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 − x + 1)/(x3 + 2x + 2).
f(x) = (3x3 − x + 1)/(x3 + 2x + 2)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e−0.5x − 1 + 0.5x)/x².
f(x) = (e−0.5x − 1 + 0.5x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = (−2x + 5)4. Find f′(−1).
f(x) = (−2x + 5)4x = −1
f′(−1) =
4.DerivativesHard
Let f(x) = x3 − 12x2 + 36x − 1. Find the x-coordinate x of the local maximum.
f(x) = x3 − 12x2 + 36x − 1
x =
5.IntegralsMedium
Evaluate I = ∫02 2x·(x² + 1)3 dx.
f(x) = 2x·(x² + 1)3[0, 2]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = x.
y = x2y = x
A =
7.Complex numbersMedium
Find the argument θ of z = −3 − j3 in degrees, as the principal value between −180° and 180°.
z = −3 − j3
θ = °
8.Complex numbersHard
Find the imaginary part of (−3 − j)⁴.
z = −3 − jn = 4
Im(z⁴) =
9.Differential equationsMedium
A forged steel part at 440 °C is left to cool in a workshop at 27 °C. After 19 min its temperature is 369 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 215 °C?
T0 = 440 °CTa = 27 °CT(19) = 369 °CT = 215 °C
t = min
10.Differential equationsHard
A tank holds 400 L of water in which 5 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. After how many minutes t does the tank contain 16 kg of salt?
V = 400 Lm0 = 5 kgc = 0.1 kg/Lr = 2 L/min
t = min
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−1 −2 −4; −3 −1 −1; 0 3 0]
det A =
12.Matrices & determinantsHard
Solve the system −2x − 6y = −3 and −x − 2y = −7. Find x.
−2x − 6y = −3−x − 2y = −7
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (0, −2, −3) and b = (6, k, 4) are perpendicular.
a = (0, −2, −3)b = (6, k, 4)
k =
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (2, −5, −5), b = (−4, 2, 5) and c = (−2, −3, 1).
a = (2, −5, −5)b = (−4, 2, 5)c = (−2, −3, 1)
V =
15.ProbabilityMedium
A bag holds 8 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
8 / 4
P =
16.ProbabilityHard
10 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?
n = 10p = 0.25
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 39, 36, 43, 46, 22, 39.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 20. What proportion P of values is above 50?
μ = 70σ = 20x = 50
P =
19.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 436 and 562 h? Give P as a decimal.
μ = 500σ = 40x1 = 436x2 = 562
P =
20.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 60% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points60%
x = points
3Solutions
1.Limits
Answer: L = 3
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 + (−1)/x2 + 1/x3)/(1 + 2/x2 + 2/x3)
Let x → ∞
L = 3/1 = 3
Answer
L = 3
For equal degrees the limit is the ratio of the leading coefficients.
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·e−0.5x + 0.5)/(2x) → 0/0
Second application
0.25·e−0.5x/2
Substitute x = 0
L = 0.25·1/2 = 0.125
Answer
L = 0.125
3.Derivatives
Answer: f′(−1) = −2744
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 + 5)3·(−2)
Substitute
f′(−1) = 4·73·(−2) = −2744
Answer
f′(−1) = −2744
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 − 24x + 36 = 3·(x2 − 8x + 12)
Solve f′(x) = 0
x = 2 or x = 6
Which one is which
f″(x) = 6x − 24: negative at a maximum, positive at a minimum.
f″(2) = −12, f″(6) = 12
x = 2
Answer
x = 2
5.Integrals
Answer: I = 156
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.
Substitution
u = x² + 1, du = 2x dx
I = ∫15 u3 du
Integrate and evaluate
I = [u4/4]15 = (625 − 1)/4 = 156
Answer
I = 156
6.Integrals
Answer: A = 0.1667
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 = x → x = 0, x = 1
Integrate upper minus lower
A = [0.5x2 − 0.3333x3]01 = 0.1667
Answer
A = 0.1667
7.Complex numbers
Answer: θ = −135 °
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 = −3 → III
Reference angle
α = atan(3/3) = 45°
Argument
θ = −(180° − α) = −135°
Answer
θ = −135°
In polar form z = 4.243∠−135°.
8.Complex numbers
Answer: Im(z⁴) = 96
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.
Powers step by step
z² = (−3 − j)(−3 − j) = 8 + j6
z³ = (8 + j6)(−3 − j) = −18 − j26
z⁴ = (−18 − j26)(−3 − j) = 28 + j96
Check with De Moivre
z = 3.162∠−161.6° → z⁴ = 100∠73.74°
Im(z⁴) = 96
Answer
Im(z⁴) = 96
9.Differential equations
Answer: t = 79.27 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(413/342)/19 = 0.009928 min⁻¹
Time to reach the target
t = ln(413/188)/0.009928 = 79.27 min
Answer
t = 79.27 min
10.Differential equations
Answer: t = 75.46 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.
Result
m(t) = 40 + (−35)·e−t/200
Solve for t
e−t/200 = (40 − 16)/(40 − 5) = 0.6857
t = 200·ln(1.458) = 75.46 min
Answer
t = 75.46 min
11.Matrices & determinants
Answer: det A = 33
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 = 3, M₁₂ = 0, M₁₃ = −9
Cofactor expansion
det A = (−1)·3 − (−2)·0 + (−4)·(−9) = 33
Answer
det A = 33
12.Matrices & determinants
Answer: x = 18
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 = (−2)·(−2) − (−6)·(−1) = −2
Replace the x column
det Aₓ = (−3)·(−2) − (−6)·(−7) = −36
x = −36/(−2) = 18
Answer
x = 18
y = −5.5
13.Vectors: dot & cross product
Answer: k = −6
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 = 0·6 + (−2)·k + (−3)·4 = 0
−12 + (−2)·k = 0
Solve for k
k = −(−12)/(−2) = −6
Answer
k = −6
14.Vectors: dot & cross product
Answer: V = 16
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 = 2·1 − 5·(−3) = 17
(b × c)y = bzcx − bxcz = 5·(−2) − (−4)·1 = −6
(b × c)z = bxcy − bycx = (−4)·(−3) − 2·(−2) = 16
Dot product with a
a·(b × c) = 2·17 + (−5)·(−6) + (−5)·16 = −16
V = |−16| = 16
Answer
V = 16
15.Probability
Answer: P = 0.4242
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 = 8/12 · 7/11 = 56/132 = 0.4242
Answer
P = 0.4242
16.Probability
Answer: P = 0.244
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) = 0.7510 = 0.05631
P(X = 1) = 10·0.25·0.759 = 0.1877
Add them
P = 0.05631 + 0.1877 = 0.244
Answer
P = 0.244
17.Descriptive statistics
Answer: s = 8.361
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) = 0.1587
P = 1 − 0.1587 = 0.8413
Answer
P = 0.8413
19.Normal distribution
Answer: P = 0.8846
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 = (436 − 500)/40 = −1.6
z₂ = (562 − 500)/40 = 1.55
Area between
P = Φ(1.55) − Φ(−1.6) = 0.9394 − 0.0548 = 0.8846
Answer
P = 0.8846
20.Normal distribution
Answer: x = 62.5 points
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.