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→∞ (−4x2 + 4x − 5)/(x2 + 5x + 6).

f(x) = (−4x2 + 4x − 5)/(x2 + 5x + 6)x → ∞
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 = 1.5

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
xy

Let f(x) = x3 + 4.5x2 + 6x − 2. Find the x-coordinate x of the local maximum.

f(x) = x3 + 4.5x2 + 6x − 2
x =
5.IntegralsMedium
xy28

Evaluate I = ∫28 4/x dx.

f(x) = 4/x[2, 8]
I =
6.IntegralsHard
xy1.5y = x²eˣ

Evaluate I = ∫₀1.5 x²·eˣ dx.

∫₀1.5 x²eˣ dx
I =
7.Complex numbersMedium
ReIm55−5−51010−10−10θz

Find the argument θ of z = −7 − j9 in degrees, as the principal value between −180° and 180°.

z = −7 − j9
θ = °
8.Complex numbersHard
ReIm11−1−122−2−2θz

Use De Moivre's theorem to find the imaginary part of z⁵ for z = 2 + j2.

z = 2 + j2n = 5
Im(z⁵) =
9.Differential equationsMedium
tyy(0) = 1y(1) = ?

Solve y″ + 4y′ + 3y = 0 with y(0) = 1 and y′(0) = 6. Find y(1).

y″ + 4y′ + 3y = 0y(0) = 1y′(0) = 6
y(1) =
10.Differential equationsHard
tyK = 5ymax​ = ?

A system obeys y″ + y′ + 9.25y = 46.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.

y″ + y′ + 9.25y = 46.25y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
A =3−204B =2−6−3−3

Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).

A = [3 −2; 0 4]B = [2 −6; −3 −3]
c21 =
12.Matrices & determinantsHard
A =3−3−60

Find the larger eigenvalue λ of the matrix A shown.

A = [3 −3; −6 0]
λ =
13.Vectors: dot & cross productMedium
xyzdFAB

A constant force F = (25, 30, 30) N acts on an object while it moves in a straight line from point A(1, 0, −5) to point B(−2, 0, −4). Coordinates are in metres. Find the work W done by the force.

F = (25, 30, 30) NA(1, 0, −5) mB(−2, 0, −4) m
W = J
14.Vectors: dot & cross productHard
xyzab

The unit vector n̂ is perpendicular to both a = (−5, −2, 1) and b = (5, −5, −3) and has a positive z-component. Find its x-component n̂x.

a = (−5, −2, 1)b = (5, −5, −3)n̂z > 0
n̂x =
15.ProbabilityMedium
xp(x)100.1200.35300.35400.2μ = 26.5

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

x: 10, 20, 30, 40p(x): 0.1, 0.35, 0.35, 0.2
Var(X) =
16.ProbabilityHard
n = 9, p = 0.25: P(X ≤ 1) = ?k0123456789

9 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?

n = 9p = 0.25
P =
17.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 52, 39, 58, 24, 26, 22, 37.

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

A quantity is normally distributed with mean 50 and standard deviation 15. What proportion P of values is below 38?

μ = 50σ = 15x = 38
P =
19.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9464

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9464. Find z to two decimals.

A = 0.9464
z =
20.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =

3Solutions

1.Limits

Answer: L = −4

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

(−4 + 4/x + (−5)/x2)/(1 + 5/x + 6/x2)

Let x → ∞

L = −4/1 = −4

Answer

L = −4

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

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′(1.5) = −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: x = −2

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 + 9x + 6 = 3·(x2 + 3x + 2)

Solve f′(x) = 0

x = −2 or x = −1

Which one is which

f″(x) = 6x + 9: negative at a maximum, positive at a minimum.

f″(−2) = −3, f″(−1) = 3
x = −2

Answer

x = −2
5.Integrals

Answer: I = 5.545

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) = 4·ln x

Evaluate at the limits

I = 4·(ln 8 − ln 2) = 4·ln 4 = 5.545

Answer

I = 5.545
6.Integrals

Answer: I = 3.602

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.

