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 − 3x) − 2x).

f(x) = √(4x2 − 3x) − 2xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (1 − cos 2x)/(3x²), with x in radians.

f(x) = (1 − cos 2x)/(3x²)x → 0
L =
3.DerivativesMedium
xyx = 1.5

Let f(x) = x·sin(x), with x in radians. Find f′(1.5).

f(x) = x·sin(x)x = 1.5 rad
f′(1.5) =
4.DerivativesHard
xyx = 2

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

f(x) = 2x2 − 3x − 2x = 2
b =
5.IntegralsMedium
xy12y = x ln x

Evaluate I = ∫₁2 x·ln x dx.

∫₁2 x ln x dx
I =
6.IntegralsHard
xyy = 0.5x2​y = 3x

Find the area A of the region enclosed by y = 0.5x2 and y = 3x.

y = 0.5x2y = 3x
A =
7.Complex numbersMedium
ReIm55−5−51010−10−10zABCD

The diagram shows z = −3 − j4. With w = 2∠90°, click the point that represents z·w.

z = −3 − j4w = 2∠90°
  1. 1A (−6 − j8)
  2. 2B (4 − j3)
  3. 3C (−8 + j6)
  4. 4D (8 − j6)
8.Complex numbersHard
ReIm11−1−122−2−2θz

Use De Moivre's theorem to find the imaginary part of z³ for z = √3 + j.

z = √3 + jn = 3
Im(z³) =
9.Differential equationsMedium
t (min)T (°C)16Ta​T = ?

A forged steel part at 460 °C is left to cool in a workshop at 30 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.05 min⁻¹, find its temperature T after 16 min.

T0 = 460 °CTa = 30 °Ck = 0.05 min⁻¹t = 16 min
T = °C
10.Differential equationsHard
ty1y(1) = ?

Solve y′ + y = 9e−0.5t with y(0) = 1. Find y(1).

y′ + y = 9e−0.5ty(0) = 1
y(1) =
11.Matrices & determinantsMedium
A =0213B =−301−3

Let C = AB for the matrices shown. Find the entry c22 (row 2, column 2).

A = [0 2; 1 3]B = [−3 0; 1 −3]
c22 =
12.Matrices & determinantsHard
A =−325−3b =0−2

Solve the system −3x + 2y = 0 and 5x − 3y = −2. Find x.

−3x + 2y = 05x − 3y = −2
x =
13.Vectors: dot & cross productMedium
xyzdFAB

A constant force F = (−15, 0, 20) N acts on an object while it moves in a straight line from point A(−1, 0, 1) to point B(2, 3, −5). Coordinates are in metres. Find the work W done by the force.

F = (−15, 0, 20) NA(−1, 0, 1) mB(2, 3, −5) m
W = J
14.Vectors: dot & cross productHard
xyzab

Find the area A of the parallelogram spanned by a = (0, 2, 5) and b = (−2, 2, −4).

a = (0, 2, 5)b = (−2, 2, −4)
A =
15.ProbabilityMedium
xp(x)10.3520.330.2540.0550.05μ = 2.15

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

x: 1, 2, 3, 4, 5p(x): 0.35, 0.3, 0.25, 0.05, 0.05
σ =
16.ProbabilityHard
xp(x)00.310.320.330.1μ

X has the distribution shown, and Y = 2X + 4. Find the standard deviation σY of Y.

x: 0, 1, 2, 3p(x): 0.3, 0.3, 0.3, 0.1Y = 2X + 4
σY =
17.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 11, 35, 18, 28, 34, 57, 47.

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

Exam scores are normally distributed with mean 60 and standard deviation 8. Find the z-score z of a score of 64.8.

μ = 60σ = 8x = 64.8
z =
19.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8611

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8611. Find z to two decimals.

A = 0.8611
z =
20.Normal distributionHard
μ = 250x = ?80%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 80% of them lie (the 80th percentile).

μ = 250σ = 5P = 80%
x = g

3Solutions

1.Limits

Answer: L = −0.75

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) = −3x/(√(4x2 − 3x) + 2x)

Divide by x

f(x) = −3/(√(4 + (−3)/x) + 2)

Let x → ∞

L = −3/(2 + 2) = −0.75

Answer

L = −0.75
2.Limits

Answer: L = 0.6667

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·sin 2x/(6x) → 0/0

Second application

4·cos 2x/6

Substitute x = 0

L = 4/6 = 0.6667

Answer

L = 0.6667
3.Derivatives

Answer: f′(1.5) = 1.104

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(x) + x·cos(x)

Substitute

f′(1.5) = 0.9975 + 1.5·0.07074 = 1.104

Answer

f′(1.5) = 1.104
4.Derivatives

Answer: b = −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.

