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)3x. Give a decimal value.

f(x) = (1 − 1/x)3xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 ln(1 + x)/sin(2x), with x in radians.

f(x) = ln(1 + x)/sin(2x)x → 0
L =
3.DerivativesMedium
xyx = 1

Let f(x) = x2·e0.5x. Find f′(1).

f(x) = x2·e0.5xx = 1
f′(1) =
4.DerivativesHard
xyx = 1

The tangent line to f(x) = 2x2 − 4x − 2 at x = 1 crosses the y-axis at y = b. Find b.

f(x) = 2x2 − 4x − 2x = 1
b =
5.IntegralsMedium
xy02

Evaluate I = ∫02 e−0.5x dx.

f(x) = e−0.5x[0, 2]
I =
6.IntegralsHard
xy−11

Find the area A of the region between the parabola y = 1 − x² and the x-axis.

y = 1 − x²
A =
7.Complex numbersMedium
ReIm55−5−51010−10−10θz

Find the argument θ of z = −5 + j9 in degrees, as the principal value between −180° and 180°.

z = −5 + j9
θ = °
8.Complex numbersHard
ReIm22−2−244−4−4ABCD

Exactly one of the points A–D is a root of z⁴ = 81∠90°. Click it.

w = 81∠90°n = 4
  1. 1A (3∠90°)
  2. 2B (3∠−112.5°)
  3. 3C (3∠−157.5°)
  4. 4D (5.4∠−112.5°)
9.Differential equationsMedium
t (min)T (°C)17Ta​33 °C

A cup of coffee at 87 °C is left in a room at 21 °C. After 17 min its temperature is 33 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 87 °CTa = 21 °CT(17) = 33 °C
k = min⁻¹
10.Differential equationsHard
ty1y(1) = ?

Solve y′ + 2y = 5e−3t with y(0) = 2. Find y(1).

y′ + 2y = 5e−3ty(0) = 2
y(1) =
11.Matrices & determinantsMedium
A =−4−23−10−2450

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

A = [−4 −2 3; −1 0 −2; 4 5 0]
det A =
12.Matrices & determinantsHard
A =24−5−7

Find the larger eigenvalue λ of the matrix A shown.

A = [2 4; −5 −7]
λ =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (4, 1, −1) onto b = (−6, −6, 2), that is, the component of a in the direction of b.

a = (4, 1, −1)b = (−6, −6, 2)
compb a =
14.Vectors: dot & cross productHard
xyzrFOA

A force F = (−20, 40, 20) N acts at point A(0, −3, −1). Coordinates are in metres. Find the z-component Mz of the moment of the force about point O(−1, −3, 3).

F = (−20, 40, 20) NA(0, −3, −1) mO(−1, −3, 3) m
Mz = N·m
15.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5P(six) = 1/6 on each roll

A fair die is rolled 5 times. What is the probability P of getting at least one six?

n = 5
P =
16.ProbabilityHard
n = 9, p = 0.5: P(X = 1) = ?k0123456789

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 1 correct answers?

n = 9p = 0.5k = 1
P =
17.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 11, 24, 57, 45, 51.

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

Exam scores are normally distributed with mean 70 and standard deviation 5. Find the z-score z of a score of 73.

μ = 70σ = 5x = 73
z =
19.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9686

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9686. Find z to two decimals.

A = 0.9686
z =
20.Normal distributionHard
μ = 500x = ?97.5%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 97.5% of them lie (the 97.5th percentile).

μ = 500σ = 40P = 97.5%
x = h

3Solutions

1.Limits

Answer: L = 0.04979

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→∞ 3·x·ln(1 − 1/x)

Evaluate

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

ln L = 3·(−1) = −3

Undo the logarithm

L = e−3 = 0.04979

Answer

L = 0.04979
2.Limits

Answer: L = 0.5

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)]/[2·cos 2x]

Substitute x = 0

L = 1/2 = 0.5

Answer

L = 0.5
3.Derivatives

Answer: f′(1) = 4.122

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 rule

f′(x) = u′v + uv′ = e0.5x·(2x + 0.5·x2)

Substitute

f′(1) = 1.649·(2 + 0.5) = 4.122

Answer

f′(1) = 4.122
4.Derivatives

Answer: b = −4

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) = −4
f′(x) = 4x − 4 → f′(1) = 0

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = −4 − 0·1 = −4

Answer

b = −4
5.Integrals

Answer: I = 1.264

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.

