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)3x. Give a decimal value.

f(x) = (1 − 2/x)3xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (e−0.5x − 1 + 0.5x)/x².

f(x) = (e−0.5x − 1 + 0.5x)/x²x → 0
L =
3.DerivativesMedium
xyx = 1.5

Let f(x) = x·e0.5x. Find f′(1.5).

f(x) = x·e0.5xx = 1.5
f′(1.5) =
4.DerivativesHard
xy

Let f(x) = x3 − 13.5x2 + 54x + 4. Find the x-coordinate x of the local minimum.

f(x) = x3 − 13.5x2 + 54x + 4
x =
5.IntegralsMedium
xy02

Evaluate I = ∫02 2x·(x² + 1)3 dx.

f(x) = 2x·(x² + 1)3[0, 2]
I =
6.IntegralsHard
xy−11

Find the area A of the region between the parabola y = 1 − x² and the x-axis.

y = 1 − x²
A =
7.Complex numbersMedium
ReIm22−2−244−4−4θz

Find the argument θ of z = 5 + j5 in degrees, as the principal value between −180° and 180°.

z = 5 + j5
θ = °
8.Complex numbersHard
ReIm11−1−122−2−2θz

Find the real part of (1 − j2)⁴.

z = 1 − j2n = 4
Re(z⁴) =
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
xy13(1, 0)y(3) = ?

Solve the initial value problem y′ + 3y/x = 6 for x > 0, with y(1) = 0. Find y(3).

y′ + 3y/x = 6y(1) = 0
y(3) =
11.Matrices & determinantsMedium
A =−3−55−42−115−5

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

A = [−3 −5 5; −4 2 −1; 1 5 −5]
det A =
12.Matrices & determinantsHard
A =56−3−6

Find the larger eigenvalue λ of the matrix A shown.

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

A constant force F = (−15, −20, 0) N acts on an object while it moves in a straight line from point A(3, 0, 4) to point B(2, −1, −3). Coordinates are in metres. Find the work W done by the force.

F = (−15, −20, 0) NA(3, 0, 4) mB(2, −1, −3) m
W = J
14.Vectors: dot & cross productHard
xyzABC

Find the area A of the triangle with corners A(−3, −3, 0), B(−4, −2, 2) and C(−4, 2, 4). Coordinates are in metres.

A(−3, −3, 0)B(−4, −2, 2)C(−4, 2, 4)
A = m²
15.ProbabilityMedium
xp(x)100.35200.25300.05400.35μ = 24

For the distribution shown, find the standard deviation σ of X.

x: 10, 20, 30, 40p(x): 0.35, 0.25, 0.05, 0.35
σ =
16.ProbabilityHard
n = 7, p = 0.6: P(X ≤ 1) = ?k01234567

7 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?

n = 7p = 0.6
P =
17.Descriptive statisticsMedium
4045505560

Find the sample standard deviation s of the data: 60, 37, 36, 41, 37, 48, 52.

n = 7
s =
18.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 60 and standard deviation 10. What proportion P of values is above 45?

μ = 60σ = 10x = 45
P =
19.Normal distributionMedium
167.15181.85P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 167.15 and 181.85 cm? Give P as a decimal.

μ = 178σ = 7x1 = 167.15x2 = 181.85
P =
20.Normal distributionHard
μ = 250x = ?25%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 25% of them lie (the 25th percentile).

μ = 250σ = 5P = 25%
x = g

3Solutions

1.Limits

Answer: L = 0.002479

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→∞ 3·x·ln(1 − 2/x)

Evaluate

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

ln L = 3·(−2) = −6

Undo the logarithm

L = e−6 = 0.002479

Answer

L = 0.002479
2.Limits

Answer: L = 0.125

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

(−0.5·e−0.5x + 0.5)/(2x) → 0/0

Second application

0.25·e−0.5x/2

Substitute x = 0

L = 0.25·1/2 = 0.125

Answer

L = 0.125
3.Derivatives

Answer: f′(1.5) = 3.705

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 rule

f′(x) = u′v + uv′ = e0.5x·(1 + 0.5·x)

Substitute

f′(1.5) = 2.117·(1 + 0.75) = 3.705

Answer

f′(1.5) = 3.705
4.Derivatives

Answer: x = 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.

