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 − 1/x)x. Give a decimal value.

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

Find the limit L = limx→0 (e2x − 1 − 2x)/x².

f(x) = (e2x − 1 − 2x)/x²x → 0
L =
3.DerivativesMedium
xyx = 1

Let f(x) = (2x + 1)3. Find f′(1).

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

The tangent line to f(x) = 2x2 − 3x − 5 at x = −1 crosses the y-axis at y = b. Find b.

f(x) = 2x2 − 3x − 5x = −1
b =
5.IntegralsMedium
xy12.5y = x ln x

Evaluate I = ∫₁2.5 x·ln x dx.

∫₁2.5 x ln x dx
I =
6.IntegralsHard
xyy = 2x2​y = 4x

Find the area A of the region enclosed by y = 2x2 and y = 4x.

y = 2x2y = 4x
A =
7.Complex numbersMedium
ReIm11−1−122−2−233−3−344−4−455−5−566−6−677−7−788−8−8zABCD

The diagram shows z = 2 + j3. With w = 0.5∠90°, click the point that represents z/w.

z = 2 + j3w = 0.5∠90°
  1. 1A (4 + j6)
  2. 2B (−6 + j4)
  3. 3C (3 − j2)
  4. 4D (6 − j4)
8.Complex numbersHard
ReIm11−1−122−2−233−3−3z₁z₂

One root of z³ + 2z² − 6z − 72 = 0 is z₁ = −3 + j3. The coefficients are real. Find the real root r.

z₁ = −3 + j3
r =
9.Differential equationsMedium
t (min)T (°C)5Ta​58 °C

A cup of coffee at 87 °C is left in a room at 22 °C. After 5 min its temperature is 58 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 87 °CTa = 22 °CT(5) = 58 °C
k = min⁻¹
10.Differential equationsHard
ty2y(2) = ?

Solve y′ + 2y = 7e−3t with y(0) = 3. Find y(2).

y′ + 2y = 7e−3ty(0) = 3
y(2) =
11.Matrices & determinantsMedium
A =1−5−105−14−2−5

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

A = [1 −5 −1; 0 5 −1; 4 −2 −5]
det A =
12.Matrices & determinantsHard
A =04−610

Find the larger eigenvalue λ of the matrix A shown.

A = [0 4; −6 10]
λ =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (5, 1, 4) onto b = (3, 1, −6), that is, the component of a in the direction of b.

a = (5, 1, 4)b = (3, 1, −6)
compb a =
14.Vectors: dot & cross productHard
xyzrFOA

A force F = (−60, −40, −10) N acts at point A(2, −1, 2). Coordinates are in metres. Find the z-component Mz of the moment of the force about point O(3, 2, 3).

F = (−60, −40, −10) NA(2, −1, 2) mO(3, 2, 3) m
Mz = N·m
15.ProbabilityMedium
xp(x)00.2510.320.2530.2μ = 1.4

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

x: 0, 1, 2, 3p(x): 0.25, 0.3, 0.25, 0.2
Var(X) =
16.ProbabilityHard
n = 11, p = 0.5: P(X ≤ 1) = ?k01234567891011

11 components are tested, and each one fails with probability 0.5, independently. What is the probability P that at most one fails?

n = 11p = 0.5
P =
17.Descriptive statisticsMedium
3035404550

Find the sample standard deviation s of the data: 48, 51, 27, 36, 34, 38.

n = 6
s =
18.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 170 and standard deviation 8. What proportion P of values is below 186?

μ = 170σ = 8x = 186
P =
19.Normal distributionMedium
438508P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 438 and 508 h? Give P as a decimal.

μ = 500σ = 40x1 = 438x2 = 508
P =
20.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.04 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.02 mm
P =

3Solutions

1.Limits

Answer: L = 0.3679

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 − 1/x)

Evaluate

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

ln L = 1·(−1) = −1

Undo the logarithm

L = e−1 = 0.3679

Answer

L = 0.3679
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

(2·e2x − 2)/(2x) → 0/0

Second application

4·e2x/2

Substitute x = 0

L = 4·1/2 = 2

Answer

L = 2
3.Derivatives

Answer: f′(1) = 54

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·(2x + 1)2·2

Substitute

f′(1) = 3·32·2 = 54

Answer

f′(1) = 54
4.Derivatives

Answer: b = −7

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) = 0
f′(x) = 4x − 3 → f′(−1) = −7

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 0 − (−7)·(−1) = −7

Answer

b = −7
5.Integrals

Answer: I = 1.551

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

(3.125·ln 2.5 − 1.5625) − (0 − 0.25) = 1.551

Answer

I = 1.551
6.Integrals

Answer: A = 2.667

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.

