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→∞ (−x3 − 3x + 3)/(x3 + 5x + 2).

f(x) = (−x3 − 3x + 3)/(x3 + 5x + 2)x → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (1 − cos 5x)/(3x²), with x in radians.

f(x) = (1 − cos 5x)/(3x²)x → 0
L =
3.DerivativesMedium
xyx = 1

Let f(x) = ln(3x2 + 1). Find f′(1).

f(x) = ln(3x2 + 1)x = 1
f′(1) =
4.DerivativesHard
xy

Let f(x) = x3 − 15x2 + 63x − 5. Find the x-coordinate x of the local minimum.

f(x) = x3 − 15x2 + 63x − 5
x =
5.IntegralsMedium
xy02

Evaluate I = ∫02 2x·(x² + 1)2 dx.

f(x) = 2x·(x² + 1)2[0, 2]
I =
6.IntegralsHard
xyy = 2x2​y = 2x

Find the area A of the region enclosed by y = 2x2 and y = 2x.

y = 2x2y = 2x
A =
7.Complex numbersMedium
ReIm55−5−5θz

Find the argument θ of z = −8 + j8 in degrees, as the principal value between −180° and 180°.

z = −8 + j8
θ = °
8.Complex numbersHard
ReIm22−2−244−4−4z₁z₂

Solve z² − 8z + 32 = 0. The roots are a ± jb; find b > 0.

p = −8q = 32
b =
9.Differential equationsMedium
tyy(0) = 4y(1.5) = ?

Solve y″ + 5y′ + 4y = 0 with y(0) = 4 and y′(0) = 1. Find y(1.5).

y″ + 5y′ + 4y = 0y(0) = 4y′(0) = 1
y(1.5) =
10.Differential equationsHard
xy12(1, −1)y(2) = ?

Solve the initial value problem y′ + y/x = 5x for x > 0, with y(1) = −1. Find y(2).

y′ + y/x = 5xy(1) = −1
y(2) =
11.Matrices & determinantsMedium
A =51−120−1−2−35

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

A = [5 1 −1; 2 0 −1; −2 −3 5]
det A =
12.Matrices & determinantsHard
A =52−12

Find the larger eigenvalue λ of the matrix A shown.

A = [5 2; −1 2]
λ =
13.Vectors: dot & cross productMedium
xyzdFAB

A constant force F = (−20, 5, 25) N acts on an object while it moves in a straight line from point A(−1, −1, −1) to point B(3, −4, −2). Coordinates are in metres. Find the work W done by the force.

F = (−20, 5, 25) NA(−1, −1, −1) mB(3, −4, −2) m
W = J
14.Vectors: dot & cross productHard
xyzrFOA

A force F = (−30, 60, 20) N acts at point A(1, 4, −4). Coordinates are in metres. Find the z-component Mz of the moment of the force about point O(2, 3, 0).

F = (−30, 60, 20) NA(1, 4, −4) mO(2, 3, 0) m
Mz = N·m
15.ProbabilityMedium
red: 3blue: 7

A bag holds 3 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

3 / 7
P =
16.ProbabilityHard
€p(x)−50.84150.152450.01

A game costs €5 to play. You win €20 with probability 0.15 and €250 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€20: p = 0.15€250: p = 0.01
E(G) = €
17.Descriptive statisticsMedium
2530354045

Find the sample standard deviation s of the data: 30, 40, 41, 23, 40, 49.

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

A quantity is normally distributed with mean 170 and standard deviation 8. What proportion P of values is above 160.4?

μ = 170σ = 8x = 160.4
P =
19.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.7823

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.7823. Find z to two decimals.

A = 0.7823
z =
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 0.015 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 = 0.015 mm
P =

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

(−1 + (−3)/x2 + 3/x3)/(1 + 5/x2 + 2/x3)

Let x → ∞

L = −1/1 = −1

Answer

L = −1

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

2.Limits

Answer: L = 4.167

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

5·sin 5x/(6x) → 0/0

Second application

25·cos 5x/6

Substitute x = 0

L = 25/6 = 4.167

Answer

L = 4.167
3.Derivatives

Answer: f′(1) = 1.5

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 for the logarithm

f′(x) = 6x/(3x2 + 1)

Substitute

f′(1) = 6/4 = 1.5

Answer

f′(1) = 1.5
4.Derivatives

Answer: x = 7

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 − 30x + 63 = 3·(x2 − 10x + 21)

Solve f′(x) = 0

x = 3 or x = 7

Which one is which

f″(x) = 6x − 30: negative at a maximum, positive at a minimum.

f″(3) = −12, f″(7) = 12
x = 7

Answer

x = 7
5.Integrals

Answer: I = 41.33

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 u2 du

Integrate and evaluate

I = [u3/3]15 = (125 − 1)/3 = 41.33

Answer

I = 41.33
6.Integrals

Answer: A = 0.3333

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.

