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

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

Find the limit L = limx→0 (e−x − 1 + x)/x².

f(x) = (e−x − 1 + x)/x²x → 0
L =
3.DerivativesMedium
xyx = 2

Let f(x) = (4x + 1)5. Find f′(2).

f(x) = (4x + 1)5x = 2
f′(2) =
4.DerivativesHard
xyx = 1

The tangent line to f(x) = x2 + x − 5 at x = 1 crosses the y-axis at y = b. Find b.

f(x) = x2 + x − 5x = 1
b =
5.IntegralsMedium
xy01

Evaluate I = ∫01 2x·(x² + 1)2 dx.

f(x) = 2x·(x² + 1)2[0, 1]
I =
6.IntegralsHard
xy4y = 2xy = 0.5x²

The region between the line y = 2x and the parabola y = 0.5x² is rotated about the x-axis. Find the volume V of the solid.

y = 2xy = 0.5x²
V =
7.Complex numbersMedium
ReIm55−5−51010−10−10θz

Find the argument θ of z = −5 + j9 in degrees, as the principal value between −180° and 180°.

z = −5 + j9
θ = °
8.Complex numbersHard
ReIm11−1−122−2−233−3−3z₁z₂

One root of z³ + 2z² − 14z − 40 = 0 is z₁ = −3 + j. The coefficients are real. Find the real root r.

z₁ = −3 + j
r =
9.Differential equationsMedium
t (min)T (°C)16Ta​21 °C15 °C

A can of soda at 6 °C is taken out of the fridge into a room at 28 °C. After 16 min its temperature is 21 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 15 °C?

T0 = 6 °CTa = 28 °CT(16) = 21 °CT = 15 °C
t = min
10.Differential equationsHard
tyK = 1y(0.25) = ?

A system obeys y″ + 3y′ + 11.25y = 11.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.25); angles in radians.

y″ + 3y′ + 11.25y = 11.25y(0) = 0y′(0) = 0
y(0.25) =
11.Matrices & determinantsMedium
A =−2555B =2−601

Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).

A = [−2 5; 5 5]B = [2 −6; 0 1]
c21 =
12.Matrices & determinantsHard
A =52−9−6

Find the larger eigenvalue λ of the matrix A shown.

A = [5 2; −9 −6]
λ =
13.Vectors: dot & cross productMedium
xyzdFAB

A constant force F = (−15, 15, −30) N acts on an object while it moves in a straight line from point A(2, −1, 4) to point B(−1, 3, −5). Coordinates are in metres. Find the work W done by the force.

F = (−15, 15, −30) NA(2, −1, 4) mB(−1, 3, −5) m
W = J
14.Vectors: dot & cross productHard
xyzab

The unit vector n̂ is perpendicular to both a = (0, 3, −2) and b = (1, 4, −1) and has a positive z-component. Find its y-component n̂y.

a = (0, 3, −2)b = (1, 4, −1)n̂z > 0
n̂y =
15.ProbabilityMedium
red: 8blue: 3

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

8 / 3
P =
16.ProbabilityHard
n = 10, p = 0.7: P(X ≤ 1) = ?k012345678910

10 components are tested, and each one fails with probability 0.7, independently. What is the probability P that at most one fails?

n = 10p = 0.7
P =
17.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 47, 57, 15, 34, 37, 39.

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 above 40?

μ = 70σ = 15x = 40
P =
19.Normal distributionMedium
55.567.5P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 55.5 and 67.5 points? Give P as a decimal.

μ = 65σ = 10x1 = 55.5x2 = 67.5
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 0.01 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 = 0.01 mm
P =

3Solutions

1.Limits

Answer: L = 20.09

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→∞ x·ln(1 + 3/x)

Evaluate

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

ln L = 1·3 = 3

Undo the logarithm

L = e3 = 20.09

Answer

L = 20.09
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

(−1·e−x + 1)/(2x) → 0/0

Second application

1·e−x/2

Substitute x = 0

L = 1·1/2 = 0.5

Answer

L = 0.5
3.Derivatives

Answer: f′(2) = 131220

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) = 5·(4x + 1)4·4

Substitute

f′(2) = 5·94·4 = 131220

Answer

f′(2) = 131220
4.Derivatives

Answer: b = −6

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(1) = −3
f′(x) = 2x + 1 → f′(1) = 3

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = −3 − 3·1 = −6

Answer

b = −6
5.Integrals

Answer: I = 2.333

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.

