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→∞ (3x2 − 3x − 5)/(x2 + 4x + 4).
f(x) = (3x2 − 3x − 5)/(x2 + 4x + 4)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 3x)/(3x²), with x in radians.
f(x) = (1 − cos 3x)/(3x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(3x), with x in radians. Find f′(1).
f(x) = x·sin(3x)x = 1 rad
f′(1) =
4.DerivativesHard
Let f(x) = x3 + 7.5x2 + 18x − 3. Find the x-coordinate x of the local minimum.
f(x) = x3 + 7.5x2 + 18x − 3
x =
5.IntegralsMedium
Evaluate I = ∫03 ex dx.
f(x) = ex[0, 3]
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
The diagram shows z = −4 + j4. With w = 0.5∠−90°, click the point that represents z/w.
z = −4 + j4w = 0.5∠−90°
1A (−8 − j8)
2B (8 + j8)
3C (−2 − j2)
4D (−8 + j8)
8.Complex numbersHard
The roots of zⁿ = w with n = 6 and w = 64∠135° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, …, 5. Find the real part of z₂.
w = 64∠135°n = 6k = 2
Re(z₂) =
9.Differential equationsMedium
A forged steel part at 380 °C is left to cool in a workshop at 29 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.04 min⁻¹, find its temperature T after 49 min.
T0 = 380 °CTa = 29 °Ck = 0.04 min⁻¹t = 49 min
T = °C
10.Differential equationsHard
Solve the initial value problem y′ + 2y/x = 6x2 for x > 0, with y(1) = −2. Find y(3).
y′ + 2y/x = 6x2y(1) = −2
y(3) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−2 −5 2; −5 3 4; −1 3 0]
det A =
12.Matrices & determinantsHard
Solve the system −3x + 5y = 7 and 2x + 2y = 11. Find x.
−3x + 5y = 72x + 2y = 11
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (5, −2, −1) and b = (−2, −5, −3), in degrees.
a = (5, −2, −1)b = (−2, −5, −3)
θ = °
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (1, 4, −3), b = (4, 4, −5) and c = (−2, 1, −5).
a = (1, 4, −3)b = (4, 4, −5)c = (−2, 1, −5)
V =
15.ProbabilityMedium
A fair die is rolled 8 times. What is the probability P of getting at least one six?
n = 8
P =
16.ProbabilityHard
X has the distribution shown, and Y = 2X + 4. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 43, 35, 22, 53, 27.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 75. Find the z-score z of a score of 612.5.
μ = 500σ = 75x = 612.5
z =
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 34.5 min. Round z to two decimals.
μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 34.5 min
P =
20.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 15% is to the left of it. Find the cut-off x.
μ = 250 gσ = 5 g15%
x = g
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 + (−3)/x + (−5)/x2)/(1 + 4/x + 4/x2)
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 = 1.5
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
3·sin 3x/(6x) → 0/0
Second application
9·cos 3x/6
Substitute x = 0
L = 9/6 = 1.5
Answer
L = 1.5
3.Derivatives
Answer: f′(1) = −2.829
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(3x) + 3x·cos(3x)
Substitute
f′(1) = 0.1411 + 3·(−0.99) = −2.829
Answer
f′(1) = −2.829
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 = −3 or x = −2
Which one is which
f″(x) = 6x + 15: negative at a maximum, positive at a minimum.
f″(−3) = −3, f″(−2) = 3
x = −2
Answer
x = −2
5.Integrals
Answer: I = 19.09
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 = (e3 − 1)/1 = (20.09 − 1)/1 = 19.09
Answer
I = 19.09
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: A (−8 − j8)
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| = |z|/0.5, arg(z/w) = arg z − (−90)°
The point
z/w = −8 − j8 → A
Answer
A (−8 − j8)
8.Complex numbers
Answer: Re(z₂) = −1.587
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.
The 6 roots lie on the circle of radius 2, 60° apart.
9.Differential equations
Answer: T = 78.44 °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 + 351·e−0.04t
Substitute t
T(49) = 29 + 351·0.1409 = 78.44 °C
Answer
T = 78.44 °C
10.Differential equations
Answer: y(3) = 32.04
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 = e∫2/x dx = x2
Integrate
(x2y)′ = 6x4 → x2y = 6x5/5 + C
Initial condition
−2 = 6/5 + C → C = −3.2
Evaluate
y(3) = (6·35/5 + (−3.2))/32 = 32.04
Answer
y(3) = 32.04
11.Matrices & determinants
Answer: det A = 20
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 = −12, M₁₂ = 4, M₁₃ = −12
Cofactor expansion
det A = (−2)·(−12) − (−5)·4 + 2·(−12) = 20
Answer
det A = 20
12.Matrices & determinants
Answer: x = 2.563
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 = (−3)·2 − 5·2 = −16
Replace the x column
det Aₓ = 7·2 − 5·11 = −41
x = −41/(−16) = 2.563
Answer
x = 2.563
y = 2.938
13.Vectors: dot & cross product
Answer: θ = 84.9 °
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 = 5·(−2) + (−2)·(−5) + (−1)·(−3) = 3
Magnitudes
|a| = √(5² + (−2)² + (−1)²) = 5.477
|b| = √((−2)² + (−5)² + (−3)²) = 6.164
Angle
cos θ = 3/(5.477·6.164) = 0.08885
θ = arccos(0.08885) = 84.9°
Answer
θ = 84.9°
14.Vectors: dot & cross product
Answer: V = 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 b × c
(b × c)x = bycz − bzcy = 4·(−5) − (−5)·1 = −15
(b × c)y = bzcx − bxcz = (−5)·(−2) − 4·(−5) = 30
(b × c)z = bxcy − bycx = 4·1 − 4·(−2) = 12
Dot product with a
a·(b × c) = 1·(−15) + 4·30 + (−3)·12 = 69
V = |69| = 69
Answer
V = 69
15.Probability
Answer: P = 0.7674
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.
Complement
P(no six) = (5/6)8 = 0.2326
P = 1 − 0.2326 = 0.7674
Answer
P = 0.7674
16.Probability
Answer: σY = 2.472
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 = 2.35, Var(X) = 1.527, σX = 1.236
Linear change
σY = |2|·1.236 = 2.472
Answer
σY = 2.472
The mean does move: E(Y) = 2·2.35 + 4 = 8.7.
17.Descriptive statistics
Answer: s = 12.41
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 = (612.5 − 500)/75 = 1.5
Answer
z = 1.5
19.Normal distribution
Answer: P = 0.1587
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 = (34.5 − 32)/2.5 = 1
Upper tail
P = 1 − Φ(1) = 1 − 0.8413 = 0.1587
Answer
P = 0.1587
20.Normal distribution
Answer: x = 244.8 g
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.