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 + 0.5/x)3x. Give a decimal value.

f(x) = (1 + 0.5/x)3xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (e2x − 1 − 2x)/x².

f(x) = (e2x − 1 − 2x)/x²x → 0
L =
3.DerivativesMedium
xyx = −1

Let f(x) = (2x + 1)5. Find f′(−1).

f(x) = (2x + 1)5x = −1
f′(−1) =
4.DerivativesHard
xy

Let f(x) = x3 − 3x − 5. Find the x-coordinate x of the local minimum.

f(x) = x3 − 3x − 5
x =
5.IntegralsMedium
xyπy = x·cos x

Evaluate I = ∫₀π x·cos x dx (x in radians).

∫₀π x·cos x dx
I =
6.IntegralsHard
xy2y = x²eˣ

Evaluate I = ∫₀2 x²·eˣ dx.

∫₀2 x²eˣ dx
I =
7.Complex numbersMedium
ReIm11−1−122−2−233−3−344−4−455−5−566−6−677−7−788−8−8zABCD

The diagram shows z = 3 + j2. With w = 0.5∠90°, click the point that represents z/w.

z = 3 + j2w = 0.5∠90°
  1. 1A (2 − j3)
  2. 2B (−4 + j6)
  3. 3C (6 + j4)
  4. 4D (4 − j6)
8.Complex numbersHard
ReIm11−1−122−2−2z₀z₁z₂

The roots of zⁿ = w with n = 3 and w = 8∠−120° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2. Find the imaginary part of z₂.

w = 8∠−120°n = 3k = 2
Im(z₂) =
9.Differential equationsMedium
t (min)T (°C)29Ta​20 °C

A can of soda at 4 °C is taken out of the fridge into a room at 22 °C. After 29 min its temperature is 20 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 4 °CTa = 22 °CT(29) = 20 °C
k = min⁻¹
10.Differential equationsHard
tyK = 5ymax​ = ?

A system obeys y″ + 2y′ + 5y = 25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.

y″ + 2y′ + 5y = 25y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
A =053−5B =6−5−6−3

Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).

A = [0 5; 3 −5]B = [6 −5; −6 −3]
c12 =
12.Matrices & determinantsHard
A =345−1b =312

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

3x + 4y = 35x − y = 12
x =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (3, −4, 5) onto b = (−3, −6, 0), that is, the component of a in the direction of b.

a = (3, −4, 5)b = (−3, −6, 0)
compb a =
14.Vectors: dot & cross productHard
xyzrFOA

A force F = (−40, 20, 40) N acts at point A(−4, −1, 4). Coordinates are in metres. Find the z-component Mz of the moment of the force about the origin O(0, 0, 0).

F = (−40, 20, 40) NA(−4, −1, 4) mO(0, 0, 0) m
Mz = N·m
15.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6P(six) = 1/6 on each roll

A fair die is rolled 6 times. What is the probability P of getting at least one six?

n = 6
P =
16.ProbabilityHard
€p(x)−100.83300.154900.02

A game costs €10 to play. You win €40 with probability 0.15 and €500 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €10€40: p = 0.15€500: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 56, 32, 22, 33, 44.

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

A quantity is normally distributed with mean 100 and standard deviation 5. What proportion P of values is above 92.5?

μ = 100σ = 5x = 92.5
P =
19.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9099

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9099. Find z to two decimals.

A = 0.9099
z =
20.Normal distributionHard
157164171178185192199cmtop 15%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 15% is to the right of it. Find the cut-off x.

μ = 178 cmσ = 7 cm15%
x = cm

3Solutions

1.Limits

Answer: L = 4.482

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 + 0.5/x)

Evaluate

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

ln L = 3·0.5 = 1.5

Undo the logarithm

L = e1.5 = 4.482

Answer

L = 4.482
2.Limits

Answer: L = 2

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

(2·e2x − 2)/(2x) → 0/0

Second application

4·e2x/2

Substitute x = 0

L = 4·1/2 = 2

Answer

L = 2
3.Derivatives

Answer: f′(−1) = 10

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.

