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→∞ (4x2 − 2x − 4)/(4x2 + 4x + 9).
f(x) = (4x2 − 2x − 4)/(4x2 + 4x + 9)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e−0.5x − 1 + 0.5x)/x².
f(x) = (e−0.5x − 1 + 0.5x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(2x), with x in radians. Find f′(1).
f(x) = x·sin(2x)x = 1 rad
f′(1) =
4.DerivativesHard
Let f(x) = x3 − 9x2 + 24x + 6. Find the x-coordinate x of the local maximum.
f(x) = x3 − 9x2 + 24x + 6
x =
5.IntegralsMedium
Evaluate I = ∫03 ex dx.
f(x) = ex[0, 3]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = 3x.
y = x2y = 3x
A =
7.Complex numbersMedium
Write z = 10∠−75° in rectangular form a + jb and give the real part a.
r = 10θ = −75°
a =
8.Complex numbersHard
Solve z² + 6z + 18 = 0. The roots are a ± jb; find b > 0.
p = 6q = 18
b =
9.Differential equationsMedium
A forged steel part at 500 °C is left to cool in a workshop at 30 °C. After 49 min its temperature is 206 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 412 °C?
T0 = 500 °CTa = 30 °CT(49) = 206 °CT = 412 °C
t = min
10.Differential equationsHard
Solve y′ + y = 2e−3t with y(0) = 4. Find y(1).
y′ + y = 2e−3ty(0) = 4
y(1) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−5 4 0; 1 −5 1; 3 −1 −2]
det A =
12.Matrices & determinantsHard
Solve the system 5x − 2y = −6 and −2x − 6y = 6. Find x.
5x − 2y = −6−2x − 6y = 6
x =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (5, −5, −3) onto b = (−3, 6, −6), that is, the component of a in the direction of b.
a = (5, −5, −3)b = (−3, 6, −6)
compb a =
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−2, −1, 2) and b = (2, −3, 1) and has a positive z-component. Find its y-component n̂y.
a = (−2, −1, 2)b = (2, −3, 1)n̂z > 0
n̂y =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4, 5p(x): 0.35, 0.15, 0.3, 0.1, 0.1
Var(X) =
16.ProbabilityHard
X has the distribution shown, and Y = 10X + 1. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 11, 56, 17, 27, 52, 10.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 5. What proportion P of values is above 60?
μ = 70σ = 5x = 60
P =
19.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 50 and 73 points? Give P as a decimal.
μ = 65σ = 10x1 = 50x2 = 73
P =
20.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 60% is to the left of it. Find the cut-off x.
μ = 500 hσ = 40 h60%
x = h
3Solutions
1.Limits
Answer: L = 1
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
(4 + (−2)/x + (−4)/x2)/(4 + 4/x + 9/x2)
Let x → ∞
L = 4/4 = 1
Answer
L = 1
For equal degrees the limit is the ratio of the leading coefficients.
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·e−0.5x + 0.5)/(2x) → 0/0
Second application
0.25·e−0.5x/2
Substitute x = 0
L = 0.25·1/2 = 0.125
Answer
L = 0.125
3.Derivatives
Answer: f′(1) = 0.077
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 and chain rule
f′(x) = sin(2x) + 2x·cos(2x)
Substitute
f′(1) = 0.9093 + 2·(−0.4161) = 0.077
Answer
f′(1) = 0.077
4.Derivatives
Answer: x = 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.
Derivative
f′(x) = 3x2 − 18x + 24 = 3·(x2 − 6x + 8)
Solve f′(x) = 0
x = 2 or x = 4
Which one is which
f″(x) = 6x − 18: negative at a maximum, positive at a minimum.
f″(2) = −6, f″(4) = 6
x = 2
Answer
x = 2
5.Integrals
Answer: I = 19.09
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) = ex/1
Evaluate at the limits
I = (e3 − 1)/1 = (20.09 − 1)/1 = 19.09
Answer
I = 19.09
6.Integrals
Answer: A = 4.5
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 = 3x → x = 0, x = 3
Integrate upper minus lower
A = [1.5x2 − 0.3333x3]03 = 4.5
Answer
A = 4.5
7.Complex numbers
Answer: a = 2.588
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.
Components
a = 10·cos(−75°) = 2.588
b = 10·sin(−75°) = −9.659
Rectangular form
z = 2.588 − j9.659
a = 2.588
Answer
a = 2.588
8.Complex numbers
Answer: b = 3
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 = 36 − 72 = −36 < 0
Quadratic formula
z = (−6 ± j√36)/2 = −3 + j3, −3 − j3
Answer
b = 3
Answer
b = 3
9.Differential equations
Answer: t = 10.34 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(470/176)/49 = 0.02005 min⁻¹
Time to reach the target
t = ln(470/382)/0.02005 = 10.34 min
Answer
t = 10.34 min
10.Differential equations
Answer: y(1) = 1.79
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 = et
(ety)′ = 2e−2t
General solution
y = 2/(−2)·e−3t + C·e−t = −1·e−3t + C·e−t
Initial condition
4 = −1 + C → C = 5
Evaluate
y(1) = −1·0.04979 + 5·0.3679 = 1.79
Answer
y(1) = 1.79
11.Matrices & determinants
Answer: det A = −35
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 = 11, M₁₂ = −5, M₁₃ = 14
Cofactor expansion
det A = (−5)·11 − 4·(−5) + 0·14 = −35
Answer
det A = −35
12.Matrices & determinants
Answer: x = −1.412
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·(−6) − (−2)·(−2) = −34
Replace the x column
det Aₓ = (−6)·(−6) − (−2)·6 = 48
x = 48/(−34) = −1.412
Answer
x = −1.412
y = −0.5294
13.Vectors: dot & cross product
Answer: compb a = −3
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·(−3) + (−5)·6 + (−3)·(−6) = −27
Length of b
|b| = √((−3)² + 6² + (−6)²) = 9
Scalar projection
compb a = a·b/|b| = −27/9 = −3
Answer
compb a = −3
Negative: a has a component pointing against b.
14.Vectors: dot & cross product
Answer: n̂y = 0.5367
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 = (−1)·1 − 2·(−3) = 5
(a × b)y = azbx − axbz = 2·2 − (−2)·1 = 6
(a × b)z = axby − aybx = (−2)·(−3) − (−1)·2 = 8
Normalise
The z-component of a × b is already positive.
|a × b| = √(5² + 6² + 8²) = 11.18
n̂ = (5, 6, 8)/11.18 = (0.4472, 0.5367, 0.7155)
n̂y = 0.5367
Answer
n̂y = 0.5367
15.Probability
Answer: Var(X) = 1.748
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 = 2.9, Var(X) = 4.19, σX = 2.047
Linear change
σY = |10|·2.047 = 20.47
Answer
σY = 20.47
The mean does move: E(Y) = 10·2.9 + 1 = 30.
17.Descriptive statistics
Answer: s = 20.45
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 = −2
Standard normal table
Φ(−2) = 0.02275
P = 1 − 0.02275 = 0.9772
Answer
P = 0.9772
19.Normal distribution
Answer: P = 0.7213
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 = (50 − 65)/10 = −1.5
z₂ = (73 − 65)/10 = 0.8
Area between
P = Φ(0.8) − Φ(−1.5) = 0.7881 − 0.06681 = 0.7213
Answer
P = 0.7213
20.Normal distribution
Answer: x = 510 h
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.