Substitution

u = x² + 1, du = 2x dx
I = ∫12 u2 du

Integrate and evaluate

I = [u3/3]12 = (8 − 1)/3 = 2.333

Answer

I = 2.333
6.Integrals

Answer: V = 107.2

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 = 0.5x² → x = 0, x = 4

Washer method

V = π∫₀4 ((2x)² − (0.5x²)²) dx

Integrate

V = π(4·4³/3 − 0.25·4⁵/5) = π·34.13 = 107.2

Answer

V = 107.2
7.Complex numbers

Answer: θ = 119.1 °

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 = −5, b = 9 → II

Reference angle

α = atan(9/5) = 60.95°

Argument

θ = 180° − α = 119.1°

Answer

θ = 119.1°

In polar form z = 10.3∠119.1°.

8.Complex numbers

Answer: r = 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.

Conjugate pair

z₂ = −3 − j
(z − z₁)(z − z₂) = z² − 2·(−3)·z + ((−3)² + 1²) = z² + 6z + 10

Factor out the quadratic

z³ + 2z² − 14z − 40 = (z² + 6z + 10)(z − 4)

Real root

r = 4

Answer

r = 4

Check with the sum of the roots: z₁ + z₂ + r = −6 + 4 = −2 = −(2).

9.Differential equations

Answer: t = 7.351 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.

Cooling constant

k = ln(-22/-7)/16 = 0.07157 min⁻¹

Time to reach the target

t = ln(-22/(−13))/0.07157 = 7.351 min

Answer

t = 7.351 min
10.Differential equations

Answer: y(0.25) = 0.2629

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.

Steady value and roots

yss = 1, r = −1.5 ± 3j

Constants from the initial values

y(0) = 1 + A = 0 → A = −1
y′(0) = −1.5A + 3B = 0 → B = −0.5

Evaluate

y(0.25) = 1·(1 − e−0.375(cos 0.75 + 0.5·sin 0.75)) = 0.2629

Answer

y(0.25) = 0.2629
11.Matrices & determinants

Answer: c21 = 10

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

c21 = 5·2 + 5·0 = 10

Answer

c21 = 10
12.Matrices & determinants

Answer: λ = 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.

Characteristic equation

λ² − (−1)λ + (−12) = 0

Solve

λ = (−1 ± √(49))/2 → λ = −4, 3
λ = 3

Answer

λ = 3
13.Vectors: dot & cross product

Answer: W = 375 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 = (−1, 3, −5) − (2, −1, 4) = (−3, 4, −9) m

Work

W = F·d = (−15)·(−3) + 15·4 + (−30)·(−9) = 375 J

Answer

W = 375 J
14.Vectors: dot & cross product

Answer: n̂y = 0.3244

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

(a × b)x = aybz − azby = 3·(−1) − (−2)·4 = 5
(a × b)y = azbx − axbz = (−2)·1 − 0·(−1) = −2
(a × b)z = axby − aybx = 0·4 − 3·1 = −3

Normalise

The z-component of a × b is negative, so use −(a × b).

|a × b| = √(5² + (−2)² + (−3)²) = 6.164
n̂ = −(5, −2, −3)/6.164 = (−0.8111, 0.3244, 0.4867)
n̂y = 0.3244

Answer

n̂y = 0.3244
15.Probability

Answer: P = 0.5091

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 = 8/11 · 7/10 = 56/110 = 0.5091

Answer

P = 0.5091
16.Probability

Answer: P = 0.0001437

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.

Zero and one failure

P(X = 0) = 0.310 = 5.905 × 10⁻⁶
P(X = 1) = 10·0.7·0.39 = 0.0001378

Add them

P = 5.905 × 10⁻⁶ + 0.0001378 = 0.0001437

Answer

P = 0.0001437
17.Descriptive statistics

Answer: s = 14.06

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 = 229/6 = 38.17

Squared deviations

Σ(x − x̄)² = 8.833² + 18.83² + (−23.17)² + (−4.167)² + (−1.167)² + 0.8333² = 988.8

Standard deviation

s = √(988.8/5) = 14.06

Answer

s = 14.06
18.Descriptive statistics

Answer: P = 0.9772

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 = −2

Standard normal table

Φ(−2) = 0.02275
P = 1 − 0.02275 = 0.9772

Answer

P = 0.9772
19.Normal distribution

Answer: P = 0.4277

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 = (55.5 − 65)/10 = −0.95
z₂ = (67.5 − 65)/10 = 0.25

Area between

P = Φ(0.25) − Φ(−0.95) = 0.5987 − 0.1711 = 0.4277

Answer

P = 0.4277
20.Normal distribution

Answer: P = 0.8665

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 = 0.01803 mm

Standardise

z = (0 − 0.02)/0.01803 = −1.11

Probability

P = 1 − Φ(−1.11) = 0.8665

Answer

P = 0.8665

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