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 + 0.5/x)3x. Give a decimal value.

f(x) = (1 + 0.5/x)3xx → ∞
L =
2.LimitsHard
xyx = 0

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

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

Let f(x) = x·sin(3x), with x in radians. Find f′(2).

f(x) = x·sin(3x)x = 2 rad
f′(2) =
4.DerivativesHard
xyx = 2

The tangent line to f(x) = −x2 − 5 at x = 2 crosses the y-axis at y = b. Find b.

f(x) = −x2 − 5x = 2
b =
5.IntegralsMedium
xy02

Evaluate I = ∫02 2x·(x² + 1)3 dx.

f(x) = 2x·(x² + 1)3[0, 2]
I =
6.IntegralsHard
xy2y = 2xy = x²

The region between the line y = 2x and the parabola y = x² is rotated about the x-axis. Find the volume V of the solid.

y = 2xy = x²
V =
7.Complex numbersMedium
ReIm55−5−5θz

Write z = 10∠−40° in rectangular form a + jb and give the imaginary part b.

r = 10θ = −40°
b =
8.Complex numbersHard
ReIm22−2−244−4−4z₁z₂

Solve z² + 3z + 27.25 = 0. The roots are a ± jb; find b > 0.

p = 3q = 27.25
b =
9.Differential equationsMedium
tyy(0) = 3y(1) = ?

Solve y″ + 6y′ + 9y = 0 with y(0) = 3 and y′(0) = 4. Find y(1).

y″ + 6y′ + 9y = 0y(0) = 3y′(0) = 4
y(1) =
10.Differential equationsHard
t (min)m (kg)t = ?202 kg

A tank holds 1000 L of water in which 10 kg of salt is dissolved. Brine with 0.25 kg of salt per litre flows in at 2 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 202 kg of salt?

V = 1000 Lm0 = 10 kgc = 0.25 kg/Lr = 2 L/min
t = min
11.Matrices & determinantsMedium
A =0−3−132−213−2

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

A = [0 −3 −1; 3 2 −2; 1 3 −2]
det A =
12.Matrices & determinantsHard
A =4652b =−12−11

Solve the system 4x + 6y = −12 and 5x + 2y = −11. Find x.

4x + 6y = −125x + 2y = −11
x =
13.Vectors: dot & cross productMedium
xyzab

Find the value of k for which the vectors a = (0, 6, −6) and b = (1, k, −5) are perpendicular.

a = (0, 6, −6)b = (1, k, −5)
k =
14.Vectors: dot & cross productHard
xyzabc

Find the volume V of the parallelepiped whose edges from one corner are a = (3, −5, 1), b = (0, 1, −5) and c = (−3, 4, 2).

a = (3, −5, 1)b = (0, 1, −5)c = (−3, 4, 2)
V =
15.ProbabilityMedium
red: 6blue: 7

A bag holds 6 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

6 / 7
P =
16.ProbabilityHard
xp(x)00.3510.3520.3μ

X has the distribution shown, and Y = 0.5X + 5. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.35, 0.35, 0.3Y = 0.5X + 5
σY =
17.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 37, 26, 51, 17, 54.

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

Exam scores are normally distributed with mean 60 and standard deviation 10. Find the z-score z of a score of 52.

μ = 60σ = 10x = 52
z =
19.Normal distributionMedium
200 mm200.5 mmX + Y?

X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 200.5 mm. Round z to two decimals.

μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 200.5 mm
P =
20.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 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 = 0.015 mmshaft: μ = 25 mm, σ = 0.02 mm
P =

3Solutions

1.Limits

Answer: L = 4.482

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→∞ 3·x·ln(1 + 0.5/x)

Evaluate

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

ln L = 3·0.5 = 1.5

Undo the logarithm

L = e1.5 = 4.482

Answer

L = 4.482
2.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).

First application

2·sin 2x/(4x) → 0/0

Second application

4·cos 2x/4

Substitute x = 0

L = 4/4 = 1

Answer

L = 1
3.Derivatives

Answer: f′(2) = 5.482

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 and chain rule

f′(x) = sin(3x) + 3x·cos(3x)

Substitute

f′(2) = −0.2794 + 6·0.9602 = 5.482

Answer

f′(2) = 5.482
4.Derivatives

Answer: b = −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.

