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 + 3/x)2x. Give a decimal value.

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

Find the limit L = limx→0 ln(1 + x)/sin(5x), with x in radians.

f(x) = ln(1 + x)/sin(5x)x → 0
L =
3.DerivativesMedium
xyx = 1.5

Let f(x) = x3·e−0.5x. Find f′(1.5).

f(x) = x3·e−0.5xx = 1.5
f′(1.5) =
4.DerivativesHard
xyx = −1

The tangent line to f(x) = 3x2 + 5x at x = −1 crosses the y-axis at y = b. Find b.

f(x) = 3x2 + 5xx = −1
b =
5.IntegralsMedium
xy4y = 3x

The region under y = 3x from x = 0 to x = 4 is rotated about the x-axis. Find the volume V of the solid (a cone).

y = 3x0 ≤ x ≤ 4
V =
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
ReIm22−2−244−4−466−6−6θz

Find the argument θ of z = −6 − j2 in degrees, as the principal value between −180° and 180°.

z = −6 − j2
θ = °
8.Complex numbersHard
ReIm11−1−122−2−2ABCD

Exactly one of the points A–D is a root of z⁴ = 16∠90°. Click it.

w = 16∠90°n = 4
  1. 1A (1∠−157.5°)
  2. 2B (2∠157.5°)
  3. 3C (2∠−157.5°)
  4. 4D (2∠−112.5°)
9.Differential equationsMedium
t (min)T (°C)18Ta​T = ?

A forged steel part at 420 °C is left to cool in a workshop at 22 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.05 min⁻¹, find its temperature T after 18 min.

T0 = 420 °CTa = 22 °Ck = 0.05 min⁻¹t = 18 min
T = °C
10.Differential equationsHard
xy12(1, 5)y(2) = ?

Solve the initial value problem y′ + 3y/x = 6x for x > 0, with y(1) = 5. Find y(2).

y′ + 3y/x = 6xy(1) = 5
y(2) =
11.Matrices & determinantsMedium
A =214−2−1410−4

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

A = [2 1 4; −2 −1 4; 1 0 −4]
det A =
12.Matrices & determinantsHard
A =525−1b =−102

Solve the system 5x + 2y = −10 and 5x − y = 2. Find x.

5x + 2y = −105x − y = 2
x =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (2, 6, 3) onto b = (0, 3, 5), that is, the component of a in the direction of b.

a = (2, 6, 3)b = (0, 3, 5)
compb a =
14.Vectors: dot & cross productHard
xyzabc

Find the volume V of the parallelepiped whose edges from one corner are a = (5, −1, −2), b = (−3, 1, 1) and c = (0, −4, 3).

a = (5, −1, −2)b = (−3, 1, 1)c = (0, −4, 3)
V =
15.ProbabilityMedium
red: 8blue: 3

A bag holds 8 red and 3 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

8 / 3
P =
16.ProbabilityHard
n = 5, p = 0.1: P(X ≤ 1) = ?k012345

5 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?

n = 5p = 0.1
P =
17.Descriptive statisticsMedium
1520253035

Find the sample standard deviation s of the data: 14, 39, 31, 28, 31.

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

A quantity is normally distributed with mean 70 and standard deviation 20. What proportion P of values is above 46?

μ = 70σ = 20x = 46
P =
19.Normal distributionMedium
400 g407.6 gT = X₁ + … + X₁₀?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 10 of them exceeds 407.6 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 10> 407.6 g
P =
20.Normal distributionHard
35455565758595pointstop 90%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 90% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points90%
x = points

3Solutions

1.Limits

Answer: L = 403.4

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 + 3/x)

Evaluate

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

ln L = 2·3 = 6

Undo the logarithm

L = e6 = 403.4

Answer

L = 403.4
2.Limits

Answer: L = 0.2

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)]/[5·cos 5x]

Substitute x = 0

L = 1/5 = 0.2

Answer

L = 0.2
3.Derivatives

Answer: f′(1.5) = 2.391

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′ = e−0.5x·(3x2 + (−0.5)·x3)

Substitute

f′(1.5) = 0.4724·(6.75 + (−1.688)) = 2.391

Answer

f′(1.5) = 2.391
4.Derivatives

Answer: b = −3

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) = −2
f′(x) = 6x + 5 → f′(−1) = −1

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = −2 − (−1)·(−1) = −3

Answer

b = −3
5.Integrals

Answer: V = 603.2

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.

