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 + 1/x)x. Give a decimal value.

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

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

f(x) = (1 − cos 3x)/(3x²)x → 0
L =
3.DerivativesMedium
xyx = 2

Let f(x) = (3x − 3)4. Find f′(2).

f(x) = (3x − 3)4x = 2
f′(2) =
4.DerivativesHard
xy

Let f(x) = x3 + 4.5x2 + 6x + 4. Find the x-coordinate x of the local maximum.

f(x) = x3 + 4.5x2 + 6x + 4
x =
5.IntegralsMedium
xy3y = 0.5x

The region under y = 0.5x from x = 0 to x = 3 is rotated about the x-axis. Find the volume V of the solid (a cone).

y = 0.5x0 ≤ x ≤ 3
V =
6.IntegralsHard
xyy = 0.5x2​y = x

Find the area A of the region enclosed by y = 0.5x2 and y = x.

y = 0.5x2y = x
A =
7.Complex numbersMedium
ReIm22−2−244−4−466−6−6θz

Find the argument θ of z = −6 + j in degrees, as the principal value between −180° and 180°.

z = −6 + j
θ = °
8.Complex numbersHard
ReIm22−2−244−4−4z₁z₂

One root of z³ − 9z² + 43z − 75 = 0 is z₁ = 3 + j4. The coefficients are real. Find the real root r.

z₁ = 3 + j4
r =
9.Differential equationsMedium
tyy(0) = 3y(1.5) = ?

Solve y″ + 6y′ + 5y = 0 with y(0) = 3 and y′(0) = -4. Find y(1.5).

y″ + 6y′ + 5y = 0y(0) = 3y′(0) = -4
y(1.5) =
10.Differential equationsHard
tyK = 10ymax​ = ?

A system obeys y″ + y′ + 25.25y = 252.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.

y″ + y′ + 25.25y = 252.5y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
A =5−551−42343

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

A = [5 −5 5; 1 −4 2; 3 4 3]
det A =
12.Matrices & determinantsHard
A =51−2−5b =0−11

Solve the system 5x + y = 0 and −2x − 5y = −11. Find x.

5x + y = 0−2x − 5y = −11
x =
13.Vectors: dot & cross productMedium
xyzdFAB

A constant force F = (−25, 10, −5) N acts on an object while it moves in a straight line from point A(3, −2, −3) to point B(−5, 2, 1). Coordinates are in metres. Find the work W done by the force.

F = (−25, 10, −5) NA(3, −2, −3) mB(−5, 2, 1) m
W = J
14.Vectors: dot & cross productHard
xyzab

For a = (6, −4, −1) and b = (−5, 6, −3), find the z-component (a × b)z of the cross product a × b.

a = (6, −4, −1)b = (−5, 6, −3)
(a × b)z =
15.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6?7?8P(six) = 1/6 on each roll

A fair die is rolled 8 times. What is the probability P of getting at least one six?

n = 8
P =
16.ProbabilityHard
€p(x)−50.7350.25950.02

A game costs €5 to play. You win €10 with probability 0.25 and €100 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€10: p = 0.25€100: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
15202530354045

Find the sample standard deviation s of the data: 26, 15, 37, 41, 47.

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

Exam scores are normally distributed with mean 50 and standard deviation 5. Find the z-score z of a score of 52.

μ = 50σ = 5x = 52
z =
19.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9660

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9660. Find z to two decimals.

A = 0.9660
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 0.01 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 = 0.01 mm
P =

3Solutions

1.Limits

Answer: L = 2.718

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→∞ x·ln(1 + 1/x)

Evaluate

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

ln L = 1·1 = 1

Undo the logarithm

L = e1 = 2.718

Answer

L = 2.718
2.Limits

Answer: L = 1.5

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/(6x) → 0/0

Second application

9·cos 3x/6

Substitute x = 0

L = 9/6 = 1.5

Answer

L = 1.5
3.Derivatives

Answer: f′(2) = 324

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) = 4·(3x − 3)3·3

Substitute

f′(2) = 4·33·3 = 324

Answer

f′(2) = 324
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 + 9x + 6 = 3·(x2 + 3x + 2)

Solve f′(x) = 0

x = −2 or x = −1

Which one is which

f″(x) = 6x + 9: negative at a maximum, positive at a minimum.

f″(−2) = −3, f″(−1) = 3
x = −2

Answer

x = −2
5.Integrals

Answer: V = 7.069

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∫₀3 0.25x² dx

Integrate

V = π·0.25·3³/3 = 7.069

Answer

V = 7.069

Check: a cone of radius 1.5 and height 3 has (1/3)πr²h = 7.069.

