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 + 2x) − x).

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

Find the limit L = limx→0 (e2x − 1 − 2x)/x².

f(x) = (e2x − 1 − 2x)/x²x → 0
L =
3.DerivativesMedium
xyx = 0

Let f(x) = (4x + 3)3. Find f′(0).

f(x) = (4x + 3)3x = 0
f′(0) =
4.DerivativesHard
xyx = 1

The tangent line to f(x) = x2 − 3x − 2 at x = 1 crosses the y-axis at y = b. Find b.

f(x) = x2 − 3x − 2x = 1
b =
5.IntegralsMedium
xy01

Evaluate I = ∫01 ex dx.

f(x) = ex[0, 1]
I =
6.IntegralsHard
xy2y = xy = 0.5x²

The region between the line y = x and the parabola y = 0.5x² is rotated about the x-axis. Find the volume V of the solid.

y = xy = 0.5x²
V =
7.Complex numbersMedium
ReIm55−5−51010−10−10θz

Find the argument θ of z = −9 − j6 in degrees, as the principal value between −180° and 180°.

z = −9 − j6
θ = °
8.Complex numbersHard
ReIm11−1−122−2−2θz

Use De Moivre's theorem to find the real part of z⁶ for z = √3 + j.

z = √3 + jn = 6
Re(z⁶) =
9.Differential equationsMedium
t (min)T (°C)8Ta​9 °C8 °C

A can of soda at 6 °C is taken out of the fridge into a room at 22 °C. After 8 min its temperature is 9 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 8 °C?

T0 = 6 °CTa = 22 °CT(8) = 9 °CT = 8 °C
t = min
10.Differential equationsHard
t (min)m (kg)25 minm = ?

A tank holds 400 L of pure water. Brine with 0.25 kg of salt per litre flows in at 10 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 25 min?

V = 400 Lm0 = 0 kgc = 0.25 kg/Lr = 10 L/min
m = kg
11.Matrices & determinantsMedium
A =0−26−2B =−60−5−1

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

A = [0 −2; 6 −2]B = [−6 0; −5 −1]
c21 =
12.Matrices & determinantsHard
A =−1−5−25b =8−2

Solve the system −x − 5y = 8 and −2x + 5y = −2. Find x.

−x − 5y = 8−2x + 5y = −2
x =
13.Vectors: dot & cross productMedium
xyzab

Find the angle θ between the vectors a = (−2, −6, −2) and b = (5, −4, 6), in degrees.

a = (−2, −6, −2)b = (5, −4, 6)
θ = °
14.Vectors: dot & cross productHard
xyzABC

Find the area A of the triangle with corners A(3, 4, −1), B(3, −2, −3) and C(−2, −3, −4). Coordinates are in metres.

A(3, 4, −1)B(3, −2, −3)C(−2, −3, −4)
A = m²
15.ProbabilityMedium
P(at least one six) = ??1?2?3?4P(six) = 1/6 on each roll

A fair die is rolled 4 times. What is the probability P of getting at least one six?

n = 4
P =
16.ProbabilityHard
xp(x)10.320.230.1540.35μ

X has the distribution shown, and Y = 2X + 5. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.3, 0.2, 0.15, 0.35Y = 2X + 5
σY =
17.Descriptive statisticsMedium
2530354045

Find the sample standard deviation s of the data: 31, 23, 38, 47, 35.

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

Exam scores are normally distributed with mean 100 and standard deviation 10. Find the z-score z of a score of 112.5.

μ = 100σ = 10x = 112.5
z =
19.Normal distributionMedium
250 kg251.6 kgT = X₁ + … + X₁₀?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 10 of them is below 251.6 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 10< 251.6 kg
P =
20.Normal distributionHard
μ = 100x = ?80%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 80% of them lie (the 80th percentile).

μ = 100σ = 15P = 80%
x = points

3Solutions

1.Limits

Answer: L = 1

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) = 2x/(√(x2 + 2x) + x)

Divide by x

f(x) = 2/(√(1 + 2/x) + 1)

Let x → ∞

L = 2/(1 + 1) = 1

Answer

L = 1
2.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).

First application

(2·e2x − 2)/(2x) → 0/0

Second application

4·e2x/2

Substitute x = 0

L = 4·1/2 = 2

Answer

L = 2
3.Derivatives

Answer: f′(0) = 108

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·(4x + 3)2·4

Substitute

f′(0) = 3·32·4 = 108

Answer

f′(0) = 108
4.Derivatives

Answer: b = −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.

