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 ln(1 + 2x)/sin(3x), with x in radians.

f(x) = ln(1 + 2x)/sin(3x)x → 0
L =
3.DerivativesMedium
xyx = 1

Let f(x) = ln(2x2 + 5). Find f′(1).

f(x) = ln(2x2 + 5)x = 1
f′(1) =
4.DerivativesHard
xy

Let f(x) = x3 + 1.5x2 − 2. Find the x-coordinate x of the local maximum.

f(x) = x3 + 1.5x2 − 2
x =
5.IntegralsMedium
xy14

Evaluate I = ∫14 4/x dx.

f(x) = 4/x[1, 4]
I =
6.IntegralsHard
xy2.5y = x²eˣ

Evaluate I = ∫₀2.5 x²·eˣ dx.

∫₀2.5 x²eˣ dx
I =
7.Complex numbersMedium
ReIm22−2−244−4−4θz

Write z = 4∠65° in rectangular form a + jb and give the imaginary part b.

r = 4θ = 65°
b =
8.Complex numbersHard
ReIm11−1−122−2−233−3−3z₁z₂

Solve z² + 3z + 11.25 = 0. The roots are a ± jb; find the real part a.

p = 3q = 11.25
a =
9.Differential equationsMedium
tyy(0) = 1y(1) = ?

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

y″ + 4y′ + 4y = 0y(0) = 1y′(0) = 4
y(1) =
10.Differential equationsHard
tyy(0) = 2y(0.5) = ?

Solve y″ + 2y′ + 26y = 0 with y(0) = 2 and y′(0) = 0 (released from rest). Find y(0.5); angles in radians.

y″ + 2y′ + 26y = 0y(0) = 2y′(0) = 0
y(0.5) =
11.Matrices & determinantsMedium
A =2−2−2−2−151−4−5

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

A = [2 −2 −2; −2 −1 5; 1 −4 −5]
det A =
12.Matrices & determinantsHard
A =3−2−5−6b =4−2

Solve the system 3x − 2y = 4 and −5x − 6y = −2. Find x.

3x − 2y = 4−5x − 6y = −2
x =
13.Vectors: dot & cross productMedium
xyzdFAB

A constant force F = (20, −25, 30) N acts on an object while it moves in a straight line from point A(−3, 5, −1) to point B(−1, 0, 0). Coordinates are in metres. Find the work W done by the force.

F = (20, −25, 30) NA(−3, 5, −1) mB(−1, 0, 0) m
W = J
14.Vectors: dot & cross productHard
xyzABC

Find the area A of the triangle with corners A(4, 0, 4), B(−2, −2, 4) and C(−3, −4, −4). Coordinates are in metres.

A(4, 0, 4)B(−2, −2, 4)C(−3, −4, −4)
A = m²
15.ProbabilityMedium
P(at least one six) = ??1?2?3P(six) = 1/6 on each roll

A fair die is rolled 3 times. What is the probability P of getting at least one six?

n = 3
P =
16.ProbabilityHard
n = 9, p = 0.25: P(X ≤ 1) = ?k0123456789

9 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?

n = 9p = 0.25
P =
17.Descriptive statisticsMedium
15202530354045

Find the sample standard deviation s of the data: 23, 28, 22, 49, 14, 45.

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

A quantity is normally distributed with mean 60 and standard deviation 20. What proportion P of values is above 30?

μ = 60σ = 20x = 30
P =
19.Normal distributionMedium
96.25115.75P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 96.25 and 115.75 points? Give P as a decimal.

μ = 100σ = 15x1 = 96.25x2 = 115.75
P =
20.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =

3Solutions

1.Limits

Answer: L = 0.3679

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 = e−1 = 0.3679

Answer

L = 0.3679
2.Limits

Answer: L = 0.6667

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).

L'Hôpital's rule

L = limx→0 [2/(1 + 2x)]/[3·cos 3x]

Substitute x = 0

L = 2/3 = 0.6667

Answer

L = 0.6667
3.Derivatives

Answer: f′(1) = 0.5714

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) = 4x/(2x2 + 5)

Substitute

f′(1) = 4/7 = 0.5714

Answer

f′(1) = 0.5714
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 + 3x = 3·(x2 + x)

Solve f′(x) = 0

x = −1 or x = 0

Which one is which

f″(x) = 6x + 3: negative at a maximum, positive at a minimum.

f″(−1) = −3, f″(0) = 3
x = −1

Answer

x = −1
5.Integrals

Answer: I = 5.545

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.

