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→∞ (x2 − 2x − 7)/(x2 + x + 8).

f(x) = (x2 − 2x − 7)/(x2 + x + 8)x → ∞
L =
2.LimitsHard
xyx = 0

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

f(x) = (1 − cos 4x)/(4x²)x → 0
L =
3.DerivativesMedium
xyx = 1.5

Let f(x) = x·sin(2x), with x in radians. Find f′(1.5).

f(x) = x·sin(2x)x = 1.5 rad
f′(1.5) =
4.DerivativesHard
xyx = −1

The tangent line to f(x) = x2 + 3x at x = −1 crosses the y-axis at y = b. Find b.

f(x) = x2 + 3xx = −1
b =
5.IntegralsMedium
xy3y = √(2x)

The region under y = √(2x) from x = 0 to x = 3 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).

y = √(2x)0 ≤ x ≤ 3
V =
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
ReIm55−5−51010−10−10θz

Find the argument θ of z = −9 − j5 in degrees, as the principal value between −180° and 180°.

z = −9 − j5
θ = °
8.Complex numbersHard
ReIm22−2−244−4−4z₁z₂

Solve z² − 3z + 27.25 = 0. The roots are a ± jb; find b > 0.

p = −3q = 27.25
b =
9.Differential equationsMedium
t (min)T (°C)52Ta​T = ?

A forged steel part at 500 °C is left to cool in a workshop at 26 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.03 min⁻¹, find its temperature T after 52 min.

T0 = 500 °CTa = 26 °Ck = 0.03 min⁻¹t = 52 min
T = °C
10.Differential equationsHard
tyK = 1y(1) = ?

A system obeys y″ + 3y′ + 6.25y = 6.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1); angles in radians.

y″ + 3y′ + 6.25y = 6.25y(0) = 0y′(0) = 0
y(1) =
11.Matrices & determinantsMedium
A =51−4−320−243

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

A = [5 1 −4; −3 2 0; −2 4 3]
det A =
12.Matrices & determinantsHard
A =33−55b =40

Solve the system 3x + 3y = 4 and −5x + 5y = 0. Find x.

3x + 3y = 4−5x + 5y = 0
x =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (5, 0, −4) onto b = (−1, 1, 6), that is, the component of a in the direction of b.

a = (5, 0, −4)b = (−1, 1, 6)
compb a =
14.Vectors: dot & cross productHard
xyzab

The unit vector n̂ is perpendicular to both a = (4, −5, 0) and b = (0, 5, 4) and has a positive z-component. Find its y-component n̂y.

a = (4, −5, 0)b = (0, 5, 4)n̂z > 0
n̂y =
15.ProbabilityMedium
xp(x)00.3520.1540.260.3μ = 2.9

For the distribution shown, find the variance Var(X) of X.

x: 0, 2, 4, 6p(x): 0.35, 0.15, 0.2, 0.3
Var(X) =
16.ProbabilityHard
xp(x)10.3520.330.0540.3μ

X has the distribution shown, and Y = 2X + 4. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.35, 0.3, 0.05, 0.3Y = 2X + 4
σY =
17.Descriptive statisticsMedium
10152025303540455055

Find the sample standard deviation s of the data: 59, 10, 55, 15, 19, 58, 20.

n = 7
s =
18.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 500 and standard deviation 75. What proportion P of values is above 612.5?

μ = 500σ = 75x = 612.5
P =
19.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.3594

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.3594. Find z to two decimals.

A = 0.3594
z =
20.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.03 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.015 mm
P =

3Solutions

1.Limits

Answer: L = 1

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

(1 + (−2)/x + (−7)/x2)/(1 + 1/x + 8/x2)

Let x → ∞

L = 1/1 = 1

Answer

L = 1

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

2.Limits

Answer: L = 2

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/(8x) → 0/0

Second application

16·cos 4x/8

Substitute x = 0

L = 16/8 = 2

Answer

L = 2
3.Derivatives

Answer: f′(1.5) = −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(2x) + 2x·cos(2x)

Substitute

f′(1.5) = 0.1411 + 3·(−0.99) = −2.829

Answer

f′(1.5) = −2.829
4.Derivatives

Answer: b = −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.

