0.06231
θ = arccos(0.06231) = 86.43°
Answer
θ = 86.43°
14.Vectors: dot & cross product
Answer: A = 37.68
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 = (−6)·6 − 3·(−2) = −30
(a × b)y = azbx − axbz = 3·4 − 1·6 = 6
(a × b)z = axby − aybx = 1·(−2) − (−6)·4 = 22
Area
A = |a × b| = √((−30)² + 6² + 22²) = 37.68
Answer
A = 37.68
15.Probability
Answer: σ = 1.049
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.
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.
Binomial formula
P(X = 4) = 5·0.74·0.31 = 5·0.2401·0.3 = 0.3601
Answer
P = 0.3601
17.Descriptive statistics
Answer: s = 14.58
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
Standard normal table
Φ(1) = 0.8413
P = 1 − 0.8413 = 0.1587
Answer
P = 0.1587
19.Normal distribution
Answer: P = 0.1075
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).
Distribution of the sum
μ = 32 min, σ = √(1.5² + 2²) = 2.5 min
Standardise
z = (35.1 − 32)/2.5 = 1.24
Upper tail
P = 1 − Φ(1.24) = 1 − 0.8925 = 0.1075
Answer
P = 0.1075
20.Normal distribution
Answer: x = 593.1 h
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.