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→∞ (4x3 − x + 2)/(x3 + 2x + 6).

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

Find the limit L = limx→0 (1 − cos 5x)/(2x²), with x in radians.

f(x) = (1 − cos 5x)/(2x²)x → 0
L =
3.DerivativesMedium
xyx = 1

Let f(x) = x·sin(3x), with x in radians. Find f′(1).

f(x) = x·sin(3x)x = 1 rad
f′(1) =
4.DerivativesHard
xy

Let f(x) = x3 + 1.5x2 − 6x − 1. Find the x-coordinate x of the local minimum.

f(x) = x3 + 1.5x2 − 6x − 1
x =
5.IntegralsMedium
xy02

Evaluate I = ∫02 2x·(x² + 1)2 dx.

f(x) = 2x·(x² + 1)2[0, 2]
I =
6.IntegralsHard
xy−22

Find the area A of the region between the parabola y = 4 − x² and the x-axis.

y = 4 − x²
A =
7.Complex numbersMedium
ReIm55−5−51010−10−10z₁z₂w

Given z₁ = 2.5∠100° and z₂ = 4∠110°, find the real part of w = z₁·z₂.

z₁ = 2.5∠100°z₂ = 4∠110°
Re(w) =
8.Complex numbersHard
ReIm22−2−244−4−4z₁z₂

One root of z³ − 11z² + 56z − 96 = 0 is z₁ = 4 + j4. The coefficients are real. Find the real root r.

z₁ = 4 + j4
r =
9.Differential equationsMedium
tyy(0) = 2y(2) = ?

Solve y″ + 4y′ + 3y = 0 with y(0) = 2 and y′(0) = -4. Find y(2).

y″ + 4y′ + 3y = 0y(0) = 2y′(0) = -4
y(2) =
10.Differential equationsHard
xy12(1, 5)y(2) = ?

Solve the initial value problem y′ + 3y/x = 1 for x > 0, with y(1) = 5. Find y(2).

y′ + 3y/x = 1y(1) = 5
y(2) =
11.Matrices & determinantsMedium
A =5−4−32201−50

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

A = [5 −4 −3; 2 2 0; 1 −5 0]
det A =
12.Matrices & determinantsHard
A =5−2−3−4b =−123

Solve the system 5x − 2y = −12 and −3x − 4y = 3. Find x.

5x − 2y = −12−3x − 4y = 3
x =
13.Vectors: dot & cross productMedium
xyzab

Find the dot product a·b of a = (−7, 6, 0) and b = (7, 6, 7).

a = (−7, 6, 0)b = (7, 6, 7)
a·b =
14.Vectors: dot & cross productHard
xyzabc

Find the volume V of the parallelepiped whose edges from one corner are a = (4, −3, 3), b = (−1, −3, 5) and c = (2, −5, 0).

a = (4, −3, 3)b = (−1, −3, 5)c = (2, −5, 0)
V =
15.ProbabilityMedium
red: 5blue: 7

A bag holds 5 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

5 / 7
P =
16.ProbabilityHard
n = 6, p = 0.5: P(X = 3) = ?k0123456

A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 3 correct answers?

n = 6p = 0.5k = 3
P =
17.Descriptive statisticsMedium
24262830323436384042

Find the sample standard deviation s of the data: 38, 42, 31, 42, 23.

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

A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is below 66?

μ = 70σ = 8x = 66
P =
19.Normal distributionMedium
200 kg200.7 kgT = X₁ + … + X₈?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 8 of them exceeds 200.7 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 8> 200.7 kg
P =
20.Normal distributionHard
235240245250255260265gbottom 70%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 70% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g70%
x = g

3Solutions

1.Limits

Answer: L = 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).

Divide by the highest power

(4 + (−1)/x2 + 2/x3)/(1 + 2/x2 + 6/x3)

Let x → ∞

L = 4/1 = 4

Answer

L = 4

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

2.Limits

Answer: L = 6.25

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).

First application

5·sin 5x/(4x) → 0/0

Second application

25·cos 5x/4

Substitute x = 0

L = 25/4 = 6.25

Answer

L = 6.25
3.Derivatives

Answer: f′(1) = −2.829

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(3x) + 3x·cos(3x)

Substitute

f′(1) = 0.1411 + 3·(−0.99) = −2.829

Answer

f′(1) = −2.829
4.Derivatives

Answer: x = 1

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.

