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 − 4x) − 3x).

f(x) = √(9x2 − 4x) − 3xx → ∞
L =
2.LimitsHard
xyx = 0

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

f(x) = ln(1 + x)/sin(4x)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 − 13.5x2 + 54x − 1. Find the x-coordinate x of the local maximum.

f(x) = x3 − 13.5x2 + 54x − 1
x =
5.IntegralsMedium
xy02

Evaluate I = ∫02 2x·(x² + 1)2 dx.

f(x) = 2x·(x² + 1)2[0, 2]
I =
6.IntegralsHard
xy2.5y = 5xy = 2x²

The region between the line y = 5x and the parabola y = 2x² is rotated about the x-axis. Find the volume V of the solid.

y = 5xy = 2x²
V =
7.Complex numbersMedium
ReIm22−2−244−4−466−6−6θz

The number z = 6·e−j0.5 has its angle in radians. Find its real part a.

r = 6θ = −0.5 rad
a =
8.Complex numbersHard
ReIm0.50.5−0.5−0.511−1−1θz

Use De Moivre's theorem to find the real part of z⁴ for z = 1 + j.

z = 1 + jn = 4
Re(z⁴) =
9.Differential equationsMedium
t (min)T (°C)30Ta​104 °C222 °C

A forged steel part at 300 °C is left to cool in a workshop at 20 °C. After 30 min its temperature is 104 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 222 °C?

T0 = 300 °CTa = 20 °CT(30) = 104 °CT = 222 °C
t = min
10.Differential equationsHard
tyK = 10y(0.75) = ?

A system obeys y″ + y′ + 9.25y = 92.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.75); angles in radians.

y″ + y′ + 9.25y = 92.5y(0) = 0y′(0) = 0
y(0.75) =
11.Matrices & determinantsMedium
A =−1−53−12−3−5−22

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

A = [−1 −5 3; −1 2 −3; −5 −2 2]
det A =
12.Matrices & determinantsHard
A =1−3−3−5b =−7−1

Solve the system x − 3y = −7 and −3x − 5y = −1. Find x.

x − 3y = −7−3x − 5y = −1
x =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (3, 6, −1) onto b = (2, 1, 3), that is, the component of a in the direction of b.

a = (3, 6, −1)b = (2, 1, 3)
compb a =
14.Vectors: dot & cross productHard
xyzrFOA

A force F = (40, −40, −20) N acts at point A(−3, 2, −3). Coordinates are in metres. Find the x-component Mx of the moment of the force about the origin O(0, 0, 0).

F = (40, −40, −20) NA(−3, 2, −3) mO(0, 0, 0) m
Mx = N·m
15.ProbabilityMedium
n = 7k = 3N = ?

In how many ways N can a committee of 3 people be chosen from a group of 7? The order does not matter.

n = 7k = 3
N =
16.ProbabilityHard
xp(x)100.3200.3300.05400.35μ

X has the distribution shown, and Y = 2X − 3. Find the standard deviation σY of Y.

x: 10, 20, 30, 40p(x): 0.3, 0.3, 0.05, 0.35Y = 2X − 3
σY =
17.Descriptive statisticsMedium
303540455055

Find the sample standard deviation s of the data: 50, 41, 55, 50, 28.

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

Exam scores are normally distributed with mean 50 and standard deviation 5. Find the z-score z of a score of 40.

μ = 50σ = 5x = 40
z =
19.Normal distributionMedium
120 kg116.2 kgT = X₁ + … + X₁₀?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 10 of them is below 116.2 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 10< 116.2 kg
P =
20.Normal distributionHard
μ = 100x = ?20%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 20% of them lie (the 20th percentile).

μ = 100σ = 15P = 20%
x = points

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).

Multiply by the conjugate

f(x) = −4x/(√(9x2 − 4x) + 3x)

Divide by x

f(x) = −4/(√(9 + (−4)/x) + 3)

Let x → ∞

L = −4/(3 + 3) = −0.6667

Answer

L = −0.6667
2.Limits

Answer: L = 0.25

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 [1/(1 + x)]/[4·cos 4x]

Substitute x = 0

L = 1/4 = 0.25

Answer

L = 0.25
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 = 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 − 27x + 54 = 3·(x2 − 9x + 18)

Solve f′(x) = 0

x = 3 or x = 6

Which one is which

f″(x) = 6x − 27: negative at a maximum, positive at a minimum.

f″(3) = −9, f″(6) = 9
x = 3

Answer

x = 3
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: V = 163.6

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

5x = 2x² → x = 0, x = 2.5

Washer method

V = π∫₀2.5 ((5x)² − (2x²)²) dx

Integrate

V = π(25·2.5³/3 − 4·2.5⁵/5) = π·52.08 = 163.6

Answer

V = 163.6
7.Complex numbers

Answer: a = 5.265

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.

