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 (1 − cos 4x)/(2x²), with x in radians.

f(x) = (1 − cos 4x)/(2x²)x → 0
L =
3.DerivativesMedium
xyx = 1.5

Let f(x) = x·e2x. Find f′(1.5).

f(x) = x·e2xx = 1.5
f′(1.5) =
4.DerivativesHard
xyx = 2

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

f(x) = x2 + 5x − 6x = 2
b =
5.IntegralsMedium
xy2y = x·sin x

Evaluate I = ∫₀2 x·sin x dx (x in radians).

∫₀2 x·sin x dx
I =
6.IntegralsHard
xy2.5y = x²eˣ

Evaluate I = ∫₀2.5 x²·eˣ dx.

∫₀2.5 x²eˣ dx
I =
7.Complex numbersMedium
ReIm11−1−122−2−2θz

The number z = 2.5·ej2.4 has its angle in radians. Find its real part a.

r = 2.5θ = 2.4 rad
a =
8.Complex numbersHard
ReIm11−1−122−2−2θz

Use De Moivre's theorem to find the imaginary part of z³ for z = −2 − j2.

z = −2 − j2n = 3
Im(z³) =
9.Differential equationsMedium
t (min)T (°C)38Ta​219 °C

A forged steel part at 440 °C is left to cool in a workshop at 24 °C. After 38 min its temperature is 219 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 440 °CTa = 24 °CT(38) = 219 °C
k = min⁻¹
10.Differential equationsHard
t (min)m (kg)40 minm = ?

A tank holds 500 L of pure water. Brine with 0.2 kg of salt per litre flows in at 10 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 40 min?

V = 500 Lm0 = 0 kgc = 0.2 kg/Lr = 10 L/min
m = kg
11.Matrices & determinantsMedium
A =4−4−2−31340−3

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

A = [4 −4 −2; −3 1 3; 4 0 −3]
det A =
12.Matrices & determinantsHard
A =4−1−56b =68

Solve the system 4x − y = 6 and −5x + 6y = 8. Find x.

4x − y = 6−5x + 6y = 8
x =
13.Vectors: dot & cross productMedium
xyzab

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

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

The unit vector n̂ is perpendicular to both a = (5, 2, −2) and b = (0, 3, 4) and has a positive z-component. Find its y-component n̂y.

a = (5, 2, −2)b = (0, 3, 4)n̂z > 0
n̂y =
15.ProbabilityMedium
xp(x)100.25200.35300.35400.05μ = 22

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

x: 10, 20, 30, 40p(x): 0.25, 0.35, 0.35, 0.05
Var(X) =
16.ProbabilityHard
n = 12, p = 0.2: P(X = 5) = ?k0123456789101112

A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 5 correct answers?

n = 12p = 0.2k = 5
P =
17.Descriptive statisticsMedium
1015202530354045505560

Find the sample standard deviation s of the data: 23, 54, 57, 10, 60, 56.

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

A quantity is normally distributed with mean 70 and standard deviation 5. What proportion P of values is above 64?

μ = 70σ = 5x = 64
P =
19.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.7154

Drag z so that the shaded area between −z and z under the standard normal curve is 0.7154. Find z to two decimals.

A = 0.7154
z =
20.Normal distributionHard
μ = 32x = ?25%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 25% of them lie (the 25th percentile).

μ = 32σ = 6P = 25%
x = min

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 = 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).

First application

4·sin 4x/(4x) → 0/0

Second application

16·cos 4x/4

Substitute x = 0

L = 16/4 = 4

Answer

L = 4
3.Derivatives

Answer: f′(1.5) = 80.34

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′ = e2x·(1 + 2·x)

Substitute

f′(1.5) = 20.09·(1 + 3) = 80.34

Answer

f′(1.5) = 80.34
4.Derivatives

Answer: b = −10

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) = 8
f′(x) = 2x + 5 → f′(2) = 9

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 8 − 9·2 = −10

Answer

b = −10
5.Integrals

Answer: I = 1.742

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 = sin x dx, v = −cos x

Antiderivative

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

Limits

(−2·cos 2 + sin 2) − (0 + 0) = 1.742

Answer

I = 1.742
6.Integrals

Answer: I = 37.59

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

12.18·(3.25) − 1·2 = 37.59

Answer

I = 37.59
7.Complex numbers

Answer: a = −1.843

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 = 2.5·(cos 2.4 + j sin 2.4)

Components (radians)

a = 2.5·cos 2.4 = −1.843
b = 2.5·sin 2.4 = 1.689
a = −1.843

Answer

a = −1.843
8.Complex numbers

Answer: Im(z³) = −16

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.