Intersection points

2x2 = 4x → x = 0, x = 2

Integrate upper minus lower

A = [2x2 − 0.6667x3]02 = 2.667

Answer

A = 2.667
7.Complex numbers

Answer: D (6 − j4)

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.

Effect of w

|z/w| = |z|/0.5, arg(z/w) = arg z − 90°

The point

z/w = 6 − j4 → D

Answer

D (6 − j4)
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 = −3 − j3
(z − z₁)(z − z₂) = z² − 2·(−3)·z + ((−3)² + 3²) = z² + 6z + 18

Factor out the quadratic

z³ + 2z² − 6z − 72 = (z² + 6z + 18)(z − 4)

Real root

r = 4

Answer

r = 4

Check with the sum of the roots: z₁ + z₂ + r = −6 + 4 = −2 = −(2).

9.Differential equations

Answer: k = 0.1182 min⁻¹

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.

Ratio of temperature differences

e−5k = 36/65

Solve for k

k = ln(65/36)/5 = 0.1182 min⁻¹

Answer

k = 0.1182 min⁻¹
10.Differential equations

Answer: y(2) = 0.1658

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 = e2t
(e2ty)′ = 7e−t

General solution

y = 7/(−1)·e−3t + C·e−2t = −7·e−3t + C·e−2t

Initial condition

3 = −7 + C → C = 10

Evaluate

y(2) = −7·0.002479 + 10·0.01832 = 0.1658

Answer

y(2) = 0.1658
11.Matrices & determinants

Answer: det A = 13

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 = −27, M₁₂ = 4, M₁₃ = −20

Cofactor expansion

det A = 1·(−27) − (−5)·4 + (−1)·(−20) = 13

Answer

det A = 13
12.Matrices & determinants

Answer: λ = 6

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.

Characteristic equation

λ² − (10)λ + (24) = 0

Solve

λ = (10 ± √(4))/2 → λ = 4, 6
λ = 6

Answer

λ = 6
13.Vectors: dot & cross product

Answer: compb a = −1.18

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 = 5·3 + 1·1 + 4·(−6) = −8

Length of b

|b| = √(3² + 1² + (−6)²) = 6.782

Scalar projection

compb a = a·b/|b| = −8/6.782 = −1.18

Answer

compb a = −1.18

Negative: a has a component pointing against b.

14.Vectors: dot & cross product

Answer: Mz = −140 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, −1, 2) − (3, 2, 3) = (−1, −3, −1) m

Moment M = r × F

(r × F)x = ryFz − rzFy = (−3)·(−10) − (−1)·(−40) = −10 N·m
(r × F)y = rzFx − rxFz = (−1)·(−60) − (−1)·(−10) = 50 N·m
(r × F)z = rxFy − ryFx = (−1)·(−40) − (−3)·(−60) = −140 N·m

Requested component

Mz = −140 N·m

Answer

Mz = −140 N·m
15.Probability

Answer: Var(X) = 1.14

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 = 0·0.25 + 1·0.3 + 2·0.25 + 3·0.2 = 1.4

Mean of the squares

E(X²) = 0·0.25 + 1·0.3 + 4·0.25 + 9·0.2 = 3.1

Variance

Var(X) = 3.1 − 1.4² = 1.14

Answer

Var(X) = 1.14
16.Probability

Answer: P = 0.005859

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) = 0.511 = 0.0004883
P(X = 1) = 11·0.5·0.510 = 0.005371

Add them

P = 0.0004883 + 0.005371 = 0.005859

Answer

P = 0.005859
17.Descriptive statistics

Answer: s = 8.989

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 = 234/6 = 39

Squared deviations

Σ(x − x̄)² = 9² + 12² + (−12)² + (−3)² + (−5)² + (−1)² = 404

Standard deviation

s = √(404/5) = 8.989

Answer

s = 8.989
18.Descriptive statistics

Answer: P = 0.9772

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 = 2

Standard normal table

Φ(2) = 0.9772
P = 0.9772

Answer

P = 0.9772
19.Normal distribution

Answer: P = 0.5187

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 = (438 − 500)/40 = −1.55
z₂ = (508 − 500)/40 = 0.2

Area between

P = Φ(0.2) − Φ(−1.55) = 0.5793 − 0.06057 = 0.5187

Answer

P = 0.5187
20.Normal distribution

Answer: P = 0.9452

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 = 25.04 − 25 = 0.04 mm, σ = 0.025 mm

Standardise

z = (0 − 0.04)/0.025 = −1.6

Probability

P = 1 − Φ(−1.6) = 0.9452

Answer

P = 0.9452

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