Euler's formula

z = 6·(cos −0.5 + j sin −0.5)

Components (radians)

a = 6·cos −0.5 = 5.265
b = 6·sin −0.5 = −2.877
a = 5.265

Answer

a = 5.265
8.Complex numbers

Answer: Re(z⁴) = −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∠45°

De Moivre

z⁴ = (√2)⁴∠180° = 4∠180°

Rectangular form

z⁴ = 4·(cos 180° + j sin 180°) = −4
Re(z⁴) = −4

Answer

Re(z⁴) = −4
9.Differential equations

Answer: t = 8.136 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(280/84)/30 = 0.04013 min⁻¹

Time to reach the target

t = ln(280/202)/0.04013 = 8.136 min

Answer

t = 8.136 min
10.Differential equations

Answer: y(0.75) = 13.43

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 = −0.5 ± 3j

Constants from the initial values

y(0) = 10 + A = 0 → A = −10
y′(0) = −0.5A + 3B = 0 → B = −1.667

Evaluate

y(0.75) = 10·(1 − e−0.375(cos 2.25 + 0.1667·sin 2.25)) = 13.43

Answer

y(0.75) = 13.43
11.Matrices & determinants

Answer: det A = −47

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 = −2, M₁₂ = −17, M₁₃ = 12

Cofactor expansion

det A = (−1)·(−2) − (−5)·(−17) + 3·12 = −47

Answer

det A = −47
12.Matrices & determinants

Answer: x = −2.286

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·(−5) − (−3)·(−3) = −14

Replace the x column

det Aₓ = (−7)·(−5) − (−3)·(−1) = 32
x = 32/(−14) = −2.286

Answer

x = −2.286
y = 1.571
13.Vectors: dot & cross product

Answer: compb a = 2.405

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 = 3·2 + 6·1 + (−1)·3 = 9

Length of b

|b| = √(2² + 1² + 3²) = 3.742

Scalar projection

compb a = a·b/|b| = 9/3.742 = 2.405

Answer

compb a = 2.405
14.Vectors: dot & cross product

Answer: Mx = −160 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 = (−3, 2, −3) − (0, 0, 0) = (−3, 2, −3) m

Moment M = r × F

(r × F)x = ryFz − rzFy = 2·(−20) − (−3)·(−40) = −160 N·m
(r × F)y = rzFx − rxFz = (−3)·40 − (−3)·(−20) = −180 N·m
(r × F)z = rxFy − ryFx = (−3)·(−40) − 2·40 = 40 N·m

Requested component

Mx = −160 N·m

Answer

Mx = −160 N·m
15.Probability

Answer: N = 35

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(7, 3) = (7·6·5)/(3·2·1) = 35

Answer

N = 35
16.Probability

Answer: σY = 24.88

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 = 24.5, Var(X) = 154.8, σX = 12.44

Linear change

σY = |2|·12.44 = 24.88

Answer

σY = 24.88

The mean does move: E(Y) = 2·24.5 − 3 = 46.

17.Descriptive statistics

Answer: s = 10.66

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 = 224/5 = 44.8

Squared deviations

Σ(x − x̄)² = 5.2² + (−3.8)² + 10.2² + 5.2² + (−16.8)² = 454.8

Standard deviation

s = √(454.8/4) = 10.66

Answer

s = 10.66
18.Descriptive statistics

Answer: z = −2

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 = (40 − 50)/5 = −2

Answer

z = −2
19.Normal distribution

Answer: P = 0.06681

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 = 120 kg, σT = 0.8·√10 = 2.53 kg

Standardise

z = (116.2 − 120)/2.53 = −1.5

Probability

P = Φ(−1.5) = 0.06681

Answer

P = 0.06681
20.Normal distribution

Answer: x = 87.38 points

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) = 0.2 → z = −0.8416

Back to x

x = 100 + −0.8416·15 = 87.38 points

Answer

x = 87.38 points

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