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)3x. Give a decimal value.

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

Find the limit L = limx→0 ln(1 + 4x)/sin(3x), with x in radians.

f(x) = ln(1 + 4x)/sin(3x)x → 0
L =
3.DerivativesMedium
xyx = 1.5

Let f(x) = x·sin(3x), with x in radians. Find f′(1.5).

f(x) = x·sin(3x)x = 1.5 rad
f′(1.5) =
4.DerivativesHard
xy

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

f(x) = x3 + 3x2 + 3
x =
5.IntegralsMedium
xy12.5y = x ln x

Evaluate I = ∫₁2.5 x·ln x dx.

∫₁2.5 x ln x dx
I =
6.IntegralsHard
xy2.5y = 5xy = 2x²

The region between the line y = 5x and the parabola y = 2x² is rotated about the x-axis. Find the volume V of the solid.

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

Write z = 8∠110° in rectangular form a + jb and give the real part a.

r = 8θ = 110°
a =
8.Complex numbersHard
ReIm11−1−122−2−233−3−3θz

Find the imaginary part of (2 − j3)⁴.

z = 2 − j3n = 4
Im(z⁴) =
9.Differential equationsMedium
t (min)T (°C)46Ta​T = ?

A forged steel part at 340 °C is left to cool in a workshop at 30 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.05 min⁻¹, find its temperature T after 46 min.

T0 = 340 °CTa = 30 °Ck = 0.05 min⁻¹t = 46 min
T = °C
10.Differential equationsHard
ty1.5y(1.5) = ?

Solve y′ + y = 3e−3t with y(0) = 1. Find y(1.5).

y′ + y = 3e−3ty(0) = 1
y(1.5) =
11.Matrices & determinantsMedium
A =−3−1−16B =−1−4−40

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

A = [−3 −1; −1 6]B = [−1 −4; −4 0]
c11 =
12.Matrices & determinantsHard
A =−1−664b =−12−11

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

−x − 6y = −126x + 4y = −11
x =
13.Vectors: dot & cross productMedium
xyzdFAB

A constant force F = (25, −30, 25) N acts on an object while it moves in a straight line from point A(3, −1, 3) to point B(4, 0, −1). Coordinates are in metres. Find the work W done by the force.

F = (25, −30, 25) NA(3, −1, 3) mB(4, 0, −1) m
W = J
14.Vectors: dot & cross productHard
xyzrFOA

A force F = (40, −50, −10) N acts at point A(−3, −3, 3). Coordinates are in metres. Find the y-component My of the moment of the force about the origin O(0, 0, 0).

F = (40, −50, −10) NA(−3, −3, 3) mO(0, 0, 0) m
My = N·m
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
€p(x)−100.7400.254900.05

A game costs €10 to play. You win €50 with probability 0.25 and €500 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €10€50: p = 0.25€500: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
303540455055

Find the sample standard deviation s of the data: 35, 30, 55, 36, 52, 41, 49.

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

Exam scores are normally distributed with mean 500 and standard deviation 25. Find the z-score z of a score of 475.

μ = 500σ = 25x = 475
z =
19.Normal distributionMedium
164179.75P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 164 and 179.75 cm? Give P as a decimal.

μ = 178σ = 7x1 = 164x2 = 179.75
P =
20.Normal distributionHard
35455565758595pointstop 95%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 95% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points95%
x = points

3Solutions

1.Limits

Answer: L = 0.04979

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 − 1/x)

Evaluate

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

ln L = 3·(−1) = −3

Undo the logarithm

L = e−3 = 0.04979

Answer

L = 0.04979
2.Limits

Answer: L = 1.333

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 [4/(1 + 4x)]/[3·cos 3x]

Substitute x = 0

L = 4/3 = 1.333

Answer

L = 1.333
3.Derivatives

Answer: f′(1.5) = −1.926

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′(1.5) = −0.9775 + 4.5·(−0.2108) = −1.926

Answer

f′(1.5) = −1.926
4.Derivatives

Answer: x = −2

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 + 6x = 3·(x2 + 2x)