Point and slope

f(2) = 0
f′(x) = 4x − 3 → f′(2) = 5

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 0 − 5·2 = −10

Answer

b = −10
5.Integrals

Answer: I = 0.6363

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.

Choose u and dv

u = ln x, du = dx/x; dv = x dx, v = x²/2

Antiderivative

∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4

Limits

(2·ln 2 − 1) − (0 − 0.25) = 0.6363

Answer

I = 0.6363
6.Integrals

Answer: A = 18

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.

Intersection points

0.5x2 = 3x → x = 0, x = 6

Integrate upper minus lower

A = [1.5x2 − 0.1667x3]06 = 18

Answer

A = 18
7.Complex numbers

Answer: D (8 − j6)

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| = 2·|z|, arg(z·w) = arg z + 90°

The point

z·w = 8 − j6 → D

Answer

D (8 − j6)
8.Complex numbers

Answer: Im(z³) = 8

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

z³ = (2)³∠90° = 8∠90°

Rectangular form

z³ = 8·(cos 90° + j sin 90°) = j8
Im(z³) = 8

Answer

Im(z³) = 8
9.Differential equations

Answer: T = 223.2 °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) = 30 + 430·e−0.05t

Substitute t

T(16) = 30 + 430·0.4493 = 223.2 °C

Answer

T = 223.2 °C
10.Differential equations

Answer: y(1) = 4.664

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.

Integrating factor

μ = et
(ety)′ = 9e0.5t

General solution

y = 9/0.5·e−0.5t + C·e−t = 18·e−0.5t + C·e−t

Initial condition

1 = 18 + C → C = −17

Evaluate

y(1) = 18·0.6065 + (−17)·0.3679 = 4.664

Answer

y(1) = 4.664
11.Matrices & determinants

Answer: c22 = −9

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

c22 = 1·0 + 3·(−3) = −9

Answer

c22 = −9
12.Matrices & determinants

Answer: x = −4

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 = (−3)·(−3) − 2·5 = −1

Replace the x column

det Aₓ = 0·(−3) − 2·(−2) = 4
x = 4/(−1) = −4

Answer

x = −4
y = −6
13.Vectors: dot & cross product

Answer: W = −165 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, 3, −5) − (−1, 0, 1) = (3, 3, −6) m

Work

W = F·d = (−15)·3 + 0·3 + 20·(−6) = −165 J

Answer

W = −165 J

Negative work: the force acts mostly against the motion.

14.Vectors: dot & cross product

Answer: A = 20.98

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·(−4) − 5·2 = −18
(a × b)y = azbx − axbz = 5·(−2) − 0·(−4) = −10
(a × b)z = axby − aybx = 0·2 − 2·(−2) = 4

Area

A = |a × b| = √((−18)² + (−10)² + 4²) = 20.98

Answer

A = 20.98
15.Probability

Answer: σ = 1.108

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 = 1·0.35 + 2·0.3 + 3·0.25 + 4·0.05 + 5·0.05 = 2.15

Mean of the squares

E(X²) = 1·0.35 + 4·0.3 + 9·0.25 + 16·0.05 + 25·0.05 = 5.85

Variance

Var(X) = 5.85 − 2.15² = 1.228
σ = √1.228 = 1.108

Answer

σ = 1.108
16.Probability

Answer: σY = 1.96

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 = 1.2, Var(X) = 0.96, σX = 0.9798

Linear change

σY = |2|·0.9798 = 1.96

Answer

σY = 1.96

The mean does move: E(Y) = 2·1.2 + 4 = 6.4.

17.Descriptive statistics

Answer: s = 15.87

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 = 230/7 = 32.86

Squared deviations

Σ(x − x̄)² = (−21.86)² + 2.143² + (−14.86)² + (−4.857)² + 1.143² + 24.14² + 14.14² = 1511

Standard deviation

s = √(1511/6) = 15.87

Answer

s = 15.87
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 = (64.8 − 60)/8 = 0.6

Answer

z = 0.6
19.Normal distribution

Answer: z = 1.48

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 + 0.8611)/2 = 0.9306

Read the z-table backwards

Find 0.9306 inside the table and read off z.

z = 1.48

Answer

z = 1.48
20.Normal distribution

Answer: x = 254.2 g

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 = 250 + 0.8416·5 = 254.21 g

Answer

x = 254.21 g

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