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→∞ (√(4x2 + 2x) − 2x).

f(x) = √(4x2 + 2x) − 2xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (e3x − 1 − 3x)/x².

f(x) = (e3x − 1 − 3x)/x²x → 0
L =
3.DerivativesMedium
xyx = 0.5

Let f(x) = x·sin(2x), with x in radians. Find f′(0.5).

f(x) = x·sin(2x)x = 0.5 rad
f′(0.5) =
4.DerivativesHard
xyx = −2

The tangent line to f(x) = 2x2 − x − 5 at x = −2 crosses the y-axis at y = b. Find b.

f(x) = 2x2 − x − 5x = −2
b =
5.IntegralsMedium
xy4y = √(2x)

The region under y = √(2x) from x = 0 to x = 4 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).

y = √(2x)0 ≤ x ≤ 4
V =
6.IntegralsHard
xy2y = 4xy = 2x²

The region between the line y = 4x and the parabola y = 2x² is rotated about the x-axis. Find the volume V of the solid.

y = 4xy = 2x²
V =
7.Complex numbersMedium
ReIm55−5−5θz

Write z = 10∠140° in rectangular form a + jb and give the imaginary part b.

r = 10θ = 140°
b =
8.Complex numbersHard
ReIm11−1−122−2−2z₀z₁z₂

The roots of zⁿ = w with n = 3 and w = 8∠135° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2. Find the real part of z₁.

w = 8∠135°n = 3k = 1
Re(z₁) =
9.Differential equationsMedium
tyy(0) = 2y(1.5) = ?

Solve y″ + 6y′ + 9y = 0 with y(0) = 2 and y′(0) = -1. Find y(1.5).

y″ + 6y′ + 9y = 0y(0) = 2y′(0) = -1
y(1.5) =
10.Differential equationsHard
tyy(0) = 10y(0.5) = ?

Solve y″ + 2y′ + 10y = 0 with y(0) = 10 and y′(0) = 0 (released from rest). Find y(0.5); angles in radians.

y″ + 2y′ + 10y = 0y(0) = 10y′(0) = 0
y(0.5) =
11.Matrices & determinantsMedium
A =−2−452−43143

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

A = [−2 −4 5; 2 −4 3; 1 4 3]
det A =
12.Matrices & determinantsHard
A =52−45b =−2−2

Solve the system 5x + 2y = −2 and −4x + 5y = −2. Find x.

5x + 2y = −2−4x + 5y = −2
x =
13.Vectors: dot & cross productMedium
xyzab

Find the angle θ between the vectors a = (1, 1, 1) and b = (3, 3, −1), in degrees.

a = (1, 1, 1)b = (3, 3, −1)
θ = °
14.Vectors: dot & cross productHard
xyzabc

Find the volume V of the parallelepiped whose edges from one corner are a = (−4, 1, 0), b = (−1, −1, 0) and c = (1, 1, −5).

a = (−4, 1, 0)b = (−1, −1, 0)c = (1, 1, −5)
V =
15.ProbabilityMedium
red: 7blue: 7

A bag holds 7 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

7 / 7
P =
16.ProbabilityHard
€p(x)−20.8920.1380.01

A game costs €2 to play. You win €4 with probability 0.1 and €40 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €2€4: p = 0.1€40: p = 0.01
E(G) = €
17.Descriptive statisticsMedium
15202530354045505560

Find the sample standard deviation s of the data: 22, 60, 40, 60, 14, 27, 23.

n = 7
s =
18.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 500 and standard deviation 50. What proportion P of values is below 400?

μ = 500σ = 50x = 400
P =
19.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.8665

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.8665. Find z to two decimals.

A = 0.8665
z =
20.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.02 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.015 mm
P =

3Solutions

1.Limits

Answer: L = 0.5

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).

Multiply by the conjugate

f(x) = 2x/(√(4x2 + 2x) + 2x)

Divide by x

f(x) = 2/(√(4 + 2/x) + 2)

Let x → ∞

L = 2/(2 + 2) = 0.5

Answer

L = 0.5
2.Limits

Answer: L = 4.5

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

(3·e3x − 3)/(2x) → 0/0

Second application

9·e3x/2

Substitute x = 0

L = 9·1/2 = 4.5

Answer

L = 4.5
3.Derivatives

Answer: f′(0.5) = 1.382

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′(0.5) = 0.8415 + 1·0.5403 = 1.382

Answer

f′(0.5) = 1.382
4.Derivatives

Answer: b = −13

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) = 5
f′(x) = 4x − 1 → f′(−2) = −9

