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 + 2/x)x. Give a decimal value.

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

Find the limit L = limx→0 (1 − cos 4x)/(x²), with x in radians.

f(x) = (1 − cos 4x)/(x²)x → 0
L =
3.DerivativesMedium
xyx = 1

Let f(x) = ln(x2 + 2). Find f′(1).

f(x) = ln(x2 + 2)x = 1
f′(1) =
4.DerivativesHard
xy

Let f(x) = x3 − 10.5x2 + 36x + 2. Find the x-coordinate x of the local minimum.

f(x) = x3 − 10.5x2 + 36x + 2
x =
5.IntegralsMedium
xy02

Evaluate I = ∫02 2x·(x² + 1)1 dx.

f(x) = 2x·(x² + 1)1[0, 2]
I =
6.IntegralsHard
xy2y = x²eˣ

Evaluate I = ∫₀2 x²·eˣ dx.

∫₀2 x²eˣ dx
I =
7.Complex numbersMedium
ReIm22−2−244−4−466−6−6θz

Write z = −1 + j6 in exponential form r·ejθ. Give θ in radians, between −π and π.

z = −1 + j6
θ = rad
8.Complex numbersHard
ReIm0.50.5−0.5−0.511−1−1z₁z₂

One root of z³ − 2z² − 6z − 8 = 0 is z₁ = −1 + j. The coefficients are real. Find the real root r.

z₁ = −1 + j
r =
9.Differential equationsMedium
t (min)T (°C)21Ta​36 °C59 °C

A cup of coffee at 82 °C is left in a room at 18 °C. After 21 min its temperature is 36 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 59 °C?

T0 = 82 °CTa = 18 °CT(21) = 36 °CT = 59 °C
t = min
10.Differential equationsHard
t (min)m (kg)55 minm = ?

A tank holds 1000 L of water in which 20 kg of salt is dissolved. Brine with 0.25 kg of salt per litre flows in at 4 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 55 min?

V = 1000 Lm0 = 20 kgc = 0.25 kg/Lr = 4 L/min
m = kg
11.Matrices & determinantsMedium
A =−2−4−63B =61−16

Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).

A = [−2 −4; −6 3]B = [6 1; −1 6]
c12 =
12.Matrices & determinantsHard
A =3−6−5−1b =−12−3

Solve the system 3x − 6y = −12 and −5x − y = −3. Find x.

3x − 6y = −12−5x − y = −3
x =
13.Vectors: dot & cross productMedium
xyzab

Find the angle θ between the vectors a = (1, 0, −5) and b = (−1, 2, −4), in degrees.

a = (1, 0, −5)b = (−1, 2, −4)
θ = °
14.Vectors: dot & cross productHard
xyzrFOA

A force F = (30, 60, 40) N acts at point A(2, −3, −4). Coordinates are in metres. Find the x-component Mx of the moment of the force about the origin O(0, 0, 0).

F = (30, 60, 40) NA(2, −3, −4) mO(0, 0, 0) m
Mx = N·m
15.ProbabilityMedium
n = 13k = 2N = ?

In how many ways N can a committee of 2 people be chosen from a group of 13? The order does not matter.

n = 13k = 2
N =
16.ProbabilityHard
n = 7, p = 0.2: P(X = 5) = ?k01234567

A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 5 correct answers?

n = 7p = 0.2k = 5
P =
17.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 13, 57, 30, 36, 20.

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 73.2?

μ = 70σ = 8x = 73.2
P =
19.Normal distributionMedium
57.573P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 57.5 and 73 points? Give P as a decimal.

μ = 65σ = 10x1 = 57.5x2 = 73
P =
20.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =

3Solutions

1.Limits

Answer: L = 7.389

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 + 2/x)

Evaluate

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

ln L = 1·2 = 2

Undo the logarithm

L = e2 = 7.389

Answer

L = 7.389
2.Limits

Answer: L = 8

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

4·sin 4x/(2x) → 0/0

Second application

16·cos 4x/2

Substitute x = 0

L = 16/2 = 8

Answer

L = 8
3.Derivatives

Answer: f′(1) = 0.6667

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 for the logarithm

f′(x) = 2x/(x2 + 2)

Substitute

f′(1) = 2/3 = 0.6667

Answer

f′(1) = 0.6667
4.Derivatives

Answer: x = 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.

