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→∞ (−2x2 + 6x − 3)/(3x2 + 4x + 8).

f(x) = (−2x2 + 6x − 3)/(3x2 + 4x + 8)x → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (1 − cos 5x)/(3x²), with x in radians.

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

Let f(x) = (2x − 3)4. Find f′(1).

f(x) = (2x − 3)4x = 1
f′(1) =
4.DerivativesHard
xy

Let f(x) = x3 − 12x2 + 36x − 6. Find the x-coordinate x of the local minimum.

f(x) = x3 − 12x2 + 36x − 6
x =
5.IntegralsMedium
xy15y = x ln x

Evaluate I = ∫₁5 x·ln x dx.

∫₁5 x ln x dx
I =
6.IntegralsHard
xy4y = 2xy = 0.5x²

The region between the line y = 2x and the parabola y = 0.5x² is rotated about the x-axis. Find the volume V of the solid.

y = 2xy = 0.5x²
V =
7.Complex numbersMedium
ReIm22−2−244−4−466−6−6z₁z₂w

Given z₁ = 3∠−60° and z₂ = 2∠45°, find the real part of w = z₁·z₂.

z₁ = 3∠−60°z₂ = 2∠45°
Re(w) =
8.Complex numbersHard
ReIm0.50.5−0.5−0.511−1−1θz

Use De Moivre's theorem to find the imaginary part of z⁶ for z = 1 − j.

z = 1 − jn = 6
Im(z⁶) =
9.Differential equationsMedium
t (min)T (°C)8Ta​T = ?

A cup of coffee at 84 °C is left in a room at 24 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.12 min⁻¹, find its temperature T after 8 min.

T0 = 84 °CTa = 24 °Ck = 0.12 min⁻¹t = 8 min
T = °C
10.Differential equationsHard
t (min)m (kg)t = ?26 kg

A tank holds 1000 L of water in which 10 kg of salt is dissolved. Brine with 0.05 kg of salt per litre flows in at 5 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 26 kg of salt?

V = 1000 Lm0 = 10 kgc = 0.05 kg/Lr = 5 L/min
t = min
11.Matrices & determinantsMedium
A =0−50403−44−4

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

A = [0 −5 0; 4 0 3; −4 4 −4]
det A =
12.Matrices & determinantsHard
A =−6−41−2b =−59

Solve the system −6x − 4y = −5 and x − 2y = 9. Find x.

−6x − 4y = −5x − 2y = 9
x =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (−4, 2, 0) onto b = (0, −2, −5), that is, the component of a in the direction of b.

a = (−4, 2, 0)b = (0, −2, −5)
compb a =
14.Vectors: dot & cross productHard
xyzABC

Find the area A of the triangle with corners A(1, 1, −4), B(−4, 4, 2) and C(−1, 0, −4). Coordinates are in metres.

A(1, 1, −4)B(−4, 4, 2)C(−1, 0, −4)
A = m²
15.ProbabilityMedium
red: 3blue: 5

A bag holds 3 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

3 / 5
P =
16.ProbabilityHard
xp(x)00.3510.0520.2530.340.05μ

X has the distribution shown, and Y = 3X + 4. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.35, 0.05, 0.25, 0.3, 0.05Y = 3X + 4
σY =
17.Descriptive statisticsMedium
202530354045

Find the sample standard deviation s of the data: 17, 29, 43, 18, 45.

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

A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is below 43.6?

μ = 50σ = 8x = 43.6
P =
19.Normal distributionMedium
32 min34.5 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 34.5 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 34.5 min
P =
20.Normal distributionHard
μ = 250x = ?97.5%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 97.5% of them lie (the 97.5th percentile).

μ = 250σ = 5P = 97.5%
x = g

3Solutions

1.Limits

Answer: L = −0.6667

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

(−2 + 6/x + (−3)/x2)/(3 + 4/x + 8/x2)

Let x → ∞

L = −2/3 = −0.6667

Answer

L = −0.6667

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

2.Limits

Answer: L = 4.167

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/(6x) → 0/0

Second application

25·cos 5x/6

Substitute x = 0

L = 25/6 = 4.167

Answer

L = 4.167
3.Derivatives

Answer: f′(1) = −8

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) = 4·(2x − 3)3·2

Substitute

f′(1) = 4·(−1)3·2 = −8

Answer

f′(1) = −8
4.Derivatives

Answer: x = 6

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 − 24x + 36 = 3·(x2 − 8x + 12)

Solve f′(x) = 0

x = 2 or x = 6

Which one is which

f″(x) = 6x − 24: negative at a maximum, positive at a minimum.

f″(2) = −12, f″(6) = 12
x = 6

Answer

x = 6
5.Integrals

Answer: I = 14.12

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.

