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→∞ (−2x2 + 3x − 4)/(4x2 + 4x + 4).

f(x) = (−2x2 + 3x − 4)/(4x2 + 4x + 4)x → ∞
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 = 1

Let f(x) = (3x + 1)3. Find f′(1).

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

The tangent line to f(x) = −x2 + 3x − 5 at x = −2 crosses the y-axis at y = b. Find b.

f(x) = −x2 + 3x − 5x = −2
b =
5.IntegralsMedium
xy13

Evaluate I = ∫13 1/x dx.

f(x) = 1/x[1, 3]
I =
6.IntegralsHard
xy4y = 4xy = x²

The region between the line y = 4x and the parabola y = x² is rotated about the x-axis. Find the volume V of the solid.

y = 4xy = x²
V =
7.Complex numbersMedium
ReIm55−5−51010−10−10θz

Write z = 12∠25° in rectangular form a + jb and give the imaginary part b.

r = 12θ = 25°
b =
8.Complex numbersHard
ReIm11−1−122−2−233−3−3θz

Find the imaginary part of (−3 − j2)⁵.

z = −3 − j2n = 5
Im(z⁵) =
9.Differential equationsMedium
tyy(0) = 5y(2) = ?

Solve y″ + 4y′ + 4y = 0 with y(0) = 5 and y′(0) = -2. Find y(2).

y″ + 4y′ + 4y = 0y(0) = 5y′(0) = -2
y(2) =
10.Differential equationsHard
xy12(1, 2)y(2) = ?

Solve the initial value problem y′ + 3y/x = 2 for x > 0, with y(1) = 2. Find y(2).

y′ + 3y/x = 2y(1) = 2
y(2) =
11.Matrices & determinantsMedium
A =3−2−6−1B =−3−5−3−1

Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).

A = [3 −2; −6 −1]B = [−3 −5; −3 −1]
c11 =
12.Matrices & determinantsHard
A =5133

Find the larger eigenvalue λ of the matrix A shown.

A = [5 1; 3 3]
λ =
13.Vectors: dot & cross productMedium
xyzdFAB

A constant force F = (−30, −10, −15) N acts on an object while it moves in a straight line from point A(−2, −4, 0) to point B(−2, −5, 2). Coordinates are in metres. Find the work W done by the force.

F = (−30, −10, −15) NA(−2, −4, 0) mB(−2, −5, 2) m
W = J
14.Vectors: dot & cross productHard
xyzrFOA

A force F = (10, 0, 50) N acts at point A(−1, −3, −4). Coordinates are in metres. Find the z-component Mz of the moment of the force about point O(2, 2, 3).

F = (10, 0, 50) NA(−1, −3, −4) mO(2, 2, 3) m
Mz = N·m
15.ProbabilityMedium
xp(x)00.3510.3520.3μ = 0.95

For the distribution shown, find the standard deviation σ of X.

x: 0, 1, 2p(x): 0.35, 0.35, 0.3
σ =
16.ProbabilityHard
€p(x)−50.75200.2450.05

A game costs €5 to play. You win €25 with probability 0.2 and €50 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€25: p = 0.2€50: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
15202530354045505560

Find the sample standard deviation s of the data: 34, 53, 52, 60, 11, 32, 18.

n = 7
s =
18.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 100 and standard deviation 12. What proportion P of values is below 90.4?

μ = 100σ = 12x = 90.4
P =
19.Normal distributionMedium
6085.5P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 60 and 85.5 points? Give P as a decimal.

μ = 65σ = 10x1 = 60x2 = 85.5
P =
20.Normal distributionHard
35455565758595pointstop 40%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points40%
x = points

3Solutions

1.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).

Divide by the highest power

(−2 + 3/x + (−4)/x2)/(4 + 4/x + 4/x2)

Let x → ∞

L = −2/4 = −0.5

Answer

L = −0.5

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

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′(1) = 144

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

f′(x) = 3·(3x + 1)2·3

Substitute

f′(1) = 3·42·3 = 144

Answer

f′(1) = 144
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) = −15
f′(x) = −2x + 3 → f′(−2) = 7

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = −15 − 7·(−2) = −1

Answer

b = −1
5.Integrals

Answer: I = 1.099

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) = ln x

Evaluate at the limits

I = (ln 3 − ln 1) = ln 3 = 1.099

Answer

I = 1.099
6.Integrals

Answer: V = 428.9

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

4x = 1x² → x = 0, x = 4

Washer method

V = π∫₀4 ((4x)² − (1x²)²) dx

Integrate

V = π(16·4³/3 − 1·4⁵/5) = π·136.5 = 428.9

Answer

V = 428.9
7.Complex numbers

Answer: b = 5.071

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 = 12·cos(25°) = 10.88
b = 12·sin(25°) = 5.071

Rectangular form

z = 10.88 + j5.071
b = 5.071

Answer

b = 5.071
8.Complex numbers

Answer: Im(z⁵) = −122

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.

