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→∞ (√(x2 + 9x) − x).

f(x) = √(x2 + 9x) − xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (1 − cos 2x)/(x²), with x in radians.

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

Let f(x) = x·e2x. Find f′(2).

f(x) = x·e2xx = 2
f′(2) =
4.DerivativesHard
xyx = −1

The tangent line to f(x) = 3x2 + 4x − 4 at x = −1 crosses the y-axis at y = b. Find b.

f(x) = 3x2 + 4x − 4x = −1
b =
5.IntegralsMedium
xy02

Evaluate I = ∫02 e−0.5x dx.

f(x) = e−0.5x[0, 2]
I =
6.IntegralsHard
xyy = x2​y = 2x

Find the area A of the region enclosed by y = x2 and y = 2x.

y = x2y = 2x
A =
7.Complex numbersMedium
ReIm11−1−122−2−2θz

The number z = 2.5·ej0.4 has its angle in radians. Find its real part a.

r = 2.5θ = 0.4 rad
a =
8.Complex numbersHard
ReIm11−1−122−2−233−3−3z₀z₁z₂

The roots of zⁿ = w with n = 3 and w = 27∠0° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2. Find the real part of z₂.

w = 27∠0°n = 3k = 2
Re(z₂) =
9.Differential equationsMedium
t (min)T (°C)20Ta​61 °C

A cup of coffee at 92 °C is left in a room at 23 °C. After 20 min its temperature is 61 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 92 °CTa = 23 °CT(20) = 61 °C
k = min⁻¹
10.Differential equationsHard
tyK = 2ymax​ = ?

A system obeys y″ + y′ + 9.25y = 18.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.

y″ + y′ + 9.25y = 18.5y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
A =−2053B =−2200

Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).

A = [−2 0; 5 3]B = [−2 2; 0 0]
c21 =
12.Matrices & determinantsHard
A =−16−26

Find the larger eigenvalue λ of the matrix A shown.

A = [−1 6; −2 6]
λ =
13.Vectors: dot & cross productMedium
xyzab

Find the value of k for which the vectors a = (6, 4, −4) and b = (k, −4, 4) are perpendicular.

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

Find the area A of the parallelogram spanned by a = (0, −2, −6) and b = (4, −4, −3).

a = (0, −2, −6)b = (4, −4, −3)
A =
15.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6P(six) = 1/6 on each roll

A fair die is rolled 6 times. What is the probability P of getting at least one six?

n = 6
P =
16.ProbabilityHard
xp(x)00.310.3520.35μ

X has the distribution shown, and Y = 3X + 5. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.3, 0.35, 0.35Y = 3X + 5
σY =
17.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 55, 22, 59, 52, 54.

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

Exam scores are normally distributed with mean 500 and standard deviation 50. Find the z-score z of a score of 450.

μ = 500σ = 50x = 450
z =
19.Normal distributionMedium
340 kg376 kgT = X₁ + … + X₄?

The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 4 of them is below 376 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 4< 376 kg
P =
20.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.03 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.02 mm
P =

3Solutions

1.Limits

Answer: L = 4.5

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).

Multiply by the conjugate

f(x) = 9x/(√(x2 + 9x) + x)

Divide by x

f(x) = 9/(√(1 + 9/x) + 1)

Let x → ∞

L = 9/(1 + 1) = 4.5

Answer

L = 4.5
2.Limits

Answer: L = 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).

First application

2·sin 2x/(2x) → 0/0

Second application

4·cos 2x/2

Substitute x = 0

L = 4/2 = 2

Answer

L = 2
3.Derivatives

Answer: f′(2) = 273

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′(2) = 54.6·(1 + 4) = 273

Answer

f′(2) = 273
4.Derivatives

Answer: b = −7

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) = −5
f′(x) = 6x + 4 → f′(−1) = −2

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = −5 − (−2)·(−1) = −7

Answer

b = −7
5.Integrals

Answer: I = 1.264

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) = e−0.5x/(−0.5)

Evaluate at the limits

I = (e−1 − 1)/(−0.5) = (0.3679 − 1)/(−0.5) = 1.264

Answer

I = 1.264
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.

