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)x. Give a decimal value.

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

Find the limit L = limx→0 ln(1 + 5x)/sin(2x), with x in radians.

f(x) = ln(1 + 5x)/sin(2x)x → 0
L =
3.DerivativesMedium
xyx = 1

Let f(x) = x·e2x. Find f′(1).

f(x) = x·e2xx = 1
f′(1) =
4.DerivativesHard
xy

Let f(x) = x3 − 13.5x2 + 54x − 2. Find the x-coordinate x of the local minimum.

f(x) = x3 − 13.5x2 + 54x − 2
x =
5.IntegralsMedium
xy12y = x ln x

Evaluate I = ∫₁2 x·ln x dx.

∫₁2 x ln x dx
I =
6.IntegralsHard
xy0.5y = x²eˣ

Evaluate I = ∫₀0.5 x²·eˣ dx.

∫₀0.5 x²eˣ dx
I =
7.Complex numbersMedium
ReIm22−2−244−4−466−6−6θz

Find the argument θ of z = −6 + j3 in degrees, as the principal value between −180° and 180°.

z = −6 + j3
θ = °
8.Complex numbersHard
ReIm11−1−122−2−2θz

Use De Moivre's theorem to find the real part of z³ for z = −2 − j2.

z = −2 − j2n = 3
Re(z³) =
9.Differential equationsMedium
tyy(0) = 5y(1) = ?

Solve y″ + 6y′ + 9y = 0 with y(0) = 5 and y′(0) = 2. Find y(1).

y″ + 6y′ + 9y = 0y(0) = 5y′(0) = 2
y(1) =
10.Differential equationsHard
xy12(1, 2)y(2) = ?

Solve the initial value problem y′ + 2y/x = x2 for x > 0, with y(1) = 2. Find y(2).

y′ + 2y/x = x2y(1) = 2
y(2) =
11.Matrices & determinantsMedium
A =022−5B =21−5−1

Let C = AB for the matrices shown. Find the entry c22 (row 2, column 2).

A = [0 2; 2 −5]B = [2 1; −5 −1]
c22 =
12.Matrices & determinantsHard
A =6−3−4−4b =92

Solve the system 6x − 3y = 9 and −4x − 4y = 2. Find x.

6x − 3y = 9−4x − 4y = 2
x =
13.Vectors: dot & cross productMedium
xyzab

Find the dot product a·b of a = (2, 5, 2) and b = (3, 7, −6).

a = (2, 5, 2)b = (3, 7, −6)
a·b =
14.Vectors: dot & cross productHard
xyzabc

Find the volume V of the parallelepiped whose edges from one corner are a = (0, −4, 2), b = (−2, 2, −2) and c = (−5, 0, 1).

a = (0, −4, 2)b = (−2, 2, −2)c = (−5, 0, 1)
V =
15.ProbabilityMedium
n = 10k = 2N = ?

In how many ways N can a committee of 2 people be chosen from a group of 10? The order does not matter.

n = 10k = 2
N =
16.ProbabilityHard
n = 6, p = 0.25: P(X ≤ 1) = ?k0123456

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

n = 6p = 0.25
P =
17.Descriptive statisticsMedium
303540455055

Find the sample standard deviation s of the data: 36, 59, 40, 30, 46, 35.

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

Exam scores are normally distributed with mean 170 and standard deviation 15. Find the z-score z of a score of 152.

μ = 170σ = 15x = 152
z =
19.Normal distributionMedium
480 g486.2 gT = X₁ + … + X₁₂?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 12 of them is below 486.2 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 12< 486.2 g
P =
20.Normal distributionHard
μ = 178x = ?80%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 80% of them lie (the 80th percentile).

μ = 178σ = 7P = 80%
x = cm

3Solutions

1.Limits

Answer: L = 0.1353

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 − 2/x)

Evaluate

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

ln L = 1·(−2) = −2

Undo the logarithm

L = e−2 = 0.1353

Answer

L = 0.1353
2.Limits

Answer: L = 2.5

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).