Polar form

z = −2 − j2 = 2√2∠−135°

De Moivre

z³ = (2√2)³∠−405° = 22.63∠−45°

Rectangular form

z³ = 22.63·(cos −45° + j sin −45°) = 16 − j16
Im(z³) = −16

Answer

Im(z³) = −16
9.Differential equations

Answer: k = 0.01994 min⁻¹

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.

Ratio of temperature differences

e−38k = 195/416

Solve for k

k = ln(416/195)/38 = 0.01994 min⁻¹

Answer

k = 0.01994 min⁻¹
10.Differential equations

Answer: m = 55.07 kg

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.

Set up the equation

dm/dt = 10·0.2 − (10/500)·m = 2 − 0.02·m

Solve

The integrating factor is ert/V; the solution approaches the equilibrium c·V.

m(t) = 100 + (−100)·e−t/50

Substitute t

m(40) = 100 + (−100)·0.4493 = 55.07 kg

Answer

m = 55.07 kg
11.Matrices & determinants

Answer: det A = −16

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 = −3, M₁₂ = −3, M₁₃ = −4

Cofactor expansion

det A = 4·(−3) − (−4)·(−3) + (−2)·(−4) = −16

Answer

det A = −16
12.Matrices & determinants

Answer: x = 2.316

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 = 4·6 − (−1)·(−5) = 19

Replace the x column

det Aₓ = 6·6 − (−1)·8 = 44
x = 44/19 = 2.316

Answer

x = 2.316
y = 3.263
13.Vectors: dot & cross product

Answer: θ = 39.76 °

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 = (−3)·(−3) + 0·3 + (−2)·(−2) = 13

Magnitudes

|a| = √((−3)² + 0² + (−2)²) = 3.606
|b| = √((−3)² + 3² + (−2)²) = 4.69

Angle

cos θ = 13/(3.606·4.69) = 0.7687
θ = arccos(0.7687) = 39.76°

Answer

θ = 39.76°
14.Vectors: dot & cross product

Answer: n̂y = −0.698

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 = 2·4 − (−2)·3 = 14
(a × b)y = azbx − axbz = (−2)·0 − 5·4 = −20
(a × b)z = axby − aybx = 5·3 − 2·0 = 15

Normalise

The z-component of a × b is already positive.

|a × b| = √(14² + (−20)² + 15²) = 28.65
n̂ = (14, −20, 15)/28.65 = (0.4886, −0.698, 0.5235)
n̂y = −0.698

Answer

n̂y = −0.698
15.Probability

Answer: Var(X) = 76

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 = 10·0.25 + 20·0.35 + 30·0.35 + 40·0.05 = 22

Mean of the squares

E(X²) = 100·0.25 + 400·0.35 + 900·0.35 + 1600·0.05 = 560

Variance

Var(X) = 560 − 22² = 76

Answer

Var(X) = 76
16.Probability

Answer: P = 0.05315

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 = 5) = 792·0.25·0.87 = 792·0.00032·0.2097 = 0.05315

Answer

P = 0.05315
17.Descriptive statistics

Answer: s = 21.28

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 = 260/6 = 43.33

Squared deviations

Σ(x − x̄)² = (−20.33)² + 10.67² + 13.67² + (−33.33)² + 16.67² + 12.67² = 2263

Standard deviation

s = √(2263/5) = 21.28

Answer

s = 21.28
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: z = 1.07

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 + 0.7154)/2 = 0.8577

Read the z-table backwards

Find 0.8577 inside the table and read off z.

z = 1.07

Answer

z = 1.07
20.Normal distribution

Answer: x = 27.95 min

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 = 32 + −0.6745·6 = 27.95 min

Answer

x = 27.95 min

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