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→∞ (√(x2 − 5x) − x).

f(x) = √(x2 − 5x) − xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (1 − cos 3x)/(x²), with x in radians.

f(x) = (1 − cos 3x)/(x²)x → 0
L =
3.DerivativesMedium
xyx = 1.5

Let f(x) = x·sin(x), with x in radians. Find f′(1.5).

f(x) = x·sin(x)x = 1.5 rad
f′(1.5) =
4.DerivativesHard
xy

Let f(x) = x3 − 10.5x2 + 36x − 1. Find the x-coordinate x of the local minimum.

f(x) = x3 − 10.5x2 + 36x − 1
x =
5.IntegralsMedium
xy12.5y = x ln x

Evaluate I = ∫₁2.5 x·ln x dx.

∫₁2.5 x ln x dx
I =
6.IntegralsHard
xy3y = x²eˣ

Evaluate I = ∫₀3 x²·eˣ dx.

∫₀3 x²eˣ dx
I =
7.Complex numbersMedium
ReIm22−2−244−4−4z₁z₂w

Given z₁ = 1.5∠−60° and z₂ = 4∠−25°, find the imaginary part of w = z₁/z₂.

z₁ = 1.5∠−60°z₂ = 4∠−25°
Im(w) =
8.Complex numbersHard
ReIm11−1−122−2−2z₀z₁z₂z₃z₄

The roots of zⁿ = w with n = 5 and w = 32∠−120° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, …, 4. Find the imaginary part of z₁.

w = 32∠−120°n = 5k = 1
Im(z₁) =
9.Differential equationsMedium
tyy(0) = 1y(2) = ?

Solve y″ + 4y′ + 4y = 0 with y(0) = 1 and y′(0) = -1. Find y(2).

y″ + 4y′ + 4y = 0y(0) = 1y′(0) = -1
y(2) =
10.Differential equationsHard
t (min)m (kg)t = ?3 kg

A tank holds 100 L of pure water. Brine with 0.05 kg of salt per litre flows in at 4 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 3 kg of salt?

V = 100 Lm0 = 0 kgc = 0.05 kg/Lr = 4 L/min
t = min
11.Matrices & determinantsMedium
A =20−4−15−15−30

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

A = [2 0 −4; −1 5 −1; 5 −3 0]
det A =
12.Matrices & determinantsHard
A =5−36−2b =−99

Solve the system 5x − 3y = −9 and 6x − 2y = 9. Find x.

5x − 3y = −96x − 2y = 9
x =
13.Vectors: dot & cross productMedium
xyzab

Find the value of k for which the vectors a = (2, −3, −6) and b = (k, 6, 4) are perpendicular.

a = (2, −3, −6)b = (k, 6, 4)
k =
14.Vectors: dot & cross productHard
xyzab

The unit vector n̂ is perpendicular to both a = (−1, −2, 3) and b = (−5, 1, 3) and has a positive z-component. Find its y-component n̂y.

a = (−1, −2, 3)b = (−5, 1, 3)n̂z > 0
n̂y =
15.ProbabilityMedium
n = 7k = 4N = ?

In how many ways N can a committee of 4 people be chosen from a group of 7? The order does not matter.

n = 7k = 4
N =
16.ProbabilityHard
€p(x)−10.8810.1490.02

A game costs €1 to play. You win €2 with probability 0.1 and €50 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €1€2: p = 0.1€50: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
354045505560

Find the sample standard deviation s of the data: 59, 60, 42, 41, 31, 32.

n = 6
s =
18.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 70 and standard deviation 12. What proportion P of values is above 82?

μ = 70σ = 12x = 82
P =
19.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.6950

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.6950. Find z to two decimals.

A = 0.6950
z =
20.Normal distributionHard
μ = 250x = ?95%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 95% of them lie (the 95th percentile).

μ = 250σ = 5P = 95%
x = g

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).

Multiply by the conjugate

f(x) = −5x/(√(x2 − 5x) + x)

Divide by x

f(x) = −5/(√(1 + (−5)/x) + 1)

Let x → ∞

L = −5/(1 + 1) = −2.5

Answer

L = −2.5
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·sin 3x/(2x) → 0/0

Second application

9·cos 3x/2

Substitute x = 0

L = 9/2 = 4.5

Answer

L = 4.5
3.Derivatives

Answer: f′(1.5) = 1.104

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′(1.5) = 0.9975 + 1.5·0.07074 = 1.104

Answer

f′(1.5) = 1.104
4.Derivatives

Answer: x = 4

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 − 21x + 36 = 3·(x2 − 7x + 12)

Solve f′(x) = 0

x = 3 or x = 4

Which one is which

f″(x) = 6x − 21: negative at a maximum, positive at a minimum.

f″(3) = −3, f″(4) = 3
x = 4

Answer

x = 4
5.Integrals

Answer: I = 1.551

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

(3.125·ln 2.5 − 1.5625) − (0 − 0.25) = 1.551

Answer

I = 1.551
6.Integrals

Answer: I = 98.43

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.

