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 + 4x) − x).

f(x) = √(x2 + 4x) − xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (1 − cos 5x)/(x²), with x in radians.

f(x) = (1 − cos 5x)/(x²)x → 0
L =
3.DerivativesMedium
xyx = −1

Let f(x) = (2x + 1)5. Find f′(−1).

f(x) = (2x + 1)5x = −1
f′(−1) =
4.DerivativesHard
xy

Let f(x) = x3 + 3x2 − 9x + 1. Find the x-coordinate x of the local maximum.

f(x) = x3 + 3x2 − 9x + 1
x =
5.IntegralsMedium
xy1y = 2x

The region under y = 2x from x = 0 to x = 1 is rotated about the x-axis. Find the volume V of the solid (a cone).

y = 2x0 ≤ x ≤ 1
V =
6.IntegralsHard
xy−33

Find the area A of the region between the parabola y = 9 − x² and the x-axis.

y = 9 − x²
A =
7.Complex numbersMedium
ReIm55−5−5θz

Write z = 6 − j8 in exponential form r·ejθ. Give θ in radians, between −π and π.

z = 6 − j8
θ = rad
8.Complex numbersHard
ReIm22−2−244−4−4ABCD

Exactly one of the points A–D is a root of z³ = 27∠−90°. Click it.

w = 27∠−90°n = 3
  1. 1A (3∠150°)
  2. 2B (3∠90°)
  3. 3C (3∠−90°)
  4. 4D (5.4∠150°)
9.Differential equationsMedium
t (min)T (°C)39Ta​T = ?

A forged steel part at 400 °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 = 0.02 min⁻¹, find its temperature T after 39 min.

T0 = 400 °CTa = 26 °Ck = 0.02 min⁻¹t = 39 min
T = °C
10.Differential equationsHard
t (min)m (kg)t = ?40 kg

A tank holds 1000 L of pure water. Brine with 0.1 kg of salt per litre flows in at 2 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 40 kg of salt?

V = 1000 Lm0 = 0 kgc = 0.1 kg/Lr = 2 L/min
t = min
11.Matrices & determinantsMedium
A =3−35−1−5−15−3−5

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

A = [3 −3 5; −1 −5 −1; 5 −3 −5]
det A =
12.Matrices & determinantsHard
A =1−247

Find the larger eigenvalue λ of the matrix A shown.

A = [1 −2; 4 7]
λ =
13.Vectors: dot & cross productMedium
xyzab

Find the value of k for which the vectors a = (−3, −6, 4) and b = (k, 0, 5) are perpendicular.

a = (−3, −6, 4)b = (k, 0, 5)
k =
14.Vectors: dot & cross productHard
xyzab

The unit vector n̂ is perpendicular to both a = (−4, 3, −1) and b = (−5, 0, 5) and has a positive z-component. Find its x-component n̂x.

a = (−4, 3, −1)b = (−5, 0, 5)n̂z > 0
n̂x =
15.ProbabilityMedium
xp(x)10.120.0530.340.350.25μ = 3.55

For the distribution shown, find the standard deviation σ of X.

x: 1, 2, 3, 4, 5p(x): 0.1, 0.05, 0.3, 0.3, 0.25
σ =
16.ProbabilityHard
€p(x)−10.8430.15190.01

A game costs €1 to play. You win €4 with probability 0.15 and €20 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €1€4: p = 0.15€20: p = 0.01
E(G) = €
17.Descriptive statisticsMedium
1520253035

Find the sample standard deviation s of the data: 25, 12, 35, 19, 28.

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

A quantity is normally distributed with mean 170 and standard deviation 5. What proportion P of values is above 173?

μ = 170σ = 5x = 173
P =
19.Normal distributionMedium
245254P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 245 and 254 g? Give P as a decimal.

μ = 250σ = 5x1 = 245x2 = 254
P =
20.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =

3Solutions

1.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).

Multiply by the conjugate

f(x) = 4x/(√(x2 + 4x) + x)

Divide by x

f(x) = 4/(√(1 + 4/x) + 1)

Let x → ∞

L = 4/(1 + 1) = 2

Answer

L = 2
2.Limits

Answer: L = 12.5

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).