Disc method

V = π∫₀4 9x² dx

Integrate

V = π·9·4³/3 = 603.2

Answer

V = 603.2

Check: a cone of radius 12 and height 4 has (1/3)πr²h = 603.2.

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: θ = −161.6 °

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 = −6, b = −2 → III

Reference angle

α = atan(2/6) = 18.43°

Argument

θ = −(180° − α) = −161.6°

Answer

θ = −161.6°

In polar form z = 6.325∠−161.6°.

8.Complex numbers

Answer: C (2∠−157.5°)

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 and arguments of the roots

|z| = 2
θ = 22.5°, 112.5°, −157.5°, −67.5°

The point

C = 2∠−157.5°

Answer

C (2∠−157.5°)
9.Differential equations

Answer: T = 183.8 °C

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.

Solution of the cooling law

T(t) = 22 + 398·e−0.05t

Substitute t

T(18) = 22 + 398·0.4066 = 183.8 °C

Answer

T = 183.8 °C
10.Differential equations

Answer: y(2) = 5.275

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)′ = 6x4 → x3y = 6x5/5 + C

Initial condition

5 = 6/5 + C → C = 3.8

Evaluate

y(2) = (6·25/5 + 3.8)/23 = 5.275

Answer

y(2) = 5.275
11.Matrices & determinants

Answer: det A = 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.

2×2 minors of the first row

M₁₁ = 4, M₁₂ = 4, M₁₃ = 1

Cofactor expansion

det A = 2·4 − 1·4 + 4·1 = 8

Answer

det A = 8
12.Matrices & determinants

Answer: x = −0.4

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 = 5·(−1) − 2·5 = −15

Replace the x column

det Aₓ = (−10)·(−1) − 2·2 = 6
x = 6/(−15) = −0.4

Answer

x = −0.4
y = −4
13.Vectors: dot & cross product

Answer: compb a = 5.659

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 = 2·0 + 6·3 + 3·5 = 33

Length of b

|b| = √(0² + 3² + 5²) = 5.831

Scalar projection

compb a = a·b/|b| = 33/5.831 = 5.659

Answer

compb a = 5.659
14.Vectors: dot & cross product

Answer: V = 2

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 = 1·3 − 1·(−4) = 7
(b × c)y = bzcx − bxcz = 1·0 − (−3)·3 = 9
(b × c)z = bxcy − bycx = (−3)·(−4) − 1·0 = 12

Dot product with a

a·(b × c) = 5·7 + (−1)·9 + (−2)·12 = 2
V = |2| = 2

Answer

V = 2
15.Probability

Answer: P = 0.5091

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.

First and second draw

P = 8/11 · 7/10 = 56/110 = 0.5091

Answer

P = 0.5091
16.Probability

Answer: P = 0.9185

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) = 0.95 = 0.5905
P(X = 1) = 5·0.1·0.94 = 0.3281

Add them

P = 0.5905 + 0.3281 = 0.9185

Answer

P = 0.9185
17.Descriptive statistics

Answer: s = 9.127

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 = 143/5 = 28.6

Squared deviations

Σ(x − x̄)² = (−14.6)² + 10.4² + 2.4² + (−0.6)² + 2.4² = 333.2

Standard deviation

s = √(333.2/4) = 9.127

Answer

s = 9.127
18.Descriptive statistics

Answer: P = 0.8849

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849

Answer

P = 0.8849
19.Normal distribution

Answer: P = 0.02275

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 total

μT = 400 g, σT = 1.2·√10 = 3.795 g

Standardise

z = (407.6 − 400)/3.795 = 2

Probability

P = 1 − Φ(2) = 0.02275

Answer

P = 0.02275
20.Normal distribution

Answer: x = 52.2 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) = 0.1

z from the table

z ≈ −1.28

Back to the original units

x = 65 − 1.28·10 = 52.2 points

Answer

x = 52.2 points

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