Engineering Mathematics: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

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1Formulas and methods

Limits

Direct substitution
try x = a first; 0/0 means simplify
Factor and cancel
xⁿ − aⁿ = (x − a)(xⁿ⁻¹ + a·xⁿ⁻² + … + aⁿ⁻¹)
Conjugate
(√x − √a)(√x + √a) = x − a
Limits at infinity
divide by the highest power of x; equal degrees give the ratio of leading coefficients
Special limit
limx→∞ (1 + a/x)bx = eab
L'Hôpital's rule
lim f/g = lim f′/g′ (0/0 or ∞/∞)

Derivatives

Power rule
(xⁿ)′ = n·xⁿ⁻¹
Product rule
(uv)′ = u′v + uv′
Quotient rule
(u/v)′ = (u′v − uv′)/v²
Chain rule
f(g(x))′ = f′(g(x))·g′(x)
Common derivatives
(eˣ)′ = eˣ, (ln x)′ = 1/x, (sin x)′ = cos x, (cos x)′ = −sin x

Method

  1. Identify the structure: sum, product, quotient or composition.
  2. Apply the power rule to each power of x.
  3. Use the product, quotient or chain rule where needed.
  4. Simplify, and evaluate at the point if asked.

Integrals

Power rule
∫xⁿ dx = xⁿ⁺¹/(n + 1) + C
Definite integral
∫ab f dx = F(b) − F(a)
Common integrals
∫eˣ dx = eˣ, ∫1/x dx = ln|x|, ∫sin x dx = −cos x
Area between curves
∫ab (f − g) dx

Method

  1. Rewrite the integrand as a sum of simple terms where possible.
  2. Apply the power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1), and standard integrals for ex, 1/x, sin and cos.
  3. Use substitution when you see a function and its derivative together.
  4. For a definite integral, evaluate F(b) − F(a).

Complex numbers

Rectangular form
z = a + jb, j² = −1, z̄ = a − jb
Modulus and argument
|z| = √(a² + b²), tan θ = b/a (check the quadrant)
Polar and exponential form
z = r∠θ = r(cos θ + j sin θ) = r·ejθ
Division
z₁/z₂ = z₁z̄₂/|z₂|²
Product and quotient
r₁∠θ₁ · r₂∠θ₂ = r₁r₂∠(θ₁ + θ₂), r₁∠θ₁ / r₂∠θ₂ = (r₁/r₂)∠(θ₁ − θ₂)
De Moivre
(r∠θ)ⁿ = rⁿ∠nθ
n-th roots
zk = R1/n∠((φ + 360°·k)/n), k = 0 … n − 1
Quadratic with p² < 4q
z = −p/2 ± j√(q − p²/4)

Method

  1. Decide which form suits the operation: rectangular a + jb for adding and subtracting, polar r∠θ for multiplying, dividing, powers and roots.
  2. Multiply out with j² = −1; divide by multiplying the top and bottom by the conjugate of the denominator.
  3. 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.
  4. 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

  1. Write every vector in components; for a force given by a magnitude and an angle, use F·cos θ and F·sin θ.
  2. For lengths and directions use the magnitude √(x² + y² + z²) and the unit vector a/|a|.
  3. Use the dot product for angles, projections, perpendicularity and work: a·b = |a||b|cos θ.
  4. 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

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. For a percentile, work backwards: find z for the area, then x = μ + z·σ.

2Practice problems

1.LimitsMedium
xyy = L

Find the limit L = limx→∞ (1 − 2/x)3x. Give a decimal value.

f(x) = (1 − 2/x)3xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 ln(1 + 4x)/sin(5x), with x in radians.

f(x) = ln(1 + 4x)/sin(5x)x → 0
L =
3.DerivativesMedium
xyx = 0.5

Let f(x) = x·sin(x), with x in radians. Find f′(0.5).

f(x) = x·sin(x)x = 0.5 rad
f′(0.5) =
4.DerivativesHard
xy

Let f(x) = x3 + 1.5x2 − 6x + 6. Find the x-coordinate x of the local maximum.

f(x) = x3 + 1.5x2 − 6x + 6
x =
5.IntegralsMedium
xy16y = x ln x

Evaluate I = ∫₁6 x·ln x dx.