Point and slope

f(2) = −9
f′(x) = −2x → f′(2) = −4

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = −9 − (−4)·2 = −1

Answer

b = −1
5.Integrals

Answer: I = 156

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 = ∫15 u3 du

Integrate and evaluate

I = [u4/4]15 = (625 − 1)/4 = 156

Answer

I = 156
6.Integrals

Answer: V = 13.4

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.

Intersection points

2x = 1x² → x = 0, x = 2

Washer method

V = π∫₀2 ((2x)² − (1x²)²) dx

Integrate

V = π(4·2³/3 − 1·2⁵/5) = π·4.267 = 13.4

Answer

V = 13.4
7.Complex numbers

Answer: b = −6.428

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 = 10·cos(−40°) = 7.66
b = 10·sin(−40°) = −6.428

Rectangular form

z = 7.66 − j6.428
b = −6.428

Answer

b = −6.428
8.Complex numbers

Answer: b = 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 − 109 = −100 < 0

Quadratic formula

z = (−3 ± j√100)/2 = −1.5 + j5, −1.5 − j5

Answer

b = 5

Answer

b = 5
9.Differential equations

Answer: y(1) = 0.7966

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) = 3
y′(0) = B − 3A → B = 4 + 9 = 13

Evaluate

y(1) = (3 + 13·1)·e−3 = 0.7966

Answer

y(1) = 0.7966
10.Differential equations

Answer: t = 804.7 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) = 250 + (−240)·e−t/500

Solve for t

e−t/500 = (250 − 202)/(250 − 10) = 0.2
t = 500·ln(5) = 804.7 min

Answer

t = 804.7 min
11.Matrices & determinants

Answer: det A = −19

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 = 2, M₁₂ = −4, M₁₃ = 7

Cofactor expansion

det A = 0·2 − (−3)·(−4) + (−1)·7 = −19

Answer

det A = −19
12.Matrices & determinants

Answer: x = −1.909

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 = 4·2 − 6·5 = −22

Replace the x column

det Aₓ = (−12)·2 − 6·(−11) = 42
x = 42/(−22) = −1.909

Answer

x = −1.909
y = −0.7273
13.Vectors: dot & cross product

Answer: k = −5

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 equals zero

a·b = 0·1 + 6·k + (−6)·(−5) = 0
30 + 6·k = 0

Solve for k

k = −30/6 = −5

Answer

k = −5
14.Vectors: dot & cross product

Answer: V = 6

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 = 1·2 − (−5)·4 = 22
(b × c)y = bzcx − bxcz = (−5)·(−3) − 0·2 = 15
(b × c)z = bxcy − bycx = 0·4 − 1·(−3) = 3

Dot product with a

a·(b × c) = 3·22 + (−5)·15 + 1·3 = −6
V = |−6| = 6

Answer

V = 6
15.Probability

Answer: P = 0.1923

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 = 6/13 · 5/12 = 30/156 = 0.1923

Answer

P = 0.1923
16.Probability

Answer: σY = 0.4023

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.

Spread of X

μ = 0.95, Var(X) = 0.6475, σX = 0.8047

Linear change

σY = |0.5|·0.8047 = 0.4023

Answer

σY = 0.4023

The mean does move: E(Y) = 0.5·0.95 + 5 = 5.475.

17.Descriptive statistics

Answer: s = 15.86

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 = 185/5 = 37

Squared deviations

Σ(x − x̄)² = 0² + (−11)² + 14² + (−20)² + 17² = 1006

Standard deviation

s = √(1006/4) = 15.86

Answer

s = 15.86
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 = (52 − 60)/10 = −0.8

Answer

z = −0.8
19.Normal distribution

Answer: P = 0.1587

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 sum

μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm

Standardise

z = (200.5 − 200)/0.5 = 1

Upper tail

P = 1 − Φ(1) = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
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 = 25.02 − 25 = 0.02 mm, σ = 0.025 mm

Standardise

z = (0 − 0.02)/0.025 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881

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