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 − 2/x)2x. Give a decimal value.
f(x) = (1 − 2/x)2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 3x)/sin(5x), with x in radians.
f(x) = ln(1 + 3x)/sin(5x)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(2x2 + 1). Find f′(1).
f(x) = ln(2x2 + 1)x = 1
f′(1) =
4.DerivativesHard
The tangent line to f(x) = −x2 + 6 at x = 2 crosses the y-axis at y = b. Find b.
f(x) = −x2 + 6x = 2
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 enclosed by y = x2 and y = 4x.
y = x2y = 4x
A =
7.Complex numbersMedium
The diagram shows z = 4 − j4. With w = 0.5∠90°, click the point that represents z/w.
z = 4 − j4w = 0.5∠90°
1A (−2 − j2)
2B (8 + j8)
3C (−8 − j8)
4D (−4 − j4)
8.Complex numbersHard
Solve z² + 8z + 32 = 0. The roots are a ± jb; find the real part a.
p = 8q = 32
a =
9.Differential equationsMedium
A forged steel part at 380 °C is left to cool in a workshop at 23 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.02 min⁻¹, find its temperature T after 58 min.
T0 = 380 °CTa = 23 °Ck = 0.02 min⁻¹t = 58 min
T = °C
10.Differential equationsHard
Solve the initial value problem y′ + y/x = 2 for x > 0, with y(1) = −1. Find y(3).
y′ + y/x = 2y(1) = −1
y(3) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [−2 −5; 6 2]B = [0 5; 6 6]
c11 =
12.Matrices & determinantsHard
Solve the system x + 6y = −11 and x + 2y = −6. Find x.
x + 6y = −11x + 2y = −6
x =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (−3, 6, 2) onto b = (−6, −6, −6), that is, the component of a in the direction of b.
a = (−3, 6, 2)b = (−6, −6, −6)
compb a =
14.Vectors: dot & cross productHard
For a = (3, −5, 6) and b = (−2, −3, −3), find the y-component (a × b)y of the cross product a × b.
a = (3, −5, 6)b = (−2, −3, −3)
(a × b)y =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4p(x): 0.35, 0.25, 0.05, 0.35
Var(X) =
16.ProbabilityHard
X has the distribution shown, and Y = 2X − 3. Find the standard deviation σY of Y.
x: 0, 2, 4, 6p(x): 0.3, 0.25, 0.3, 0.15Y = 2X − 3
σY =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 45, 54, 46, 38, 48, 53.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is above 73.2?
μ = 70σ = 8x = 73.2
P =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.9500. Find z to two decimals.
A = 0.9500
z =
20.Normal distributionHard
Hole diameters D are normal with mean 25.05 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.05 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
3Solutions
1.Limits
Answer: L = 0.01832
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 − 2/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 2·(−2) = −4
Undo the logarithm
L = e−4 = 0.01832
Answer
L = 0.01832
2.Limits
Answer: L = 0.6
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 [3/(1 + 3x)]/[5·cos 5x]
Substitute x = 0
L = 3/5 = 0.6
Answer
L = 0.6
3.Derivatives
Answer: f′(1) = 1.333
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) = 4x/(2x2 + 1)
Substitute
f′(1) = 4/3 = 1.333
Answer
f′(1) = 1.333
4.Derivatives
Answer: b = 10
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) = 2
f′(x) = −2x → f′(2) = −4
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 2 − (−4)·2 = 10
Answer
b = 10
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 = 10.67
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
x2 = 4x → x = 0, x = 4
Integrate upper minus lower
A = [2x2 − 0.3333x3]04 = 10.67
Answer
A = 10.67
7.Complex numbers
Answer: C (−8 − j8)
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 = −8 − j8 → C
Answer
C (−8 − j8)
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 − 128 = −64 < 0
Quadratic formula
z = (−8 ± j√64)/2 = −4 + j4, −4 − j4
Answer
a = −4
Answer
a = −4
9.Differential equations
Answer: T = 134.9 °C
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.
Solution of the cooling law
T(t) = 23 + 357·e−0.02t
Substitute t
T(58) = 23 + 357·0.3135 = 134.9 °C
Answer
T = 134.9 °C
10.Differential equations
Answer: y(3) = 2.333
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)′ = 2x → x1y = 2x2/2 + C
Initial condition
−1 = 2/2 + C → C = −2
Evaluate
y(3) = (2·32/2 + (−2))/31 = 2.333
Answer
y(3) = 2.333
11.Matrices & determinants
Answer: c11 = −30
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
c11 = (−2)·0 + (−5)·6 = −30
Answer
c11 = −30
12.Matrices & determinants
Answer: x = −3.5
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 = 1·2 − 6·1 = −4
Replace the x column
det Aₓ = (−11)·2 − 6·(−6) = 14
x = 14/(−4) = −3.5
Answer
x = −3.5
y = −1.25
13.Vectors: dot & cross product
Answer: compb a = −2.887
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)·(−6) + 6·(−6) + 2·(−6) = −30
Length of b
|b| = √((−6)² + (−6)² + (−6)²) = 10.39
Scalar projection
compb a = a·b/|b| = −30/10.39 = −2.887
Answer
compb a = −2.887
Negative: a has a component pointing against b.
14.Vectors: dot & cross product
Answer: (a × b)y = −3
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 = (−5)·(−3) − 6·(−3) = 33
(a × b)y = azbx − axbz = 6·(−2) − 3·(−3) = −3
(a × b)z = axby − aybx = 3·(−3) − (−5)·(−2) = −19
Requested component
(a × b)y = −3
Answer
(a × b)y = −3
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: Var(X) = 1.64
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 = 1·0.35 + 2·0.25 + 3·0.05 + 4·0.35 = 2.4
Mean of the squares
E(X²) = 1·0.35 + 4·0.25 + 9·0.05 + 16·0.35 = 7.4
Variance
Var(X) = 7.4 − 2.4² = 1.64
Answer
Var(X) = 1.64
16.Probability
Answer: σY = 4.214
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 = 2.6, Var(X) = 4.44, σX = 2.107
Linear change
σY = |2|·2.107 = 4.214
Answer
σY = 4.214
The mean does move: E(Y) = 2·2.6 − 3 = 2.2.
17.Descriptive statistics
Answer: s = 5.854
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 = 0.4
Standard normal table
Φ(0.4) = 0.6554
P = 1 − 0.6554 = 0.3446
Answer
P = 0.3446
19.Normal distribution
Answer: z = 1.96
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 + 0.9500)/2 = 0.9750
Read the z-table backwards
Find 0.9750 inside the table and read off z.
z = 1.96
Answer
z = 1.96
20.Normal distribution
Answer: P = 0.9772
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).