∫₁6 x ln x dx
I =
6.IntegralsHard
xy1.333y = 4xy = 3x²

The region between the line y = 4x and the parabola y = 3x² is rotated about the x-axis. Find the volume V of the solid.

y = 4xy = 3x²
V =
7.Complex numbersMedium
ReIm22−2−244−4−466−6−6z₁z₂w

Given z₁ = 4∠70° and z₂ = 1.5∠−95°, find the real part of w = z₁·z₂.

z₁ = 4∠70°z₂ = 1.5∠−95°
Re(w) =
8.Complex numbersHard
ReIm11−1−122−2−233−3−3θz

Find the imaginary part of (−3 + j3)³.

z = −3 + j3n = 3
Im(z³) =
9.Differential equationsMedium
tyy(0) = 3y(2) = ?

Solve y″ + 5y′ + 6y = 0 with y(0) = 3 and y′(0) = 4. Find y(2).

y″ + 5y′ + 6y = 0y(0) = 3y′(0) = 4
y(2) =
10.Differential equationsHard
tyK = 1ymax​ = ?

A system obeys y″ + 4y′ + 29y = 29 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.

y″ + 4y′ + 29y = 29y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
A =1121−4−5323

Find the determinant det A of the 3×3 matrix A shown.

A = [1 1 2; 1 −4 −5; 3 2 3]
det A =
12.Matrices & determinantsHard
A =−14−53b =−12−7

Solve the system −x + 4y = −12 and −5x + 3y = −7. Find x.

−x + 4y = −12−5x + 3y = −7
x =
13.Vectors: dot & cross productMedium
xyzab

Find the dot product a·b of a = (−4, −3, −1) and b = (−1, 3, 5).

a = (−4, −3, −1)b = (−1, 3, 5)
a·b =
14.Vectors: dot & cross productHard
xyzabc

Find the volume V of the parallelepiped whose edges from one corner are a = (−2, 2, 3), b = (1, −3, 3) and c = (−1, −1, 2).

a = (−2, 2, 3)b = (1, −3, 3)c = (−1, −1, 2)
V =
15.ProbabilityMedium
n = 6k = 2N = ?

In how many ways N can a committee of 2 people be chosen from a group of 6? The order does not matter.

n = 6k = 2
N =
16.ProbabilityHard
xp(x)00.210.2520.3530.1540.05μ

X has the distribution shown, and Y = 2X + 1. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.2, 0.25, 0.35, 0.15, 0.05Y = 2X + 1
σY =
17.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 33, 55, 57, 38, 16.

n = 5
s =
18.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 70 and standard deviation 10. What proportion P of values is below 62?

μ = 70σ = 10x = 62
P =
19.Normal distributionMedium
244254P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 244 and 254 g? Give P as a decimal.

μ = 250σ = 5x1 = 244x2 = 254
P =
20.Normal distributionHard
μ = 250x = ?10%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 10% of them lie (the 10th percentile).

μ = 250σ = 5P = 10%
x = g

3Solutions

1.Limits

Answer: L = 0.002479

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→∞ 3·x·ln(1 − 2/x)

Evaluate

ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.

ln L = 3·(−2) = −6

Undo the logarithm

L = e−6 = 0.002479

Answer

L = 0.002479
2.Limits

Answer: L = 0.8

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 [4/(1 + 4x)]/[5·cos 5x]

Substitute x = 0

L = 4/5 = 0.8

Answer

L = 0.8
3.Derivatives

Answer: f′(0.5) = 0.9182

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(x) + x·cos(x)

Substitute

f′(0.5) = 0.4794 + 0.5·0.8776 = 0.9182

Answer

f′(0.5) = 0.9182
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 + 3x − 6 = 3·(x2 + x − 2)

Solve f′(x) = 0

x = −2 or x = 1

Which one is which

f″(x) = 6x + 3: negative at a maximum, positive at a minimum.

f″(−2) = −9, f″(1) = 9
x = −2

Answer

x = −2
5.Integrals

Answer: I = 23.5

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 = ln x, du = dx/x; dv = x dx, v = x²/2

Antiderivative

∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4

Limits

(18·ln 6 − 9) − (0 − 0.25) = 23.5

Answer

I = 23.5
6.Integrals

Answer: V = 15.89

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.