Antiderivative

F(x) = 4·ln x

Evaluate at the limits

I = 4·(ln 4 − ln 1) = 4·ln 4 = 5.545

Answer

I = 5.545
6.Integrals

Answer: I = 37.59

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

12.18·(3.25) − 1·2 = 37.59

Answer

I = 37.59
7.Complex numbers

Answer: b = 3.625

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 = 4·cos(65°) = 1.69
b = 4·sin(65°) = 3.625

Rectangular form

z = 1.69 + j3.625
b = 3.625

Answer

b = 3.625
8.Complex numbers

Answer: a = −1.5

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 = 9 − 45 = −36 < 0

Quadratic formula

z = (−3 ± j√36)/2 = −1.5 + j3, −1.5 − j3

Answer

a = −1.5

Answer

a = −1.5
9.Differential equations

Answer: y(1) = 0.9473

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 = 4 + 2 = 6

Evaluate

y(1) = (1 + 6·1)·e−2 = 0.9473

Answer

y(1) = 0.9473
10.Differential equations

Answer: y(0.5) = −0.8266

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.

Complex roots

r = −1 ± 5j

Constants

A = 2, B = 1·2/5 = 0.4

Evaluate

y(0.5) = e−0.5(2·cos 2.5 + 0.4·sin 2.5) = −0.8266

Answer

y(0.5) = −0.8266
11.Matrices & determinants

Answer: det A = 42

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 = 25, M₁₂ = 5, M₁₃ = 9

Cofactor expansion

det A = 2·25 − (−2)·5 + (−2)·9 = 42

Answer

det A = 42
12.Matrices & determinants

Answer: x = 1

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·(−6) − (−2)·(−5) = −28

Replace the x column

det Aₓ = 4·(−6) − (−2)·(−2) = −28
x = −28/(−28) = 1

Answer

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

Answer: W = 195 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 = (−1, 0, 0) − (−3, 5, −1) = (2, −5, 1) m

Work

W = F·d = 20·2 + (−25)·(−5) + 30·1 = 195 J

Answer

W = 195 J
14.Vectors: dot & cross product

Answer: A = 25.79 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 = (−6, −2, 0)
AC = C − A = (−7, −4, −8)

Cross product

(AB × AC)x = AByACz − ABzACy = (−2)·(−8) − 0·(−4) = 16
(AB × AC)y = ABzACx − ABxACz = 0·(−7) − (−6)·(−8) = −48
(AB × AC)z = ABxACy − AByACx = (−6)·(−4) − (−2)·(−7) = 10
|AB × AC| = √(16² + (−48)² + 10²) = 51.58

Area of the triangle

A = ½·51.58 = 25.79 m²

Answer

A = 25.79 m²
15.Probability

Answer: P = 0.4213

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)3 = 0.5787
P = 1 − 0.5787 = 0.4213

Answer

P = 0.4213
16.Probability

Answer: P = 0.3003

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.

Zero and one failure

P(X = 0) = 0.759 = 0.07508
P(X = 1) = 9·0.25·0.758 = 0.2253

Add them

P = 0.07508 + 0.2253 = 0.3003

Answer

P = 0.3003
17.Descriptive statistics

Answer: s = 13.85

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 = 181/6 = 30.17

Squared deviations

Σ(x − x̄)² = (−7.167)² + (−2.167)² + (−8.167)² + 18.83² + (−16.17)² + 14.83² = 958.8

Standard deviation

s = √(958.8/5) = 13.85

Answer

s = 13.85
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: P = 0.4518

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 = (96.25 − 100)/15 = −0.25
z₂ = (115.75 − 100)/15 = 1.05

Area between

P = Φ(1.05) − Φ(−0.25) = 0.8531 − 0.4013 = 0.4518

Answer

P = 0.4518
20.Normal distribution

Answer: P = 0.7881

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 = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881

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