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→∞ (√(9x2 − 10x) − 3x).

f(x) = √(9x2 − 10x) − 3xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 ln(1 + 2x)/sin(5x), with x in radians.

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

Let f(x) = x·sin(2x), with x in radians. Find f′(1).

f(x) = x·sin(2x)x = 1 rad
f′(1) =
4.DerivativesHard
xyx = 2

The tangent line to f(x) = 0.5x2 − 5 at x = 2 crosses the y-axis at y = b. Find b.

f(x) = 0.5x2 − 5x = 2
b =
5.IntegralsMedium
xy24

Evaluate I = ∫24 2/x dx.

f(x) = 2/x[2, 4]
I =
6.IntegralsHard
xyy = 2x2​y = 2x

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

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

Write z = 2.5∠160° in rectangular form a + jb and give the imaginary part b.

r = 2.5θ = 160°
b =
8.Complex numbersHard
ReIm11−1−122−2−2θz

Use De Moivre's theorem to find the imaginary part of z⁵ for z = −√3 + j.

z = −√3 + jn = 5
Im(z⁵) =
9.Differential equationsMedium
tyy(0) = 1y(1) = ?

Solve y″ + 4y′ + 3y = 0 with y(0) = 1 and y′(0) = 0. Find y(1).

y″ + 4y′ + 3y = 0y(0) = 1y′(0) = 0
y(1) =
10.Differential equationsHard
tyy(0) = 1y(1) = ?

Solve y″ + y′ + 25.25y = 0 with y(0) = 1 and y′(0) = 0 (released from rest). Find y(1); angles in radians.

y″ + y′ + 25.25y = 0y(0) = 1y′(0) = 0
y(1) =
11.Matrices & determinantsMedium
A =144−5−41−4−5−5

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

A = [1 4 4; −5 −4 1; −4 −5 −5]
det A =
12.Matrices & determinantsHard
A =2263

Find the larger eigenvalue λ of the matrix A shown.

A = [2 2; 6 3]
λ =
13.Vectors: dot & cross productMedium
xyzab

Find the dot product a·b of a = (−6, −3, −5) and b = (−3, 4, 6).

a = (−6, −3, −5)b = (−3, 4, 6)
a·b =
14.Vectors: dot & cross productHard
xyzrFOA

A force F = (−30, 20, 30) N acts at point A(−3, −4, −1). Coordinates are in metres. Find the magnitude |M| of the moment of the force about point O(1, −3, 0).

F = (−30, 20, 30) NA(−3, −4, −1) mO(1, −3, 0) m
|M| = N·m
15.ProbabilityMedium
red: 4blue: 4

A bag holds 4 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

4 / 4
P =
16.ProbabilityHard
n = 8, p = 0.1: P(X ≤ 1) = ?k012345678

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

n = 8p = 0.1
P =
17.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 28, 38, 11, 52, 44, 59, 36.

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

A quantity is normally distributed with mean 100 and standard deviation 10. What proportion P of values is above 112.5?

μ = 100σ = 10x = 112.5
P =
19.Normal distributionMedium
750 g754.9 gX + Y?

X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 754.9 g. Round z to two decimals.

μX = 250, σX = 4μY = 500, σY = 9X + Y > 754.9 g
P =
20.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.01 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 = 25.04 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.02 mm
P =

3Solutions

1.Limits

Answer: L = −1.667

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) = −10x/(√(9x2 − 10x) + 3x)

Divide by x

f(x) = −10/(√(9 + (−10)/x) + 3)

Let x → ∞

L = −10/(3 + 3) = −1.667

Answer

L = −1.667
2.Limits

Answer: L = 0.4

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).

L'Hôpital's rule

L = limx→0 [2/(1 + 2x)]/[5·cos 5x]

Substitute x = 0

L = 2/5 = 0.4

Answer

L = 0.4
3.Derivatives

Answer: f′(1) = 0.077

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 and chain rule

f′(x) = sin(2x) + 2x·cos(2x)

Substitute

f′(1) = 0.9093 + 2·(−0.4161) = 0.077

Answer

f′(1) = 0.077
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(2) = −3
f′(x) = x → f′(2) = 2

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = −3 − 2·2 = −7

Answer

b = −7
5.Integrals

Answer: I = 1.386

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) = 2·ln x

Evaluate at the limits

I = 2·(ln 4 − ln 2) = 2·ln 2 = 1.386

Answer

I = 1.386
6.Integrals

Answer: A = 0.3333

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

2x2 = 2x → x = 0, x = 1

Integrate upper minus lower

A = [x2 − 0.6667x3]01 = 0.3333

Answer

A = 0.3333
7.Complex numbers

Answer: b = 0.8551

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.