Derivative

f′(x) = 3x2 + 3x − 6 = 3·(x2 + x − 2)

Solve f′(x) = 0

x = −2 or x = 1

Which one is which

f″(x) = 6x + 3: negative at a maximum, positive at a minimum.

f″(−2) = −9, f″(1) = 9
x = 1

Answer

x = 1
5.Integrals

Answer: I = 41.33

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.

Substitution

u = x² + 1, du = 2x dx
I = ∫15 u2 du

Integrate and evaluate

I = [u3/3]15 = (125 − 1)/3 = 41.33

Answer

I = 41.33
6.Integrals

Answer: A = 10.67

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.

Zeros

4 − x² = 0 → x = ±2

Integrate

A = [4x − x³/3]−22 = 5.333 − (−5.333) = 10.67

Answer

A = 10.67
7.Complex numbers

Answer: Re(w) = −8.66

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 and argument of w

|w| = 2.5 · 4 = 10
θ = 100° + 110° = 210° → −150°

Rectangular form

Re(w) = 10·cos(−150°) = −8.66
Im(w) = 10·sin(−150°) = −5
Re(w) = −8.66

Answer

Re(w) = −8.66
8.Complex numbers

Answer: r = 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.

Conjugate pair

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

Factor out the quadratic

z³ − 11z² + 56z − 96 = (z² − 8z + 32)(z − 3)

Real root

r = 3

Answer

r = 3

Check with the sum of the roots: z₁ + z₂ + r = 8 + 3 = 11 = −(−11).

9.Differential equations

Answer: y(2) = 0.1378

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 = 2, −1A − 3B = -4
A = 1, B = 1

Evaluate

y(2) = 1·e−2 + 1·e−6 = 0.1378

Answer

y(2) = 0.1378
10.Differential equations

Answer: y(2) = 1.094

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.

Integrating factor

μ = e∫3/x dx = x3

Integrate

(x3y)′ = x3 → x3y = 1x4/4 + C

Initial condition

5 = 1/4 + C → C = 4.75

Evaluate

y(2) = (1·24/4 + 4.75)/23 = 1.094

Answer

y(2) = 1.094
11.Matrices & determinants

Answer: det A = 36

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 = 0, M₁₂ = 0, M₁₃ = −12

Cofactor expansion

det A = 5·0 − (−4)·0 + (−3)·(−12) = 36

Answer

det A = 36
12.Matrices & determinants

Answer: x = −2.077

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·(−4) − (−2)·(−3) = −26

Replace the x column

det Aₓ = (−12)·(−4) − (−2)·3 = 54
x = 54/(−26) = −2.077

Answer

x = −2.077
y = 0.8077
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 = (−7)·7 + 6·6 + 0·7 = (−49) + 36 + 0 = −13

Answer

a·b = −13

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

14.Vectors: dot & cross product

Answer: V = 103

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 b × c

(b × c)x = bycz − bzcy = (−3)·0 − 5·(−5) = 25
(b × c)y = bzcx − bxcz = 5·2 − (−1)·0 = 10
(b × c)z = bxcy − bycx = (−1)·(−5) − (−3)·2 = 11

Dot product with a

a·(b × c) = 4·25 + (−3)·10 + 3·11 = 103
V = |103| = 103

Answer

V = 103
15.Probability

Answer: P = 0.1515

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 = 5/12 · 4/11 = 20/132 = 0.1515

Answer

P = 0.1515
16.Probability

Answer: P = 0.3125

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.

Binomial formula

P(X = 3) = 20·0.53·0.53 = 20·0.125·0.125 = 0.3125

Answer

P = 0.3125
17.Descriptive statistics

Answer: s = 8.167

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 = 176/5 = 35.2

Squared deviations

Σ(x − x̄)² = 2.8² + 6.8² + (−4.2)² + 6.8² + (−12.2)² = 266.8

Standard deviation

s = √(266.8/4) = 8.167

Answer

s = 8.167
18.Descriptive statistics

Answer: P = 0.3085

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 = −0.5

Standard normal table

Φ(−0.5) = 0.3085
P = 0.3085

Answer

P = 0.3085
19.Normal distribution

Answer: P = 0.3121

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 = 200 kg, σT = 0.5·√8 = 1.414 kg

Standardise

z = (200.7 − 200)/1.414 = 0.49

Probability

P = 1 − Φ(0.49) = 0.3121

Answer

P = 0.3121
20.Normal distribution

Answer: x = 252.6 g

From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.

Area to the left

Φ(z) = 0.7

z from the table

z ≈ 0.52

Back to the original units

x = 250 + 0.52·5 = 252.6 g

Answer

x = 252.6 g

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