Antiderivative

F(x) = e−0.5x/(−0.5)

Evaluate at the limits

I = (e−1 − 1)/(−0.5) = (0.3679 − 1)/(−0.5) = 1.264

Answer

I = 1.264
6.Integrals

Answer: A = 1.333

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

1 − x² = 0 → x = ±1

Integrate

A = [1x − x³/3]−11 = 0.6667 − (−0.6667) = 1.333

Answer

A = 1.333
7.Complex numbers

Answer: θ = 119.1 °

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 = −5, b = 9 → II

Reference angle

α = atan(9/5) = 60.95°

Argument

θ = 180° − α = 119.1°

Answer

θ = 119.1°

In polar form z = 10.3∠119.1°.

8.Complex numbers

Answer: C (3∠−157.5°)

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.

Modulus and arguments of the roots

|z| = 3
θ = 22.5°, 112.5°, −157.5°, −67.5°

The point

C = 3∠−157.5°

Answer

C (3∠−157.5°)
9.Differential equations

Answer: k = 0.1003 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−17k = 12/66

Solve for k

k = ln(66/12)/17 = 0.1003 min⁻¹

Answer

k = 0.1003 min⁻¹
10.Differential equations

Answer: y(1) = 0.6984

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)′ = 5e−t

General solution

y = 5/(−1)·e−3t + C·e−2t = −5·e−3t + C·e−2t

Initial condition

2 = −5 + C → C = 7

Evaluate

y(1) = −5·0.04979 + 7·0.1353 = 0.6984

Answer

y(1) = 0.6984
11.Matrices & determinants

Answer: det A = −39

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 = 10, M₁₂ = 8, M₁₃ = −5

Cofactor expansion

det A = (−4)·10 − (−2)·8 + 3·(−5) = −39

Answer

det A = −39
12.Matrices & determinants

Answer: λ = −2

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 − (−5)λ + (6) = 0

Solve

λ = (−5 ± √(1))/2 → λ = −3, −2
λ = −2

Answer

λ = −2
13.Vectors: dot & cross product

Answer: compb a = −3.671

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·(−6) + 1·(−6) + (−1)·2 = −32

Length of b

|b| = √((−6)² + (−6)² + 2²) = 8.718

Scalar projection

compb a = a·b/|b| = −32/8.718 = −3.671

Answer

compb a = −3.671

Negative: a has a component pointing against b.

14.Vectors: dot & cross product

Answer: Mz = 40 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 = (0, −3, −1) − (−1, −3, 3) = (1, 0, −4) m

Moment M = r × F

(r × F)x = ryFz − rzFy = 0·20 − (−4)·40 = 160 N·m
(r × F)y = rzFx − rxFz = (−4)·(−20) − 1·20 = 60 N·m
(r × F)z = rxFy − ryFx = 1·40 − 0·(−20) = 40 N·m

Requested component

Mz = 40 N·m

Answer

Mz = 40 N·m
15.Probability

Answer: P = 0.5981

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.

Complement

P(no six) = (5/6)5 = 0.4019
P = 1 − 0.4019 = 0.5981

Answer

P = 0.5981
16.Probability

Answer: P = 0.01758

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.

Binomial formula

P(X = 1) = 9·0.51·0.58 = 9·0.5·0.003906 = 0.01758

Answer

P = 0.01758
17.Descriptive statistics

Answer: s = 19.39

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 = 188/5 = 37.6

Squared deviations

Σ(x − x̄)² = (−26.6)² + (−13.6)² + 19.4² + 7.4² + 13.4² = 1503

Standard deviation

s = √(1503/4) = 19.39

Answer

s = 19.39
18.Descriptive statistics

Answer: z = 0.6

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 = (73 − 70)/5 = 0.6

Answer

z = 0.6
19.Normal distribution

Answer: z = −1.86

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 − 0.9686 = 0.0314

Read the z-table backwards

Find 0.0314 inside the table and read off z.

z = −1.86

Answer

z = −1.86
20.Normal distribution

Answer: x = 578.4 h

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 = 500 + 1.96·40 = 578.4 h

Answer

x = 578.4 h

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