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→∞ (5x2 − 3x + 3)/(5x2 + 2x + 1).

f(x) = (5x2 − 3x + 3)/(5x2 + 2x + 1)x → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (1 − cos 2x)/(4x²), with x in radians.

f(x) = (1 − cos 2x)/(4x²)x → 0
L =
3.DerivativesMedium
xyx = 1

Let f(x) = x2·ex. Find f′(1).

f(x) = x2·exx = 1
f′(1) =
4.DerivativesHard
xy

Let f(x) = x3 − 6x2 + 3. Find the x-coordinate x of the local minimum.

f(x) = x3 − 6x2 + 3
x =
5.IntegralsMedium
xy1y = √(2x)

The region under y = √(2x) from x = 0 to x = 1 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).

y = √(2x)0 ≤ x ≤ 1
V =
6.IntegralsHard
xy−11

Find the area A of the region between the parabola y = 1 − x² and the x-axis.

y = 1 − x²
A =
7.Complex numbersMedium
ReIm22−2−244−4−4z₁z₂w

Given z₁ = 2∠70° and z₂ = 2∠−95°, find the real part of w = z₁·z₂.

z₁ = 2∠70°z₂ = 2∠−95°
Re(w) =
8.Complex numbersHard
ReIm0.50.5−0.5−0.511−1−1θz

Use De Moivre's theorem to find the imaginary part of z⁷ for z = 1 + j.

z = 1 + jn = 7
Im(z⁷) =
9.Differential equationsMedium
tyy(0) = 1y(1) = ?

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

y″ + 2y′ + y = 0y(0) = 1y′(0) = 0
y(1) =
10.Differential equationsHard
t (min)m (kg)t = ?23 kg

A tank holds 400 L of water in which 5 kg of salt is dissolved. Brine with 0.1 kg of salt per litre flows in at 4 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 23 kg of salt?

V = 400 Lm0 = 5 kgc = 0.1 kg/Lr = 4 L/min
t = min
11.Matrices & determinantsMedium
A =1−1−11−43451

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

A = [1 −1 −1; 1 −4 3; 4 5 1]
det A =
12.Matrices & determinantsHard
A =−6−42−1b =−33

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

−6x − 4y = −32x − y = 3
x =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (4, −4, −2) onto b = (5, −6, 2), that is, the component of a in the direction of b.

a = (4, −4, −2)b = (5, −6, 2)
compb a =
14.Vectors: dot & cross productHard
xyzab

For a = (6, −3, 2) and b = (4, 4, −3), find the y-component (a × b)y of the cross product a × b.

a = (6, −3, 2)b = (4, 4, −3)
(a × b)y =
15.ProbabilityMedium
red: 3blue: 8

A bag holds 3 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

3 / 8
P =
16.ProbabilityHard
€p(x)−20.6560.3980.05

A game costs €2 to play. You win €8 with probability 0.3 and €100 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €2€8: p = 0.3€100: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
253035404550

Find the sample standard deviation s of the data: 34, 38, 28, 21, 32, 51.

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

A quantity is normally distributed with mean 70 and standard deviation 15. What proportion P of values is below 79?

μ = 70σ = 15x = 79
P =
19.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.7887

Drag z so that the shaded area between −z and z under the standard normal curve is 0.7887. Find z to two decimals.

A = 0.7887
z =
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 = 1

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).

Divide by the highest power

(5 + (−3)/x + 3/x2)/(5 + 2/x + 1/x2)

Let x → ∞

L = 5/5 = 1

Answer

L = 1

For equal degrees the limit is the ratio of the leading coefficients.

2.Limits

Answer: L = 0.5

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

2·sin 2x/(8x) → 0/0

Second application

4·cos 2x/8

Substitute x = 0

L = 4/8 = 0.5

Answer

L = 0.5
3.Derivatives

Answer: f′(1) = 8.155

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′ = ex·(2x + 1·x2)

Substitute

f′(1) = 2.718·(2 + 1) = 8.155

Answer

f′(1) = 8.155
4.Derivatives

Answer: x = 4

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 − 12x = 3·(x2 − 4x)

Solve f′(x) = 0

x = 0 or x = 4

Which one is which

f″(x) = 6x − 12: negative at a maximum, positive at a minimum.

f″(0) = −12, f″(4) = 12
x = 4

Answer

x = 4
5.Integrals

Answer: V = 3.142

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∫₀1 2x dx

Integrate

V = π·2·1²/2 = 3.142

Answer

V = 3.142
6.Integrals

Answer: A = 1.333

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.