First round

∫x²eˣ dx = x²eˣ − ∫2x eˣ dx

Second round

∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)

Limits

4.482·(1.25) − 1·2 = 3.602

Answer

I = 3.602
7.Complex numbers

Answer: θ = −127.9 °

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 = −7, b = −9 → III

Reference angle

α = atan(9/7) = 52.13°

Argument

θ = −(180° − α) = −127.9°

Answer

θ = −127.9°

In polar form z = 11.4∠−127.9°.

8.Complex numbers

Answer: Im(z⁵) = −128

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 = 2 + j2 = 2√2∠45°

De Moivre

z⁵ = (2√2)⁵∠225° = 181∠−135°

Rectangular form

z⁵ = 181·(cos −135° + j sin −135°) = −128 − j128
Im(z⁵) = −128

Answer

Im(z⁵) = −128
9.Differential equations

Answer: y(1) = 1.481

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 + 3) = 0 → r = −1, −3

Constants from the initial values

A + B = 1, −1A − 3B = 6
A = 4.5, B = −3.5

Evaluate

y(1) = 4.5·e−1 + (−3.5)·e−3 = 1.481

Answer

y(1) = 1.481
10.Differential equations

Answer: ymax = 7.962

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 = 5, r = −0.5 ± 3j

Constants from the initial values

y(0) = 5 + A = 0 → A = −5
y′(0) = −0.5A + 3B = 0 → B = −0.8333

Peak

t = π/3 = 1.047
ymax = 5·(1 + e−0.5·π/3) = 7.962

Answer

ymax = 7.962
11.Matrices & determinants

Answer: c21 = −12

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.

Row times column

c21 = 0·2 + 4·(−3) = −12

Answer

c21 = −12
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 − (3)λ + (−18) = 0

Solve

λ = (3 ± √(81))/2 → λ = −3, 6
λ = 6

Answer

λ = 6
13.Vectors: dot & cross product

Answer: W = −45 J

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.

Displacement

d = B − A = (−2, 0, −4) − (1, 0, −5) = (−3, 0, 1) m

Work

W = F·d = 25·(−3) + 30·0 + 30·1 = −45 J

Answer

W = −45 J

Negative work: the force acts mostly against the motion.

14.Vectors: dot & cross product

Answer: n̂x = 0.2893

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 = (−2)·(−3) − 1·(−5) = 11
(a × b)y = azbx − axbz = 1·5 − (−5)·(−3) = −10
(a × b)z = axby − aybx = (−5)·(−5) − (−2)·5 = 35

Normalise

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

|a × b| = √(11² + (−10)² + 35²) = 38.03
n̂ = (11, −10, 35)/38.03 = (0.2893, −0.263, 0.9204)
n̂x = 0.2893

Answer

n̂x = 0.2893
15.Probability

Answer: Var(X) = 82.75

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 = 10·0.1 + 20·0.35 + 30·0.35 + 40·0.2 = 26.5

Mean of the squares

E(X²) = 100·0.1 + 400·0.35 + 900·0.35 + 1600·0.2 = 785

Variance

Var(X) = 785 − 26.5² = 82.75

Answer

Var(X) = 82.75
16.Probability

Answer: P = 0.3003

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) = 0.759 = 0.07508
P(X = 1) = 9·0.25·0.758 = 0.2253

Add them

P = 0.07508 + 0.2253 = 0.3003

Answer

P = 0.3003
17.Descriptive statistics

Answer: s = 14.05

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 = 258/7 = 36.86

Squared deviations

Σ(x − x̄)² = 15.14² + 2.143² + 21.14² + (−12.86)² + (−10.86)² + (−14.86)² + 0.1429² = 1185

Standard deviation

s = √(1185/6) = 14.05

Answer

s = 14.05
18.Descriptive statistics

Answer: P = 0.2119

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 = −0.8

Standard normal table

Φ(−0.8) = 0.2119
P = 0.2119

Answer

P = 0.2119
19.Normal distribution

Answer: z = 1.93

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 + 0.9464)/2 = 0.9732

Read the z-table backwards

Find 0.9732 inside the table and read off z.

z = 1.93

Answer

z = 1.93
20.Normal distribution

Answer: P = 0.7389

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 = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389

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