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 + 2/x)2x. Give a decimal value.

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

Find the limit L = limx→0 (1 − cos 5x)/(x²), with x in radians.

f(x) = (1 − cos 5x)/(x²)x → 0
L =
3.DerivativesMedium
xyx = −2

Let f(x) = ln(3x2 + 1). Find f′(−2).

f(x) = ln(3x2 + 1)x = −2
f′(−2) =
4.DerivativesHard
xyx = 3

The tangent line to f(x) = 2x2 + 4x − 4 at x = 3 crosses the y-axis at y = b. Find b.

f(x) = 2x2 + 4x − 4x = 3
b =
5.IntegralsMedium
xy01

Evaluate I = ∫01 2x·(x² + 1)1 dx.

f(x) = 2x·(x² + 1)1[0, 1]
I =
6.IntegralsHard
xy1.5y = x²eˣ

Evaluate I = ∫₀1.5 x²·eˣ dx.

∫₀1.5 x²eˣ dx
I =
7.Complex numbersMedium
ReIm22−2−244−4−4θz

Write z = −3 − j5 in exponential form r·ejθ. Give θ in radians, between −π and π.

z = −3 − j5
θ = rad
8.Complex numbersHard
ReIm22−2−244−4−4z₁z₂

Solve z² − 4z + 20 = 0. The roots are a ± jb; find the real part a.

p = −4q = 20
a =
9.Differential equationsMedium
t (min)T (°C)29Ta​30 °C

A cup of coffee at 80 °C is left in a room at 19 °C. After 29 min its temperature is 30 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 80 °CTa = 19 °CT(29) = 30 °C
k = min⁻¹
10.Differential equationsHard
tyK = 1ymax​ = ?

A system obeys y″ + y′ + 16.25y = 16.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.

y″ + y′ + 16.25y = 16.25y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
A =5−5−5−5−2−1−20−4

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

A = [5 −5 −5; −5 −2 −1; −2 0 −4]
det A =
12.Matrices & determinantsHard
A =−4−2−43b =05

Solve the system −4x − 2y = 0 and −4x + 3y = 5. Find x.

−4x − 2y = 0−4x + 3y = 5
x =
13.Vectors: dot & cross productMedium
xyzab

Find the angle θ between the vectors a = (−3, −3, 6) and b = (−4, 0, −1), in degrees.

a = (−3, −3, 6)b = (−4, 0, −1)
θ = °
14.Vectors: dot & cross productHard
xyzABC

Find the area A of the triangle with corners A(−2, 3, 3), B(−3, 2, −1) and C(0, 2, 2). Coordinates are in metres.

A(−2, 3, 3)B(−3, 2, −1)C(0, 2, 2)
A = m²
15.ProbabilityMedium
red: 5blue: 6

A bag holds 5 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

5 / 6
P =
16.ProbabilityHard
€p(x)−100.78100.21900.02

A game costs €10 to play. You win €20 with probability 0.2 and €200 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €10€20: p = 0.2€200: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 57, 38, 31, 22, 36, 46.

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

Exam scores are normally distributed with mean 100 and standard deviation 15. Find the z-score z of a score of 88.

μ = 100σ = 15x = 88
z =
19.Normal distributionMedium
5785.5P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 57 and 85.5 points? Give P as a decimal.

μ = 65σ = 10x1 = 57x2 = 85.5
P =
20.Normal distributionHard
μ = 500x = ?5%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 5% of them lie (the 5th percentile).

μ = 500σ = 40P = 5%
x = h

3Solutions

1.Limits

Answer: L = 54.6

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→∞ 2·x·ln(1 + 2/x)

Evaluate

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

ln L = 2·2 = 4

Undo the logarithm

L = e4 = 54.6

Answer

L = 54.6
2.Limits

Answer: L = 12.5

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

5·sin 5x/(2x) → 0/0

Second application

25·cos 5x/2

Substitute x = 0

L = 25/2 = 12.5

Answer

L = 12.5
3.Derivatives

Answer: f′(−2) = −0.9231

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 for the logarithm

f′(x) = 6x/(3x2 + 1)

Substitute

f′(−2) = −12/13 = −0.9231

Answer

f′(−2) = −0.9231
4.Derivatives

Answer: b = −22

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(3) = 26
f′(x) = 4x + 4 → f′(3) = 16

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 26 − 16·3 = −22

Answer

b = −22
5.Integrals

Answer: I = 1.5

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.

