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→∞ (6x3 + 6x − 4)/(4x3 + 2x + 7).

f(x) = (6x3 + 6x − 4)/(4x3 + 2x + 7)x → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 ln(1 + 3x)/sin(5x), with x in radians.

f(x) = ln(1 + 3x)/sin(5x)x → 0
L =
3.DerivativesMedium
xyx = −2

Let f(x) = ln(x2 + 2). Find f′(−2).

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

The tangent line to f(x) = −x2 + 5x + 4 at x = 3 crosses the y-axis at y = b. Find b.

f(x) = −x2 + 5x + 4x = 3
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
xy2.5y = x²eˣ

Evaluate I = ∫₀2.5 x²·eˣ dx.

∫₀2.5 x²eˣ dx
I =
7.Complex numbersMedium
ReIm0.50.5−0.5−0.511−1−1θz

Write z = 1 − j in exponential form r·ejθ. Give θ in radians, between −π and π.

z = 1 − j
θ = rad
8.Complex numbersHard
ReIm11−1−122−2−233−3−3z₁z₂

Solve z² − 6z + 18 = 0. The roots are a ± jb; find the real part a.

p = −6q = 18
a =
9.Differential equationsMedium
t (min)T (°C)8Ta​58 °C

A cup of coffee at 94 °C is left in a room at 23 °C. After 8 min its temperature is 58 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 94 °CTa = 23 °CT(8) = 58 °C
k = min⁻¹
10.Differential equationsHard
tyK = 5y(1.5) = ?

A system obeys y″ + 3y′ + 18.25y = 91.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1.5); angles in radians.

y″ + 3y′ + 18.25y = 91.25y(0) = 0y′(0) = 0
y(1.5) =
11.Matrices & determinantsMedium
A =35−50−2−42−51

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

A = [3 5 −5; 0 −2 −4; 2 −5 1]
det A =
12.Matrices & determinantsHard
A =−54−35b =512

Solve the system −5x + 4y = 5 and −3x + 5y = 12. Find x.

−5x + 4y = 5−3x + 5y = 12
x =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (0, −6, −6) onto b = (−1, 6, 4), that is, the component of a in the direction of b.

a = (0, −6, −6)b = (−1, 6, 4)
compb a =
14.Vectors: dot & cross productHard
xyzab

For a = (4, −1, −1) and b = (−2, 3, −4), find the y-component (a × b)y of the cross product a × b.

a = (4, −1, −1)b = (−2, 3, −4)
(a × b)y =
15.ProbabilityMedium
xp(x)00.3510.320.0530.1540.15μ = 1.45

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

x: 0, 1, 2, 3, 4p(x): 0.35, 0.3, 0.05, 0.15, 0.15
Var(X) =
16.ProbabilityHard
n = 12, p = 0.6: P(X ≤ 1) = ?k0123456789101112

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

n = 12p = 0.6
P =
17.Descriptive statisticsMedium
15202530354045

Find the sample standard deviation s of the data: 47, 47, 15, 28, 43, 38, 29.

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

A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is below 58?

μ = 50σ = 8x = 58
P =
19.Normal distributionMedium
200 mm200.8 mmX + Y?

X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 200.8 mm. Round z to two decimals.

μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 200.8 mm
P =
20.Normal distributionHard
μ = 100x = ?10%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 10% of them lie (the 10th percentile).

μ = 100σ = 15P = 10%
x = points

3Solutions

1.Limits

Answer: L = 1.5

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).

Divide by the highest power

(6 + 6/x2 + (−4)/x3)/(4 + 2/x2 + 7/x3)

Let x → ∞

L = 6/4 = 1.5

Answer

L = 1.5

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

2.Limits

Answer: L = 0.6

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).

