Engineering Mathematics: cram pack

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

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 + 8x) − 2x).

f(x) = √(4x2 + 8x) − 2xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (1 − cos 3x)/(2x²), with x in radians.

f(x) = (1 − cos 3x)/(2x²)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 + 7.5x2 + 12x + 3. Find the x-coordinate x of the local minimum.

f(x) = x3 + 7.5x2 + 12x + 3
x =
5.IntegralsMedium
xy2y = x·sin x

Evaluate I = ∫₀2 x·sin x dx (x in radians).

∫₀2 x·sin x dx
I =
6.IntegralsHard
xy1.667y = 5xy = 3x²

The region between the line y = 5x and the parabola y = 3x² is rotated about the x-axis. Find the volume V of the solid.

y = 5xy = 3x²
V =
7.Complex numbersMedium
ReIm22−2−244−4−4θz

Write z = 6∠−130° in rectangular form a + jb and give the real part a.

r = 6θ = −130°
a =
8.Complex numbersHard
ReIm11−1−122−2−233−3−3θz

Find the real part of (3 − j3)³.

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

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

y″ + 2y′ + y = 0y(0) = 3y′(0) = 0
y(2) =
10.Differential equationsHard
tyK = 1y(0.5) = ?

A system obeys y″ + 4y′ + 13y = 13 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.5); angles in radians.

y″ + 4y′ + 13y = 13y(0) = 0y′(0) = 0
y(0.5) =
11.Matrices & determinantsMedium
A =435451−2−3−5

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

A = [4 3 5; 4 5 1; −2 −3 −5]
det A =
12.Matrices & determinantsHard
A =−2−43−4b =−12−9

Solve the system −2x − 4y = −12 and 3x − 4y = −9. Find x.

−2x − 4y = −123x − 4y = −9
x =
13.Vectors: dot & cross productMedium
xyzdFAB

A constant force F = (0, 15, 25) N acts on an object while it moves in a straight line from point A(1, 3, 4) to point B(0, −3, 2). Coordinates are in metres. Find the work W done by the force.

F = (0, 15, 25) NA(1, 3, 4) mB(0, −3, 2) m
W = J
14.Vectors: dot & cross productHard
xyzab

The unit vector n̂ is perpendicular to both a = (4, −2, −4) and b = (2, 5, −4) and has a positive z-component. Find its y-component n̂y.

a = (4, −2, −4)b = (2, 5, −4)n̂z > 0
n̂y =
15.ProbabilityMedium
red: 8blue: 4

A bag holds 8 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

8 / 4
P =
16.ProbabilityHard
€p(x)−20.8940.1380.01

A game costs €2 to play. You win €6 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€6: p = 0.1€40: p = 0.01
E(G) = €
17.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 15, 26, 33, 26, 20, 58, 42.

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

Exam scores are normally distributed with mean 170 and standard deviation 12. Find the z-score z of a score of 185.

μ = 170σ = 12x = 185
z =
19.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8764

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8764. Find z to two decimals.

A = 0.8764
z =
20.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =

3Solutions

1.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).

Multiply by the conjugate

f(x) = 8x/(√(4x2 + 8x) + 2x)

Divide by x

f(x) = 8/(√(4 + 8/x) + 2)

Let x → ∞

L = 8/(2 + 2) = 2

Answer

L = 2
2.Limits

Answer: L = 2.25

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/(4x) → 0/0

Second application

9·cos 3x/4

Substitute x = 0

L = 9/4 = 2.25

Answer

L = 2.25
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 = −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 + 15x + 12 = 3·(x2 + 5x + 4)

Solve f′(x) = 0

x = −4 or x = −1

Which one is which

f″(x) = 6x + 15: negative at a maximum, positive at a minimum.

