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→∞ (√(x2 − 5x) − x).
f(x) = √(x2 − 5x) − xx → ∞
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) = ln(2x2 + 2). Find f′(−2).
f(x) = ln(2x2 + 2)x = −2
f′(−2) =
4.DerivativesHard
The tangent line to f(x) = −x2 + 3x − 3 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = −x2 + 3x − 3x = 1
b =
5.IntegralsMedium
Evaluate I = ∫02 2x·(x² + 1)3 dx.
f(x) = 2x·(x² + 1)3[0, 2]
I =
6.IntegralsHard
The region between the line y = 3x and the parabola y = x² is rotated about the x-axis. Find the volume V of the solid.
y = 3xy = x²
V =
7.Complex numbersMedium
The number z = 4·ej0.6 has its angle in radians. Find its real part a.
r = 4θ = 0.6 rad
a =
8.Complex numbersHard
Find the real part of (2 − j2)³.
z = 2 − j2n = 3
Re(z³) =
9.Differential equationsMedium
A can of soda at 3 °C is taken out of the fridge into a room at 24 °C. After 24 min its temperature is 20 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 22 °C?
T0 = 3 °CTa = 24 °CT(24) = 20 °CT = 22 °C
t = min
10.Differential equationsHard
Solve y′ + 2y = 9e−t with y(0) = 5. Find y(1.5).
y′ + 2y = 9e−ty(0) = 5
y(1.5) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c22 (row 2, column 2).
A = [−4 −4; 1 6]B = [4 4; 0 4]
c22 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−2 4; −2 7]
λ =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (−5, −6, 3) and b = (k, 4, −1) are perpendicular.
a = (−5, −6, 3)b = (k, 4, −1)
k =
14.Vectors: dot & cross productHard
For a = (−4, −4, 1) and b = (−6, 5, −2), find the y-component (a × b)y of the cross product a × b.
a = (−4, −4, 1)b = (−6, 5, −2)
(a × b)y =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 10, 20, 30, 40p(x): 0.3, 0.25, 0.1, 0.35
Var(X) =
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: 20, 18, 19, 41, 21, 26, 36.
n = 7
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 25. Find the z-score z of a score of 480.
μ = 500σ = 25x = 480
z =
19.Normal distributionMedium
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 247.5 and 252.5 g? Give P as a decimal.
μ = 250σ = 5x1 = 247.5x2 = 252.5
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 = −2.5
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).
Multiply by the conjugate
f(x) = −5x/(√(x2 − 5x) + x)
Divide by x
f(x) = −5/(√(1 + (−5)/x) + 1)
Let x → ∞
L = −5/(1 + 1) = −2.5
Answer
L = −2.5
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′(−2) = −0.8
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 + 2)
Substitute
f′(−2) = −8/10 = −0.8
Answer
f′(−2) = −0.8
4.Derivatives
Answer: b = −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.
Point and slope
f(1) = −1
f′(x) = −2x + 3 → f′(1) = 1
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −1 − 1·1 = −2
Answer
b = −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: V = 101.8
Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.
Intersection points
3x = 1x² → x = 0, x = 3
Washer method
V = π∫₀3 ((3x)² − (1x²)²) dx
Integrate
V = π(9·3³/3 − 1·3⁵/5) = π·32.4 = 101.8
Answer
V = 101.8
7.Complex numbers
Answer: a = 3.301
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.
Euler's formula
z = 4·(cos 0.6 + j sin 0.6)
Components (radians)
a = 4·cos 0.6 = 3.301
b = 4·sin 0.6 = 2.259
a = 3.301
Answer
a = 3.301
8.Complex numbers
Answer: Re(z³) = −16
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 = (2 − j2)(2 − j2) = −j8
z³ = (−j8)(2 − j2) = −16 − j16
Check with De Moivre
z = 2.828∠−45° → z³ = 22.63∠−135°
Re(z³) = −16
Answer
Re(z³) = −16
9.Differential equations
Answer: t = 34.03 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(-21/-4)/24 = 0.06909 min⁻¹
Time to reach the target
t = ln(-21/(−2))/0.06909 = 34.03 min
Answer
t = 34.03 min
10.Differential equations
Answer: y(1.5) = 1.809
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 = e2t
(e2ty)′ = 9et
General solution
y = 9/1·e−t + C·e−2t = 9·e−t + C·e−2t
Initial condition
5 = 9 + C → C = −4
Evaluate
y(1.5) = 9·0.2231 + (−4)·0.04979 = 1.809
Answer
y(1.5) = 1.809
11.Matrices & determinants
Answer: c22 = 28
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
c22 = 1·4 + 6·4 = 28
Answer
c22 = 28
12.Matrices & determinants
Answer: λ = 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.
Characteristic equation
λ² − (5)λ + (−6) = 0
Solve
λ = (5 ± √(49))/2 → λ = −1, 6
λ = 6
Answer
λ = 6
13.Vectors: dot & cross product
Answer: k = −5.4
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 = (−5)·k + (−6)·4 + 3·(−1) = 0
−27 + (−5)·k = 0
Solve for k
k = −(−27)/(−5) = −5.4
Answer
k = −5.4
14.Vectors: dot & cross product
Answer: (a × b)y = −14
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, component by component
(a × b)x = aybz − azby = (−4)·(−2) − 1·5 = 3
(a × b)y = azbx − axbz = 1·(−6) − (−4)·(−2) = −14
(a × b)z = axby − aybx = (−4)·5 − (−4)·(−6) = −44
Requested component
(a × b)y = −14
Answer
(a × b)y = −14
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: Var(X) = 155
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.
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 = 9.118
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 = (480 − 500)/25 = −0.8
Answer
z = −0.8
19.Normal distribution
Answer: P = 0.3829
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 = (247.5 − 250)/5 = −0.5
z₂ = (252.5 − 250)/5 = 0.5
Area between
P = Φ(0.5) − Φ(−0.5) = 0.6915 − 0.3085 = 0.3829
Answer
P = 0.3829
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.