Point and slope

f(1) = −4
f′(x) = 2x − 3 → f′(1) = −1

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = −4 − (−1)·1 = −3

Answer

b = −3
5.Integrals

Answer: I = 1.718

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 = (e1 − 1)/1 = (2.718 − 1)/1 = 1.718

Answer

I = 1.718
6.Integrals

Answer: V = 3.351

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

1x = 0.5x² → x = 0, x = 2

Washer method

V = π∫₀2 ((1x)² − (0.5x²)²) dx

Integrate

V = π(1·2³/3 − 0.25·2⁵/5) = π·1.067 = 3.351

Answer

V = 3.351
7.Complex numbers

Answer: θ = −146.3 °

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 = −9, b = −6 → III

Reference angle

α = atan(6/9) = 33.69°

Argument

θ = −(180° − α) = −146.3°

Answer

θ = −146.3°

In polar form z = 10.82∠−146.3°.

8.Complex numbers

Answer: Re(z⁶) = −64

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 = √3 + j = 2∠30°

De Moivre

z⁶ = (2)⁶∠180° = 64∠180°

Rectangular form

z⁶ = 64·(cos 180° + j sin 180°) = −64
Re(z⁶) = −64

Answer

Re(z⁶) = −64
9.Differential equations

Answer: t = 5.145 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.

Cooling constant

k = ln(-16/-13)/8 = 0.02595 min⁻¹

Time to reach the target

t = ln(-16/(−14))/0.02595 = 5.145 min

Answer

t = 5.145 min
10.Differential equations

Answer: m = 46.47 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 = 10·0.25 − (10/400)·m = 2.5 − 0.025·m

Solve

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

m(t) = 100 + (−100)·e−t/40

Substitute t

m(25) = 100 + (−100)·0.5353 = 46.47 kg

Answer

m = 46.47 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 = 6·(−6) + (−2)·(−5) = −26

Answer

c21 = −26
12.Matrices & determinants

Answer: x = −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.

Determinant of the coefficient matrix

det A = (−1)·5 − (−5)·(−2) = −15

Replace the x column

det Aₓ = 8·5 − (−5)·(−2) = 30
x = 30/(−15) = −2

Answer

x = −2
y = −1.2
13.Vectors: dot & cross product

Answer: θ = 88.03 °

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 = (−2)·5 + (−6)·(−4) + (−2)·6 = 2

Magnitudes

|a| = √((−2)² + (−6)² + (−2)²) = 6.633
|b| = √(5² + (−4)² + 6²) = 8.775

Angle

cos θ = 2/(6.633·8.775) = 0.03436
θ = arccos(0.03436) = 88.03°

Answer

θ = 88.03°
14.Vectors: dot & cross product

Answer: A = 15.94 m²

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.

Edge vectors

AB = B − A = (0, −6, −2)
AC = C − A = (−5, −7, −3)

Cross product

(AB × AC)x = AByACz − ABzACy = (−6)·(−3) − (−2)·(−7) = 4
(AB × AC)y = ABzACx − ABxACz = (−2)·(−5) − 0·(−3) = 10
(AB × AC)z = ABxACy − AByACx = 0·(−7) − (−6)·(−5) = −30
|AB × AC| = √(4² + 10² + (−30)²) = 31.87

Area of the triangle

A = ½·31.87 = 15.94 m²

Answer

A = 15.94 m²
15.Probability

Answer: P = 0.5177

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)4 = 0.4823
P = 1 − 0.4823 = 0.5177

Answer

P = 0.5177
16.Probability

Answer: σY = 2.488

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 = 2.55, Var(X) = 1.548, σX = 1.244

Linear change

σY = |2|·1.244 = 2.488

Answer

σY = 2.488

The mean does move: E(Y) = 2·2.55 + 5 = 10.1.

17.Descriptive statistics

Answer: s = 8.843

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 = 174/5 = 34.8

Squared deviations

Σ(x − x̄)² = (−3.8)² + (−11.8)² + 3.2² + 12.2² + 0.2² = 312.8

Standard deviation

s = √(312.8/4) = 8.843

Answer

s = 8.843
18.Descriptive statistics

Answer: z = 1.25

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 = (112.5 − 100)/10 = 1.25

Answer

z = 1.25
19.Normal distribution

Answer: P = 0.8438

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 = 250 kg, σT = 0.5·√10 = 1.581 kg

Standardise

z = (251.6 − 250)/1.581 = 1.01

Probability

P = Φ(1.01) = 0.8438

Answer

P = 0.8438
20.Normal distribution

Answer: x = 112.6 points

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) = 0.8 → z = 0.8416

Back to x

x = 100 + 0.8416·15 = 112.62 points

Answer

x = 112.62 points

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