Intersection points

x2 = 2x → x = 0, x = 2

Integrate upper minus lower

A = [x2 − 0.3333x3]02 = 1.333

Answer

A = 1.333
7.Complex numbers

Answer: a = 2.303

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 0.4 + j sin 0.4)

Components (radians)

a = 2.5·cos 0.4 = 2.303
b = 2.5·sin 0.4 = 0.9735
a = 2.303

Answer

a = 2.303
8.Complex numbers

Answer: Re(z₂) = −1.5

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 of the roots

271/3 = 3

Argument of the root

θ₂ = (0° + 360°·2)/3 = 240° → −120°

Rectangular form

z₂ = 3·(cos −120° + j sin −120°) = −1.5 − j2.598
Re(z₂) = −1.5

Answer

Re(z₂) = −1.5

The 3 roots lie on the circle of radius 3, 120° apart.

9.Differential equations

Answer: k = 0.02983 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−20k = 38/69

Solve for k

k = ln(69/38)/20 = 0.02983 min⁻¹

Answer

k = 0.02983 min⁻¹
10.Differential equations

Answer: ymax = 3.185

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 = 2, r = −0.5 ± 3j

Constants from the initial values

y(0) = 2 + A = 0 → A = −2
y′(0) = −0.5A + 3B = 0 → B = −0.3333

Peak

t = π/3 = 1.047
ymax = 2·(1 + e−0.5·π/3) = 3.185

Answer

ymax = 3.185
11.Matrices & determinants

Answer: c21 = −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.

Row times column

c21 = 5·(−2) + 3·0 = −10

Answer

c21 = −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 − (5)λ + (6) = 0

Solve

λ = (5 ± √(1))/2 → λ = 2, 3
λ = 3

Answer

λ = 3
13.Vectors: dot & cross product

Answer: k = 5.333

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 equals zero

a·b = 6·k + 4·(−4) + (−4)·4 = 0
−32 + 6·k = 0

Solve for k

k = −(−32)/6 = 5.333

Answer

k = 5.333
14.Vectors: dot & cross product

Answer: A = 31.05

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)·(−3) − (−6)·(−4) = −18
(a × b)y = azbx − axbz = (−6)·4 − 0·(−3) = −24
(a × b)z = axby − aybx = 0·(−4) − (−2)·4 = 8

Area

A = |a × b| = √((−18)² + (−24)² + 8²) = 31.05

Answer

A = 31.05
15.Probability

Answer: P = 0.6651

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.

Complement

P(no six) = (5/6)6 = 0.3349
P = 1 − 0.3349 = 0.6651

Answer

P = 0.6651
16.Probability

Answer: σY = 2.414

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 = 1.05, Var(X) = 0.6475, σX = 0.8047

Linear change

σY = |3|·0.8047 = 2.414

Answer

σY = 2.414

The mean does move: E(Y) = 3·1.05 + 5 = 8.15.

17.Descriptive statistics

Answer: s = 14.98

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 = 242/5 = 48.4

Squared deviations

Σ(x − x̄)² = 6.6² + (−26.4)² + 10.6² + 3.6² + 5.6² = 897.2

Standard deviation

s = √(897.2/4) = 14.98

Answer

s = 14.98
18.Descriptive statistics

Answer: z = −1

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 = (450 − 500)/50 = −1

Answer

z = −1
19.Normal distribution

Answer: P = 0.9332

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 = 340 kg, σT = 12·√4 = 24 kg

Standardise

z = (376 − 340)/24 = 1.5

Probability

P = Φ(1.5) = 0.9332

Answer

P = 0.9332
20.Normal distribution

Answer: P = 0.8849

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 = 25.03 − 25 = 0.03 mm, σ = 0.025 mm

Standardise

z = (0 − 0.03)/0.025 = −1.2

Probability

P = 1 − Φ(−1.2) = 0.8849

Answer

P = 0.8849

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