Derivative

f′(x) = 3x2 − 21x + 36 = 3·(x2 − 7x + 12)

Solve f′(x) = 0

x = 3 or x = 4

Which one is which

f″(x) = 6x − 21: negative at a maximum, positive at a minimum.

f″(3) = −3, f″(4) = 3
x = 4

Answer

x = 4
5.Integrals

Answer: I = 12

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 u1 du

Integrate and evaluate

I = [u2/2]15 = (25 − 1)/2 = 12

Answer

I = 12
6.Integrals

Answer: I = 12.78

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

7.389·(2) − 1·2 = 12.78

Answer

I = 12.78
7.Complex numbers

Answer: θ = 1.736 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 = √(1 + 36) = 6.083

Argument in radians

θ = 99.46° = 99.46·π/180 = 1.736 rad

Exponential form

z = 6.083·ej1.736

Answer

θ = 1.736 rad
8.Complex numbers

Answer: r = 4

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 = −1 − j
(z − z₁)(z − z₂) = z² − 2·(−1)·z + ((−1)² + 1²) = z² + 2z + 2

Factor out the quadratic

z³ − 2z² − 6z − 8 = (z² + 2z + 2)(z − 4)

Real root

r = 4

Answer

r = 4

Check with the sum of the roots: z₁ + z₂ + r = −2 + 4 = 2 = −(−2).

9.Differential equations

Answer: t = 7.372 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(64/18)/21 = 0.06041 min⁻¹

Time to reach the target

t = ln(64/41)/0.06041 = 7.372 min

Answer

t = 7.372 min
10.Differential equations

Answer: m = 65.42 kg

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.

Set up the equation

dm/dt = 4·0.25 − (4/1000)·m = 1 − 0.004·m

Solve

The integrating factor is ert/V; the solution approaches the equilibrium c·V.

m(t) = 250 + (−230)·e−t/250

Substitute t

m(55) = 250 + (−230)·0.8025 = 65.42 kg

Answer

m = 65.42 kg
11.Matrices & determinants

Answer: c12 = −26

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

c12 = (−2)·1 + (−4)·6 = −26

Answer

c12 = −26
12.Matrices & determinants

Answer: x = 0.1818

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 = 3·(−1) − (−6)·(−5) = −33

Replace the x column

det Aₓ = (−12)·(−1) − (−6)·(−3) = −6
x = −6/(−33) = 0.1818

Answer

x = 0.1818
y = 2.091
13.Vectors: dot & cross product

Answer: θ = 35.6 °

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

a·b = 1·(−1) + 0·2 + (−5)·(−4) = 19

Magnitudes

|a| = √(1² + 0² + (−5)²) = 5.099
|b| = √((−1)² + 2² + (−4)²) = 4.583

Angle

cos θ = 19/(5.099·4.583) = 0.8131
θ = arccos(0.8131) = 35.6°

Answer

θ = 35.6°
14.Vectors: dot & cross product

Answer: Mx = 120 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 = (2, −3, −4) − (0, 0, 0) = (2, −3, −4) m

Moment M = r × F

(r × F)x = ryFz − rzFy = (−3)·40 − (−4)·60 = 120 N·m
(r × F)y = rzFx − rxFz = (−4)·30 − 2·40 = −200 N·m
(r × F)z = rxFy − ryFx = 2·60 − (−3)·30 = 210 N·m

Requested component

Mx = 120 N·m

Answer

Mx = 120 N·m
15.Probability

Answer: N = 78

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(13, 2) = (13·12)/(2·1) = 78

Answer

N = 78
16.Probability

Answer: P = 0.004301

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 = 5) = 21·0.25·0.82 = 21·0.00032·0.64 = 0.004301

Answer

P = 0.004301
17.Descriptive statistics

Answer: s = 16.93

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 = 156/5 = 31.2

Squared deviations

Σ(x − x̄)² = (−18.2)² + 25.8² + (−1.2)² + 4.8² + (−11.2)² = 1147

Standard deviation

s = √(1147/4) = 16.93

Answer

s = 16.93
18.Descriptive statistics

Answer: P = 0.6554

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 = 0.6554

Answer

P = 0.6554
19.Normal distribution

Answer: P = 0.5615

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 = (57.5 − 65)/10 = −0.75
z₂ = (73 − 65)/10 = 0.8

Area between

P = Φ(0.8) − Φ(−0.75) = 0.7881 − 0.2266 = 0.5615

Answer

P = 0.5615
20.Normal distribution

Answer: P = 0.8413

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 = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413

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