Chain rule

f′(x) = 5·(2x + 1)4·2

Substitute

f′(−1) = 5·(−1)4·2 = 10

Answer

f′(−1) = 10
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 − 3 = 3·(x2 − 1)

Solve f′(x) = 0

x = −1 or x = 1

Which one is which

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

f″(−1) = −6, f″(1) = 6
x = 1

Answer

x = 1
5.Integrals

Answer: I = −2

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·sin π + cos π) − (0 + 1) = −2

Answer

I = −2
6.Integrals

Answer: I = 12.78

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

7.389·(2) − 1·2 = 12.78

Answer

I = 12.78
7.Complex numbers

Answer: D (4 − j6)

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| = |z|/0.5, arg(z/w) = arg z − 90°

The point

z/w = 4 − j6 → D

Answer

D (4 − j6)
8.Complex numbers

Answer: Im(z₂) = −0.684

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

θ₂ = (−120° + 360°·2)/3 = 200° → −160°

Rectangular form

z₂ = 2·(cos −160° + j sin −160°) = −1.879 − j0.684
Im(z₂) = −0.684

Answer

Im(z₂) = −0.684

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

9.Differential equations

Answer: k = 0.07577 min⁻¹

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.

Ratio of temperature differences

e−29k = -2/-18

Solve for k

k = ln(-18/-2)/29 = 0.07577 min⁻¹

Answer

k = 0.07577 min⁻¹
10.Differential equations

Answer: ymax = 6.039

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 = 5, r = −1 ± 2j

Constants from the initial values

y(0) = 5 + A = 0 → A = −5
y′(0) = −1A + 2B = 0 → B = −2.5

Peak

t = π/2 = 1.571
ymax = 5·(1 + e−1·π/2) = 6.039

Answer

ymax = 6.039
11.Matrices & determinants

Answer: c12 = −15

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.

Row times column

c12 = 0·(−5) + 5·(−3) = −15

Answer

c12 = −15
12.Matrices & determinants

Answer: x = 2.217

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 = 3·(−1) − 4·5 = −23

Replace the x column

det Aₓ = 3·(−1) − 4·12 = −51
x = −51/(−23) = 2.217

Answer

x = 2.217
y = −0.913
13.Vectors: dot & cross product

Answer: compb a = 2.236

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 = 3·(−3) + (−4)·(−6) + 5·0 = 15

Length of b

|b| = √((−3)² + (−6)² + 0²) = 6.708

Scalar projection

compb a = a·b/|b| = 15/6.708 = 2.236

Answer

compb a = 2.236
14.Vectors: dot & cross product

Answer: Mz = −120 N·m

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.

Position vector from O to A

r = A − O = (−4, −1, 4) − (0, 0, 0) = (−4, −1, 4) m

Moment M = r × F

(r × F)x = ryFz − rzFy = (−1)·40 − 4·20 = −120 N·m
(r × F)y = rzFx − rxFz = 4·(−40) − (−4)·40 = 0 N·m
(r × F)z = rxFy − ryFx = (−4)·20 − (−1)·(−40) = −120 N·m

Requested component

Mz = −120 N·m

Answer

Mz = −120 N·m
15.Probability

Answer: P = 0.6651

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.

Complement

P(no six) = (5/6)6 = 0.3349
P = 1 − 0.3349 = 0.6651

Answer

P = 0.6651
16.Probability

Answer: E(G) = 6 €

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: 30 (p = 0.15), 490 (p = 0.02), −10 (p = 0.83)

Expected value

E(G) = 30·0.15 + 490·0.02 − 10·0.83 = 6 €

Answer

E(G) = 6 €

A positive value means you win money on average.

17.Descriptive statistics

Answer: s = 12.99

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 = 187/5 = 37.4

Squared deviations

Σ(x − x̄)² = 18.6² + (−5.4)² + (−15.4)² + (−4.4)² + 6.6² = 675.2

Standard deviation

s = √(675.2/4) = 12.99

Answer

s = 12.99
18.Descriptive statistics

Answer: P = 0.9332

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 = −1.5

Standard normal table

Φ(−1.5) = 0.06681
P = 1 − 0.06681 = 0.9332

Answer

P = 0.9332
19.Normal distribution

Answer: z = −1.34

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 − 0.9099 = 0.0901

Read the z-table backwards

Find 0.0901 inside the table and read off z.

z = −1.34

Answer

z = −1.34
20.Normal distribution

Answer: x = 185.3 cm

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

Φ(z) = 0.85

z from the table

z ≈ 1.04

Back to the original units

x = 178 + 1.04·7 = 185.3 cm

Answer

x = 185.3 cm

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