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 + 1/x)x. Give a decimal value.

f(x) = (1 + 1/x)xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 ln(1 + 5x)/sin(4x), with x in radians.

f(x) = ln(1 + 5x)/sin(4x)x → 0
L =
3.DerivativesMedium
xyx = 0

Let f(x) = (3x − 1)3. Find f′(0).

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

The tangent line to f(x) = 2x2 + 4x + 2 at x = 3 crosses the y-axis at y = b. Find b.

f(x) = 2x2 + 4x + 2x = 3
b =
5.IntegralsMedium
xy01

Evaluate I = ∫01 e2x dx.

f(x) = e2x[0, 1]
I =
6.IntegralsHard
xy1y = x²eˣ

Evaluate I = ∫₀1 x²·eˣ dx.

∫₀1 x²eˣ dx
I =
7.Complex numbersMedium
ReIm55−5−51010−10−10θz

Write z = −2 + j9 in exponential form r·ejθ. Give θ in radians, between −π and π.

z = −2 + j9
θ = rad
8.Complex numbersHard
ReIm22−2−244−4−4ABCD

Exactly one of the points A–D is a root of z⁶ = 729∠180°. Click it.

w = 729∠180°n = 6
  1. 1A (3∠180°)
  2. 2B (3∠30°)
  3. 3C (5.4∠60°)
  4. 4D (3∠60°)
9.Differential equationsMedium
t (min)T (°C)52Ta​200 °C248 °C

A forged steel part at 320 °C is left to cool in a workshop at 25 °C. After 52 min its temperature is 200 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 248 °C?

T0 = 320 °CTa = 25 °CT(52) = 200 °CT = 248 °C
t = min
10.Differential equationsHard
tyK = 10y(0.75) = ?

A system obeys y″ + 3y′ + 18.25y = 182.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.75); angles in radians.

y″ + 3y′ + 18.25y = 182.5y(0) = 0y′(0) = 0
y(0.75) =
11.Matrices & determinantsMedium
A =−4035B =4−233

Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).

A = [−4 0; 3 5]B = [4 −2; 3 3]
c11 =
12.Matrices & determinantsHard
A =46−4−7

Find the larger eigenvalue λ of the matrix A shown.

A = [4 6; −4 −7]
λ =
13.Vectors: dot & cross productMedium
xyzdFAB

A constant force F = (10, 0, −30) N acts on an object while it moves in a straight line from point A(−5, −3, 4) to point B(2, −5, −5). Coordinates are in metres. Find the work W done by the force.

F = (10, 0, −30) NA(−5, −3, 4) mB(2, −5, −5) m
W = J
14.Vectors: dot & cross productHard
xyzABC

Find the area A of the triangle with corners A(3, 4, 2), B(−2, −4, −2) and C(−4, −4, −2). Coordinates are in metres.

A(3, 4, 2)B(−2, −4, −2)C(−4, −4, −2)
A = m²
15.ProbabilityMedium
n = 14k = 5N = ?

In how many ways N can a committee of 5 people be chosen from a group of 14? The order does not matter.

n = 14k = 5
N =
16.ProbabilityHard
n = 5, p = 0.2: P(X ≤ 1) = ?k012345

5 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?

n = 5p = 0.2
P =
17.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 26, 53, 56, 17, 35.

n = 5
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 72?

μ = 70σ = 5x = 72
P =
19.Normal distributionMedium
246252.75P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 246 and 252.75 g? Give P as a decimal.

μ = 250σ = 5x1 = 246x2 = 252.75
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 = 2.718

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→∞ x·ln(1 + 1/x)

Evaluate

ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.

ln L = 1·1 = 1

Undo the logarithm

L = e1 = 2.718

Answer

L = 2.718
2.Limits

Answer: L = 1.25

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 [5/(1 + 5x)]/[4·cos 4x]

Substitute x = 0

L = 5/4 = 1.25

Answer

L = 1.25
3.Derivatives

Answer: f′(0) = 9

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) = 3·(3x − 1)2·3

Substitute

f′(0) = 3·(−1)2·3 = 9

Answer

f′(0) = 9
4.Derivatives

Answer: b = −16

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) = 32
f′(x) = 4x + 4 → f′(3) = 16