Components

a = 2.5·cos(160°) = −2.349
b = 2.5·sin(160°) = 0.8551

Rectangular form

z = −2.349 + j0.8551
b = 0.8551

Answer

b = 0.8551
8.Complex numbers

Answer: Im(z⁵) = 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 = −√3 + j = 2∠150°

De Moivre

z⁵ = (2)⁵∠750° = 32∠30°

Rectangular form

z⁵ = 32·(cos 30° + j sin 30°) = 27.71 + j16
Im(z⁵) = 16

Answer

Im(z⁵) = 16
9.Differential equations

Answer: y(1) = 0.5269

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.

Characteristic roots

(r + 1)(r + 3) = 0 → r = −1, −3

Constants from the initial values

A + B = 1, −1A − 3B = 0
A = 1.5, B = −0.5

Evaluate

y(1) = 1.5·e−1 + (−0.5)·e−3 = 0.5269

Answer

y(1) = 0.5269
10.Differential equations

Answer: y(1) = 0.1139

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.

Complex roots

r = −0.5 ± 5j

Constants

A = 1, B = 0.5·1/5 = 0.1

Evaluate

y(1) = e−0.5(1·cos 5 + 0.1·sin 5) = 0.1139

Answer

y(1) = 0.1139
11.Matrices & determinants

Answer: det A = −55

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 = 25, M₁₂ = 29, M₁₃ = 9

Cofactor expansion

det A = 1·25 − 4·29 + 4·9 = −55

Answer

det A = −55
12.Matrices & determinants

Answer: λ = 6

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 ± √(49))/2 → λ = −1, 6
λ = 6

Answer

λ = 6
13.Vectors: dot & cross product

Answer: a·b = −24

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 = (−6)·(−3) + (−3)·4 + (−5)·6 = 18 + (−12) + (−30) = −24

Answer

a·b = −24

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

14.Vectors: dot & cross product

Answer: |M| = 186.3 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 = (−3, −4, −1) − (1, −3, 0) = (−4, −1, −1) m

Moment M = r × F

(r × F)x = ryFz − rzFy = (−1)·30 − (−1)·20 = −10 N·m
(r × F)y = rzFx − rxFz = (−1)·(−30) − (−4)·30 = 150 N·m
(r × F)z = rxFy − ryFx = (−4)·20 − (−1)·(−30) = −110 N·m

Magnitude

|M| = √((−10)² + 150² + (−110)²) = 186.3 N·m

Answer

|M| = 186.3 N·m
15.Probability

Answer: P = 0.2143

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 = 4/8 · 3/7 = 12/56 = 0.2143

Answer

P = 0.2143
16.Probability

Answer: P = 0.8131

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) = 0.98 = 0.4305
P(X = 1) = 8·0.1·0.97 = 0.3826

Add them

P = 0.4305 + 0.3826 = 0.8131

Answer

P = 0.8131
17.Descriptive statistics

Answer: s = 15.84

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 = 268/7 = 38.29

Squared deviations

Σ(x − x̄)² = (−10.29)² + (−0.2857)² + (−27.29)² + 13.71² + 5.714² + 20.71² + (−2.286)² = 1505

Standard deviation

s = √(1505/6) = 15.84

Answer

s = 15.84
18.Descriptive statistics

Answer: P = 0.1056

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 = 1.25

Standard normal table

Φ(1.25) = 0.8944
P = 1 − 0.8944 = 0.1056

Answer

P = 0.1056
19.Normal distribution

Answer: P = 0.3085

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 = 750 g, σ = √(4² + 9²) = 9.849 g

Standardise

z = (754.9 − 750)/9.849 = 0.5

Upper tail

P = 1 − Φ(0.5) = 1 − 0.6915 = 0.3085

Answer

P = 0.3085
20.Normal distribution

Answer: P = 0.9633

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 = 25.04 − 25 = 0.04 mm, σ = 0.02236 mm

Standardise

z = (0 − 0.04)/0.02236 = −1.79

Probability

P = 1 − Φ(−1.79) = 0.9633

Answer

P = 0.9633

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