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→∞ (5x2 + 6x + 1)/(2x2 + x + 4).
f(x) = (5x2 + 6x + 1)/(2x2 + x + 4)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 5x)/(x²), with x in radians.
f(x) = (1 − cos 5x)/(x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(2x), with x in radians. Find f′(2).
f(x) = x·sin(2x)x = 2 rad
f′(2) =
4.DerivativesHard
Let f(x) = x3 + 1.5x2 − 6. Find the x-coordinate x of the local maximum.
f(x) = x3 + 1.5x2 − 6
x =
5.IntegralsMedium
Evaluate I = ∫₀π/2 x·cos x dx (x in radians).
∫₀π/2 x·cos x dx
I =
6.IntegralsHard
Find the area A of the region enclosed by y = 2x2 and y = x.
y = 2x2y = x
A =
7.Complex numbersMedium
The diagram shows z = 2 + j. With w = 2∠−90°, click the point that represents z·w.
z = 2 + jw = 2∠−90°
1A (−2 + j4)
2B (2 − j4)
3C (1 − j2)
4D (4 + j2)
8.Complex numbersHard
The roots of zⁿ = w with n = 5 and w = 243∠−120° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, …, 4. Find the real part of z₄.
w = 243∠−120°n = 5k = 4
Re(z₄) =
9.Differential equationsMedium
A cup of coffee at 83 °C is left in a room at 22 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.08 min⁻¹, find its temperature T after 19 min.
T0 = 83 °CTa = 22 °Ck = 0.08 min⁻¹t = 19 min
T = °C
10.Differential equationsHard
Solve y″ + 2y′ + 26y = 0 with y(0) = 1 and y′(0) = 0 (released from rest). Find y(1); angles in radians.
y″ + 2y′ + 26y = 0y(0) = 1y′(0) = 0
y(1) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−3 −2 4; −4 5 −5; 1 −4 −3]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−1 2; −4 5]
λ =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (1, 4, −1) and b = (5, 1, 7).
a = (1, 4, −1)b = (5, 1, 7)
a·b =
14.Vectors: dot & cross productHard
For a = (3, −1, −2) and b = (3, 3, 3), find the y-component (a × b)y of the cross product a × b.
a = (3, −1, −2)b = (3, 3, 3)
(a × b)y =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2p(x): 0.35, 0.35, 0.3
σ =
16.ProbabilityHard
11 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?
n = 11p = 0.2
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 58, 54, 21, 39, 13.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 8. Find the z-score z of a score of 63.6.
μ = 70σ = 8x = 63.6
z =
19.Normal distributionMedium
X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 742.1 g. Round z to two decimals.
μX = 250, σX = 4μY = 500, σY = 9X + Y > 742.1 g
P =
20.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 80% of them lie (the 80th percentile).
μ = 100σ = 15P = 80%
x = points
3Solutions
1.Limits
Answer: L = 2.5
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
(5 + 6/x + 1/x2)/(2 + 1/x + 4/x2)
Let x → ∞
L = 5/2 = 2.5
Answer
L = 2.5
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 12.5
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/(2x) → 0/0
Second application
25·cos 5x/2
Substitute x = 0
L = 25/2 = 12.5
Answer
L = 12.5
3.Derivatives
Answer: f′(2) = −3.371
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′(2) = −0.7568 + 4·(−0.6536) = −3.371
Answer
f′(2) = −3.371
4.Derivatives
Answer: x = −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.
Derivative
f′(x) = 3x2 + 3x = 3·(x2 + x)
Solve f′(x) = 0
x = −1 or x = 0
Which one is which
f″(x) = 6x + 3: negative at a maximum, positive at a minimum.
f″(−1) = −3, f″(0) = 3
x = −1
Answer
x = −1
5.Integrals
Answer: I = 0.5708
Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.
Choose u and dv
u = x, du = dx; dv = cos x dx, v = sin x
Antiderivative
∫x cos x dx = x sin x − ∫sin x dx = x sin x + cos x
Limits
(π/2·sin π/2 + cos π/2) − (0 + 1) = 0.5708
Answer
I = 0.5708
6.Integrals
Answer: A = 0.04167
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
2x2 = x → x = 0, x = 0.5
Integrate upper minus lower
A = [0.5x2 − 0.6667x3]00.5 = 0.04167
Answer
A = 0.04167
7.Complex numbers
Answer: B (2 − 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 = 2 − j4 → B
Answer
B (2 − j4)
8.Complex numbers
Answer: Re(z₄) = −0.3136
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 of the roots
2431/5 = 3
Argument of the root
θ₄ = (−120° + 360°·4)/5 = 264° → −96°
Rectangular form
z₄ = 3·(cos −96° + j sin −96°) = −0.3136 − j2.984
Re(z₄) = −0.3136
Answer
Re(z₄) = −0.3136
The 5 roots lie on the circle of radius 3, 72° apart.
9.Differential equations
Answer: T = 35.34 °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) = 22 + 61·e−0.08t
Substitute t
T(19) = 22 + 61·0.2187 = 35.34 °C
Answer
T = 35.34 °C
10.Differential equations
Answer: y(1) = 0.0338
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.
Complex roots
r = −1 ± 5j
Constants
A = 1, B = 1·1/5 = 0.2
Evaluate
y(1) = e−1(1·cos 5 + 0.2·sin 5) = 0.0338
Answer
y(1) = 0.0338
11.Matrices & determinants
Answer: det A = 183
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 = −35, M₁₂ = 17, M₁₃ = 11
Cofactor expansion
det A = (−3)·(−35) − (−2)·17 + 4·11 = 183
Answer
det A = 183
12.Matrices & determinants
Answer: λ = 3
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.
Characteristic equation
λ² − (4)λ + (3) = 0
Solve
λ = (4 ± √(4))/2 → λ = 1, 3
λ = 3
Answer
λ = 3
13.Vectors: dot & cross product
Answer: a·b = 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.
Multiply matching components and add
a·b = 1·5 + 4·1 + (−1)·7 = 5 + 4 + (−7) = 2
Answer
a·b = 2
Positive: the angle between the vectors is less than 90°.
14.Vectors: dot & cross product
Answer: (a × b)y = −15
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 = (−1)·3 − (−2)·3 = 3
(a × b)y = azbx − axbz = (−2)·3 − 3·3 = −15
(a × b)z = axby − aybx = 3·3 − (−1)·3 = 12
Requested component
(a × b)y = −15
Answer
(a × b)y = −15
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: σ = 0.8047
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 = 0·0.35 + 1·0.35 + 2·0.3 = 0.95
Mean of the squares
E(X²) = 0·0.35 + 1·0.35 + 4·0.3 = 1.55
Variance
Var(X) = 1.55 − 0.95² = 0.6475
σ = √0.6475 = 0.8047
Answer
σ = 0.8047
16.Probability
Answer: P = 0.3221
Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.
Zero and one failure
P(X = 0) = 0.811 = 0.0859
P(X = 1) = 11·0.2·0.810 = 0.2362
Add them
P = 0.0859 + 0.2362 = 0.3221
Answer
P = 0.3221
17.Descriptive statistics
Answer: s = 19.79
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 = (63.6 − 70)/8 = −0.8
Answer
z = −0.8
19.Normal distribution
Answer: P = 0.7881
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 = 750 g, σ = √(4² + 9²) = 9.849 g
Standardise
z = (742.1 − 750)/9.849 = −0.8
Upper tail
P = 1 − Φ(−0.8) = 1 − 0.2119 = 0.7881
Answer
P = 0.7881
20.Normal distribution
Answer: x = 112.6 points
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.