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→∞ (2x3 + x − 2)/(3x3 + 4x + 4).

f(x) = (2x3 + x − 2)/(3x3 + 4x + 4)x → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (e1.5x − 1 − 1.5x)/x².

f(x) = (e1.5x − 1 − 1.5x)/x²x → 0
L =
3.DerivativesMedium
xyx = 0

Let f(x) = (3x + 1)4. Find f′(0).

f(x) = (3x + 1)4x = 0
f′(0) =
4.DerivativesHard
xyx = 2

The tangent line to f(x) = 0.5x2 + x + 6 at x = 2 crosses the y-axis at y = b. Find b.

f(x) = 0.5x2 + x + 6x = 2
b =
5.IntegralsMedium
xy26

Evaluate I = ∫26 1/x dx.

f(x) = 1/x[2, 6]
I =
6.IntegralsHard
xy2.5y = x²eˣ

Evaluate I = ∫₀2.5 x²·eˣ dx.

∫₀2.5 x²eˣ dx
I =
7.Complex numbersMedium
ReIm22−2−244−4−466−6−6θz

The number z = 6·e−j1.8 has its angle in radians. Find its imaginary part b.

r = 6θ = −1.8 rad
b =
8.Complex numbersHard
ReIm22−2−244−4−4z₁z₂

One root of z³ − 6z² + 16z + 64 = 0 is z₁ = 4 + j4. The coefficients are real. Find the real root r.

z₁ = 4 + j4
r =
9.Differential equationsMedium
t (min)T (°C)15Ta​T = ?

A cup of coffee at 81 °C is left in a room at 23 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.11 min⁻¹, find its temperature T after 15 min.

T0 = 81 °CTa = 23 °Ck = 0.11 min⁻¹t = 15 min
T = °C
10.Differential equationsHard
tyK = 5ymax​ = ?

A system obeys y″ + 2y′ + 17y = 85 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.

y″ + 2y′ + 17y = 85y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
A =5−54−43−5−510

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

A = [5 −5 4; −4 3 −5; −5 1 0]
det A =
12.Matrices & determinantsHard
A =1−42−4b =8−2

Solve the system x − 4y = 8 and 2x − 4y = −2. Find x.

x − 4y = 82x − 4y = −2
x =
13.Vectors: dot & cross productMedium
xyzab

Find the dot product a·b of a = (−4, −7, −7) and b = (−2, −3, 6).

a = (−4, −7, −7)b = (−2, −3, 6)
a·b =
14.Vectors: dot & cross productHard
xyzrFOA

A force F = (20, 10, 30) N acts at point A(−1, 2, −3). Coordinates are in metres. Find the y-component My of the moment of the force about the origin O(0, 0, 0).

F = (20, 10, 30) NA(−1, 2, −3) mO(0, 0, 0) m
My = N·m
15.ProbabilityMedium
red: 3blue: 5

A bag holds 3 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

3 / 5
P =
16.ProbabilityHard
xp(x)10.2520.330.140.35μ

X has the distribution shown, and Y = 5X + 10. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.25, 0.3, 0.1, 0.35Y = 5X + 10
σY =
17.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 38, 30, 20, 57, 46, 37, 27.

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

A quantity is normally distributed with mean 170 and standard deviation 5. What proportion P of values is above 180?

μ = 170σ = 5x = 180
P =
19.Normal distributionMedium
480 g486.2 gT = X₁ + … + X₁₂?

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 12 of them exceeds 486.2 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 12> 486.2 g
P =
20.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =

3Solutions

1.Limits

Answer: L = 0.6667

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

(2 + 1/x2 + (−2)/x3)/(3 + 4/x2 + 4/x3)

Let x → ∞

L = 2/3 = 0.6667

Answer

L = 0.6667

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

2.Limits

Answer: L = 1.125

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

(1.5·e1.5x − 1.5)/(2x) → 0/0

Second application

2.25·e1.5x/2

Substitute x = 0

L = 2.25·1/2 = 1.125

Answer

L = 1.125
3.Derivatives

Answer: f′(0) = 12

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

f′(x) = 4·(3x + 1)3·3

Substitute

f′(0) = 4·13·3 = 12

Answer

f′(0) = 12
4.Derivatives

Answer: b = 4

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) = 10
f′(x) = x + 1 → f′(2) = 3

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 10 − 3·2 = 4

Answer

b = 4
5.Integrals

Answer: I = 1.099

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) = ln x

Evaluate at the limits

I = (ln 6 − ln 2) = ln 3 = 1.099

Answer

I = 1.099
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: b = −5.843

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 = 6·(cos −1.8 + j sin −1.8)

Components (radians)

a = 6·cos −1.8 = −1.363
b = 6·sin −1.8 = −5.843
b = −5.843

Answer

b = −5.843
8.Complex numbers

Answer: r = −2

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.