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 5 − (−9)·(−2) = −13

Answer

b = −13
5.Integrals

Answer: V = 50.27

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 2x dx

Integrate

V = π·2·4²/2 = 50.27

Answer

V = 50.27
6.Integrals

Answer: V = 53.62

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 = 2x² → x = 0, x = 2

Washer method

V = π∫₀2 ((4x)² − (2x²)²) dx

Integrate

V = π(16·2³/3 − 4·2⁵/5) = π·17.07 = 53.62

Answer

V = 53.62
7.Complex numbers

Answer: b = 6.428

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(140°) = −7.66
b = 10·sin(140°) = 6.428

Rectangular form

z = −7.66 + j6.428
b = 6.428

Answer

b = 6.428
8.Complex numbers

Answer: Re(z₁) = −1.932

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

81/3 = 2

Argument of the root

θ₁ = (135° + 360°·1)/3 = 165°

Rectangular form

z₁ = 2·(cos 165° + j sin 165°) = −1.932 + j0.5176
Re(z₁) = −1.932

Answer

Re(z₁) = −1.932

The 3 roots lie on the circle of radius 2, 120° apart.

9.Differential equations

Answer: y(1.5) = 0.1055

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.

Double root

(r + 3)² = 0 → r = −3

Constants

A = y(0) = 2
y′(0) = B − 3A → B = -1 + 6 = 5

Evaluate

y(1.5) = (2 + 5·1.5)·e−4.5 = 0.1055

Answer

y(1.5) = 0.1055
10.Differential equations

Answer: y(0.5) = 2.446

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 ± 3j

Constants

A = 10, B = 1·10/3 = 3.333

Evaluate

y(0.5) = e−0.5(10·cos 1.5 + 3.333·sin 1.5) = 2.446

Answer

y(0.5) = 2.446
11.Matrices & determinants

Answer: det A = 120

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 = −24, M₁₂ = 3, M₁₃ = 12

Cofactor expansion

det A = (−2)·(−24) − (−4)·3 + 5·12 = 120

Answer

det A = 120
12.Matrices & determinants

Answer: x = −0.1818

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 = 5·5 − 2·(−4) = 33

Replace the x column

det Aₓ = (−2)·5 − 2·(−2) = −6
x = −6/33 = −0.1818

Answer

x = −0.1818
y = −0.5455
13.Vectors: dot & cross product

Answer: θ = 48.53 °

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 = 1·3 + 1·3 + 1·(−1) = 5

Magnitudes

|a| = √(1² + 1² + 1²) = 1.732
|b| = √(3² + 3² + (−1)²) = 4.359

Angle

cos θ = 5/(1.732·4.359) = 0.6623
θ = arccos(0.6623) = 48.53°

Answer

θ = 48.53°
14.Vectors: dot & cross product

Answer: V = 25

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 = (−1)·(−5) − 0·1 = 5
(b × c)y = bzcx − bxcz = 0·1 − (−1)·(−5) = −5
(b × c)z = bxcy − bycx = (−1)·1 − (−1)·1 = 0

Dot product with a

a·(b × c) = (−4)·5 + 1·(−5) + 0·0 = −25
V = |−25| = 25

Answer

V = 25
15.Probability

Answer: P = 0.2308

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.

First and second draw

P = 7/14 · 6/13 = 42/182 = 0.2308

Answer

P = 0.2308
16.Probability

Answer: E(G) = −1.2 €

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.

Net outcomes

G: 2 (p = 0.1), 38 (p = 0.01), −2 (p = 0.89)

Expected value

E(G) = 2·0.1 + 38·0.01 − 2·0.89 = −1.2 €

Answer

E(G) = −1.2 €

A negative value means you lose money on average.

17.Descriptive statistics

Answer: s = 18.68

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 = 246/7 = 35.14

Squared deviations

Σ(x − x̄)² = (−13.14)² + 24.86² + 4.857² + 24.86² + (−21.14)² + (−8.143)² + (−12.14)² = 2093

Standard deviation

s = √(2093/6) = 18.68

Answer

s = 18.68
18.Descriptive statistics

Answer: P = 0.02275

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 = 0.02275

Answer

P = 0.02275
19.Normal distribution

Answer: z = −1.11

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 − 0.8665 = 0.1335

Read the z-table backwards

Find 0.1335 inside the table and read off z.

z = −1.11

Answer

z = −1.11
20.Normal distribution

Answer: P = 0.8665

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 difference

μ = 25.02 − 25 = 0.02 mm, σ = 0.01803 mm

Standardise

z = (0 − 0.02)/0.01803 = −1.11

Probability

P = 1 − Φ(−1.11) = 0.8665

Answer

P = 0.8665

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