Point and slope

f(−1) = −2
f′(x) = 2x + 3 → f′(−1) = 1

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = −2 − 1·(−1) = −1

Answer

b = −1
5.Integrals

Answer: V = 28.27

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∫₀3 2x dx

Integrate

V = π·2·3²/2 = 28.27

Answer

V = 28.27
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: θ = −150.9 °

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 = −9, b = −5 → III

Reference angle

α = atan(5/9) = 29.05°

Argument

θ = −(180° − α) = −150.9°

Answer

θ = −150.9°

In polar form z = 10.3∠−150.9°.

8.Complex numbers

Answer: b = 5

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.

Discriminant

p² − 4q = 9 − 109 = −100 < 0

Quadratic formula

z = (3 ± j√100)/2 = 1.5 + j5, 1.5 − j5

Answer

b = 5

Answer

b = 5
9.Differential equations

Answer: T = 125.6 °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) = 26 + 474·e−0.03t

Substitute t

T(52) = 26 + 474·0.2101 = 125.6 °C

Answer

T = 125.6 °C
10.Differential equations

Answer: y(1) = 0.9407

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 = 1, r = −1.5 ± 2j

Constants from the initial values

y(0) = 1 + A = 0 → A = −1
y′(0) = −1.5A + 2B = 0 → B = −0.75

Evaluate

y(1) = 1·(1 − e−1.5(cos 2 + 0.75·sin 2)) = 0.9407

Answer

y(1) = 0.9407
11.Matrices & determinants

Answer: det A = 71

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 = 6, M₁₂ = −9, M₁₃ = −8

Cofactor expansion

det A = 5·6 − 1·(−9) + (−4)·(−8) = 71

Answer

det A = 71
12.Matrices & determinants

Answer: x = 0.6667

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 = 3·5 − 3·(−5) = 30

Replace the x column

det Aₓ = 4·5 − 3·0 = 20
x = 20/30 = 0.6667

Answer

x = 0.6667
y = 0.6667
13.Vectors: dot & cross product

Answer: compb a = −4.704

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 = 5·(−1) + 0·1 + (−4)·6 = −29

Length of b

|b| = √((−1)² + 1² + 6²) = 6.164

Scalar projection

compb a = a·b/|b| = −29/6.164 = −4.704

Answer

compb a = −4.704

Negative: a has a component pointing against b.

14.Vectors: dot & cross product

Answer: n̂y = −0.4924

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

(a × b)x = aybz − azby = (−5)·4 − 0·5 = −20
(a × b)y = azbx − axbz = 0·0 − 4·4 = −16
(a × b)z = axby − aybx = 4·5 − (−5)·0 = 20

Normalise

The z-component of a × b is already positive.

|a × b| = √((−20)² + (−16)² + 20²) = 32.5
n̂ = (−20, −16, 20)/32.5 = (−0.6155, −0.4924, 0.6155)
n̂y = −0.4924

Answer

n̂y = −0.4924
15.Probability

Answer: Var(X) = 6.19

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.

Mean

μ = 0·0.35 + 2·0.15 + 4·0.2 + 6·0.3 = 2.9

Mean of the squares

E(X²) = 0·0.35 + 4·0.15 + 16·0.2 + 36·0.3 = 14.6

Variance

Var(X) = 14.6 − 2.9² = 6.19

Answer

Var(X) = 6.19
16.Probability

Answer: σY = 2.458

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 = 2.3, Var(X) = 1.51, σX = 1.229

Linear change

σY = |2|·1.229 = 2.458

Answer

σY = 2.458

The mean does move: E(Y) = 2·2.3 + 4 = 8.6.

17.Descriptive statistics

Answer: s = 22.36

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 = 236/7 = 33.71

Squared deviations

Σ(x − x̄)² = 25.29² + (−23.71)² + 21.29² + (−18.71)² + (−14.71)² + 24.29² + (−13.71)² = 2999

Standard deviation

s = √(2999/6) = 22.36

Answer

s = 22.36
18.Descriptive statistics

Answer: P = 0.06681

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 = 1.5

Standard normal table

Φ(1.5) = 0.9332
P = 1 − 0.9332 = 0.06681

Answer

P = 0.06681
19.Normal distribution

Answer: z = 0.36

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 of z

Φ(z) = 1 − 0.3594 = 0.6406

Read the z-table backwards

Find 0.6406 inside the table and read off z.

z = 0.36

Answer

z = 0.36
20.Normal distribution

Answer: P = 0.9207

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 = 25.03 − 25 = 0.03 mm, σ = 0.02121 mm

Standardise

z = (0 − 0.03)/0.02121 = −1.41

Probability

P = 1 − Φ(−1.41) = 0.9207

Answer

P = 0.9207

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