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 + 8x) − 3x).

f(x) = √(9x2 + 8x) − 3xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 ln(1 + 5x)/sin(x), with x in radians.

f(x) = ln(1 + 5x)/sin(x)x → 0
L =
3.DerivativesMedium
xyx = 1

Let f(x) = (3x + 3)3. Find f′(1).

f(x) = (3x + 3)3x = 1
f′(1) =
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
xy03

Evaluate I = ∫03 ex dx.

f(x) = ex[0, 3]
I =
6.IntegralsHard
xy−22

Find the area A of the region between the parabola y = 4 − x² and the x-axis.

y = 4 − x²
A =
7.Complex numbersMedium
ReIm22−2−244−4−4θz

Find the argument θ of z = −5 + j5 in degrees, as the principal value between −180° and 180°.

z = −5 + j5
θ = °
8.Complex numbersHard
ReIm22−2−244−4−466−6−6ABCD

Exactly one of the points A–D is a root of z³ = 27∠180°. Click it.

w = 27∠180°n = 3
  1. 1A (3∠−30°)
  2. 2B (5.4∠0°)
  3. 3C (3∠−90°)
  4. 4D (3∠−60°)
9.Differential equationsMedium
tyy(0) = 3y(2) = ?

Solve y″ + 8y′ + 16y = 0 with y(0) = 3 and y′(0) = 1. Find y(2).

y″ + 8y′ + 16y = 0y(0) = 3y′(0) = 1
y(2) =
10.Differential equationsHard
t (min)m (kg)35 minm = ?

A tank holds 400 L of pure water. Brine with 0.25 kg of salt per litre flows in at 5 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 35 min?

V = 400 Lm0 = 0 kgc = 0.25 kg/Lr = 5 L/min
m = kg
11.Matrices & determinantsMedium
A =66−23B =4−1−6−6

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

A = [6 6; −2 3]B = [4 −1; −6 −6]
c21 =
12.Matrices & determinantsHard
A =5−37−5

Find the larger eigenvalue λ of the matrix A shown.

A = [5 −3; 7 −5]
λ =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (3, 0, 3) onto b = (1, 4, 4), that is, the component of a in the direction of b.

a = (3, 0, 3)b = (1, 4, 4)
compb a =
14.Vectors: dot & cross productHard
xyzabc

Find the volume V of the parallelepiped whose edges from one corner are a = (3, 4, 1), b = (1, 2, 2) and c = (2, −1, −2).

a = (3, 4, 1)b = (1, 2, 2)c = (2, −1, −2)
V =
15.ProbabilityMedium
n = 14k = 3N = ?

In how many ways N can a committee of 3 people be chosen from a group of 14? The order does not matter.

n = 14k = 3
N =
16.ProbabilityHard
n = 12, p = 0.5: P(X ≤ 1) = ?k0123456789101112

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

n = 12p = 0.5
P =
17.Descriptive statisticsMedium
2530354045

Find the sample standard deviation s of the data: 24, 32, 24, 25, 45.

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

A quantity is normally distributed with mean 50 and standard deviation 12. What proportion P of values is above 35.6?

μ = 50σ = 12x = 35.6
P =
19.Normal distributionMedium
360 g358.2 gT = X₁ + … + X₉?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 9 of them is below 358.2 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 9< 358.2 g
P =
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 = 1.333

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) = 8x/(√(9x2 + 8x) + 3x)

Divide by x

f(x) = 8/(√(9 + 8/x) + 3)

Let x → ∞

L = 8/(3 + 3) = 1.333

Answer

L = 1.333
2.Limits

Answer: L = 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 [5/(1 + 5x)]/[cos x]

Substitute x = 0

L = 5/1 = 5

Answer

L = 5
3.Derivatives

Answer: f′(1) = 324

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·(3x + 3)2·3

Substitute

f′(1) = 3·62·3 = 324

Answer

f′(1) = 324
4.Derivatives

Answer: x = 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.

