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 + 10x) − 3x).

f(x) = √(9x2 + 10x) − 3xx → ∞
L =
2.LimitsHard
xyx = 0

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

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

Let f(x) = x·sin(2x), with x in radians. Find f′(1).

f(x) = x·sin(2x)x = 1 rad
f′(1) =
4.DerivativesHard
xyx = −1

The tangent line to f(x) = 2x2 − 5x − 4 at x = −1 crosses the y-axis at y = b. Find b.

f(x) = 2x2 − 5x − 4x = −1
b =
5.IntegralsMedium
xy02

Evaluate I = ∫02 e2x dx.

f(x) = e2x[0, 2]
I =
6.IntegralsHard
xy1y = 2xy = 2x²

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

y = 2xy = 2x²
V =
7.Complex numbersMedium
ReIm22−2−244−4−4θz

The number z = 5·ej2 has its angle in radians. Find its imaginary part b.

r = 5θ = 2 rad
b =
8.Complex numbersHard
ReIm11−1−122−2−2z₁z₂

One root of z³ − 3z² + 4z + 8 = 0 is z₁ = 2 + j2. The coefficients are real. Find the real root r.

z₁ = 2 + j2
r =
9.Differential equationsMedium
tyy(0) = 1y(0.5) = ?

Solve y″ + 6y′ + 8y = 0 with y(0) = 1 and y′(0) = -2. Find y(0.5).

y″ + 6y′ + 8y = 0y(0) = 1y′(0) = -2
y(0.5) =
10.Differential equationsHard
t (min)m (kg)t = ?35 kg

A tank holds 200 L of water in which 20 kg of salt is dissolved. Brine with 0.25 kg of salt per litre flows in at 4 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 35 kg of salt?

V = 200 Lm0 = 20 kgc = 0.25 kg/Lr = 4 L/min
t = min
11.Matrices & determinantsMedium
A =0−54−5B =0−366

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

A = [0 −5; 4 −5]B = [0 −3; 6 6]
c11 =
12.Matrices & determinantsHard
A =−6−4−2−2b =5−9

Solve the system −6x − 4y = 5 and −2x − 2y = −9. Find x.

−6x − 4y = 5−2x − 2y = −9
x =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (0, −3, 3) onto b = (−1, −2, 0), that is, the component of a in the direction of b.

a = (0, −3, 3)b = (−1, −2, 0)
compb a =
14.Vectors: dot & cross productHard
xyzabc

Find the volume V of the parallelepiped whose edges from one corner are a = (5, −3, −1), b = (−3, 1, 3) and c = (3, −4, 4).

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

A fair die is rolled 2 times. What is the probability P of getting at least one six?

n = 2
P =
16.ProbabilityHard
€p(x)−30.7490.251470.01

A game costs €3 to play. You win €12 with probability 0.25 and €150 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€12: p = 0.25€150: p = 0.01
E(G) = €
17.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 51, 43, 44, 17, 35, 23.

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

A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is above 70?

μ = 100σ = 20x = 70
P =
19.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.3035

Drag z so that the shaded area between −z and z under the standard normal curve is 0.3035. Find z to two decimals.

A = 0.3035
z =
20.Normal distributionHard
μ = 100x = ?10%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 10% of them lie (the 10th percentile).

μ = 100σ = 15P = 10%
x = points

3Solutions

1.Limits

Answer: L = 1.667

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) = 10x/(√(9x2 + 10x) + 3x)

Divide by x

f(x) = 10/(√(9 + 10/x) + 3)

Let x → ∞

L = 10/(3 + 3) = 1.667

Answer

L = 1.667
2.Limits

Answer: L = 0.3333

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 [1/(1 + x)]/[3·cos 3x]

Substitute x = 0

L = 1/3 = 0.3333

Answer

L = 0.3333
3.Derivatives

Answer: f′(1) = 0.077

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(2x) + 2x·cos(2x)

Substitute

f′(1) = 0.9093 + 2·(−0.4161) = 0.077

Answer

f′(1) = 0.077
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) = 4x − 5 → f′(−1) = −9

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 3 − (−9)·(−1) = −6

Answer

b = −6
5.Integrals

Answer: I = 26.8

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) = e2x/2

Evaluate at the limits

I = (e4 − 1)/2 = (54.6 − 1)/2 = 26.8

Answer

I = 26.8
6.Integrals

Answer: V = 1.676

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 = 2x² → x = 0, x = 1

Washer method

V = π∫₀1 ((2x)² − (2x²)²) dx

Integrate

V = π(4·1³/3 − 4·1⁵/5) = π·0.5333 = 1.676

Answer

V = 1.676
7.Complex numbers

Answer: b = 4.546

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.