Powers step by step

z² = (−3 − j2)(−3 − j2) = 5 + j12
z³ = (5 + j12)(−3 − j2) = 9 − j46
z⁴ = (9 − j46)(−3 − j2) = −119 + j120
z⁵ = (−119 + j120)(−3 − j2) = 597 − j122

Check with De Moivre

z = 3.606∠−146.3° → z⁵ = 609.3∠−11.55°
Im(z⁵) = −122

Answer

Im(z⁵) = −122
9.Differential equations

Answer: y(2) = 0.3846

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) = 5
y′(0) = B − 2A → B = -2 + 10 = 8

Evaluate

y(2) = (5 + 8·2)·e−4 = 0.3846

Answer

y(2) = 0.3846
10.Differential equations

Answer: y(2) = 1.188

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∫3/x dx = x3

Integrate

(x3y)′ = 2x3 → x3y = 2x4/4 + C

Initial condition

2 = 2/4 + C → C = 1.5

Evaluate

y(2) = (2·24/4 + 1.5)/23 = 1.188

Answer

y(2) = 1.188
11.Matrices & determinants

Answer: c11 = −3

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

c11 = 3·(−3) + (−2)·(−3) = −3

Answer

c11 = −3
12.Matrices & determinants

Answer: λ = 6

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.

Characteristic equation

λ² − (8)λ + (12) = 0

Solve

λ = (8 ± √(16))/2 → λ = 2, 6
λ = 6

Answer

λ = 6
13.Vectors: dot & cross product

Answer: W = −20 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 = (−2, −5, 2) − (−2, −4, 0) = (0, −1, 2) m

Work

W = F·d = (−30)·0 + (−10)·(−1) + (−15)·2 = −20 J

Answer

W = −20 J

Negative work: the force acts mostly against the motion.

14.Vectors: dot & cross product

Answer: Mz = 50 N·m

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.

Position vector from O to A

r = A − O = (−1, −3, −4) − (2, 2, 3) = (−3, −5, −7) m

Moment M = r × F

(r × F)x = ryFz − rzFy = (−5)·50 − (−7)·0 = −250 N·m
(r × F)y = rzFx − rxFz = (−7)·10 − (−3)·50 = 80 N·m
(r × F)z = rxFy − ryFx = (−3)·0 − (−5)·10 = 50 N·m

Requested component

Mz = 50 N·m

Answer

Mz = 50 N·m
15.Probability

Answer: σ = 0.8047

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.

Mean

μ = 0·0.35 + 1·0.35 + 2·0.3 = 0.95

Mean of the squares

E(X²) = 0·0.35 + 1·0.35 + 4·0.3 = 1.55

Variance

Var(X) = 1.55 − 0.95² = 0.6475
σ = √0.6475 = 0.8047

Answer

σ = 0.8047
16.Probability

Answer: E(G) = 2.5 €

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: 20 (p = 0.2), 45 (p = 0.05), −5 (p = 0.75)

Expected value

E(G) = 20·0.2 + 45·0.05 − 5·0.75 = 2.5 €

Answer

E(G) = 2.5 €

A positive value means you win money on average.

17.Descriptive statistics

Answer: s = 18.62

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 = 260/7 = 37.14

Squared deviations

Σ(x − x̄)² = (−3.143)² + 15.86² + 14.86² + 22.86² + (−26.14)² + (−5.143)² + (−19.14)² = 2081

Standard deviation

s = √(2081/6) = 18.62

Answer

s = 18.62
18.Descriptive statistics

Answer: P = 0.2119

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 = −0.8

Standard normal table

Φ(−0.8) = 0.2119
P = 0.2119

Answer

P = 0.2119
19.Normal distribution

Answer: P = 0.6713

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 = (60 − 65)/10 = −0.5
z₂ = (85.5 − 65)/10 = 2.05

Area between

P = Φ(2.05) − Φ(−0.5) = 0.9798 − 0.3085 = 0.6713

Answer

P = 0.6713
20.Normal distribution

Answer: x = 67.5 points

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

Φ(z) = 0.6

z from the table

z ≈ 0.25

Back to the original units

x = 65 + 0.25·10 = 67.5 points

Answer

x = 67.5 points

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