Zeros

1 − x² = 0 → x = ±1

Integrate

A = [1x − x³/3]−11 = 0.6667 − (−0.6667) = 1.333

Answer

A = 1.333
7.Complex numbers

Answer: Re(w) = 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.

Modulus and argument of w

|w| = 2 · 2 = 4
θ = 70° + (−95)° = −25°

Rectangular form

Re(w) = 4·cos(−25°) = 3.625
Im(w) = 4·sin(−25°) = −1.69
Re(w) = 3.625

Answer

Re(w) = 3.625
8.Complex numbers

Answer: Im(z⁷) = −8

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 = 1 + j = √2∠45°

De Moivre

z⁷ = (√2)⁷∠315° = 11.31∠−45°

Rectangular form

z⁷ = 11.31·(cos −45° + j sin −45°) = 8 − j8
Im(z⁷) = −8

Answer

Im(z⁷) = −8
9.Differential equations

Answer: y(1) = 0.7358

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) = 1
y′(0) = B − 1A → B = 0 + 1 = 1

Evaluate

y(1) = (1 + 1·1)·e−1 = 0.7358

Answer

y(1) = 0.7358
10.Differential equations

Answer: t = 72.21 min

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.

Result

m(t) = 40 + (−35)·e−t/100

Solve for t

e−t/100 = (40 − 23)/(40 − 5) = 0.4857
t = 100·ln(2.059) = 72.21 min

Answer

t = 72.21 min
11.Matrices & determinants

Answer: det A = −51

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 = −19, M₁₂ = −11, M₁₃ = 21

Cofactor expansion

det A = 1·(−19) − (−1)·(−11) + (−1)·21 = −51

Answer

det A = −51
12.Matrices & determinants

Answer: x = 1.071

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)·(−1) − (−4)·2 = 14

Replace the x column

det Aₓ = (−3)·(−1) − (−4)·3 = 15
x = 15/14 = 1.071

Answer

x = 1.071
y = −0.8571
13.Vectors: dot & cross product

Answer: compb a = 4.961

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 = 4·5 + (−4)·(−6) + (−2)·2 = 40

Length of b

|b| = √(5² + (−6)² + 2²) = 8.062

Scalar projection

compb a = a·b/|b| = 40/8.062 = 4.961

Answer

compb a = 4.961
14.Vectors: dot & cross product

Answer: (a × b)y = 26

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 = (−3)·(−3) − 2·4 = 1
(a × b)y = azbx − axbz = 2·4 − 6·(−3) = 26
(a × b)z = axby − aybx = 6·4 − (−3)·4 = 36

Requested component

(a × b)y = 26

Answer

(a × b)y = 26

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

15.Probability

Answer: P = 0.05455

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 = 3/11 · 2/10 = 6/110 = 0.05455

Answer

P = 0.05455
16.Probability

Answer: E(G) = 5.4 €

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: 6 (p = 0.3), 98 (p = 0.05), −2 (p = 0.65)

Expected value

E(G) = 6·0.3 + 98·0.05 − 2·0.65 = 5.4 €

Answer

E(G) = 5.4 €

A positive value means you win money on average.

17.Descriptive statistics

Answer: s = 10.14

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 = 204/6 = 34

Squared deviations

Σ(x − x̄)² = 0² + 4² + (−6)² + (−13)² + (−2)² + 17² = 514

Standard deviation

s = √(514/5) = 10.14

Answer

s = 10.14
18.Descriptive statistics

Answer: P = 0.7257

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 0.7257

Answer

P = 0.7257
19.Normal distribution

Answer: z = 1.25

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 + 0.7887)/2 = 0.8944

Read the z-table backwards

Find 0.8944 inside the table and read off z.

z = 1.25

Answer

z = 1.25
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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