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 − 10x) − 3x).
f(x) = √(9x2 − 10x) − 3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e0.5x − 1 − 0.5x)/x².
f(x) = (e0.5x − 1 − 0.5x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = ln(x2 + 4). Find f′(−2).
f(x) = ln(x2 + 4)x = −2
f′(−2) =
4.DerivativesHard
The tangent line to f(x) = 2x2 − 5x − 3 at x = −2 crosses the y-axis at y = b. Find b.
f(x) = 2x2 − 5x − 3x = −2
b =
5.IntegralsMedium
Evaluate I = ∫28 5/x dx.
f(x) = 5/x[2, 8]
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 = 1 + j2. With w = 0.5∠90°, click the point that represents z/w.
z = 1 + j2w = 0.5∠90°
1A (−4 + j2)
2B (4 − j2)
3C (2 + j4)
4D (2 − j)
8.Complex numbersHard
Exactly one of the points A–D is a root of z³ = 8∠120°. Click it.
w = 8∠120°n = 3
1A (2∠70°)
2B (2∠40°)
3C (3.6∠40°)
4D (2∠−40°)
9.Differential equationsMedium
A can of soda at 3 °C is taken out of the fridge into a room at 25 °C. After 7 min its temperature is 11 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 3 °CTa = 25 °CT(7) = 11 °C
k = min⁻¹
10.Differential equationsHard
A system obeys y″ + 2y′ + 17y = 170 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.5); angles in radians.
y″ + 2y′ + 17y = 170y(0) = 0y′(0) = 0
y(0.5) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [3 2 3; −3 0 1; 2 −1 5]
det A =
12.Matrices & determinantsHard
Solve the system −4x − 2y = 9 and 5x + 2y = −7. Find x.
−4x − 2y = 95x + 2y = −7
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (1, 2, −4) and b = (6, 6, k) are perpendicular.
a = (1, 2, −4)b = (6, 6, k)
k =
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (4, 2, −3), b = (5, 0, −1) and c = (3, −5, 2).
a = (4, 2, −3)b = (5, 0, −1)c = (3, −5, 2)
V =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4p(x): 0.1, 0.35, 0.25, 0.3
σ =
16.ProbabilityHard
X has the distribution shown, and Y = −2X + 5. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 16, 44, 47, 20, 27.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 60 and standard deviation 8. Find the z-score z of a score of 63.2.
μ = 60σ = 8x = 63.2
z =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.9312. Find z to two decimals.
A = 0.9312
z =
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 top 30% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points30%
x = points
3Solutions
1.Limits
Answer: L = −1.667
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/(√(9x2 − 10x) + 3x)
Divide by x
f(x) = −10/(√(9 + (−10)/x) + 3)
Let x → ∞
L = −10/(3 + 3) = −1.667
Answer
L = −1.667
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·e0.5x − 0.5)/(2x) → 0/0
Second application
0.25·e0.5x/2
Substitute x = 0
L = 0.25·1/2 = 0.125
Answer
L = 0.125
3.Derivatives
Answer: f′(−2) = −0.5
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 + 4)
Substitute
f′(−2) = −4/8 = −0.5
Answer
f′(−2) = −0.5
4.Derivatives
Answer: b = −11
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(−2) = 15
f′(x) = 4x − 5 → f′(−2) = −13
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 15 − (−13)·(−2) = −11
Answer
b = −11
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 8 − ln 2) = 5·ln 4 = 6.931
Answer
I = 6.931
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: B (4 − j2)
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 = 4 − j2 → B
Answer
B (4 − j2)
8.Complex numbers
Answer: B (2∠40°)
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 and arguments of the roots
|z| = 2
θ = 40°, 160°, −80°
The point
B = 2∠40°
Answer
B (2∠40°)
9.Differential equations
Answer: k = 0.06457 min⁻¹
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.
Ratio of temperature differences
e−7k = -14/-22
Solve for k
k = ln(-22/-14)/7 = 0.06457 min⁻¹
Answer
k = 0.06457 min⁻¹
10.Differential equations
Answer: y(0.5) = 11.15
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.
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 = 1, M₁₂ = −17, M₁₃ = 3
Cofactor expansion
det A = 3·1 − 2·(−17) + 3·3 = 46
Answer
det A = 46
12.Matrices & determinants
Answer: x = 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 = (−4)·2 − (−2)·5 = 2
Replace the x column
det Aₓ = 9·2 − (−2)·(−7) = 4
x = 4/2 = 2
Answer
x = 2
y = −8.5
13.Vectors: dot & cross product
Answer: k = 4.5
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 = 1·6 + 2·6 + (−4)·k = 0
18 + (−4)·k = 0
Solve for k
k = −18/(−4) = 4.5
Answer
k = 4.5
14.Vectors: dot & cross product
Answer: V = 29
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 = 0·2 − (−1)·(−5) = −5
(b × c)y = bzcx − bxcz = (−1)·3 − 5·2 = −13
(b × c)z = bxcy − bycx = 5·(−5) − 0·3 = −25
Dot product with a
a·(b × c) = 4·(−5) + 2·(−13) + (−3)·(−25) = 29
V = |29| = 29
Answer
V = 29
15.Probability
Answer: σ = 0.9937
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 = 1·0.1 + 2·0.35 + 3·0.25 + 4·0.3 = 2.75
Mean of the squares
E(X²) = 1·0.1 + 4·0.35 + 9·0.25 + 16·0.3 = 8.55
Variance
Var(X) = 8.55 − 2.75² = 0.9875
σ = √0.9875 = 0.9937
Answer
σ = 0.9937
16.Probability
Answer: σY = 4.247
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 = 3.7, Var(X) = 4.51, σX = 2.124
Linear change
σY = |−2|·2.124 = 4.247
Answer
σY = 4.247
The mean does move: E(Y) = −2·3.7 + 5 = −2.4.
17.Descriptive statistics
Answer: s = 14.02
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 = (63.2 − 60)/8 = 0.4
Answer
z = 0.4
19.Normal distribution
Answer: z = 1.82
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 + 0.9312)/2 = 0.9656
Read the z-table backwards
Find 0.9656 inside the table and read off z.
z = 1.82
Answer
z = 1.82
20.Normal distribution
Answer: x = 70.2 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.