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→∞ (√(9x2 − 8x) − 3x).

f(x) = √(9x2 − 8x) − 3xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 ln(1 + 2x)/sin(x), with x in radians.

f(x) = ln(1 + 2x)/sin(x)x → 0
L =
3.DerivativesMedium
xyx = 0

Let f(x) = (3x + 3)4. Find f′(0).

f(x) = (3x + 3)4x = 0
f′(0) =
4.DerivativesHard
xyx = 3

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

f(x) = −x2 + 3x + 5x = 3
b =
5.IntegralsMedium
xy01

Evaluate I = ∫01 2x·(x² + 1)2 dx.

f(x) = 2x·(x² + 1)2[0, 1]
I =
6.IntegralsHard
xy1.667y = 5xy = 3x²

The region between the line y = 5x and the parabola y = 3x² is rotated about the x-axis. Find the volume V of the solid.

y = 5xy = 3x²
V =
7.Complex numbersMedium
ReIm55−5−5z₁z₂w

Given z₁ = 5∠55° and z₂ = 0.5∠110°, find the argument θ of w = z₁/z₂ in degrees, between −180° and 180°.

z₁ = 5∠55°z₂ = 0.5∠110°
θ = °
8.Complex numbersHard
ReIm11−1−122−2−2θz

Use De Moivre's theorem to find the real part of z⁶ for z = 1 + j√3.

z = 1 + j√3n = 6
Re(z⁶) =
9.Differential equationsMedium
tyy(0) = 2y(2) = ?

Solve y″ + 4y′ + 3y = 0 with y(0) = 2 and y′(0) = -4. Find y(2).

y″ + 4y′ + 3y = 0y(0) = 2y′(0) = -4
y(2) =
10.Differential equationsHard
ty2y(2) = ?

Solve y′ + 0.5y = 6e−2t with y(0) = 4. Find y(2).

y′ + 0.5y = 6e−2ty(0) = 4
y(2) =
11.Matrices & determinantsMedium
A =4−25510−4−42

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

A = [4 −2 5; 5 1 0; −4 −4 2]
det A =
12.Matrices & determinantsHard
A =53−6−6

Find the larger eigenvalue λ of the matrix A shown.

A = [5 3; −6 −6]
λ =
13.Vectors: dot & cross productMedium
xyzab

Find the angle θ between the vectors a = (−3, 4, −2) and b = (6, 4, 4), in degrees.

a = (−3, 4, −2)b = (6, 4, 4)
θ = °
14.Vectors: dot & cross productHard
xyzab

For a = (0, −4, 4) and b = (−5, −2, 5), find the x-component (a × b)x of the cross product a × b.

a = (0, −4, 4)b = (−5, −2, 5)
(a × b)x =
15.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5P(six) = 1/6 on each roll

A fair die is rolled 5 times. What is the probability P of getting at least one six?

n = 5
P =
16.ProbabilityHard
n = 12, p = 0.5: P(X = 4) = ?k0123456789101112

A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 4 correct answers?

n = 12p = 0.5k = 4
P =
17.Descriptive statisticsMedium
253035404550

Find the sample standard deviation s of the data: 52, 21, 47, 23, 34.

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

A quantity is normally distributed with mean 500 and standard deviation 40. What proportion P of values is above 516?

μ = 500σ = 40x = 516
P =
19.Normal distributionMedium
850 kg793.1 kgT = X₁ + … + X₁₀?

The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 10 of them is below 793.1 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 10< 793.1 kg
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 = −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).