Intersection points

4x = 3x² → x = 0, x = 1.333

Washer method

V = π∫₀1.333 ((4x)² − (3x²)²) dx

Integrate

V = π(16·1.333³/3 − 9·1.333⁵/5) = π·5.057 = 15.89

Answer

V = 15.89
7.Complex numbers

Answer: Re(w) = 5.438

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.

Modulus and argument of w

|w| = 4 · 1.5 = 6
θ = 70° + (−95)° = −25°

Rectangular form

Re(w) = 6·cos(−25°) = 5.438
Im(w) = 6·sin(−25°) = −2.536
Re(w) = 5.438

Answer

Re(w) = 5.438
8.Complex numbers

Answer: Im(z³) = 54

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.

Powers step by step

z² = (−3 + j3)(−3 + j3) = −j18
z³ = (−j18)(−3 + j3) = 54 + j54

Check with De Moivre

z = 4.243∠135° → z³ = 76.37∠45°
Im(z³) = 54

Answer

Im(z³) = 54
9.Differential equations

Answer: y(2) = 0.2133

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.

Characteristic roots

(r + 2)(r + 3) = 0 → r = −2, −3

Constants from the initial values

A + B = 3, −2A − 3B = 4
A = 13, B = −10

Evaluate

y(2) = 13·e−4 + (−10)·e−6 = 0.2133

Answer

y(2) = 0.2133
10.Differential equations

Answer: ymax = 1.285

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 = −2 ± 5j

Constants from the initial values

y(0) = 1 + A = 0 → A = −1
y′(0) = −2A + 5B = 0 → B = −0.4

Peak

t = π/5 = 0.6283
ymax = 1·(1 + e−2·π/5) = 1.285

Answer

ymax = 1.285
11.Matrices & determinants

Answer: det A = 8

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 = −2, M₁₂ = 18, M₁₃ = 14

Cofactor expansion

det A = 1·(−2) − 1·18 + 2·14 = 8

Answer

det A = 8
12.Matrices & determinants

Answer: x = −0.4706

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)·3 − 4·(−5) = 17

Replace the x column

det Aₓ = (−12)·3 − 4·(−7) = −8
x = −8/17 = −0.4706

Answer

x = −0.4706
y = −3.118
13.Vectors: dot & cross product

Answer: a·b = −10

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 = (−4)·(−1) + (−3)·3 + (−1)·5 = 4 + (−9) + (−5) = −10

Answer

a·b = −10

Negative: the angle between the vectors is more than 90°.

14.Vectors: dot & cross product

Answer: V = 16

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 = (−3)·2 − 3·(−1) = −3
(b × c)y = bzcx − bxcz = 3·(−1) − 1·2 = −5
(b × c)z = bxcy − bycx = 1·(−1) − (−3)·(−1) = −4

Dot product with a

a·(b × c) = (−2)·(−3) + 2·(−5) + 3·(−4) = −16
V = |−16| = 16

Answer

V = 16
15.Probability

Answer: N = 15

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.

Combinations

C(6, 2) = (6·5)/(2·1) = 15

Answer

N = 15
16.Probability

Answer: σY = 2.227

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 = 1.6, Var(X) = 1.24, σX = 1.114

Linear change

σY = |2|·1.114 = 2.227

Answer

σY = 2.227

The mean does move: E(Y) = 2·1.6 + 1 = 4.2.

17.Descriptive statistics

Answer: s = 16.9

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.

Mean

x̄ = 199/5 = 39.8

Squared deviations

Σ(x − x̄)² = (−6.8)² + 15.2² + 17.2² + (−1.8)² + (−23.8)² = 1143

Standard deviation

s = √(1143/4) = 16.9

Answer

s = 16.9
18.Descriptive statistics

Answer: P = 0.2119

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 = −0.8

Standard normal table

Φ(−0.8) = 0.2119
P = 0.2119

Answer

P = 0.2119
19.Normal distribution

Answer: P = 0.6731

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 = (244 − 250)/5 = −1.2
z₂ = (254 − 250)/5 = 0.8

Area between

P = Φ(0.8) − Φ(−1.2) = 0.7881 − 0.1151 = 0.6731

Answer

P = 0.6731
20.Normal distribution

Answer: x = 243.6 g

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 for the percentile

Φ(z) = 0.1 → z = −1.282

Back to x

x = 250 + −1.282·5 = 243.59 g

Answer

x = 243.59 g

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