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 − x + 7)/(2x2 + x + 1).

f(x) = (x2 − x + 7)/(2x2 + x + 1)x → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (1 − cos 4x)/(2x²), with x in radians.

f(x) = (1 − cos 4x)/(2x²)x → 0
L =
3.DerivativesMedium
xyx = 1

Let f(x) = x·ex. Find f′(1).

f(x) = x·exx = 1
f′(1) =
4.DerivativesHard
xyx = 3

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

f(x) = x2 − 2x + 2x = 3
b =
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
xy4y = 4xy = x²

The region between the line y = 4x and the parabola y = x² is rotated about the x-axis. Find the volume V of the solid.

y = 4xy = x²
V =
7.Complex numbersMedium
ReIm11−1−122−2−233−3−3zABCD

The diagram shows z = 1 + j. With w = 0.5∠90°, click the point that represents z/w.

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

One root of z³ + 2z² − 11z − 52 = 0 is z₁ = −3 + j2. The coefficients are real. Find the real root r.

z₁ = −3 + j2
r =
9.Differential equationsMedium
t (min)T (°C)5Ta​T = ?

A can of soda at 3 °C is taken out of the fridge into a room at 27 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.07 min⁻¹, find its temperature T after 5 min.

T0 = 3 °CTa = 27 °Ck = 0.07 min⁻¹t = 5 min
T = °C
10.Differential equationsHard
t (min)m (kg)15 minm = ?

A tank holds 400 L of pure water. Brine with 0.05 kg of salt per litre flows in at 2 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 15 min?

V = 400 Lm0 = 0 kgc = 0.05 kg/Lr = 2 L/min
m = kg
11.Matrices & determinantsMedium
A =−512−3B =−42−3−4

Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).

A = [−5 1; 2 −3]B = [−4 2; −3 −4]
c11 =
12.Matrices & determinantsHard
A =5−6−25b =712

Solve the system 5x − 6y = 7 and −2x + 5y = 12. Find x.

5x − 6y = 7−2x + 5y = 12
x =
13.Vectors: dot & cross productMedium
xyzab

Find the angle θ between the vectors a = (3, 3, 0) and b = (1, −6, 2), in degrees.

a = (3, 3, 0)b = (1, −6, 2)
θ = °
14.Vectors: dot & cross productHard
xyzab

For a = (1, 6, 6) and b = (3, −5, 5), find the y-component (a × b)y of the cross product a × b.

a = (1, 6, 6)b = (3, −5, 5)
(a × b)y =
15.ProbabilityMedium
red: 7blue: 5

A bag holds 7 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

7 / 5
P =
16.ProbabilityHard
xp(x)100.25200.35300.3400.1μ

X has the distribution shown, and Y = 3X + 20. Find the standard deviation σY of Y.

x: 10, 20, 30, 40p(x): 0.25, 0.35, 0.3, 0.1Y = 3X + 20
σY =
17.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 23, 55, 23, 32, 24, 22, 48.

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

Exam scores are normally distributed with mean 500 and standard deviation 75. Find the z-score z of a score of 545.

μ = 500σ = 75x = 545
z =
19.Normal distributionMedium
5777.5P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 57 and 77.5 points? Give P as a decimal.

μ = 65σ = 10x1 = 57x2 = 77.5
P =
20.Normal distributionHard
0.05 mm0D − d?

Hole diameters D are normal with mean 25.05 mm and standard deviation 0.02 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 = 25.05 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.02 mm
P =

3Solutions

1.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).

Divide by the highest power

(1 + (−1)/x + 7/x2)/(2 + 1/x + 1/x2)

Let x → ∞

L = 1/2 = 0.5

Answer

L = 0.5

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

2.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).

First application

4·sin 4x/(4x) → 0/0

Second application

16·cos 4x/4

Substitute x = 0

L = 16/4 = 4

Answer

L = 4
3.Derivatives

Answer: f′(1) = 5.437

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′ = ex·(1 + 1·x)

Substitute

f′(1) = 2.718·(1 + 1) = 5.437

Answer

f′(1) = 5.437
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(3) = 5
f′(x) = 2x − 2 → f′(3) = 4

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 5 − 4·3 = −7

Answer

b = −7
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: V = 428.9

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.