Substitution

u = x² + 1, du = 2x dx
I = ∫12 u1 du

Integrate and evaluate

I = [u2/2]12 = (4 − 1)/2 = 1.5

Answer

I = 1.5
6.Integrals

Answer: I = 3.602

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

4.482·(1.25) − 1·2 = 3.602

Answer

I = 3.602
7.Complex numbers

Answer: θ = −2.111 rad

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

r = √(9 + 25) = 5.831

Argument in radians

θ = −121° = −121·π/180 = −2.111 rad

Exponential form

z = 5.831·e−j2.111

Answer

θ = −2.111 rad
8.Complex numbers

Answer: a = 2

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.

Discriminant

p² − 4q = 16 − 80 = −64 < 0

Quadratic formula

z = (4 ± j√64)/2 = 2 + j4, 2 − j4

Answer

a = 2

Answer

a = 2
9.Differential equations

Answer: k = 0.05907 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 = 11/61

Solve for k

k = ln(61/11)/29 = 0.05907 min⁻¹

Answer

k = 0.05907 min⁻¹
10.Differential equations

Answer: ymax = 1.675

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 = −0.5 ± 4j

Constants from the initial values

y(0) = 1 + A = 0 → A = −1
y′(0) = −0.5A + 4B = 0 → B = −0.125

Peak

t = π/4 = 0.7854
ymax = 1·(1 + e−0.5·π/4) = 1.675

Answer

ymax = 1.675
11.Matrices & determinants

Answer: det A = 150

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 = 8, M₁₂ = 18, M₁₃ = −4

Cofactor expansion

det A = 5·8 − (−5)·18 + (−5)·(−4) = 150

Answer

det A = 150
12.Matrices & determinants

Answer: x = −0.5

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 = (−4)·3 − (−2)·(−4) = −20

Replace the x column

det Aₓ = 0·3 − (−2)·5 = 10
x = 10/(−20) = −0.5

Answer

x = −0.5
y = 1
13.Vectors: dot & cross product

Answer: θ = 78.58 °

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)·(−4) + (−3)·0 + 6·(−1) = 6

Magnitudes

|a| = √((−3)² + (−3)² + 6²) = 7.348
|b| = √((−4)² + 0² + (−1)²) = 4.123

Angle

cos θ = 6/(7.348·4.123) = 0.198
θ = arccos(0.198) = 78.58°

Answer

θ = 78.58°
14.Vectors: dot & cross product

Answer: A = 4.975 m²

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.

Edge vectors

AB = B − A = (−1, −1, −4)
AC = C − A = (2, −1, −1)

Cross product

(AB × AC)x = AByACz − ABzACy = (−1)·(−1) − (−4)·(−1) = −3
(AB × AC)y = ABzACx − ABxACz = (−4)·2 − (−1)·(−1) = −9
(AB × AC)z = ABxACy − AByACx = (−1)·(−1) − (−1)·2 = 3
|AB × AC| = √((−3)² + (−9)² + 3²) = 9.95

Area of the triangle

A = ½·9.95 = 4.975 m²

Answer

A = 4.975 m²
15.Probability

Answer: P = 0.1818

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 = 5/11 · 4/10 = 20/110 = 0.1818

Answer

P = 0.1818
16.Probability

Answer: E(G) = −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: 10 (p = 0.2), 190 (p = 0.02), −10 (p = 0.78)

Expected value

E(G) = 10·0.2 + 190·0.02 − 10·0.78 = −2 €

Answer

E(G) = −2 €

A negative value means you lose money on average.

17.Descriptive statistics

Answer: s = 12.11

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 = 230/6 = 38.33

Squared deviations

Σ(x − x̄)² = 18.67² + (−0.3333)² + (−7.333)² + (−16.33)² + (−2.333)² + 7.667² = 733.3

Standard deviation

s = √(733.3/5) = 12.11

Answer

s = 12.11
18.Descriptive statistics

Answer: z = −0.8

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 = (88 − 100)/15 = −0.8

Answer

z = −0.8
19.Normal distribution

Answer: P = 0.768

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-scores

z₁ = (57 − 65)/10 = −0.8
z₂ = (85.5 − 65)/10 = 2.05

Area between

P = Φ(2.05) − Φ(−0.8) = 0.9798 − 0.2119 = 0.768

Answer

P = 0.768
20.Normal distribution

Answer: x = 434.2 h

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) = 0.05 → z = −1.645

Back to x

x = 500 + −1.645·40 = 434.2 h

Answer

x = 434.2 h

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