Intersection points

2x2 = 2x → x = 0, x = 1

Integrate upper minus lower

A = [x2 − 0.6667x3]01 = 0.3333

Answer

A = 0.3333
7.Complex numbers

Answer: θ = 135 °

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 = −8, b = 8 → II

Reference angle

α = atan(8/8) = 45°

Argument

θ = 180° − α = 135°

Answer

θ = 135°

In polar form z = 11.31∠135°.

8.Complex numbers

Answer: b = 4

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 = 64 − 128 = −64 < 0

Quadratic formula

z = (8 ± j√64)/2 = 4 + j4, 4 − j4

Answer

b = 4

Answer

b = 4
9.Differential equations

Answer: y(1.5) = 1.26

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.

Characteristic roots

(r + 1)(r + 4) = 0 → r = −1, −4

Constants from the initial values

A + B = 4, −1A − 4B = 1
A = 5.667, B = −1.667

Evaluate

y(1.5) = 5.667·e−1.5 + (−1.667)·e−6 = 1.26

Answer

y(1.5) = 1.26
10.Differential equations

Answer: y(2) = 5.333

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∫1/x dx = x1

Integrate

(x1y)′ = 5x2 → x1y = 5x3/3 + C

Initial condition

−1 = 5/3 + C → C = −2.667

Evaluate

y(2) = (5·23/3 + (−2.667))/21 = 5.333

Answer

y(2) = 5.333
11.Matrices & determinants

Answer: det A = −17

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 = −3, M₁₂ = 8, M₁₃ = −6

Cofactor expansion

det A = 5·(−3) − 1·8 + (−1)·(−6) = −17

Answer

det A = −17
12.Matrices & determinants

Answer: λ = 4

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 − (7)λ + (12) = 0

Solve

λ = (7 ± √(1))/2 → λ = 3, 4
λ = 4

Answer

λ = 4
13.Vectors: dot & cross product

Answer: W = −120 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 = (3, −4, −2) − (−1, −1, −1) = (4, −3, −1) m

Work

W = F·d = (−20)·4 + 5·(−3) + 25·(−1) = −120 J

Answer

W = −120 J

Negative work: the force acts mostly against the motion.

14.Vectors: dot & cross product

Answer: Mz = −30 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, 4, −4) − (2, 3, 0) = (−1, 1, −4) m

Moment M = r × F

(r × F)x = ryFz − rzFy = 1·20 − (−4)·60 = 260 N·m
(r × F)y = rzFx − rxFz = (−4)·(−30) − (−1)·20 = 140 N·m
(r × F)z = rxFy − ryFx = (−1)·60 − 1·(−30) = −30 N·m

Requested component

Mz = −30 N·m

Answer

Mz = −30 N·m
15.Probability

Answer: P = 0.06667

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 = 3/10 · 2/9 = 6/90 = 0.06667

Answer

P = 0.06667
16.Probability

Answer: E(G) = 0.5 €

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: 15 (p = 0.15), 245 (p = 0.01), −5 (p = 0.84)

Expected value

E(G) = 15·0.15 + 245·0.01 − 5·0.84 = 0.5 €

Answer

E(G) = 0.5 €

A positive value means you win money on average.

17.Descriptive statistics

Answer: s = 9.196

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 = 223/6 = 37.17

Squared deviations

Σ(x − x̄)² = (−7.167)² + 2.833² + 3.833² + (−14.17)² + 2.833² + 11.83² = 422.8

Standard deviation

s = √(422.8/5) = 9.196

Answer

s = 9.196
18.Descriptive statistics

Answer: P = 0.8849

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849

Answer

P = 0.8849
19.Normal distribution

Answer: z = −0.78

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 − 0.7823 = 0.2177

Read the z-table backwards

Find 0.2177 inside the table and read off z.

z = −0.78

Answer

z = −0.78
20.Normal distribution

Answer: P = 0.8264

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 = 0.02121 mm

Standardise

z = (0 − 0.02)/0.02121 = −0.94

Probability

P = 1 − Φ(−0.94) = 0.8264

Answer

P = 0.8264

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