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 + 2/x)2x. Give a decimal value.

f(x) = (1 + 2/x)2xx → ∞
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 = 0.5

Let f(x) = x·sin(3x), with x in radians. Find f′(0.5).

f(x) = x·sin(3x)x = 0.5 rad
f′(0.5) =
4.DerivativesHard
xyx = 3

The tangent line to f(x) = 3x2 + 1 at x = 3 crosses the y-axis at y = b. Find b.

f(x) = 3x2 + 1x = 3
b =
5.IntegralsMedium
xy03

Evaluate I = ∫03 e−x dx.

f(x) = e−x[0, 3]
I =
6.IntegralsHard
xy−22

Find the area A of the region between the parabola y = 4 − x² and the x-axis.

y = 4 − x²
A =
7.Complex numbersMedium
ReIm0.50.5−0.5−0.511−1−1θz

The number z = 1.5·ej0.6 has its angle in radians. Find its imaginary part b.

r = 1.5θ = 0.6 rad
b =
8.Complex numbersHard
ReIm22−2−244−4−4z₀z₁z₂

The roots of zⁿ = w with n = 3 and w = 64∠180° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2. Find the argument of z₁ in degrees, between −180° and 180°.

w = 64∠180°n = 3k = 1
θ₁ = °
9.Differential equationsMedium
tyy(0) = 3y(1) = ?

Solve y″ + 4y′ + 4y = 0 with y(0) = 3 and y′(0) = 1. Find y(1).

y″ + 4y′ + 4y = 0y(0) = 3y′(0) = 1
y(1) =
10.Differential equationsHard
tyK = 5ymax​ = ?

A system obeys y″ + y′ + 9.25y = 46.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.

y″ + y′ + 9.25y = 46.25y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
A =5−4−4−1−1−4422

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

A = [5 −4 −4; −1 −1 −4; 4 2 2]
det A =
12.Matrices & determinantsHard
A =−6−2−36b =22

Solve the system −6x − 2y = 2 and −3x + 6y = 2. Find x.

−6x − 2y = 2−3x + 6y = 2
x =
13.Vectors: dot & cross productMedium
xyzab

Find the dot product a·b of a = (4, −1, 3) and b = (3, −2, −4).

a = (4, −1, 3)b = (3, −2, −4)
a·b =
14.Vectors: dot & cross productHard
xyzab

Find the area A of the parallelogram spanned by a = (5, −3, 0) and b = (−4, −4, 5).

a = (5, −3, 0)b = (−4, −4, 5)
A =
15.ProbabilityMedium
n = 9k = 2N = ?

In how many ways N can a committee of 2 people be chosen from a group of 9? The order does not matter.

n = 9k = 2
N =
16.ProbabilityHard
xp(x)10.1520.3530.2540.25μ

X has the distribution shown, and Y = 0.5X + 5. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.15, 0.35, 0.25, 0.25Y = 0.5X + 5
σY =
17.Descriptive statisticsMedium
15202530354045

Find the sample standard deviation s of the data: 18, 27, 46, 18, 13, 32.

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

A quantity is normally distributed with mean 170 and standard deviation 20. What proportion P of values is above 182?

μ = 170σ = 20x = 182
P =
19.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.2886

Drag z so that the shaded area between −z and z under the standard normal curve is 0.2886. Find z to two decimals.

A = 0.2886
z =
20.Normal distributionHard
557085100115130145pointstop 10%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 10% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points10%
x = points

3Solutions

1.Limits

Answer: L = 54.6

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→∞ 2·x·ln(1 + 2/x)

Evaluate

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

ln L = 2·2 = 4

Undo the logarithm

L = e4 = 54.6

Answer

L = 54.6
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′(0.5) = 1.104

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(3x) + 3x·cos(3x)

Substitute

f′(0.5) = 0.9975 + 1.5·0.07074 = 1.104

Answer

f′(0.5) = 1.104
4.Derivatives

Answer: b = −26

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) = 28
f′(x) = 6x → f′(3) = 18

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 28 − 18·3 = −26

Answer

b = −26
5.Integrals

Answer: I = 0.9502

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) = e−x/(−1)

Evaluate at the limits

I = (e−3 − 1)/(−1) = (0.04979 − 1)/(−1) = 0.9502

Answer

I = 0.9502
6.Integrals

Answer: A = 10.67

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.

