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 (e−0.5x − 1 + 0.5x)/x².

f(x) = (e−0.5x − 1 + 0.5x)/x²x → 0
L =
3.DerivativesMedium
xyx = 2

Let f(x) = (3x − 3)3. Find f′(2).

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

Let f(x) = x3 + 3x2 − 9x − 4. Find the x-coordinate x of the local maximum.

f(x) = x3 + 3x2 − 9x − 4
x =
5.IntegralsMedium
xy1y = 3x

The region under y = 3x from x = 0 to x = 1 is rotated about the x-axis. Find the volume V of the solid (a cone).

y = 3x0 ≤ x ≤ 1
V =
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
ReIm55−5−51010−10−10θz

Find the argument θ of z = −4 + j9 in degrees, as the principal value between −180° and 180°.

z = −4 + j9
θ = °
8.Complex numbersHard
ReIm0.50.5−0.5−0.511−1−1θz

Use De Moivre's theorem to find the real part of z⁵ for z = −1 + j.

z = −1 + jn = 5
Re(z⁵) =
9.Differential equationsMedium
tyy(0) = 3y(2) = ?

Solve y″ + 6y′ + 5y = 0 with y(0) = 3 and y′(0) = 6. Find y(2).

y″ + 6y′ + 5y = 0y(0) = 3y′(0) = 6
y(2) =
10.Differential equationsHard
ty1y(1) = ?

Solve y′ + 3y = 9e−0.5t with y(0) = 5. Find y(1).

y′ + 3y = 9e−0.5ty(0) = 5
y(1) =
11.Matrices & determinantsMedium
A =5−40543−3−5−2

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

A = [5 −4 0; 5 4 3; −3 −5 −2]
det A =
12.Matrices & determinantsHard
A =−1−6−6−1b =121

Solve the system −x − 6y = 12 and −6x − y = 1. Find x.

−x − 6y = 12−6x − y = 1
x =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (4, 0, 1) onto b = (5, 1, −2), that is, the component of a in the direction of b.

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

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

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

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

7 / 7
P =
16.ProbabilityHard
n = 8, p = 0.6: P(X ≤ 1) = ?k012345678

8 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?

n = 8p = 0.6
P =
17.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 34, 16, 53, 41, 22.

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

Exam scores are normally distributed with mean 70 and standard deviation 20. Find the z-score z of a score of 50.

μ = 70σ = 20x = 50
z =
19.Normal distributionMedium
680 kg730.9 kgT = X₁ + … + X₈?

The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 8 of them is below 730.9 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 8< 730.9 kg
P =
20.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
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 = 0.125

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

(−0.5·e−0.5x + 0.5)/(2x) → 0/0

Second application

0.25·e−0.5x/2

Substitute x = 0

L = 0.25·1/2 = 0.125

Answer

L = 0.125
3.Derivatives

Answer: f′(2) = 81

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 − 3)2·3

Substitute

f′(2) = 3·32·3 = 81

Answer

f′(2) = 81
4.Derivatives

Answer: x = −3

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 + 6x − 9 = 3·(x2 + 2x − 3)

Solve f′(x) = 0

x = −3 or x = 1

Which one is which

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

f″(−3) = −12, f″(1) = 12
x = −3

Answer

x = −3
5.Integrals

Answer: V = 9.425

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.

Disc method

V = π∫₀1 9x² dx

Integrate

V = π·9·1³/3 = 9.425

Answer

V = 9.425

Check: a cone of radius 3 and height 1 has (1/3)πr²h = 9.425.

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: θ = 114 °

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.

Quadrant

a = −4, b = 9 → II

Reference angle

α = atan(9/4) = 66.04°

Argument

θ = 180° − α = 114°

Answer

θ = 114°

In polar form z = 9.849∠114°.

8.Complex numbers

Answer: Re(z⁵) = 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.

Polar form

z = −1 + j = √2∠135°

De Moivre

z⁵ = (√2)⁵∠675° = 5.657∠−45°

Rectangular form

z⁵ = 5.657·(cos −45° + j sin −45°) = 4 − j4
Re(z⁵) = 4

Answer

Re(z⁵) = 4
9.Differential equations

Answer: y(2) = 0.7104

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 + 5) = 0 → r = −1, −5

Constants from the initial values

A + B = 3, −1A − 5B = 6
A = 5.25, B = −2.25

Evaluate

y(2) = 5.25·e−2 + (−2.25)·e−10 = 0.7104

Answer

y(2) = 0.7104
10.Differential equations

Answer: y(1) = 2.253

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 = e3t
(e3ty)′ = 9e2.5t

General solution

y = 9/2.5·e−0.5t + C·e−3t = 3.6·e−0.5t + C·e−3t

Initial condition

5 = 3.6 + C → C = 1.4

Evaluate

y(1) = 3.6·0.6065 + 1.4·0.04979 = 2.253

Answer

y(1) = 2.253
11.Matrices & determinants

Answer: det A = 31

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 = 7, M₁₂ = −1, M₁₃ = −13

Cofactor expansion

det A = 5·7 − (−4)·(−1) + 0·(−13) = 31

Answer

det A = 31
12.Matrices & determinants

Answer: x = 0.1714

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 = (−1)·(−1) − (−6)·(−6) = −35

Replace the x column

det Aₓ = 12·(−1) − (−6)·1 = −6
x = −6/(−35) = 0.1714

Answer

x = 0.1714
y = −2.029
13.Vectors: dot & cross product

Answer: compb a = 3.286

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 = 4·5 + 0·1 + 1·(−2) = 18

Length of b

|b| = √(5² + 1² + (−2)²) = 5.477

Scalar projection

compb a = a·b/|b| = 18/5.477 = 3.286

Answer

compb a = 3.286
14.Vectors: dot & cross product

Answer: V = 3

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 = 0·1 − 2·0 = 0
(b × c)y = bzcx − bxcz = 2·2 − 3·1 = 1
(b × c)z = bxcy − bycx = 3·0 − 0·2 = 0

Dot product with a

a·(b × c) = 0·0 + 3·1 + 3·0 = 3
V = |3| = 3

Answer

V = 3
15.Probability

Answer: P = 0.2308

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 = 7/14 · 6/13 = 42/182 = 0.2308

Answer

P = 0.2308
16.Probability

Answer: P = 0.00852

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) = 0.48 = 0.0006554
P(X = 1) = 8·0.6·0.47 = 0.007864

Add them

P = 0.0006554 + 0.007864 = 0.00852

Answer

P = 0.00852
17.Descriptive statistics

Answer: s = 14.79

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 = 166/5 = 33.2

Squared deviations

Σ(x − x̄)² = 0.8² + (−17.2)² + 19.8² + 7.8² + (−11.2)² = 874.8

Standard deviation

s = √(874.8/4) = 14.79

Answer

s = 14.79
18.Descriptive statistics

Answer: z = −1

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 = (50 − 70)/20 = −1

Answer

z = −1
19.Normal distribution

Answer: P = 0.9332

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 = 680 kg, σT = 12·√8 = 33.94 kg

Standardise

z = (730.9 − 680)/33.94 = 1.5

Probability

P = Φ(1.5) = 0.9332

Answer

P = 0.9332
20.Normal distribution

Answer: P = 0.7881

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 = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881

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