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→∞ (1 + 1/x)2x. Give a decimal value.
f(x) = (1 + 1/x)2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 5x)/(3x²), with x in radians.
f(x) = (1 − cos 5x)/(3x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x3·e−x. Find f′(1).
f(x) = x3·e−xx = 1
f′(1) =
4.DerivativesHard
The tangent line to f(x) = 2x2 − 4 at x = −1 crosses the y-axis at y = b. Find b.
f(x) = 2x2 − 4x = −1
b =
5.IntegralsMedium
The region under y = x from x = 0 to x = 4 is rotated about the x-axis. Find the volume V of the solid (a cone).
y = x0 ≤ x ≤ 4
V =
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
Given z₁ = 5∠−60° and z₂ = 2∠−25°, find the real part of w = z₁/z₂.
z₁ = 5∠−60°z₂ = 2∠−25°
Re(w) =
8.Complex numbersHard
The roots of zⁿ = w with n = 3 and w = 64∠45° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2. Find the argument of z₂ in degrees, between −180° and 180°.
w = 64∠45°n = 3k = 2
θ₂ = °
9.Differential equationsMedium
A forged steel part at 360 °C is left to cool in a workshop at 27 °C. After 27 min its temperature is 221 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 291 °C?
T0 = 360 °CTa = 27 °CT(27) = 221 °CT = 291 °C
t = min
10.Differential equationsHard
A tank holds 200 L of water in which 20 kg of salt is dissolved. Brine with 0.05 kg of salt per litre flows in at 5 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 55 min?
V = 200 Lm0 = 20 kgc = 0.05 kg/Lr = 5 L/min
m = kg
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [3 −4; −1 6]B = [−2 2; −2 4]
c21 =
12.Matrices & determinantsHard
Solve the system −5x − 3y = 11 and 5x + 2y = −2. Find x.
−5x − 3y = 115x + 2y = −2
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (1, 4, 4) and b = (5, 3, 2), in degrees.
a = (1, 4, 4)b = (5, 3, 2)
θ = °
14.Vectors: dot & cross productHard
A force F = (−60, −50, −30) N acts at point A(3, 3, 3). Coordinates are in metres. Find the z-component Mz of the moment of the force about point O(−1, 0, 3).
F = (−60, −50, −30) NA(3, 3, 3) mO(−1, 0, 3) m
Mz = N·m
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3p(x): 0.1, 0.35, 0.35, 0.2
Var(X) =
16.ProbabilityHard
X has the distribution shown, and Y = 3X + 1. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 12, 39, 54, 52, 15.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 15. Find the z-score z of a score of 58.
μ = 70σ = 15x = 58
z =
19.Normal distributionMedium
The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 5 of them exceeds 127.2 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 5> 127.2 kg
P =
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 90% is to the left of it. Find the cut-off x.
μ = 100 pointsσ = 15 points90%
x = points
3Solutions
1.Limits
Answer: L = 7.389
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).
Take the logarithm
ln L = limx→∞ 2·x·ln(1 + 1/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 2·1 = 2
Undo the logarithm
L = e2 = 7.389
Answer
L = 7.389
2.Limits
Answer: L = 4.167
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/(6x) → 0/0
Second application
25·cos 5x/6
Substitute x = 0
L = 25/6 = 4.167
Answer
L = 4.167
3.Derivatives
Answer: f′(1) = 0.7358
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 rule
f′(x) = u′v + uv′ = e−x·(3x2 + (−1)·x3)
Substitute
f′(1) = 0.3679·(3 + (−1)) = 0.7358
Answer
f′(1) = 0.7358
4.Derivatives
Answer: b = −6
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) = −2
f′(x) = 4x → f′(−1) = −4
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −2 − (−4)·(−1) = −6
Answer
b = −6
5.Integrals
Answer: V = 67.02
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.
Disc method
V = π∫₀4 1x² dx
Integrate
V = π·1·4³/3 = 67.02
Answer
V = 67.02
Check: a cone of radius 4 and height 4 has (1/3)πr²h = 67.02.
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: Re(w) = 2.048
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.
Modulus and argument of w
|w| = 5 / 2 = 2.5
θ = −60° − (−25)° = −35°
Rectangular form
Re(w) = 2.5·cos(−35°) = 2.048
Im(w) = 2.5·sin(−35°) = −1.434
Re(w) = 2.048
Answer
Re(w) = 2.048
8.Complex numbers
Answer: θ₂ = −105 °
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 of the roots
641/3 = 4
Argument of the root
θ₂ = (45° + 360°·2)/3 = 255° → −105°
Argument
θ₂ = −105°
Answer
θ₂ = −105°
The 3 roots lie on the circle of radius 4, 120° apart.
9.Differential equations
Answer: t = 11.6 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(333/194)/27 = 0.02001 min⁻¹
Time to reach the target
t = ln(333/264)/0.02001 = 11.6 min
Answer
t = 11.6 min
10.Differential equations
Answer: m = 12.53 kg
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.
Set up the equation
dm/dt = 5·0.05 − (5/200)·m = 0.25 − 0.025·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 10 + 10·e−t/40
Substitute t
m(55) = 10 + 10·0.2528 = 12.53 kg
Answer
m = 12.53 kg
11.Matrices & determinants
Answer: c21 = −10
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
c21 = (−1)·(−2) + 6·(−2) = −10
Answer
c21 = −10
12.Matrices & determinants
Answer: x = 3.2
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 − (−3)·5 = 5
Replace the x column
det Aₓ = 11·2 − (−3)·(−2) = 16
x = 16/5 = 3.2
Answer
x = 3.2
y = −9
13.Vectors: dot & cross product
Answer: θ = 45.09 °
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 = 1·5 + 4·3 + 4·2 = 25
Magnitudes
|a| = √(1² + 4² + 4²) = 5.745
|b| = √(5² + 3² + 2²) = 6.164
Angle
cos θ = 25/(5.745·6.164) = 0.706
θ = arccos(0.706) = 45.09°
Answer
θ = 45.09°
14.Vectors: dot & cross product
Answer: Mz = −20 N·m
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.
Mean
μ = 0·0.1 + 1·0.35 + 2·0.35 + 3·0.2 = 1.65
Mean of the squares
E(X²) = 0·0.1 + 1·0.35 + 4·0.35 + 9·0.2 = 3.55
Variance
Var(X) = 3.55 − 1.65² = 0.8275
Answer
Var(X) = 0.8275
16.Probability
Answer: σY = 33.67
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 = 28, Var(X) = 126, σX = 11.22
Linear change
σY = |3|·11.22 = 33.67
Answer
σY = 33.67
The mean does move: E(Y) = 3·28 + 1 = 85.
17.Descriptive statistics
Answer: s = 19.96
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 = (58 − 70)/15 = −0.8
Answer
z = −0.8
19.Normal distribution
Answer: P = 0.02442
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 = 125 kg, σT = 0.5·√5 = 1.118 kg
Standardise
z = (127.2 − 125)/1.118 = 1.97
Probability
P = 1 − Φ(1.97) = 0.02442
Answer
P = 0.02442
20.Normal distribution
Answer: x = 119.2 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.