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→∞ (√(9x2 − 8x) − 3x).
f(x) = √(9x2 − 8x) − 3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 4x)/(2x²), with x in radians.
f(x) = (1 − cos 4x)/(2x²)x → 0
L =
3.DerivativesMedium
Let f(x) = (−2x − 3)3. Find f′(−1).
f(x) = (−2x − 3)3x = −1
f′(−1) =
4.DerivativesHard
Let f(x) = x3 + 9x2 + 24x + 5. Find the x-coordinate x of the local maximum.
f(x) = x3 + 9x2 + 24x + 5
x =
5.IntegralsMedium
The region under y = √(2x) from x = 0 to x = 1 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).
y = √(2x)0 ≤ x ≤ 1
V =
6.IntegralsHard
Find the area A of the region enclosed by y = 0.5x2 and y = 3x.
y = 0.5x2y = 3x
A =
7.Complex numbersMedium
The diagram shows z = −2 − j4. With w = 2∠−90°, click the point that represents z·w.
z = −2 − j4w = 2∠−90°
1A (−4 + j2)
2B (−4 − j8)
3C (−2 + j)
4D (−8 + j4)
8.Complex numbersHard
One root of z³ + 7z² + 12z − 20 = 0 is z₁ = −4 + j2. The coefficients are real. Find the real root r.
z₁ = −4 + j2
r =
9.Differential equationsMedium
Solve y″ + 6y′ + 9y = 0 with y(0) = 5 and y′(0) = 1. Find y(1).
y″ + 6y′ + 9y = 0y(0) = 5y′(0) = 1
y(1) =
10.Differential equationsHard
Solve the initial value problem y′ + 2y/x = 2x for x > 0, with y(1) = −3. Find y(3).
y′ + 2y/x = 2xy(1) = −3
y(3) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−3 1 −3; −3 −2 3; 2 5 −1]
det A =
12.Matrices & determinantsHard
Solve the system x − 6y = 0 and 4x + 4y = −8. Find x.
x − 6y = 04x + 4y = −8
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (−5, 3, 2) and b = (−2, 1, −1), in degrees.
a = (−5, 3, 2)b = (−2, 1, −1)
θ = °
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−5, 4, 1) and b = (−4, 0, 1) and has a positive z-component. Find its y-component n̂y.
a = (−5, 4, 1)b = (−4, 0, 1)n̂z > 0
n̂y =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 2, 4, 6p(x): 0.3, 0.3, 0.1, 0.3
Var(X) =
16.ProbabilityHard
X has the distribution shown, and Y = 2X + 1. Find the standard deviation σY of Y.
x: 0, 1, 2, 3p(x): 0.2, 0.1, 0.35, 0.35Y = 2X + 1
σY =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 40, 58, 54, 43, 26, 35, 57.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 5. What proportion P of values is above 47.5?
μ = 50σ = 5x = 47.5
P =
19.Normal distributionMedium
X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 200.4 mm. Round z to two decimals.
μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 200.4 mm
P =
20.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 20% is to the left of it. Find the cut-off x.
μ = 65 pointsσ = 10 points20%
x = points
3Solutions
1.Limits
Answer: L = −1.333
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) = −8x/(√(9x2 − 8x) + 3x)
Divide by x
f(x) = −8/(√(9 + (−8)/x) + 3)
Let x → ∞
L = −8/(3 + 3) = −1.333
Answer
L = −1.333
2.Limits
Answer: L = 4
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
4·sin 4x/(4x) → 0/0
Second application
16·cos 4x/4
Substitute x = 0
L = 16/4 = 4
Answer
L = 4
3.Derivatives
Answer: f′(−1) = −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.
Chain rule
f′(x) = 3·(−2x − 3)2·(−2)
Substitute
f′(−1) = 3·(−1)2·(−2) = −6
Answer
f′(−1) = −6
4.Derivatives
Answer: x = −4
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 + 18x + 24 = 3·(x2 + 6x + 8)
Solve f′(x) = 0
x = −4 or x = −2
Which one is which
f″(x) = 6x + 18: negative at a maximum, positive at a minimum.
f″(−4) = −6, f″(−2) = 6
x = −4
Answer
x = −4
5.Integrals
Answer: V = 3.142
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∫₀1 2x dx
Integrate
V = π·2·1²/2 = 3.142
Answer
V = 3.142
6.Integrals
Answer: A = 18
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.