Solve f′(x) = 0

x = −2 or x = 0

Which one is which

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

f″(−2) = −6, f″(0) = 6
x = −2

Answer

x = −2
5.Integrals

Answer: I = 1.551

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

(3.125·ln 2.5 − 1.5625) − (0 − 0.25) = 1.551

Answer

I = 1.551
6.Integrals

Answer: V = 163.6

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

5x = 2x² → x = 0, x = 2.5

Washer method

V = π∫₀2.5 ((5x)² − (2x²)²) dx

Integrate

V = π(25·2.5³/3 − 4·2.5⁵/5) = π·52.08 = 163.6

Answer

V = 163.6
7.Complex numbers

Answer: a = −2.736

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 = 8·cos(110°) = −2.736
b = 8·sin(110°) = 7.518

Rectangular form

z = −2.736 + j7.518
a = −2.736

Answer

a = −2.736
8.Complex numbers

Answer: Im(z⁴) = 120

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 = (2 − j3)(2 − j3) = −5 − j12
z³ = (−5 − j12)(2 − j3) = −46 − j9
z⁴ = (−46 − j9)(2 − j3) = −119 + j120

Check with De Moivre

z = 3.606∠−56.31° → z⁴ = 169∠134.8°
Im(z⁴) = 120

Answer

Im(z⁴) = 120
9.Differential equations

Answer: T = 61.08 °C

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.

Solution of the cooling law

T(t) = 30 + 310·e−0.05t

Substitute t

T(46) = 30 + 310·0.1003 = 61.08 °C

Answer

T = 61.08 °C
10.Differential equations

Answer: y(1.5) = 0.5412

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 = et
(ety)′ = 3e−2t

General solution

y = 3/(−2)·e−3t + C·e−t = −1.5·e−3t + C·e−t

Initial condition

1 = −1.5 + C → C = 2.5

Evaluate

y(1.5) = −1.5·0.01111 + 2.5·0.2231 = 0.5412

Answer

y(1.5) = 0.5412
11.Matrices & determinants

Answer: c11 = 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

c11 = (−3)·(−1) + (−1)·(−4) = 7

Answer

c11 = 7
12.Matrices & determinants

Answer: x = −3.563

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 = (−1)·4 − (−6)·6 = 32

Replace the x column

det Aₓ = (−12)·4 − (−6)·(−11) = −114
x = −114/32 = −3.563

Answer

x = −3.563
y = 2.594
13.Vectors: dot & cross product

Answer: W = −105 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 = (4, 0, −1) − (3, −1, 3) = (1, 1, −4) m

Work

W = F·d = 25·1 + (−30)·1 + 25·(−4) = −105 J

Answer

W = −105 J

Negative work: the force acts mostly against the motion.

14.Vectors: dot & cross product

Answer: My = 90 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 = (−3, −3, 3) − (0, 0, 0) = (−3, −3, 3) m

Moment M = r × F

(r × F)x = ryFz − rzFy = (−3)·(−10) − 3·(−50) = 180 N·m
(r × F)y = rzFx − rxFz = 3·40 − (−3)·(−10) = 90 N·m
(r × F)z = rxFy − ryFx = (−3)·(−50) − (−3)·40 = 270 N·m

Requested component

My = 90 N·m

Answer

My = 90 N·m
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: E(G) = 27.5 €

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: 40 (p = 0.25), 490 (p = 0.05), −10 (p = 0.7)

Expected value

E(G) = 40·0.25 + 490·0.05 − 10·0.7 = 27.5 €

Answer

E(G) = 27.5 €

A positive value means you win money on average.

17.Descriptive statistics

Answer: s = 9.537

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 = 298/7 = 42.57

Squared deviations

Σ(x − x̄)² = (−7.571)² + (−12.57)² + 12.43² + (−6.571)² + 9.429² + (−1.571)² + 6.429² = 545.7

Standard deviation

s = √(545.7/6) = 9.537

Answer

s = 9.537
18.Descriptive statistics

Answer: z = −1

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 = (475 − 500)/25 = −1

Answer

z = −1
19.Normal distribution

Answer: P = 0.576

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 = (164 − 178)/7 = −2
z₂ = (179.75 − 178)/7 = 0.25

Area between

P = Φ(0.25) − Φ(−2) = 0.5987 − 0.02275 = 0.576

Answer

P = 0.576
20.Normal distribution

Answer: x = 48.6 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) = 0.05

z from the table

z ≈ −1.64

Back to the original units

x = 65 − 1.64·10 = 48.6 points

Answer

x = 48.6 points

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