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 − 10x) − x).
f(x) = √(x2 − 10x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 4x)/sin(5x), with x in radians.
f(x) = ln(1 + 4x)/sin(5x)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(x2 + 1). Find f′(2).
f(x) = ln(x2 + 1)x = 2
f′(2) =
4.DerivativesHard
The tangent line to f(x) = −x2 + 3x + 6 at x = 3 crosses the y-axis at y = b. Find b.
f(x) = −x2 + 3x + 6x = 3
b =
5.IntegralsMedium
Evaluate I = ∫14 5/x dx.
f(x) = 5/x[1, 4]
I =
6.IntegralsHard
Evaluate I = ∫₀2 x²·eˣ dx.
∫₀2 x²eˣ dx
I =
7.Complex numbersMedium
The number z = 4·e−j0.5 has its angle in radians. Find its imaginary part b.
r = 4θ = −0.5 rad
b =
8.Complex numbersHard
Solve z² + 8z + 17 = 0. The roots are a ± jb; find the real part a.
p = 8q = 17
a =
9.Differential equationsMedium
A forged steel part at 500 °C is left to cool in a workshop at 22 °C. After 15 min its temperature is 327 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 159 °C?
T0 = 500 °CTa = 22 °CT(15) = 327 °CT = 159 °C
t = min
10.Differential equationsHard
Solve the initial value problem y′ + y/x = 4x for x > 0, with y(1) = −2. Find y(3).
y′ + y/x = 4xy(1) = −2
y(3) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [5 −1 −3; −1 4 3; −2 0 −3]
det A =
12.Matrices & determinantsHard
Solve the system 5x − 5y = −7 and −3x + 2y = −10. Find x.
5x − 5y = −7−3x + 2y = −10
x =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (−3, 6, −6) onto b = (2, −1, 6), that is, the component of a in the direction of b.
a = (−3, 6, −6)b = (2, −1, 6)
compb a =
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(−3, 4, −2), B(−4, −2, 3) and C(3, −4, 4). Coordinates are in metres.
A(−3, 4, −2)B(−4, −2, 3)C(3, −4, 4)
A = m²
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 10, 20, 30, 40p(x): 0.1, 0.3, 0.35, 0.25
σ =
16.ProbabilityHard
X has the distribution shown, and Y = 10X + 4. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 12, 36, 18, 15, 11, 23.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 8. What proportion P of values is above 92?
μ = 100σ = 8x = 92
P =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.8230. Find z to two decimals.
A = 0.8230
z =
20.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 40% is to the left of it. Find the cut-off x.
μ = 100 pointsσ = 15 points40%
x = points
3Solutions
1.Limits
Answer: L = −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) = −10x/(√(x2 − 10x) + x)
Divide by x
f(x) = −10/(√(1 + (−10)/x) + 1)
Let x → ∞
L = −10/(1 + 1) = −5
Answer
L = −5
2.Limits
Answer: L = 0.8
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).
L'Hôpital's rule
L = limx→0 [4/(1 + 4x)]/[5·cos 5x]
Substitute x = 0
L = 4/5 = 0.8
Answer
L = 0.8
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) = 2x/(x2 + 1)
Substitute
f′(2) = 4/5 = 0.8
Answer
f′(2) = 0.8
4.Derivatives
Answer: b = 15
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(3) = 6
f′(x) = −2x + 3 → f′(3) = −3
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 6 − (−3)·3 = 15
Answer
b = 15
5.Integrals
Answer: I = 6.931
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) = 5·ln x
Evaluate at the limits
I = 5·(ln 4 − ln 1) = 5·ln 4 = 6.931
Answer
I = 6.931
6.Integrals
Answer: I = 12.78
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.
First round
∫x²eˣ dx = x²eˣ − ∫2x eˣ dx
Second round
∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)
Limits
7.389·(2) − 1·2 = 12.78
Answer
I = 12.78
7.Complex numbers
Answer: b = −1.918
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 −0.5 + j sin −0.5)
Components (radians)
a = 4·cos −0.5 = 3.51
b = 4·sin −0.5 = −1.918
b = −1.918
Answer
b = −1.918
8.Complex numbers
Answer: a = −4
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 = 64 − 68 = −4 < 0
Quadratic formula
z = (−8 ± j√4)/2 = −4 + j, −4 − j
Answer
a = −4
Answer
a = −4
9.Differential equations
Answer: t = 41.72 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(478/305)/15 = 0.02995 min⁻¹
Time to reach the target
t = ln(478/137)/0.02995 = 41.72 min
Answer
t = 41.72 min
10.Differential equations
Answer: y(3) = 10.89
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∫1/x dx = x1
Integrate
(x1y)′ = 4x2 → x1y = 4x3/3 + C
Initial condition
−2 = 4/3 + C → C = −3.333
Evaluate
y(3) = (4·33/3 + (−3.333))/31 = 10.89
Answer
y(3) = 10.89
11.Matrices & determinants
Answer: det A = −75
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 = 9, M₁₃ = 8
Cofactor expansion
det A = 5·(−12) − (−1)·9 + (−3)·8 = −75
Answer
det A = −75
12.Matrices & determinants
Answer: x = 12.8
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 = 5·2 − (−5)·(−3) = −5
Replace the x column
det Aₓ = (−7)·2 − (−5)·(−10) = −64
x = −64/(−5) = 12.8
Answer
x = 12.8
y = 14.2
13.Vectors: dot & cross product
Answer: compb a = −7.496
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 = (−3)·2 + 6·(−1) + (−6)·6 = −48
Length of b
|b| = √(2² + (−1)² + 6²) = 6.403
Scalar projection
compb a = a·b/|b| = −48/6.403 = −7.496
Answer
compb a = −7.496
Negative: a has a component pointing against b.
14.Vectors: dot & cross product
Answer: A = 28.5 m²
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.
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.
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 = 2.7, Var(X) = 3.31, σX = 1.819
Linear change
σY = |10|·1.819 = 18.19
Answer
σY = 18.19
The mean does move: E(Y) = 10·2.7 + 4 = 31.
17.Descriptive statistics
Answer: s = 9.326
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: z = 1.35
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.
Area to the left of z
Φ(z) = (1 + 0.8230)/2 = 0.9115
Read the z-table backwards
Find 0.9115 inside the table and read off z.
z = 1.35
Answer
z = 1.35
20.Normal distribution
Answer: x = 96.3 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.