Euler's formula

z = 5·(cos 2 + j sin 2)

Components (radians)

a = 5·cos 2 = −2.081
b = 5·sin 2 = 4.546
b = 4.546

Answer

b = 4.546
8.Complex numbers

Answer: r = −1

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 = 2 − j2
(z − z₁)(z − z₂) = z² − 2·2·z + (2² + 2²) = z² − 4z + 8

Factor out the quadratic

z³ − 3z² + 4z + 8 = (z² − 4z + 8)(z + 1)

Real root

r = −1

Answer

r = −1

Check with the sum of the roots: z₁ + z₂ + r = 4 + (−1) = 3 = −(−3).

9.Differential equations

Answer: y(0.5) = 0.3679

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 + 2)(r + 4) = 0 → r = −2, −4

Constants from the initial values

A + B = 1, −2A − 4B = -2
A = 1, B = 0

Evaluate

y(0.5) = 1·e−1 + 0·e−2 = 0.3679

Answer

y(0.5) = 0.3679
10.Differential equations

Answer: t = 34.66 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.

Result

m(t) = 50 + (−30)·e−t/50

Solve for t

e−t/50 = (50 − 35)/(50 − 20) = 0.5
t = 50·ln(2) = 34.66 min

Answer

t = 34.66 min
11.Matrices & determinants

Answer: c11 = −30

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 = 0·0 + (−5)·6 = −30

Answer

c11 = −30
12.Matrices & determinants

Answer: x = −11.5

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 = (−6)·(−2) − (−4)·(−2) = 4

Replace the x column

det Aₓ = 5·(−2) − (−4)·(−9) = −46
x = −46/4 = −11.5

Answer

x = −11.5
y = 16
13.Vectors: dot & cross product

Answer: compb a = 2.683

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 = 0·(−1) + (−3)·(−2) + 3·0 = 6

Length of b

|b| = √((−1)² + (−2)² + 0²) = 2.236

Scalar projection

compb a = a·b/|b| = 6/2.236 = 2.683

Answer

compb a = 2.683
14.Vectors: dot & cross product

Answer: V = 8

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 b × c

(b × c)x = bycz − bzcy = 1·4 − 3·(−4) = 16
(b × c)y = bzcx − bxcz = 3·3 − (−3)·4 = 21
(b × c)z = bxcy − bycx = (−3)·(−4) − 1·3 = 9

Dot product with a

a·(b × c) = 5·16 + (−3)·21 + (−1)·9 = 8
V = |8| = 8

Answer

V = 8
15.Probability

Answer: P = 0.3056

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)2 = 0.6944
P = 1 − 0.6944 = 0.3056

Answer

P = 0.3056
16.Probability

Answer: E(G) = 1.5 €

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: 9 (p = 0.25), 147 (p = 0.01), −3 (p = 0.74)

Expected value

E(G) = 9·0.25 + 147·0.01 − 3·0.74 = 1.5 €

Answer

E(G) = 1.5 €

A positive value means you win money on average.

17.Descriptive statistics

Answer: s = 13.17

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 = 213/6 = 35.5

Squared deviations

Σ(x − x̄)² = 15.5² + 7.5² + 8.5² + (−18.5)² + (−0.5)² + (−12.5)² = 867.5

Standard deviation

s = √(867.5/5) = 13.17

Answer

s = 13.17
18.Descriptive statistics

Answer: P = 0.9332

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 = −1.5

Standard normal table

Φ(−1.5) = 0.06681
P = 1 − 0.06681 = 0.9332

Answer

P = 0.9332
19.Normal distribution

Answer: z = 0.39

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 + 0.3035)/2 = 0.6517

Read the z-table backwards

Find 0.6517 inside the table and read off z.

z = 0.39

Answer

z = 0.39
20.Normal distribution

Answer: x = 80.78 points

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 for the percentile

Φ(z) = 0.1 → z = −1.282

Back to x

x = 100 + −1.282·15 = 80.78 points

Answer

x = 80.78 points

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