Derivative

f′(x) = 3x2 − 27x + 54 = 3·(x2 − 9x + 18)

Solve f′(x) = 0

x = 3 or x = 6

Which one is which

f″(x) = 6x − 27: negative at a maximum, positive at a minimum.

f″(3) = −9, f″(6) = 9
x = 6

Answer

x = 6
5.Integrals

Answer: I = 156

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 = ∫15 u3 du

Integrate and evaluate

I = [u4/4]15 = (625 − 1)/4 = 156

Answer

I = 156
6.Integrals

Answer: A = 1.333

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.

Zeros

1 − x² = 0 → x = ±1

Integrate

A = [1x − x³/3]−11 = 0.6667 − (−0.6667) = 1.333

Answer

A = 1.333
7.Complex numbers

Answer: θ = 45 °

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.

Quadrant

a = 5, b = 5 → I

Reference angle

α = atan(5/5) = 45°

Argument

θ = α = 45°

Answer

θ = 45°

In polar form z = 7.071∠45°.

8.Complex numbers

Answer: Re(z⁴) = −7

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.

Powers step by step

z² = (1 − j2)(1 − j2) = −3 − j4
z³ = (−3 − j4)(1 − j2) = −11 + j2
z⁴ = (−11 + j2)(1 − j2) = −7 + j24

Check with De Moivre

z = 2.236∠−63.43° → z⁴ = 25∠106.3°
Re(z⁴) = −7

Answer

Re(z⁴) = −7
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(3) = 4.444

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 = e∫3/x dx = x3

Integrate

(x3y)′ = 6x3 → x3y = 6x4/4 + C

Initial condition

0 = 6/4 + C → C = −1.5

Evaluate

y(3) = (6·34/4 + (−1.5))/33 = 4.444

Answer

y(3) = 4.444
11.Matrices & determinants

Answer: det A = 10

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 = −5, M₁₂ = 21, M₁₃ = −22

Cofactor expansion

det A = (−3)·(−5) − (−5)·21 + 5·(−22) = 10

Answer

det A = 10
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: W = 35 J

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.

Displacement

d = B − A = (2, −1, −3) − (3, 0, 4) = (−1, −1, −7) m

Work

W = F·d = (−15)·(−1) + (−20)·(−1) + 0·(−7) = 35 J

Answer

W = 35 J
14.Vectors: dot & cross product

Answer: A = 3.742 m²

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.

Edge vectors

AB = B − A = (−1, 1, 2)
AC = C − A = (−1, 5, 4)

Cross product

(AB × AC)x = AByACz − ABzACy = 1·4 − 2·5 = −6
(AB × AC)y = ABzACx − ABxACz = 2·(−1) − (−1)·4 = 2
(AB × AC)z = ABxACy − AByACx = (−1)·5 − 1·(−1) = −4
|AB × AC| = √((−6)² + 2² + (−4)²) = 7.483

Area of the triangle

A = ½·7.483 = 3.742 m²

Answer

A = 3.742 m²
15.Probability

Answer: σ = 12.81

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 = 10·0.35 + 20·0.25 + 30·0.05 + 40·0.35 = 24

Mean of the squares

E(X²) = 100·0.35 + 400·0.25 + 900·0.05 + 1600·0.35 = 740

Variance

Var(X) = 740 − 24² = 164
σ = √164 = 12.81

Answer

σ = 12.81
16.Probability

Answer: P = 0.01884

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.

Zero and one failure

P(X = 0) = 0.47 = 0.001638
P(X = 1) = 7·0.6·0.46 = 0.0172

Add them

P = 0.001638 + 0.0172 = 0.01884

Answer

P = 0.01884
17.Descriptive statistics

Answer: s = 9.181

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 = 311/7 = 44.43

Squared deviations

Σ(x − x̄)² = 15.57² + (−7.429)² + (−8.429)² + (−3.429)² + (−7.429)² + 3.571² + 7.571² = 505.7

Standard deviation

s = √(505.7/6) = 9.181

Answer

s = 9.181
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: P = 0.6483

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 = (167.15 − 178)/7 = −1.55
z₂ = (181.85 − 178)/7 = 0.55

Area between

P = Φ(0.55) − Φ(−1.55) = 0.7088 − 0.06057 = 0.6483

Answer

P = 0.6483
20.Normal distribution

Answer: x = 246.6 g

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) = 0.25 → z = −0.6745

Back to x

x = 250 + −0.6745·5 = 246.63 g

Answer

x = 246.63 g

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