L'Hôpital's rule

L = limx→0 [5/(1 + 5x)]/[2·cos 2x]

Substitute x = 0

L = 5/2 = 2.5

Answer

L = 2.5
3.Derivatives

Answer: f′(1) = 22.17

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 rule

f′(x) = u′v + uv′ = e2x·(1 + 2·x)

Substitute

f′(1) = 7.389·(1 + 2) = 22.17

Answer

f′(1) = 22.17
4.Derivatives

Answer: x = 6

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 − 27x + 54 = 3·(x2 − 9x + 18)

Solve f′(x) = 0

x = 3 or x = 6

Which one is which

f″(x) = 6x − 27: negative at a maximum, positive at a minimum.

f″(3) = −9, f″(6) = 9
x = 6

Answer

x = 6
5.Integrals

Answer: I = 0.6363

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 = ln x, du = dx/x; dv = x dx, v = x²/2

Antiderivative

∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4

Limits

(2·ln 2 − 1) − (0 − 0.25) = 0.6363

Answer

I = 0.6363
6.Integrals

Answer: I = 0.0609

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

1.649·(1.25) − 1·2 = 0.0609

Answer

I = 0.0609
7.Complex numbers

Answer: θ = 153.4 °

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 = 3 → II

Reference angle

α = atan(3/6) = 26.57°

Argument

θ = 180° − α = 153.4°

Answer

θ = 153.4°

In polar form z = 6.708∠153.4°.

8.Complex numbers

Answer: Re(z³) = 16

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.

Polar form

z = −2 − j2 = 2√2∠−135°

De Moivre

z³ = (2√2)³∠−405° = 22.63∠−45°

Rectangular form

z³ = 22.63·(cos −45° + j sin −45°) = 16 − j16
Re(z³) = 16

Answer

Re(z³) = 16
9.Differential equations

Answer: y(1) = 1.095

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 + 3)² = 0 → r = −3

Constants

A = y(0) = 5
y′(0) = B − 3A → B = 2 + 15 = 17

Evaluate

y(1) = (5 + 17·1)·e−3 = 1.095

Answer

y(1) = 1.095
10.Differential equations

Answer: y(2) = 2.05

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.

Integrating factor

μ = e∫2/x dx = x2

Integrate

(x2y)′ = x4 → x2y = 1x5/5 + C

Initial condition

2 = 1/5 + C → C = 1.8

Evaluate

y(2) = (1·25/5 + 1.8)/22 = 2.05

Answer

y(2) = 2.05
11.Matrices & determinants

Answer: c22 = 7

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

c22 = 2·1 + (−5)·(−1) = 7

Answer

c22 = 7
12.Matrices & determinants

Answer: x = 0.8333

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 = 6·(−4) − (−3)·(−4) = −36

Replace the x column

det Aₓ = 9·(−4) − (−3)·2 = −30
x = −30/(−36) = 0.8333

Answer

x = 0.8333
y = −1.333
13.Vectors: dot & cross product

Answer: a·b = 29

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.

Multiply matching components and add

a·b = 2·3 + 5·7 + 2·(−6) = 6 + 35 + (−12) = 29

Answer

a·b = 29

Positive: the angle between the vectors is less than 90°.

14.Vectors: dot & cross product

Answer: V = 28

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 b × c

(b × c)x = bycz − bzcy = 2·1 − (−2)·0 = 2
(b × c)y = bzcx − bxcz = (−2)·(−5) − (−2)·1 = 12
(b × c)z = bxcy − bycx = (−2)·0 − 2·(−5) = 10

Dot product with a

a·(b × c) = 0·2 + (−4)·12 + 2·10 = −28
V = |−28| = 28

Answer

V = 28
15.Probability

Answer: N = 45

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.

Combinations

C(10, 2) = (10·9)/(2·1) = 45

Answer

N = 45
16.Probability

Answer: P = 0.5339

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) = 0.756 = 0.178
P(X = 1) = 6·0.25·0.755 = 0.356

Add them

P = 0.178 + 0.356 = 0.5339

Answer

P = 0.5339
17.Descriptive statistics

Answer: s = 10.32

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 = 246/6 = 41

Squared deviations

Σ(x − x̄)² = (−5)² + 18² + (−1)² + (−11)² + 5² + (−6)² = 532

Standard deviation

s = √(532/5) = 10.32

Answer

s = 10.32
18.Descriptive statistics

Answer: z = −1.2

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 = (152 − 170)/15 = −1.2

Answer

z = −1.2
19.Normal distribution

Answer: P = 0.9319

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 total

μT = 480 g, σT = 1.2·√12 = 4.157 g

Standardise

z = (486.2 − 480)/4.157 = 1.49

Probability

P = Φ(1.49) = 0.9319

Answer

P = 0.9319
20.Normal distribution

Answer: x = 183.9 cm

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) = 0.8 → z = 0.8416

Back to x

x = 178 + 0.8416·7 = 183.89 cm

Answer

x = 183.89 cm

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