Multiply by the conjugate

f(x) = −8x/(√(9x2 − 8x) + 3x)

Divide by x

f(x) = −8/(√(9 + (−8)/x) + 3)

Let x → ∞

L = −8/(3 + 3) = −1.333

Answer

L = −1.333
2.Limits

Answer: L = 2

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 [2/(1 + 2x)]/[cos x]

Substitute x = 0

L = 2/1 = 2

Answer

L = 2
3.Derivatives

Answer: f′(0) = 324

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) = 4·(3x + 3)3·3

Substitute

f′(0) = 4·33·3 = 324

Answer

f′(0) = 324
4.Derivatives

Answer: b = 14

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(3) = 5
f′(x) = −2x + 3 → f′(3) = −3

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 5 − (−3)·3 = 14

Answer

b = 14
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 = 48.48

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 = 3x² → x = 0, x = 1.667

Washer method

V = π∫₀1.667 ((5x)² − (3x²)²) dx

Integrate

V = π(25·1.667³/3 − 9·1.667⁵/5) = π·15.43 = 48.48

Answer

V = 48.48
7.Complex numbers

Answer: θ = −55 °

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.

Modulus and argument of w

|w| = 5 / 0.5 = 10
θ = 55° − 110° = −55°

Argument

θ = −55°

Answer

θ = −55°
8.Complex numbers

Answer: Re(z⁶) = 64

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.

Polar form

z = 1 + j√3 = 2∠60°

De Moivre

z⁶ = (2)⁶∠360° = 64∠0°

Rectangular form

z⁶ = 64·(cos 0° + j sin 0°) = 64
Re(z⁶) = 64

Answer

Re(z⁶) = 64
9.Differential equations

Answer: y(2) = 0.1378

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 + 3) = 0 → r = −1, −3

Constants from the initial values

A + B = 2, −1A − 3B = -4
A = 1, B = 1

Evaluate

y(2) = 1·e−2 + 1·e−6 = 0.1378

Answer

y(2) = 0.1378
10.Differential equations

Answer: y(2) = 2.87

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 = e0.5t
(e0.5ty)′ = 6e−1.5t

General solution

y = 6/(−1.5)·e−2t + C·e−0.5t = −4·e−2t + C·e−0.5t

Initial condition

4 = −4 + C → C = 8

Evaluate

y(2) = −4·0.01832 + 8·0.3679 = 2.87

Answer

y(2) = 2.87
11.Matrices & determinants

Answer: det A = −52

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 = 10, M₁₃ = −16

Cofactor expansion

det A = 4·2 − (−2)·10 + 5·(−16) = −52

Answer

det A = −52
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: θ = 103 °

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

a·b = (−3)·6 + 4·4 + (−2)·4 = −10

Magnitudes

|a| = √((−3)² + 4² + (−2)²) = 5.385
|b| = √(6² + 4² + 4²) = 8.246

Angle

cos θ = −10/(5.385·8.246) = −0.2252
θ = arccos(−0.2252) = 103°

Answer

θ = 103°
14.Vectors: dot & cross product

Answer: (a × b)x = −12

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, component by component

(a × b)x = aybz − azby = (−4)·5 − 4·(−2) = −12
(a × b)y = azbx − axbz = 4·(−5) − 0·5 = −20
(a × b)z = axby − aybx = 0·(−2) − (−4)·(−5) = −20

Requested component

(a × b)x = −12

Answer

(a × b)x = −12

Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.

15.Probability

Answer: P = 0.5981

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)5 = 0.4019
P = 1 − 0.4019 = 0.5981

Answer

P = 0.5981
16.Probability

Answer: P = 0.1208

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.

Binomial formula

P(X = 4) = 495·0.54·0.58 = 495·0.0625·0.003906 = 0.1208

Answer

P = 0.1208
17.Descriptive statistics

Answer: s = 13.9

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 = 177/5 = 35.4

Squared deviations

Σ(x − x̄)² = 16.6² + (−14.4)² + 11.6² + (−12.4)² + (−1.4)² = 773.2

Standard deviation

s = √(773.2/4) = 13.9

Answer

s = 13.9
18.Descriptive statistics

Answer: P = 0.3446

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 = 0.4

Standard normal table

Φ(0.4) = 0.6554
P = 1 − 0.6554 = 0.3446

Answer

P = 0.3446
19.Normal distribution

Answer: P = 0.06681

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 total

μT = 850 kg, σT = 12·√10 = 37.95 kg

Standardise

z = (793.1 − 850)/37.95 = −1.5

Probability

P = Φ(−1.5) = 0.06681

Answer

P = 0.06681
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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