Choose u and dv

u = ln x, du = dx/x; dv = x dx, v = x²/2

Antiderivative

∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4

Limits

(12.5·ln 5 − 6.25) − (0 − 0.25) = 14.12

Answer

I = 14.12
6.Integrals

Answer: V = 107.2

Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.

Intersection points

2x = 0.5x² → x = 0, x = 4

Washer method

V = π∫₀4 ((2x)² − (0.5x²)²) dx

Integrate

V = π(4·4³/3 − 0.25·4⁵/5) = π·34.13 = 107.2

Answer

V = 107.2
7.Complex numbers

Answer: Re(w) = 5.796

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| = 3 · 2 = 6
θ = −60° + 45° = −15°

Rectangular form

Re(w) = 6·cos(−15°) = 5.796
Im(w) = 6·sin(−15°) = −1.553
Re(w) = 5.796

Answer

Re(w) = 5.796
8.Complex numbers

Answer: Im(z⁶) = 8

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 = 1 − j = √2∠−45°

De Moivre

z⁶ = (√2)⁶∠−270° = 8∠90°

Rectangular form

z⁶ = 8·(cos 90° + j sin 90°) = j8
Im(z⁶) = 8

Answer

Im(z⁶) = 8
9.Differential equations

Answer: T = 46.97 °C

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.

Solution of the cooling law

T(t) = 24 + 60·e−0.12t

Substitute t

T(8) = 24 + 60·0.3829 = 46.97 °C

Answer

T = 46.97 °C
10.Differential equations

Answer: t = 102.2 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.

Result

m(t) = 50 + (−40)·e−t/200

Solve for t

e−t/200 = (50 − 26)/(50 − 10) = 0.6
t = 200·ln(1.667) = 102.2 min

Answer

t = 102.2 min
11.Matrices & determinants

Answer: det A = −20

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 = −12, M₁₂ = −4, M₁₃ = 16

Cofactor expansion

det A = 0·(−12) − (−5)·(−4) + 0·16 = −20

Answer

det A = −20
12.Matrices & determinants

Answer: x = 2.875

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 = (−6)·(−2) − (−4)·1 = 16

Replace the x column

det Aₓ = (−5)·(−2) − (−4)·9 = 46
x = 46/16 = 2.875

Answer

x = 2.875
y = −3.063
13.Vectors: dot & cross product

Answer: compb a = −0.7428

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 = (−4)·0 + 2·(−2) + 0·(−5) = −4

Length of b

|b| = √(0² + (−2)² + (−5)²) = 5.385

Scalar projection

compb a = a·b/|b| = −4/5.385 = −0.7428

Answer

compb a = −0.7428

Negative: a has a component pointing against b.

14.Vectors: dot & cross product

Answer: A = 8.675 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, 3, 6)
AC = C − A = (−2, −1, 0)

Cross product

(AB × AC)x = AByACz − ABzACy = 3·0 − 6·(−1) = 6
(AB × AC)y = ABzACx − ABxACz = 6·(−2) − (−5)·0 = −12
(AB × AC)z = ABxACy − AByACx = (−5)·(−1) − 3·(−2) = 11
|AB × AC| = √(6² + (−12)² + 11²) = 17.35

Area of the triangle

A = ½·17.35 = 8.675 m²

Answer

A = 8.675 m²
15.Probability

Answer: P = 0.1071

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 = 3/8 · 2/7 = 6/56 = 0.1071

Answer

P = 0.1071
16.Probability

Answer: σY = 4.056

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 = 1.65, Var(X) = 1.828, σX = 1.352

Linear change

σY = |3|·1.352 = 4.056

Answer

σY = 4.056

The mean does move: E(Y) = 3·1.65 + 4 = 8.95.

17.Descriptive statistics

Answer: s = 13.3

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 = 152/5 = 30.4

Squared deviations

Σ(x − x̄)² = (−13.4)² + (−1.4)² + 12.6² + (−12.4)² + 14.6² = 707.2

Standard deviation

s = √(707.2/4) = 13.3

Answer

s = 13.3
18.Descriptive statistics

Answer: P = 0.2119

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 = −0.8

Standard normal table

Φ(−0.8) = 0.2119
P = 0.2119

Answer

P = 0.2119
19.Normal distribution

Answer: P = 0.1587

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 = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (34.5 − 32)/2.5 = 1

Upper tail

P = 1 − Φ(1) = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
20.Normal distribution

Answer: x = 259.8 g

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 for the percentile

Φ(z) = 0.975 → z = 1.96

Back to x

x = 250 + 1.96·5 = 259.8 g

Answer

x = 259.8 g

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