First application

5·sin 5x/(2x) → 0/0

Second application

25·cos 5x/2

Substitute x = 0

L = 25/2 = 12.5

Answer

L = 12.5
3.Derivatives

Answer: f′(−1) = 10

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) = 5·(2x + 1)4·2

Substitute

f′(−1) = 5·(−1)4·2 = 10

Answer

f′(−1) = 10
4.Derivatives

Answer: x = −3

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 + 6x − 9 = 3·(x2 + 2x − 3)

Solve f′(x) = 0

x = −3 or x = 1

Which one is which

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

f″(−3) = −12, f″(1) = 12
x = −3

Answer

x = −3
5.Integrals

Answer: V = 4.189

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∫₀1 4x² dx

Integrate

V = π·4·1³/3 = 4.189

Answer

V = 4.189

Check: a cone of radius 2 and height 1 has (1/3)πr²h = 4.189.

6.Integrals

Answer: A = 36

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

9 − x² = 0 → x = ±3

Integrate

A = [9x − x³/3]−33 = 18 − (−18) = 36

Answer

A = 36
7.Complex numbers

Answer: θ = −0.9273 rad

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.

Modulus

r = √(36 + 64) = 10

Argument in radians

θ = −53.13° = −53.13·π/180 = −0.9273 rad

Exponential form

z = 10·e−j0.9273

Answer

θ = −0.9273 rad
8.Complex numbers

Answer: B (3∠90°)

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 = −30°, 90°, −150°

The point

B = 3∠90°

Answer

B (3∠90°)
9.Differential equations

Answer: T = 197.4 °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 + 374·e−0.02t

Substitute t

T(39) = 26 + 374·0.4584 = 197.4 °C

Answer

T = 197.4 °C
10.Differential equations

Answer: t = 255.4 min

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.

Result

m(t) = 100 + (−100)·e−t/500

Solve for t

e−t/500 = (100 − 40)/(100 − 0) = 0.6
t = 500·ln(1.667) = 255.4 min

Answer

t = 255.4 min
11.Matrices & determinants

Answer: det A = 236

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 = 22, M₁₂ = 10, M₁₃ = 28

Cofactor expansion

det A = 3·22 − (−3)·10 + 5·28 = 236

Answer

det A = 236
12.Matrices & determinants

Answer: λ = 5

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 − (8)λ + (15) = 0

Solve

λ = (8 ± √(4))/2 → λ = 3, 5
λ = 5

Answer

λ = 5
13.Vectors: dot & cross product

Answer: k = 6.667

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 equals zero

a·b = (−3)·k + (−6)·0 + 4·5 = 0
20 + (−3)·k = 0

Solve for k

k = −20/(−3) = 6.667

Answer

k = 6.667
14.Vectors: dot & cross product

Answer: n̂x = 0.4575

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 = 3·5 − (−1)·0 = 15
(a × b)y = azbx − axbz = (−1)·(−5) − (−4)·5 = 25
(a × b)z = axby − aybx = (−4)·0 − 3·(−5) = 15

Normalise

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

|a × b| = √(15² + 25² + 15²) = 32.79
n̂ = (15, 25, 15)/32.79 = (0.4575, 0.7625, 0.4575)
n̂x = 0.4575

Answer

n̂x = 0.4575
15.Probability

Answer: σ = 1.203

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 = 1·0.1 + 2·0.05 + 3·0.3 + 4·0.3 + 5·0.25 = 3.55

Mean of the squares

E(X²) = 1·0.1 + 4·0.05 + 9·0.3 + 16·0.3 + 25·0.25 = 14.05

Variance

Var(X) = 14.05 − 3.55² = 1.448
σ = √1.448 = 1.203

Answer

σ = 1.203
16.Probability

Answer: E(G) = −0.2 €

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.

Net outcomes

G: 3 (p = 0.15), 19 (p = 0.01), −1 (p = 0.84)

Expected value

E(G) = 3·0.15 + 19·0.01 − 1·0.84 = −0.2 €

Answer

E(G) = −0.2 €

A negative value means you lose money on average.

17.Descriptive statistics

Answer: s = 8.758

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 = 119/5 = 23.8

Squared deviations

Σ(x − x̄)² = 1.2² + (−11.8)² + 11.2² + (−4.8)² + 4.2² = 306.8

Standard deviation

s = √(306.8/4) = 8.758

Answer

s = 8.758
18.Descriptive statistics

Answer: P = 0.2743

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743

Answer

P = 0.2743
19.Normal distribution

Answer: P = 0.6295

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 = (245 − 250)/5 = −1
z₂ = (254 − 250)/5 = 0.8

Area between

P = Φ(0.8) − Φ(−1) = 0.7881 − 0.1587 = 0.6295

Answer

P = 0.6295
20.Normal distribution

Answer: P = 0.8413

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 = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413

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