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 32 − 16·3 = −16

Answer

b = −16
5.Integrals

Answer: I = 3.195

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) = e2x/2

Evaluate at the limits

I = (e2 − 1)/2 = (7.389 − 1)/2 = 3.195

Answer

I = 3.195
6.Integrals

Answer: I = 0.7183

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

2.718·(1) − 1·2 = 0.7183

Answer

I = 0.7183
7.Complex numbers

Answer: θ = 1.789 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 = √(4 + 81) = 9.22

Argument in radians

θ = 102.5° = 102.5·π/180 = 1.789 rad

Exponential form

z = 9.22·ej1.789

Answer

θ = 1.789 rad
8.Complex numbers

Answer: B (3∠30°)

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 and arguments of the roots

|z| = 3
θ = 30°, 90°, 150°, −150°, −90°, −30°

The point

B = 3∠30°

Answer

B (3∠30°)
9.Differential equations

Answer: t = 27.86 min

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.

Cooling constant

k = ln(295/175)/52 = 0.01004 min⁻¹

Time to reach the target

t = ln(295/223)/0.01004 = 27.86 min

Answer

t = 27.86 min
10.Differential equations

Answer: y(0.75) = 13.04

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 = 10, r = −1.5 ± 4j

Constants from the initial values

y(0) = 10 + A = 0 → A = −10
y′(0) = −1.5A + 4B = 0 → B = −3.75

Evaluate

y(0.75) = 10·(1 − e−1.125(cos 3 + 0.375·sin 3)) = 13.04

Answer

y(0.75) = 13.04
11.Matrices & determinants

Answer: c11 = −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.

Row times column

c11 = (−4)·4 + 0·3 = −16

Answer

c11 = −16
12.Matrices & determinants

Answer: λ = 1

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 − (−3)λ + (−4) = 0

Solve

λ = (−3 ± √(25))/2 → λ = −4, 1
λ = 1

Answer

λ = 1
13.Vectors: dot & cross product

Answer: W = 340 J

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.

Displacement

d = B − A = (2, −5, −5) − (−5, −3, 4) = (7, −2, −9) m

Work

W = F·d = 10·7 + 0·(−2) + (−30)·(−9) = 340 J

Answer

W = 340 J
14.Vectors: dot & cross product

Answer: A = 8.944 m²

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.

Edge vectors

AB = B − A = (−5, −8, −4)
AC = C − A = (−7, −8, −4)

Cross product

(AB × AC)x = AByACz − ABzACy = (−8)·(−4) − (−4)·(−8) = 0
(AB × AC)y = ABzACx − ABxACz = (−4)·(−7) − (−5)·(−4) = 8
(AB × AC)z = ABxACy − AByACx = (−5)·(−8) − (−8)·(−7) = −16
|AB × AC| = √(0² + 8² + (−16)²) = 17.89

Area of the triangle

A = ½·17.89 = 8.944 m²

Answer

A = 8.944 m²
15.Probability

Answer: N = 2002

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.

Combinations

C(14, 5) = (14·13·12·11·10)/(5·4·3·2·1) = 2002

Answer

N = 2002
16.Probability

Answer: P = 0.7373

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) = 0.85 = 0.3277
P(X = 1) = 5·0.2·0.84 = 0.4096

Add them

P = 0.3277 + 0.4096 = 0.7373

Answer

P = 0.7373
17.Descriptive statistics

Answer: s = 16.89

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 = 187/5 = 37.4

Squared deviations

Σ(x − x̄)² = (−11.4)² + 15.6² + 18.6² + (−20.4)² + (−2.4)² = 1141

Standard deviation

s = √(1141/4) = 16.89

Answer

s = 16.89
18.Descriptive statistics

Answer: P = 0.3446

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 = 0.4

Standard normal table

Φ(0.4) = 0.6554
P = 1 − 0.6554 = 0.3446

Answer

P = 0.3446
19.Normal distribution

Answer: P = 0.497

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-scores

z₁ = (246 − 250)/5 = −0.8
z₂ = (252.75 − 250)/5 = 0.55

Area between

P = Φ(0.55) − Φ(−0.8) = 0.7088 − 0.2119 = 0.497

Answer

P = 0.497
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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