L'Hôpital's rule

L = limx→0 [3/(1 + 3x)]/[5·cos 5x]

Substitute x = 0

L = 3/5 = 0.6

Answer

L = 0.6
3.Derivatives

Answer: f′(−2) = −0.6667

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) = 2x/(x2 + 2)

Substitute

f′(−2) = −4/6 = −0.6667

Answer

f′(−2) = −0.6667
4.Derivatives

Answer: b = 13

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) = 10
f′(x) = −2x + 5 → f′(3) = −1

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 10 − (−1)·3 = 13

Answer

b = 13
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 = 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: θ = −0.7854 rad

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

r = √(1 + 1) = 1.414

Argument in radians

θ = −45° = −45·π/180 = −0.7854 rad

Exponential form

z = 1.414·e−j0.7854

Answer

θ = −0.7854 rad
8.Complex numbers

Answer: a = 3

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 = 36 − 72 = −36 < 0

Quadratic formula

z = (6 ± j√36)/2 = 3 + j3, 3 − j3

Answer

a = 3

Answer

a = 3
9.Differential equations

Answer: k = 0.08842 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−8k = 35/71

Solve for k

k = ln(71/35)/8 = 0.08842 min⁻¹

Answer

k = 0.08842 min⁻¹
10.Differential equations

Answer: y(1.5) = 4.549

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 = −1.5 ± 4j

Constants from the initial values

y(0) = 5 + A = 0 → A = −5
y′(0) = −1.5A + 4B = 0 → B = −1.875

Evaluate

y(1.5) = 5·(1 − e−2.25(cos 6 + 0.375·sin 6)) = 4.549

Answer

y(1.5) = 4.549
11.Matrices & determinants

Answer: det A = −126

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 = −22, M₁₂ = 8, M₁₃ = 4

Cofactor expansion

det A = 3·(−22) − 5·8 + (−5)·4 = −126

Answer

det A = −126
12.Matrices & determinants

Answer: x = 1.769

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)·5 − 4·(−3) = −13

Replace the x column

det Aₓ = 5·5 − 4·12 = −23
x = −23/(−13) = 1.769

Answer

x = 1.769
y = 3.462
13.Vectors: dot & cross product

Answer: compb a = −8.242

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·(−1) + (−6)·6 + (−6)·4 = −60

Length of b

|b| = √((−1)² + 6² + 4²) = 7.28

Scalar projection

compb a = a·b/|b| = −60/7.28 = −8.242

Answer

compb a = −8.242

Negative: a has a component pointing against b.

14.Vectors: dot & cross product

Answer: (a × b)y = 18

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, component by component

(a × b)x = aybz − azby = (−1)·(−4) − (−1)·3 = 7
(a × b)y = azbx − axbz = (−1)·(−2) − 4·(−4) = 18
(a × b)z = axby − aybx = 4·3 − (−1)·(−2) = 10

Requested component

(a × b)y = 18

Answer

(a × b)y = 18

Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.

15.Probability

Answer: Var(X) = 2.147

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 = 0·0.35 + 1·0.3 + 2·0.05 + 3·0.15 + 4·0.15 = 1.45

Mean of the squares

E(X²) = 0·0.35 + 1·0.3 + 4·0.05 + 9·0.15 + 16·0.15 = 4.25

Variance

Var(X) = 4.25 − 1.45² = 2.147

Answer

Var(X) = 2.147
16.Probability

Answer: P = 0.0003188

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) = 0.412 = 1.678 × 10⁻⁵
P(X = 1) = 12·0.6·0.411 = 0.000302

Add them

P = 1.678 × 10⁻⁵ + 0.000302 = 0.0003188

Answer

P = 0.0003188
17.Descriptive statistics

Answer: s = 11.87

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 = 247/7 = 35.29

Squared deviations

Σ(x − x̄)² = 11.71² + 11.71² + (−20.29)² + (−7.286)² + 7.714² + 2.714² + (−6.286)² = 845.4

Standard deviation

s = √(845.4/6) = 11.87

Answer

s = 11.87
18.Descriptive statistics

Answer: P = 0.8413

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

Standard normal table

Φ(1) = 0.8413
P = 0.8413

Answer

P = 0.8413
19.Normal distribution

Answer: P = 0.0548

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 sum

μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm

Standardise

z = (200.8 − 200)/0.5 = 1.6

Upper tail

P = 1 − Φ(1.6) = 1 − 0.9452 = 0.0548

Answer

P = 0.0548
20.Normal distribution

Answer: x = 80.78 points

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) = 0.1 → z = −1.282

Back to x

x = 100 + −1.282·15 = 80.78 points

Answer

x = 80.78 points

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