Intersection points
0.5x2 = 3x → x = 0, x = 6
Integrate upper minus lower
A = [1.5x2 − 0.1667x3]06 = 18
Answer
A = 18
7.Complex numbers
Answer: D (−8 + 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 = −8 + j4 → D
Answer
D (−8 + j4)
8.Complex numbers
Answer: r = 1
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.
Check with the sum of the roots: z₁ + z₂ + r = −8 + 1 = −7 = −(7).
9.Differential equations
Answer: y(1) = 1.046
Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.
Double root
(r + 3)² = 0 → r = −3
Constants
A = y(0) = 5
y′(0) = B − 3A → B = 1 + 15 = 16
Evaluate
y(1) = (5 + 16·1)·e−3 = 1.046
Answer
y(1) = 1.046
10.Differential equations
Answer: y(3) = 4.111
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)′ = 2x3 → x2y = 2x4/4 + C
Initial condition
−3 = 2/4 + C → C = −3.5
Evaluate
y(3) = (2·34/4 + (−3.5))/32 = 4.111
Answer
y(3) = 4.111
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 = −13, M₁₂ = −3, M₁₃ = −11
Cofactor expansion
det A = (−3)·(−13) − 1·(−3) + (−3)·(−11) = 75
Answer
det A = 75
12.Matrices & determinants
Answer: x = −1.714
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 = 1·4 − (−6)·4 = 28
Replace the x column
det Aₓ = 0·4 − (−6)·(−8) = −48
x = −48/28 = −1.714
Answer
x = −1.714
y = −0.2857
13.Vectors: dot & cross product
Answer: θ = 43.24 °
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) + 3·1 + 2·(−1) = 11
Magnitudes
|a| = √((−5)² + 3² + 2²) = 6.164
|b| = √((−2)² + 1² + (−1)²) = 2.449
Angle
cos θ = 11/(6.164·2.449) = 0.7285
θ = arccos(0.7285) = 43.24°
Answer
θ = 43.24°
14.Vectors: dot & cross product
Answer: n̂y = 0.06052
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
(a × b)x = aybz − azby = 4·1 − 1·0 = 4
(a × b)y = azbx − axbz = 1·(−4) − (−5)·1 = 1
(a × b)z = axby − aybx = (−5)·0 − 4·(−4) = 16
Normalise
The z-component of a × b is already positive.
|a × b| = √(4² + 1² + 16²) = 16.52
n̂ = (4, 1, 16)/16.52 = (0.2421, 0.06052, 0.9684)
n̂y = 0.06052
Answer
n̂y = 0.06052
15.Probability
Answer: Var(X) = 5.76
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 = 0·0.3 + 2·0.3 + 4·0.1 + 6·0.3 = 2.8
Mean of the squares
E(X²) = 0·0.3 + 4·0.3 + 16·0.1 + 36·0.3 = 13.6
Variance
Var(X) = 13.6 − 2.8² = 5.76
Answer
Var(X) = 5.76
16.Probability
Answer: σY = 2.216
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 = 1.85, Var(X) = 1.228, σX = 1.108
Linear change
σY = |2|·1.108 = 2.216
Answer
σY = 2.216
The mean does move: E(Y) = 2·1.85 + 1 = 4.7.
17.Descriptive statistics
Answer: s = 12.13
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 = −0.5
Standard normal table
Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915
Answer
P = 0.6915
19.Normal distribution
Answer: P = 0.2119
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 = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm
Standardise
z = (200.4 − 200)/0.5 = 0.8
Upper tail
P = 1 − Φ(0.8) = 1 − 0.7881 = 0.2119
Answer
P = 0.2119
20.Normal distribution
Answer: x = 56.6 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.