Intersection points

4x = 1x² → x = 0, x = 4

Washer method

V = π∫₀4 ((4x)² − (1x²)²) dx

Integrate

V = π(16·4³/3 − 1·4⁵/5) = π·136.5 = 428.9

Answer

V = 428.9
7.Complex numbers

Answer: B (2 − j2)

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 = 2 − j2 → B

Answer

B (2 − j2)
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 − j2
(z − z₁)(z − z₂) = z² − 2·(−3)·z + ((−3)² + 2²) = z² + 6z + 13

Factor out the quadratic

z³ + 2z² − 11z − 52 = (z² + 6z + 13)(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: T = 10.09 °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) = 27 + (−24)·e−0.07t

Substitute t

T(5) = 27 + (−24)·0.7047 = 10.09 °C

Answer

T = 10.09 °C
10.Differential equations

Answer: m = 1.445 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 = 2·0.05 − (2/400)·m = 0.1 − 0.005·m

Solve

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

m(t) = 20 + (−20)·e−t/200

Substitute t

m(15) = 20 + (−20)·0.9277 = 1.445 kg

Answer

m = 1.445 kg
11.Matrices & determinants

Answer: c11 = 17

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

c11 = (−5)·(−4) + 1·(−3) = 17

Answer

c11 = 17
12.Matrices & determinants

Answer: x = 8.231

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.

Determinant of the coefficient matrix

det A = 5·5 − (−6)·(−2) = 13

Replace the x column

det Aₓ = 7·5 − (−6)·12 = 107
x = 107/13 = 8.231

Answer

x = 8.231
y = 5.692
13.Vectors: dot & cross product

Answer: θ = 123.5 °

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 + 3·(−6) + 0·2 = −15

Magnitudes

|a| = √(3² + 3² + 0²) = 4.243
|b| = √(1² + (−6)² + 2²) = 6.403

Angle

cos θ = −15/(4.243·6.403) = −0.5522
θ = arccos(−0.5522) = 123.5°

Answer

θ = 123.5°
14.Vectors: dot & cross product

Answer: (a × b)y = 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, component by component

(a × b)x = aybz − azby = 6·5 − 6·(−5) = 60
(a × b)y = azbx − axbz = 6·3 − 1·5 = 13
(a × b)z = axby − aybx = 1·(−5) − 6·3 = −23

Requested component

(a × b)y = 13

Answer

(a × b)y = 13

Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.

15.Probability

Answer: P = 0.3182

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.

First and second draw

P = 7/12 · 6/11 = 42/132 = 0.3182

Answer

P = 0.3182
16.Probability

Answer: σY = 28.26

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.

Spread of X

μ = 22.5, Var(X) = 88.75, σX = 9.421

Linear change

σY = |3|·9.421 = 28.26

Answer

σY = 28.26

The mean does move: E(Y) = 3·22.5 + 20 = 87.5.

17.Descriptive statistics

Answer: s = 13.6

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.

Mean

x̄ = 227/7 = 32.43

Squared deviations

Σ(x − x̄)² = (−9.429)² + 22.57² + (−9.429)² + (−0.4286)² + (−8.429)² + (−10.43)² + 15.57² = 1110

Standard deviation

s = √(1110/6) = 13.6

Answer

s = 13.6
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 = (545 − 500)/75 = 0.6

Answer

z = 0.6
19.Normal distribution

Answer: P = 0.6825

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 = (57 − 65)/10 = −0.8
z₂ = (77.5 − 65)/10 = 1.25

Area between

P = Φ(1.25) − Φ(−0.8) = 0.8944 − 0.2119 = 0.6825

Answer

P = 0.6825
20.Normal distribution

Answer: P = 0.9616

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 = 25.05 − 25 = 0.05 mm, σ = 0.02828 mm

Standardise

z = (0 − 0.05)/0.02828 = −1.77

Probability

P = 1 − Φ(−1.77) = 0.9616

Answer

P = 0.9616

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