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→∞ (−6x3 − 5x + 9)/(5x3 + 3x + 9).
f(x) = (−6x3 − 5x + 9)/(5x3 + 3x + 9)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 5x)/(2x²), with x in radians.
f(x) = (1 − cos 5x)/(2x²)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(2x2 + 4). Find f′(−1).
f(x) = ln(2x2 + 4)x = −1
f′(−1) =
4.DerivativesHard
Let f(x) = x3 + 3x2 − 9x + 6. Find the x-coordinate x of the local maximum.
f(x) = x3 + 3x2 − 9x + 6
x =
5.IntegralsMedium
Evaluate I = ∫01 ex dx.
f(x) = ex[0, 1]
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 4 − x² and the x-axis.
y = 4 − x²
A =
7.Complex numbersMedium
The diagram shows z = 2 + j2. With w = 2∠−90°, click the point that represents z·w.
z = 2 + j2w = 2∠−90°
1A (1 − j)
2B (−4 + j4)
3C (2 − j2)
4D (4 − j4)
8.Complex numbersHard
Exactly one of the points A–D is a root of z⁴ = 81∠90°. Click it.
w = 81∠90°n = 4
1A (3∠67.5°)
2B (5.4∠−22.5°)
3C (3∠−22.5°)
4D (3∠−67.5°)
9.Differential equationsMedium
A forged steel part at 420 °C is left to cool in a workshop at 24 °C. After 40 min its temperature is 104 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 56 °C?
T0 = 420 °CTa = 24 °CT(40) = 104 °CT = 56 °C
t = min
10.Differential equationsHard
Solve y′ + 3y = 4e−t with y(0) = 5. Find y(0.5).
y′ + 3y = 4e−ty(0) = 5
y(0.5) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [3 5 −4; −1 3 3; −4 3 1]
det A =
12.Matrices & determinantsHard
Solve the system −2x + 6y = −11 and 4x − 4y = 5. Find x.
−2x + 6y = −114x − 4y = 5
x =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (−6, 1, 7) onto b = (6, −6, −5), that is, the component of a in the direction of b.
a = (−6, 1, 7)b = (6, −6, −5)
compb a =
14.Vectors: dot & cross productHard
For a = (−6, −4, 5) and b = (−6, −3, 4), find the z-component (a × b)z of the cross product a × b.
a = (−6, −4, 5)b = (−6, −3, 4)
(a × b)z =
15.ProbabilityMedium
A bag holds 4 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
4 / 5
P =
16.ProbabilityHard
A test has 8 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 2 correct answers?
n = 8p = 0.25k = 2
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 36, 42, 14, 24, 18, 29.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 50. What proportion P of values is below 450?
μ = 500σ = 50x = 450
P =
19.Normal distributionMedium
The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 9 of them exceeds 356.4 g. Round z to two decimals.
μ = 40 gσ = 1.2 gn = 9> 356.4 g
P =
20.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 80% of them lie (the 80th percentile).
μ = 250σ = 5P = 80%
x = g
3Solutions
1.Limits
Answer: L = −1.2
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
(−6 + (−5)/x2 + 9/x3)/(5 + 3/x2 + 9/x3)
Let x → ∞
L = −6/5 = −1.2
Answer
L = −1.2
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 6.25
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/(4x) → 0/0
Second application
25·cos 5x/4
Substitute x = 0
L = 25/4 = 6.25
Answer
L = 6.25
3.Derivatives
Answer: f′(−1) = −0.6667
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 for the logarithm
f′(x) = 4x/(2x2 + 4)
Substitute
f′(−1) = −4/6 = −0.6667
Answer
f′(−1) = −0.6667
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 + 6x − 9 = 3·(x2 + 2x − 3)
Solve f′(x) = 0
x = −3 or x = 1
Which one is which
f″(x) = 6x + 6: negative at a maximum, positive at a minimum.
f″(−3) = −12, f″(1) = 12
x = −3
Answer
x = −3
5.Integrals
Answer: I = 1.718
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 = (e1 − 1)/1 = (2.718 − 1)/1 = 1.718
Answer
I = 1.718
6.Integrals
Answer: A = 10.67
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
4 − x² = 0 → x = ±2
Integrate
A = [4x − x³/3]−22 = 5.333 − (−5.333) = 10.67
Answer
A = 10.67
7.Complex numbers
Answer: D (4 − j4)
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 = 4 − j4 → D
Answer
D (4 − j4)
8.Complex numbers
Answer: D (3∠−67.5°)
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 and arguments of the roots
|z| = 3
θ = 22.5°, 112.5°, −157.5°, −67.5°
The point
D = 3∠−67.5°
Answer
D (3∠−67.5°)
9.Differential equations
Answer: t = 62.92 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(396/80)/40 = 0.03998 min⁻¹
Time to reach the target
t = ln(396/32)/0.03998 = 62.92 min
Answer
t = 62.92 min
10.Differential equations
Answer: y(0.5) = 1.882
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.
Integrating factor
μ = e3t
(e3ty)′ = 4e2t
General solution
y = 4/2·e−t + C·e−3t = 2·e−t + C·e−3t
Initial condition
5 = 2 + C → C = 3
Evaluate
y(0.5) = 2·0.6065 + 3·0.2231 = 1.882
Answer
y(0.5) = 1.882
11.Matrices & determinants
Answer: det A = −109
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 = −6, M₁₂ = 11, M₁₃ = 9
Cofactor expansion
det A = 3·(−6) − 5·11 + (−4)·9 = −109
Answer
det A = −109
12.Matrices & determinants
Answer: x = −0.875
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)·(−4) − 6·4 = −16
Replace the x column
det Aₓ = (−11)·(−4) − 6·5 = 14
x = 14/(−16) = −0.875
Answer
x = −0.875
y = −2.125
13.Vectors: dot & cross product
Answer: compb a = −7.818
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 = (−6)·6 + 1·(−6) + 7·(−5) = −77
Length of b
|b| = √(6² + (−6)² + (−5)²) = 9.849
Scalar projection
compb a = a·b/|b| = −77/9.849 = −7.818
Answer
compb a = −7.818
Negative: a has a component pointing against b.
14.Vectors: dot & cross product
Answer: (a × b)z = −6
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.
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: P = 0.1667
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 = 4/9 · 3/8 = 12/72 = 0.1667
Answer
P = 0.1667
16.Probability
Answer: P = 0.3115
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.
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 = 0.1587
Answer
P = 0.1587
19.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).
Distribution of the total
μT = 360 g, σT = 1.2·√9 = 3.6 g
Standardise
z = (356.4 − 360)/3.6 = −1
Probability
P = 1 − Φ(−1) = 0.8413
Answer
P = 0.8413
20.Normal distribution
Answer: x = 254.2 g
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.