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→∞ (−3x2 + 2x − 1)/(4x2 + 6x + 6).

f(x) = (−3x2 + 2x − 1)/(4x2 + 6x + 6)x → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (e1.5x − 1 − 1.5x)/x².

f(x) = (e1.5x − 1 − 1.5x)/x²x → 0
L =
3.DerivativesMedium
xyx = 3

Let f(x) = ln(3x2 + 4). Find f′(3).

f(x) = ln(3x2 + 4)x = 3
f′(3) =
4.DerivativesHard
xyx = −2

The tangent line to f(x) = 0.5x2 − 5x − 6 at x = −2 crosses the y-axis at y = b. Find b.

f(x) = 0.5x2 − 5x − 6x = −2
b =
5.IntegralsMedium
xyπ/3y = x·cos x

Evaluate I = ∫₀π/3 x·cos x dx (x in radians).

∫₀π/3 x·cos x dx
I =
6.IntegralsHard
xy0.5y = x²eˣ

Evaluate I = ∫₀0.5 x²·eˣ dx.

∫₀0.5 x²eˣ dx
I =
7.Complex numbersMedium
ReIm22−2−244−4−4z₁z₂w

Given z₁ = 2∠125° and z₂ = 2∠−25°, find the argument θ of w = z₁·z₂ in degrees, between −180° and 180°.

z₁ = 2∠125°z₂ = 2∠−25°
θ = °
8.Complex numbersHard
ReIm11−1−122−2−2z₁z₂

Solve z² − 5z + 10.25 = 0. The roots are a ± jb; find b > 0.

p = −5q = 10.25
b =
9.Differential equationsMedium
tyy(0) = 2y(2) = ?

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

y″ + 3y′ + 2y = 0y(0) = 2y′(0) = 4
y(2) =
10.Differential equationsHard
tyy(0) = 10y(1.5) = ?

Solve y″ + 3y′ + 6.25y = 0 with y(0) = 10 and y′(0) = 0 (released from rest). Find y(1.5); angles in radians.

y″ + 3y′ + 6.25y = 0y(0) = 10y′(0) = 0
y(1.5) =
11.Matrices & determinantsMedium
A =2−202B =5−5−46

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

A = [2 −2; 0 2]B = [5 −5; −4 6]
c21 =
12.Matrices & determinantsHard
A =5−160

Find the larger eigenvalue λ of the matrix A shown.

A = [5 −1; 6 0]
λ =
13.Vectors: dot & cross productMedium
xyzab

Find the angle θ between the vectors a = (0, −2, 4) and b = (−3, 2, −2), in degrees.

a = (0, −2, 4)b = (−3, 2, −2)
θ = °
14.Vectors: dot & cross productHard
xyzrFOA

A force F = (−30, 10, 30) N acts at point A(4, 4, 0). Coordinates are in metres. Find the y-component My of the moment of the force about the origin O(0, 0, 0).

F = (−30, 10, 30) NA(4, 4, 0) mO(0, 0, 0) m
My = N·m
15.ProbabilityMedium
xp(x)00.120.3540.260.35μ = 3.6

For the distribution shown, find the variance Var(X) of X.

x: 0, 2, 4, 6p(x): 0.1, 0.35, 0.2, 0.35
Var(X) =
16.ProbabilityHard
n = 11, p = 0.1: P(X = 3) = ?k01234567891011

A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 3 correct answers?

n = 11p = 0.1k = 3
P =
17.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 11, 45, 51, 32, 45, 58.

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

A quantity is normally distributed with mean 50 and standard deviation 5. What proportion P of values is above 47.5?

μ = 50σ = 5x = 47.5
P =
19.Normal distributionMedium
92.5103.75P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 92.5 and 103.75 points? Give P as a decimal.

μ = 100σ = 15x1 = 92.5x2 = 103.75
P =
20.Normal distributionHard
157164171178185192199cmtop 70%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 70% is to the right of it. Find the cut-off x.

μ = 178 cmσ = 7 cm70%
x = cm

3Solutions

1.Limits

Answer: L = −0.75

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

(−3 + 2/x + (−1)/x2)/(4 + 6/x + 6/x2)

Let x → ∞

L = −3/4 = −0.75

Answer

L = −0.75

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

2.Limits

Answer: L = 1.125

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

(1.5·e1.5x − 1.5)/(2x) → 0/0

Second application

2.25·e1.5x/2

Substitute x = 0

L = 2.25·1/2 = 1.125

Answer

L = 1.125
3.Derivatives

Answer: f′(3) = 0.5806

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 + 4)

Substitute

f′(3) = 18/31 = 0.5806

Answer

f′(3) = 0.5806
4.Derivatives

Answer: b = −8

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) = x − 5 → f′(−2) = −7

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 6 − (−7)·(−2) = −8

Answer

b = −8
5.Integrals

Answer: I = 0.4069

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.