Derivative

f′(x) = 3x2 − 9x + 6 = 3·(x2 − 3x + 2)

Solve f′(x) = 0

x = 1 or x = 2

Which one is which

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

f″(1) = −3, f″(2) = 3
x = 1

Answer

x = 1
5.Integrals

Answer: I = 19.09

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) = ex/1

Evaluate at the limits

I = (e3 − 1)/1 = (20.09 − 1)/1 = 19.09

Answer

I = 19.09
6.Integrals

Answer: A = 10.67

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

4 − x² = 0 → x = ±2

Integrate

A = [4x − x³/3]−22 = 5.333 − (−5.333) = 10.67

Answer

A = 10.67
7.Complex numbers

Answer: θ = 135 °

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 = 5 → II

Reference angle

α = atan(5/5) = 45°

Argument

θ = 180° − α = 135°

Answer

θ = 135°

In polar form z = 7.071∠135°.

8.Complex numbers

Answer: D (3∠−60°)

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 = 60°, 180°, −60°

The point

D = 3∠−60°

Answer

D (3∠−60°)
9.Differential equations

Answer: y(2) = 0.009728

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.

Double root

(r + 4)² = 0 → r = −4

Constants

A = y(0) = 3
y′(0) = B − 4A → B = 1 + 12 = 13

Evaluate

y(2) = (3 + 13·2)·e−8 = 0.009728

Answer

y(2) = 0.009728
10.Differential equations

Answer: m = 35.44 kg

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.

Set up the equation

dm/dt = 5·0.25 − (5/400)·m = 1.25 − 0.0125·m

Solve

The integrating factor is ert/V; the solution approaches the equilibrium c·V.

m(t) = 100 + (−100)·e−t/80

Substitute t

m(35) = 100 + (−100)·0.6456 = 35.44 kg

Answer

m = 35.44 kg
11.Matrices & determinants

Answer: c21 = −26

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 = (−2)·4 + 3·(−6) = −26

Answer

c21 = −26
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 − (0)λ + (−4) = 0

Solve

λ = (0 ± √(16))/2 → λ = −2, 2
λ = 2

Answer

λ = 2
13.Vectors: dot & cross product

Answer: compb a = 2.611

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 = 3·1 + 0·4 + 3·4 = 15

Length of b

|b| = √(1² + 4² + 4²) = 5.745

Scalar projection

compb a = a·b/|b| = 15/5.745 = 2.611

Answer

compb a = 2.611
14.Vectors: dot & cross product

Answer: V = 13

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 b × c

(b × c)x = bycz − bzcy = 2·(−2) − 2·(−1) = −2
(b × c)y = bzcx − bxcz = 2·2 − 1·(−2) = 6
(b × c)z = bxcy − bycx = 1·(−1) − 2·2 = −5

Dot product with a

a·(b × c) = 3·(−2) + 4·6 + 1·(−5) = 13
V = |13| = 13

Answer

V = 13
15.Probability

Answer: N = 364

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(14, 3) = (14·13·12)/(3·2·1) = 364

Answer

N = 364
16.Probability

Answer: P = 0.003174

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) = 0.512 = 0.0002441
P(X = 1) = 12·0.5·0.511 = 0.00293

Add them

P = 0.0002441 + 0.00293 = 0.003174

Answer

P = 0.003174
17.Descriptive statistics

Answer: s = 9.028

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 = 150/5 = 30

Squared deviations

Σ(x − x̄)² = (−6)² + 2² + (−6)² + (−5)² + 15² = 326

Standard deviation

s = √(326/4) = 9.028

Answer

s = 9.028
18.Descriptive statistics

Answer: P = 0.8849

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849

Answer

P = 0.8849
19.Normal distribution

Answer: P = 0.3085

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 = 360 g, σT = 1.2·√9 = 3.6 g

Standardise

z = (358.2 − 360)/3.6 = −0.5

Probability

P = Φ(−0.5) = 0.3085

Answer

P = 0.3085
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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