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→∞ (6x2 + 2x + 3)/(x2 + 2x + 6).
f(x) = (6x2 + 2x + 3)/(x2 + 2x + 6)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 5x)/(x²), with x in radians.
f(x) = (1 − cos 5x)/(x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(x), with x in radians. Find f′(2).
f(x) = x·sin(x)x = 2 rad
f′(2) =
4.DerivativesHard
Let f(x) = x3 − 3x2 − 9x − 4. Find the x-coordinate x of the local maximum.
f(x) = x3 − 3x2 − 9x − 4
x =
5.IntegralsMedium
Evaluate I = ∫₀π/3 x·cos x dx (x in radians).
∫₀π/3 x·cos x dx
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
Write z = 4∠35° in rectangular form a + jb and give the imaginary part b.
r = 4θ = 35°
b =
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 forged steel part at 500 °C is left to cool in a workshop at 21 °C. After 60 min its temperature is 45 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 31 °C?
T0 = 500 °CTa = 21 °CT(60) = 45 °CT = 31 °C
t = min
10.Differential equationsHard
A tank holds 500 L of water in which 10 kg of salt is dissolved. 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 75 min?
V = 500 Lm0 = 10 kgc = 0.1 kg/Lr = 2 L/min
m = kg
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−4 4 −4; 1 −4 4; −3 5 −1]
det A =
12.Matrices & determinantsHard
Solve the system −x + 2y = −3 and 6x − 2y = 6. Find x.
−x + 2y = −36x − 2y = 6
x =
13.Vectors: dot & cross productMedium
A constant force F = (−15, 5, 20) N acts on an object while it moves in a straight line from point A(−2, 0, 4) to point B(0, −3, 1). Coordinates are in metres. Find the work W done by the force.
F = (−15, 5, 20) NA(−2, 0, 4) mB(0, −3, 1) m
W = J
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (2, 3, −4), b = (5, −3, 0) and c = (−1, −5, −4).
a = (2, 3, −4)b = (5, −3, 0)c = (−1, −5, −4)
V =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4p(x): 0.15, 0.25, 0.25, 0.35
Var(X) =
16.ProbabilityHard
A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 1 correct answers?
n = 7p = 0.3k = 1
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 58, 44, 54, 57, 17.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 5. What proportion P of values is above 97.5?
μ = 100σ = 5x = 97.5
P =
19.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 442 and 580 h? Give P as a decimal.
μ = 500σ = 40x1 = 442x2 = 580
P =
20.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 95% of them lie (the 95th percentile).
μ = 178σ = 7P = 95%
x = cm
3Solutions
1.Limits
Answer: L = 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
(6 + 2/x + 3/x2)/(1 + 2/x + 6/x2)
Let x → ∞
L = 6/1 = 6
Answer
L = 6
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 12.5
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
5·sin 5x/(2x) → 0/0
Second application
25·cos 5x/2
Substitute x = 0
L = 25/2 = 12.5
Answer
L = 12.5
3.Derivatives
Answer: f′(2) = 0.077
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(x) + x·cos(x)
Substitute
f′(2) = 0.9093 + 2·(−0.4161) = 0.077
Answer
f′(2) = 0.077
4.Derivatives
Answer: x = −1
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 − 6x − 9 = 3·(x2 − 2x − 3)
Solve f′(x) = 0
x = −1 or x = 3
Which one is which
f″(x) = 6x − 6: negative at a maximum, positive at a minimum.
f″(−1) = −12, f″(3) = 12
x = −1
Answer
x = −1
5.Integrals
Answer: I = 0.4069
Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.
Choose u and dv
u = x, du = dx; dv = cos x dx, v = sin x
Antiderivative
∫x cos x dx = x sin x − ∫sin x dx = x sin x + cos x
Limits
(π/3·sin π/3 + cos π/3) − (0 + 1) = 0.4069
Answer
I = 0.4069
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: b = 2.294
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.
Components
a = 4·cos(35°) = 3.277
b = 4·sin(35°) = 2.294
Rectangular form
z = 3.277 + j2.294
b = 2.294
Answer
b = 2.294
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 = 77.55 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(479/24)/60 = 0.04989 min⁻¹
Time to reach the target
t = ln(479/10)/0.04989 = 77.55 min
Answer
t = 77.55 min
10.Differential equations
Answer: m = 20.37 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 − (2/500)·m = 0.2 − 0.004·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 50 + (−40)·e−t/250
Substitute t
m(75) = 50 + (−40)·0.7408 = 20.37 kg
Answer
m = 20.37 kg
11.Matrices & determinants
Answer: det A = 48
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.
2×2 minors of the first row
M₁₁ = −16, M₁₂ = 11, M₁₃ = −7
Cofactor expansion
det A = (−4)·(−16) − 4·11 + (−4)·(−7) = 48
Answer
det A = 48
12.Matrices & determinants
Answer: x = 0.6
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)·(−2) − 2·6 = −10
Replace the x column
det Aₓ = (−3)·(−2) − 2·6 = −6
x = −6/(−10) = 0.6
Answer
x = 0.6
y = −1.2
13.Vectors: dot & cross product
Answer: W = −105 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 = (0, −3, 1) − (−2, 0, 4) = (2, −3, −3) m
Work
W = F·d = (−15)·2 + 5·(−3) + 20·(−3) = −105 J
Answer
W = −105 J
Negative work: the force acts mostly against the motion.
14.Vectors: dot & cross product
Answer: V = 196
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 = (−3)·(−4) − 0·(−5) = 12
(b × c)y = bzcx − bxcz = 0·(−1) − 5·(−4) = 20
(b × c)z = bxcy − bycx = 5·(−5) − (−3)·(−1) = −28
Dot product with a
a·(b × c) = 2·12 + 3·20 + (−4)·(−28) = 196
V = |196| = 196
Answer
V = 196
15.Probability
Answer: Var(X) = 1.16
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 = 1·0.15 + 2·0.25 + 3·0.25 + 4·0.35 = 2.8
Mean of the squares
E(X²) = 1·0.15 + 4·0.25 + 9·0.25 + 16·0.35 = 9
Variance
Var(X) = 9 − 2.8² = 1.16
Answer
Var(X) = 1.16
16.Probability
Answer: P = 0.2471
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 = 1) = 7·0.31·0.76 = 7·0.3·0.1176 = 0.2471
Answer
P = 0.2471
17.Descriptive statistics
Answer: s = 17.13
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: P = 0.9037
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 = (442 − 500)/40 = −1.45
z₂ = (580 − 500)/40 = 2
Area between
P = Φ(2) − Φ(−1.45) = 0.9772 − 0.07353 = 0.9037
Answer
P = 0.9037
20.Normal distribution
Answer: x = 189.5 cm
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.