First round

∫x²eˣ dx = x²eˣ − ∫2x eˣ dx

Second round

∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)

Limits

20.09·(5) − 1·2 = 98.43

Answer

I = 98.43
7.Complex numbers

Answer: Im(w) = −0.2151

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| = 1.5 / 4 = 0.375
θ = −60° − (−25)° = −35°

Rectangular form

Re(w) = 0.375·cos(−35°) = 0.3072
Im(w) = 0.375·sin(−35°) = −0.2151
Im(w) = −0.2151

Answer

Im(w) = −0.2151
8.Complex numbers

Answer: Im(z₁) = 1.486

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

321/5 = 2

Argument of the root

θ₁ = (−120° + 360°·1)/5 = 48°

Rectangular form

z₁ = 2·(cos 48° + j sin 48°) = 1.338 + j1.486
Im(z₁) = 1.486

Answer

Im(z₁) = 1.486

The 5 roots lie on the circle of radius 2, 72° apart.

9.Differential equations

Answer: y(2) = 0.05495

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 + 2)² = 0 → r = −2

Constants

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

Evaluate

y(2) = (1 + 1·2)·e−4 = 0.05495

Answer

y(2) = 0.05495
10.Differential equations

Answer: t = 22.91 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.

Result

m(t) = 5 + (−5)·e−t/25

Solve for t

e−t/25 = (5 − 3)/(5 − 0) = 0.4
t = 25·ln(2.5) = 22.91 min

Answer

t = 22.91 min
11.Matrices & determinants

Answer: det A = 82

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 = −3, M₁₂ = 5, M₁₃ = −22

Cofactor expansion

det A = 2·(−3) − 0·5 + (−4)·(−22) = 82

Answer

det A = 82
12.Matrices & determinants

Answer: x = 5.625

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·(−2) − (−3)·6 = 8

Replace the x column

det Aₓ = (−9)·(−2) − (−3)·9 = 45
x = 45/8 = 5.625

Answer

x = 5.625
y = 12.38
13.Vectors: dot & cross product

Answer: k = 21

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 equals zero

a·b = 2·k + (−3)·6 + (−6)·4 = 0
−42 + 2·k = 0

Solve for k

k = −(−42)/2 = 21

Answer

k = 21
14.Vectors: dot & cross product

Answer: n̂y = 0.6451

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 = (−2)·3 − 3·1 = −9
(a × b)y = azbx − axbz = 3·(−5) − (−1)·3 = −12
(a × b)z = axby − aybx = (−1)·1 − (−2)·(−5) = −11

Normalise

The z-component of a × b is negative, so use −(a × b).

|a × b| = √((−9)² + (−12)² + (−11)²) = 18.6
n̂ = −(−9, −12, −11)/18.6 = (0.4838, 0.6451, 0.5914)
n̂y = 0.6451

Answer

n̂y = 0.6451
15.Probability

Answer: N = 35

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(7, 4) = (7·6·5·4)/(4·3·2·1) = 35

Answer

N = 35
16.Probability

Answer: E(G) = 0.2 €

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: 1 (p = 0.1), 49 (p = 0.02), −1 (p = 0.88)

Expected value

E(G) = 1·0.1 + 49·0.02 − 1·0.88 = 0.2 €

Answer

E(G) = 0.2 €

A positive value means you win money on average.

17.Descriptive statistics

Answer: s = 12.7

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 = 265/6 = 44.17

Squared deviations

Σ(x − x̄)² = 14.83² + 15.83² + (−2.167)² + (−3.167)² + (−13.17)² + (−12.17)² = 806.8

Standard deviation

s = √(806.8/5) = 12.7

Answer

s = 12.7
18.Descriptive statistics

Answer: P = 0.1587

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 = 1

Standard normal table

Φ(1) = 0.8413
P = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
19.Normal distribution

Answer: z = −0.51

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 − 0.6950 = 0.3050

Read the z-table backwards

Find 0.3050 inside the table and read off z.

z = −0.51

Answer

z = −0.51
20.Normal distribution

Answer: x = 258.2 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) = 0.95 → z = 1.645

Back to x

x = 250 + 1.645·5 = 258.22 g

Answer

x = 258.22 g

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