f″(−4) = −9, f″(−1) = 9
x = −1

Answer

x = −1
5.Integrals

Answer: I = 1.742

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 = sin x dx, v = −cos x

Antiderivative

∫x sin x dx = −x cos x + ∫cos x dx = −x cos x + sin x

Limits

(−2·cos 2 + sin 2) − (0 + 0) = 1.742

Answer

I = 1.742
6.Integrals

Answer: V = 48.48

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

5x = 3x² → x = 0, x = 1.667

Washer method

V = π∫₀1.667 ((5x)² − (3x²)²) dx

Integrate

V = π(25·1.667³/3 − 9·1.667⁵/5) = π·15.43 = 48.48

Answer

V = 48.48
7.Complex numbers

Answer: a = −3.857

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 = 6·cos(−130°) = −3.857
b = 6·sin(−130°) = −4.596

Rectangular form

z = −3.857 − j4.596
a = −3.857

Answer

a = −3.857
8.Complex numbers

Answer: Re(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∠−45° → z³ = 76.37∠−135°
Re(z³) = −54

Answer

Re(z³) = −54
9.Differential equations

Answer: y(2) = 1.218

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 + 1)² = 0 → r = −1

Constants

A = y(0) = 3
y′(0) = B − 1A → B = 0 + 3 = 3

Evaluate

y(2) = (3 + 3·2)·e−2 = 1.218

Answer

y(2) = 1.218
10.Differential equations

Answer: y(0.5) = 0.7293

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 ± 3j

Constants from the initial values

y(0) = 1 + A = 0 → A = −1
y′(0) = −2A + 3B = 0 → B = −0.6667

Evaluate

y(0.5) = 1·(1 − e−1(cos 1.5 + 0.6667·sin 1.5)) = 0.7293

Answer

y(0.5) = 0.7293
11.Matrices & determinants

Answer: det A = −44

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 = −22, M₁₂ = −18, M₁₃ = −2

Cofactor expansion

det A = 4·(−22) − 3·(−18) + 5·(−2) = −44

Answer

det A = −44
12.Matrices & determinants

Answer: x = 0.6

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 = (−2)·(−4) − (−4)·3 = 20

Replace the x column

det Aₓ = (−12)·(−4) − (−4)·(−9) = 12
x = 12/20 = 0.6

Answer

x = 0.6
y = 2.7
13.Vectors: dot & cross product

Answer: W = −140 J

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.

Displacement

d = B − A = (0, −3, 2) − (1, 3, 4) = (−1, −6, −2) m

Work

W = F·d = 0·(−1) + 15·(−6) + 25·(−2) = −140 J

Answer

W = −140 J

Negative work: the force acts mostly against the motion.

14.Vectors: dot & cross product

Answer: n̂y = 0.212

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)·(−4) − (−4)·5 = 28
(a × b)y = azbx − axbz = (−4)·2 − 4·(−4) = 8
(a × b)z = axby − aybx = 4·5 − (−2)·2 = 24

Normalise

The z-component of a × b is already positive.

|a × b| = √(28² + 8² + 24²) = 37.74
n̂ = (28, 8, 24)/37.74 = (0.742, 0.212, 0.636)
n̂y = 0.212

Answer

n̂y = 0.212
15.Probability

Answer: P = 0.4242

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 = 8/12 · 7/11 = 56/132 = 0.4242

Answer

P = 0.4242
16.Probability

Answer: E(G) = −1 €

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

Expected value

E(G) = 4·0.1 + 38·0.01 − 2·0.89 = −1 €

Answer

E(G) = −1 €

A negative value means you lose money on average.

17.Descriptive statistics

Answer: s = 14.6

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 = 220/7 = 31.43

Squared deviations

Σ(x − x̄)² = (−16.43)² + (−5.429)² + 1.571² + (−5.429)² + (−11.43)² + 26.57² + 10.57² = 1280

Standard deviation

s = √(1280/6) = 14.6

Answer

s = 14.6
18.Descriptive statistics

Answer: z = 1.25

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 = (185 − 170)/12 = 1.25

Answer

z = 1.25
19.Normal distribution

Answer: z = 1.54

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 + 0.8764)/2 = 0.9382

Read the z-table backwards

Find 0.9382 inside the table and read off z.

z = 1.54

Answer

z = 1.54
20.Normal distribution

Answer: P = 0.7389

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 = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389

ExamProblems · examproblems.com/courses/engineering-mathematics