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→∞ (−2x2 − 2x + 1)/(4x2 + x + 5).

f(x) = (−2x2 − 2x + 1)/(4x2 + x + 5)x → ∞
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

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

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

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

f(x) = x2 + 5xx = −2
b =
5.IntegralsMedium
xy01

Evaluate I = ∫01 2x·(x² + 1)2 dx.

f(x) = 2x·(x² + 1)2[0, 1]
I =
6.IntegralsHard
xy1y = x²eˣ

Evaluate I = ∫₀1 x²·eˣ dx.

∫₀1 x²eˣ dx
I =
7.Complex numbersMedium
ReIm22−2−244−4−466−6−6θz

The number z = 6·e−j0.5 has its angle in radians. Find its imaginary part b.

r = 6θ = −0.5 rad
b =
8.Complex numbersHard
ReIm11−1−122−2−233−3−3z₁z₂

One root of z³ − 4z² + 14z − 20 = 0 is z₁ = 1 + j3. The coefficients are real. Find the real root r.

z₁ = 1 + j3
r =
9.Differential equationsMedium
tyy(0) = 1y(1) = ?

Solve y″ + 8y′ + 16y = 0 with y(0) = 1 and y′(0) = 1. Find y(1).

y″ + 8y′ + 16y = 0y(0) = 1y′(0) = 1
y(1) =
10.Differential equationsHard
tyK = 2y(0.5) = ?

A system obeys y″ + 3y′ + 6.25y = 12.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.5); angles in radians.

y″ + 3y′ + 6.25y = 12.5y(0) = 0y′(0) = 0
y(0.5) =
11.Matrices & determinantsMedium
A =−33−60B =−5−2−5−1

Let C = AB for the matrices shown. Find the entry c22 (row 2, column 2).

A = [−3 3; −6 0]B = [−5 −2; −5 −1]
c22 =
12.Matrices & determinantsHard
A =−43−2−5b =−4−2

Solve the system −4x + 3y = −4 and −2x − 5y = −2. Find x.

−4x + 3y = −4−2x − 5y = −2
x =
13.Vectors: dot & cross productMedium
xyzab

Find the dot product a·b of a = (−3, −4, −2) and b = (3, 5, 6).

a = (−3, −4, −2)b = (3, 5, 6)
a·b =
14.Vectors: dot & cross productHard
xyzrFOA

A force F = (−60, 10, 50) N acts at point A(−3, −2, 2). Coordinates are in metres. Find the x-component Mx of the moment of the force about point O(2, −1, 2).

F = (−60, 10, 50) NA(−3, −2, 2) mO(2, −1, 2) m
Mx = 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)−100.79100.2900.01

A game costs €10 to play. You win €20 with probability 0.2 and €100 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €10€20: p = 0.2€100: p = 0.01
E(G) = €
17.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 15, 53, 51, 50, 25.

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

Exam scores are normally distributed with mean 60 and standard deviation 10. Find the z-score z of a score of 80.

μ = 60σ = 10x = 80
z =
19.Normal distributionMedium
200 mm199.5 mmX + Y?

X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 199.5 mm. Round z to two decimals.

μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 199.5 mm
P =
20.Normal distributionHard
380420460500540580620hbottom 80%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 80% is to the left of it. Find the cut-off x.

μ = 500 hσ = 40 h80%
x = h

3Solutions

1.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).

Divide by the highest power

(−2 + (−2)/x + 1/x2)/(4 + 1/x + 5/x2)

Let x → ∞

L = −2/4 = −0.5

Answer

L = −0.5

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

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) = 1.382

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(x) + x·cos(x)

Substitute

f′(1) = 0.8415 + 1·0.5403 = 1.382

Answer

f′(1) = 1.382
4.Derivatives

Answer: b = −4

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(−2) = −6
f′(x) = 2x + 5 → f′(−2) = 1

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = −6 − 1·(−2) = −4

Answer

b = −4
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: I = 0.7183

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.