Zeros

4 − x² = 0 → x = ±2

Integrate

A = [4x − x³/3]−22 = 5.333 − (−5.333) = 10.67

Answer

A = 10.67
7.Complex numbers

Answer: b = 0.847

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 = 1.5·(cos 0.6 + j sin 0.6)

Components (radians)

a = 1.5·cos 0.6 = 1.238
b = 1.5·sin 0.6 = 0.847
b = 0.847

Answer

b = 0.847
8.Complex numbers

Answer: θ₁ = 180 °

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.

Modulus of the roots

641/3 = 4

Argument of the root

θ₁ = (180° + 360°·1)/3 = 180°

Argument

θ₁ = 180°

Answer

θ₁ = 180°

The 3 roots lie on the circle of radius 4, 120° apart.

9.Differential equations

Answer: y(1) = 1.353

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 + 2)² = 0 → r = −2

Constants

A = y(0) = 3
y′(0) = B − 2A → B = 1 + 6 = 7

Evaluate

y(1) = (3 + 7·1)·e−2 = 1.353

Answer

y(1) = 1.353
10.Differential equations

Answer: ymax = 7.962

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 = 5, r = −0.5 ± 3j

Constants from the initial values

y(0) = 5 + A = 0 → A = −5
y′(0) = −0.5A + 3B = 0 → B = −0.8333

Peak

t = π/3 = 1.047
ymax = 5·(1 + e−0.5·π/3) = 7.962

Answer

ymax = 7.962
11.Matrices & determinants

Answer: det A = 78

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 = 6, M₁₂ = 14, M₁₃ = 2

Cofactor expansion

det A = 5·6 − (−4)·14 + (−4)·2 = 78

Answer

det A = 78
12.Matrices & determinants

Answer: x = −0.381

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)·6 − (−2)·(−3) = −42

Replace the x column

det Aₓ = 2·6 − (−2)·2 = 16
x = 16/(−42) = −0.381

Answer

x = −0.381
y = 0.1429
13.Vectors: dot & cross product

Answer: a·b = 2

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 = 4·3 + (−1)·(−2) + 3·(−4) = 12 + 2 + (−12) = 2

Answer

a·b = 2

Positive: the angle between the vectors is less than 90°.

14.Vectors: dot & cross product

Answer: A = 43.29

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)·5 − 0·(−4) = −15
(a × b)y = azbx − axbz = 0·(−4) − 5·5 = −25
(a × b)z = axby − aybx = 5·(−4) − (−3)·(−4) = −32

Area

A = |a × b| = √((−15)² + (−25)² + (−32)²) = 43.29

Answer

A = 43.29
15.Probability

Answer: N = 36

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.

Combinations

C(9, 2) = (9·8)/(2·1) = 36

Answer

N = 36
16.Probability

Answer: σY = 0.5099

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.

Spread of X

μ = 2.6, Var(X) = 1.04, σX = 1.02

Linear change

σY = |0.5|·1.02 = 0.5099

Answer

σY = 0.5099

The mean does move: E(Y) = 0.5·2.6 + 5 = 6.3.

17.Descriptive statistics

Answer: s = 12.11

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 = 154/6 = 25.67

Squared deviations

Σ(x − x̄)² = (−7.667)² + 1.333² + 20.33² + (−7.667)² + (−12.67)² + 6.333² = 733.3

Standard deviation

s = √(733.3/5) = 12.11

Answer

s = 12.11
18.Descriptive statistics

Answer: P = 0.2743

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743

Answer

P = 0.2743
19.Normal distribution

Answer: z = 0.37

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 + 0.2886)/2 = 0.6443

Read the z-table backwards

Find 0.6443 inside the table and read off z.

z = 0.37

Answer

z = 0.37
20.Normal distribution

Answer: x = 119.2 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) = 0.9

z from the table

z ≈ 1.28

Back to the original units

x = 100 + 1.28·15 = 119.2 points

Answer

x = 119.2 points

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