Conjugate pair

z₂ = 4 − j4
(z − z₁)(z − z₂) = z² − 2·4·z + (4² + 4²) = z² − 8z + 32

Factor out the quadratic

z³ − 6z² + 16z + 64 = (z² − 8z + 32)(z + 2)

Real root

r = −2

Answer

r = −2

Check with the sum of the roots: z₁ + z₂ + r = 8 + (−2) = 6 = −(−6).

9.Differential equations

Answer: T = 34.14 °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) = 23 + 58·e−0.11t

Substitute t

T(15) = 23 + 58·0.192 = 34.14 °C

Answer

T = 34.14 °C
10.Differential equations

Answer: ymax = 7.28

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 ± 4j

Constants from the initial values

y(0) = 5 + A = 0 → A = −5
y′(0) = −1A + 4B = 0 → B = −1.25

Peak

t = π/4 = 0.7854
ymax = 5·(1 + e−1·π/4) = 7.28

Answer

ymax = 7.28
11.Matrices & determinants

Answer: det A = −56

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 = −25, M₁₃ = 11

Cofactor expansion

det A = 5·5 − (−5)·(−25) + 4·11 = −56

Answer

det A = −56
12.Matrices & determinants

Answer: x = −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.

Determinant of the coefficient matrix

det A = 1·(−4) − (−4)·2 = 4

Replace the x column

det Aₓ = 8·(−4) − (−4)·(−2) = −40
x = −40/4 = −10

Answer

x = −10
y = −4.5
13.Vectors: dot & cross product

Answer: a·b = −13

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)·(−2) + (−7)·(−3) + (−7)·6 = 8 + 21 + (−42) = −13

Answer

a·b = −13

Negative: the angle between the vectors is more than 90°.

14.Vectors: dot & cross product

Answer: My = −30 N·m

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.

Position vector from O to A

r = A − O = (−1, 2, −3) − (0, 0, 0) = (−1, 2, −3) m

Moment M = r × F

(r × F)x = ryFz − rzFy = 2·30 − (−3)·10 = 90 N·m
(r × F)y = rzFx − rxFz = (−3)·20 − (−1)·30 = −30 N·m
(r × F)z = rxFy − ryFx = (−1)·10 − 2·20 = −50 N·m

Requested component

My = −30 N·m

Answer

My = −30 N·m
15.Probability

Answer: P = 0.1071

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 = 3/8 · 2/7 = 6/56 = 0.1071

Answer

P = 0.1071
16.Probability

Answer: σY = 6.016

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 = 2.55, Var(X) = 1.447, σX = 1.203

Linear change

σY = |5|·1.203 = 6.016

Answer

σY = 6.016

The mean does move: E(Y) = 5·2.55 + 10 = 22.75.

17.Descriptive statistics

Answer: s = 12.37

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 = 255/7 = 36.43

Squared deviations

Σ(x − x̄)² = 1.571² + (−6.429)² + (−16.43)² + 20.57² + 9.571² + 0.5714² + (−9.429)² = 917.7

Standard deviation

s = √(917.7/6) = 12.37

Answer

s = 12.37
18.Descriptive statistics

Answer: P = 0.02275

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 = 2

Standard normal table

Φ(2) = 0.9772
P = 1 − 0.9772 = 0.02275

Answer

P = 0.02275
19.Normal distribution

Answer: P = 0.06811

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 = 480 g, σT = 1.2·√12 = 4.157 g

Standardise

z = (486.2 − 480)/4.157 = 1.49

Probability

P = 1 − Φ(1.49) = 0.06811

Answer

P = 0.06811
20.Normal distribution

Answer: P = 0.7389

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 difference

μ = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389

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