6.Integrals

Answer: A = 0.6667

Definite integrals: ∫ₐᵇ f(x) dx is the signed area between the graph and the x-axis from a to b. Find an antiderivative F (with F′ = f) and evaluate F(b) − F(a). Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1). The area between two curves is ∫(upper − lower) dx.

Intersection points

0.5x2 = x → x = 0, x = 2

Integrate upper minus lower

A = [0.5x2 − 0.1667x3]02 = 0.6667

Answer

A = 0.6667
7.Complex numbers

Answer: θ = 170.5 °

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.

Quadrant

a = −6, b = 1 → II

Reference angle

α = atan(1/6) = 9.462°

Argument

θ = 180° − α = 170.5°

Answer

θ = 170.5°

In polar form z = 6.083∠170.5°.

8.Complex numbers

Answer: r = 3

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.

Conjugate pair

z₂ = 3 − j4
(z − z₁)(z − z₂) = z² − 2·3·z + (3² + 4²) = z² − 6z + 25

Factor out the quadratic

z³ − 9z² + 43z − 75 = (z² − 6z + 25)(z − 3)

Real root

r = 3

Answer

r = 3

Check with the sum of the roots: z₁ + z₂ + r = 6 + 3 = 9 = −(−9).

9.Differential equations

Answer: y(1.5) = 0.6137

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 + 1)(r + 5) = 0 → r = −1, −5

Constants from the initial values

A + B = 3, −1A − 5B = -4
A = 2.75, B = 0.25

Evaluate

y(1.5) = 2.75·e−1.5 + 0.25·e−7.5 = 0.6137

Answer

y(1.5) = 0.6137
10.Differential equations

Answer: ymax = 17.3

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 = 10, r = −0.5 ± 5j

Constants from the initial values

y(0) = 10 + A = 0 → A = −10
y′(0) = −0.5A + 5B = 0 → B = −1

Peak

t = π/5 = 0.6283
ymax = 10·(1 + e−0.5·π/5) = 17.3

Answer

ymax = 17.3
11.Matrices & determinants

Answer: det A = −35

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 = −20, M₁₂ = −3, M₁₃ = 16

Cofactor expansion

det A = 5·(−20) − (−5)·(−3) + 5·16 = −35

Answer

det A = −35
12.Matrices & determinants

Answer: x = −0.4783

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) − 1·(−2) = −23

Replace the x column

det Aₓ = 0·(−5) − 1·(−11) = 11
x = 11/(−23) = −0.4783

Answer

x = −0.4783
y = 2.391
13.Vectors: dot & cross product

Answer: W = 220 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 = (−5, 2, 1) − (3, −2, −3) = (−8, 4, 4) m

Work

W = F·d = (−25)·(−8) + 10·4 + (−5)·4 = 220 J

Answer

W = 220 J
14.Vectors: dot & cross product

Answer: (a × b)z = 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, component by component

(a × b)x = aybz − azby = (−4)·(−3) − (−1)·6 = 18
(a × b)y = azbx − axbz = (−1)·(−5) − 6·(−3) = 23
(a × b)z = axby − aybx = 6·6 − (−4)·(−5) = 16

Requested component

(a × b)z = 16

Answer

(a × b)z = 16

Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.

15.Probability

Answer: P = 0.7674

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)8 = 0.2326
P = 1 − 0.2326 = 0.7674

Answer

P = 0.7674
16.Probability

Answer: E(G) = −0.5 €

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: 5 (p = 0.25), 95 (p = 0.02), −5 (p = 0.73)

Expected value

E(G) = 5·0.25 + 95·0.02 − 5·0.73 = −0.5 €

Answer

E(G) = −0.5 €

A negative value means you lose money on average.

17.Descriptive statistics

Answer: s = 12.74

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 = 166/5 = 33.2

Squared deviations

Σ(x − x̄)² = (−7.2)² + (−18.2)² + 3.8² + 7.8² + 13.8² = 648.8

Standard deviation

s = √(648.8/4) = 12.74

Answer

s = 12.74
18.Descriptive statistics

Answer: z = 0.4

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 = (52 − 50)/5 = 0.4

Answer

z = 0.4
19.Normal distribution

Answer: z = 2.12

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 + 0.9660)/2 = 0.9830

Read the z-table backwards

Find 0.9830 inside the table and read off z.

z = 2.12

Answer

z = 2.12
20.Normal distribution

Answer: P = 0.9207

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 = 0.01414 mm

Standardise

z = (0 − 0.02)/0.01414 = −1.41

Probability

P = 1 − Φ(−1.41) = 0.9207

Answer

P = 0.9207

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