Choose u and dv

u = x, du = dx; dv = cos x dx, v = sin x

Antiderivative

∫x cos x dx = x sin x − ∫sin x dx = x sin x + cos x

Limits

(π/3·sin π/3 + cos π/3) − (0 + 1) = 0.4069

Answer

I = 0.4069
6.Integrals

Answer: I = 0.0609

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

1.649·(1.25) − 1·2 = 0.0609

Answer

I = 0.0609
7.Complex numbers

Answer: θ = 100 °

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| = 2 · 2 = 4
θ = 125° + (−25)° = 100°

Argument

θ = 100°

Answer

θ = 100°
8.Complex numbers

Answer: b = 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.

Discriminant

p² − 4q = 25 − 41 = −16 < 0

Quadratic formula

z = (5 ± j√16)/2 = 2.5 + j2, 2.5 − j2

Answer

b = 2

Answer

b = 2
9.Differential equations

Answer: y(2) = 0.9728

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 + 2) = 0 → r = −1, −2

Constants from the initial values

A + B = 2, −1A − 2B = 4
A = 8, B = −6

Evaluate

y(2) = 8·e−2 + (−6)·e−4 = 0.9728

Answer

y(2) = 0.9728
10.Differential equations

Answer: y(1.5) = −0.9319

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.

Complex roots

r = −1.5 ± 2j

Constants

A = 10, B = 1.5·10/2 = 7.5

Evaluate

y(1.5) = e−2.25(10·cos 3 + 7.5·sin 3) = −0.9319

Answer

y(1.5) = −0.9319
11.Matrices & determinants

Answer: c21 = −8

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 = 0·5 + 2·(−4) = −8

Answer

c21 = −8
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 − (5)λ + (6) = 0

Solve

λ = (5 ± √(1))/2 → λ = 2, 3
λ = 3

Answer

λ = 3
13.Vectors: dot & cross product

Answer: θ = 130.6 °

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·(−3) + (−2)·2 + 4·(−2) = −12

Magnitudes

|a| = √(0² + (−2)² + 4²) = 4.472
|b| = √((−3)² + 2² + (−2)²) = 4.123

Angle

cos θ = −12/(4.472·4.123) = −0.6508
θ = arccos(−0.6508) = 130.6°

Answer

θ = 130.6°
14.Vectors: dot & cross product

Answer: My = −120 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 = (4, 4, 0) − (0, 0, 0) = (4, 4, 0) m

Moment M = r × F

(r × F)x = ryFz − rzFy = 4·30 − 0·10 = 120 N·m
(r × F)y = rzFx − rxFz = 0·(−30) − 4·30 = −120 N·m
(r × F)z = rxFy − ryFx = 4·10 − 4·(−30) = 160 N·m

Requested component

My = −120 N·m

Answer

My = −120 N·m
15.Probability

Answer: Var(X) = 4.24

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.

Mean

μ = 0·0.1 + 2·0.35 + 4·0.2 + 6·0.35 = 3.6

Mean of the squares

E(X²) = 0·0.1 + 4·0.35 + 16·0.2 + 36·0.35 = 17.2

Variance

Var(X) = 17.2 − 3.6² = 4.24

Answer

Var(X) = 4.24
16.Probability

Answer: P = 0.07103

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 = 3) = 165·0.13·0.98 = 165·0.001·0.4305 = 0.07103

Answer

P = 0.07103
17.Descriptive statistics

Answer: s = 16.73

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 = 242/6 = 40.33

Squared deviations

Σ(x − x̄)² = (−29.33)² + 4.667² + 10.67² + (−8.333)² + 4.667² + 17.67² = 1399

Standard deviation

s = √(1399/5) = 16.73

Answer

s = 16.73
18.Descriptive statistics

Answer: P = 0.6915

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 = −0.5

Standard normal table

Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915

Answer

P = 0.6915
19.Normal distribution

Answer: P = 0.2902

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 = (92.5 − 100)/15 = −0.5
z₂ = (103.75 − 100)/15 = 0.25

Area between

P = Φ(0.25) − Φ(−0.5) = 0.5987 − 0.3085 = 0.2902

Answer

P = 0.2902
20.Normal distribution

Answer: x = 174.4 cm

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) = 0.3

z from the table

z ≈ −0.52

Back to the original units

x = 178 − 0.52·7 = 174.4 cm

Answer

x = 174.4 cm

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