First round

∫x²eˣ dx = x²eˣ − ∫2x eˣ dx

Second round

∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)

Limits

2.718·(1) − 1·2 = 0.7183

Answer

I = 0.7183
7.Complex numbers

Answer: b = −2.877

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 = 6·(cos −0.5 + j sin −0.5)

Components (radians)

a = 6·cos −0.5 = 5.265
b = 6·sin −0.5 = −2.877
b = −2.877

Answer

b = −2.877
8.Complex numbers

Answer: r = 2

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 = 1 − j3
(z − z₁)(z − z₂) = z² − 2·1·z + (1² + 3²) = z² − 2z + 10

Factor out the quadratic

z³ − 4z² + 14z − 20 = (z² − 2z + 10)(z − 2)

Real root

r = 2

Answer

r = 2

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

9.Differential equations

Answer: y(1) = 0.1099

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.

Double root

(r + 4)² = 0 → r = −4

Constants

A = y(0) = 1
y′(0) = B − 4A → B = 1 + 4 = 5

Evaluate

y(1) = (1 + 5·1)·e−4 = 0.1099

Answer

y(1) = 0.1099
10.Differential equations

Answer: y(0.5) = 0.8933

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 = 2, r = −1.5 ± 2j

Constants from the initial values

y(0) = 2 + A = 0 → A = −2
y′(0) = −1.5A + 2B = 0 → B = −1.5

Evaluate

y(0.5) = 2·(1 − e−0.75(cos 1 + 0.75·sin 1)) = 0.8933

Answer

y(0.5) = 0.8933
11.Matrices & determinants

Answer: c22 = 12

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

c22 = (−6)·(−2) + 0·(−1) = 12

Answer

c22 = 12
12.Matrices & determinants

Answer: x = 1

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 = (−4)·(−5) − 3·(−2) = 26

Replace the x column

det Aₓ = (−4)·(−5) − 3·(−2) = 26
x = 26/26 = 1

Answer

x = 1
y = 0
13.Vectors: dot & cross product

Answer: a·b = −41

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.

Multiply matching components and add

a·b = (−3)·3 + (−4)·5 + (−2)·6 = (−9) + (−20) + (−12) = −41

Answer

a·b = −41

Negative: the angle between the vectors is more than 90°.

14.Vectors: dot & cross product

Answer: Mx = −50 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, −2, 2) − (2, −1, 2) = (−5, −1, 0) m

Moment M = r × F

(r × F)x = ryFz − rzFy = (−1)·50 − 0·10 = −50 N·m
(r × F)y = rzFx − rxFz = 0·(−60) − (−5)·50 = 250 N·m
(r × F)z = rxFy − ryFx = (−5)·10 − (−1)·(−60) = −110 N·m

Requested component

Mx = −50 N·m

Answer

Mx = −50 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) = −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: 10 (p = 0.2), 90 (p = 0.01), −10 (p = 0.79)

Expected value

E(G) = 10·0.2 + 90·0.01 − 10·0.79 = −5 €

Answer

E(G) = −5 €

A negative value means you lose money on average.

17.Descriptive statistics

Answer: s = 17.56

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 = 194/5 = 38.8

Squared deviations

Σ(x − x̄)² = (−23.8)² + 14.2² + 12.2² + 11.2² + (−13.8)² = 1233

Standard deviation

s = √(1233/4) = 17.56

Answer

s = 17.56
18.Descriptive statistics

Answer: z = 2

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 = (80 − 60)/10 = 2

Answer

z = 2
19.Normal distribution

Answer: P = 0.8413

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 sum

μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm

Standardise

z = (199.5 − 200)/0.5 = −1

Upper tail

P = 1 − Φ(−1) = 1 − 0.1587 = 0.8413

Answer

P = 0.8413
20.Normal distribution

Answer: x = 534 h

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) = 0.8

z from the table

z ≈ 0.84

Back to the original units

x = 500 + 0.84·40 = 534 h

Answer

x = 534 h

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