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→∞ (−6x2 + 3x + 9)/(2x2 + 2x + 9).
f(x) = (−6x2 + 3x + 9)/(2x2 + 2x + 9)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 5x)/(4x²), with x in radians.
f(x) = (1 − cos 5x)/(4x²)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(3x2 + 2). Find f′(1).
f(x) = ln(3x2 + 2)x = 1
f′(1) =
4.DerivativesHard
The tangent line to f(x) = 2x2 + 2x + 3 at x = −1 crosses the y-axis at y = b. Find b.
f(x) = 2x2 + 2x + 3x = −1
b =
5.IntegralsMedium
Evaluate I = ∫14 1/x dx.
f(x) = 1/x[1, 4]
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 number z = 5·ej0.6 has its angle in radians. Find its real part a.
r = 5θ = 0.6 rad
a =
8.Complex numbersHard
One root of z³ + 9z² + 25z + 25 = 0 is z₁ = −2 + j. The coefficients are real. Find the real root r.
z₁ = −2 + j
r =
9.Differential equationsMedium
A forged steel part at 480 °C is left to cool in a workshop at 26 °C. After 15 min its temperature is 362 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 409 °C?
T0 = 480 °CTa = 26 °CT(15) = 362 °CT = 409 °C
t = min
10.Differential equationsHard
A system obeys y″ + 3y′ + 11.25y = 11.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 3y′ + 11.25y = 11.25y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c22 (row 2, column 2).
A = [4 0; −2 2]B = [5 6; 1 4]
c22 =
12.Matrices & determinantsHard
Solve the system 5x − 4y = −5 and 5x + 4y = 3. Find x.
5x − 4y = −55x + 4y = 3
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (−5, −2, −2) and b = (k, 4, −1) are perpendicular.
a = (−5, −2, −2)b = (k, 4, −1)
k =
14.Vectors: dot & cross productHard
For a = (2, 4, 0) and b = (−2, 2, 3), find the z-component (a × b)z of the cross product a × b.
a = (2, 4, 0)b = (−2, 2, 3)
(a × b)z =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 10, 20, 30, 40p(x): 0.35, 0.05, 0.35, 0.25
Var(X) =
16.ProbabilityHard
X has the distribution shown, and Y = 5X − 3. Find the standard deviation σY of Y.
x: 0, 1, 2p(x): 0.35, 0.3, 0.35Y = 5X − 3
σY =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 58, 38, 21, 44, 15.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 5. Find the z-score z of a score of 72.
μ = 70σ = 5x = 72
z =
19.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 53.5 and 70.5 points? Give P as a decimal.
μ = 65σ = 10x1 = 53.5x2 = 70.5
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 top 15% is to the right of it. Find the cut-off x.
μ = 100 pointsσ = 15 points15%
x = points
3Solutions
1.Limits
Answer: L = −3
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).
Divide by the highest power
(−6 + 3/x + 9/x2)/(2 + 2/x + 9/x2)
Let x → ∞
L = −6/2 = −3
Answer
L = −3
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 3.125
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
5·sin 5x/(8x) → 0/0
Second application
25·cos 5x/8
Substitute x = 0
L = 25/8 = 3.125
Answer
L = 3.125
3.Derivatives
Answer: f′(1) = 1.2
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.
Chain rule for the logarithm
f′(x) = 6x/(3x2 + 2)
Substitute
f′(1) = 6/5 = 1.2
Answer
f′(1) = 1.2
4.Derivatives
Answer: b = 1
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) = 3
f′(x) = 4x + 2 → f′(−1) = −2
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 3 − (−2)·(−1) = 1
Answer
b = 1
5.Integrals
Answer: I = 1.386
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) = ln x
Evaluate at the limits
I = (ln 4 − ln 1) = ln 4 = 1.386
Answer
I = 1.386
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: a = 4.127
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 = 5·(cos 0.6 + j sin 0.6)
Components (radians)
a = 5·cos 0.6 = 4.127
b = 5·sin 0.6 = 2.823
a = 4.127
Answer
a = 4.127
8.Complex numbers
Answer: r = −5
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 = −4 + (−5) = −9 = −(9).
9.Differential equations
Answer: t = 8.475 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(454/336)/15 = 0.02007 min⁻¹
Time to reach the target
t = ln(454/383)/0.02007 = 8.475 min
Answer
t = 8.475 min
10.Differential equations
Answer: ymax = 1.208
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.
Steady value and roots
yss = 1, r = −1.5 ± 3j
Constants from the initial values
y(0) = 1 + A = 0 → A = −1
y′(0) = −1.5A + 3B = 0 → B = −0.5
Peak
t = π/3 = 1.047
ymax = 1·(1 + e−1.5·π/3) = 1.208
Answer
ymax = 1.208
11.Matrices & determinants
Answer: c22 = −4
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 = (−2)·6 + 2·4 = −4
Answer
c22 = −4
12.Matrices & determinants
Answer: x = −0.2
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 = 5·4 − (−4)·5 = 40
Replace the x column
det Aₓ = (−5)·4 − (−4)·3 = −8
x = −8/40 = −0.2
Answer
x = −0.2
y = 1
13.Vectors: dot & cross product
Answer: k = −1.2
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 + (−2)·4 + (−2)·(−1) = 0
−6 + (−5)·k = 0
Solve for k
k = −(−6)/(−5) = −1.2
Answer
k = −1.2
14.Vectors: dot & cross product
Answer: (a × b)z = 12
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 = 4·3 − 0·2 = 12
(a × b)y = azbx − axbz = 0·(−2) − 2·3 = −6
(a × b)z = axby − aybx = 2·2 − 4·(−2) = 12
Requested component
(a × b)z = 12
Answer
(a × b)z = 12
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: Var(X) = 145
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 = 1, Var(X) = 0.7, σX = 0.8367
Linear change
σY = |5|·0.8367 = 4.183
Answer
σY = 4.183
The mean does move: E(Y) = 5·1 − 3 = 2.
17.Descriptive statistics
Answer: s = 17.43
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 = (72 − 70)/5 = 0.4
Answer
z = 0.4
19.Normal distribution
Answer: P = 0.5838
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 = (53.5 − 65)/10 = −1.15
z₂ = (70.5 − 65)/10 = 0.55
Area between
P = Φ(0.55) − Φ(−1.15) = 0.7088 − 0.1251 = 0.5838
Answer
P = 0.5838
20.Normal distribution
Answer: x = 115.6 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.