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→∞ (1 − 1/x)2x. Give a decimal value.

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

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

f(x) = (1 − cos 4x)/(x²)x → 0
L =
3.DerivativesMedium
xyx = 1.5

Let f(x) = x2·e2x. Find f′(1.5).

f(x) = x2·e2xx = 1.5
f′(1.5) =
4.DerivativesHard
xy

Let f(x) = x3 − 7.5x2 + 12x + 3. Find the x-coordinate x of the local minimum.

f(x) = x3 − 7.5x2 + 12x + 3
x =
5.IntegralsMedium
xy01

Evaluate I = ∫01 e−x dx.

f(x) = e−x[0, 1]
I =
6.IntegralsHard
xyy = 0.5x2​y = x

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

y = 0.5x2y = x
A =
7.Complex numbersMedium
ReIm22−2−244−4−466−6−6θz

Write z = 6∠25° in rectangular form a + jb and give the real part a.

r = 6θ = 25°
a =
8.Complex numbersHard
ReIm11−1−122−2−2z₁z₂

The equation z² − 4z + 10.25 = 0 has two complex roots. Find their modulus |z|.

p = −4q = 10.25
|z| =
9.Differential equationsMedium
t (min)T (°C)27Ta​24 °C17 °C

A can of soda at 6 °C is taken out of the fridge into a room at 26 °C. After 27 min its temperature is 24 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 17 °C?

T0 = 6 °CTa = 26 °CT(27) = 24 °CT = 17 °C
t = min
10.Differential equationsHard
ty0.5y(0.5) = ?

Solve y′ + 3y = 2e−0.5t with y(0) = 4. Find y(0.5).

y′ + 3y = 2e−0.5ty(0) = 4
y(0.5) =
11.Matrices & determinantsMedium
A =−63−6−5B =4−31−6

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

A = [−6 3; −6 −5]B = [4 −3; 1 −6]
c22 =
12.Matrices & determinantsHard
A =−616−2b =−103

Solve the system −6x + y = −10 and 6x − 2y = 3. Find x.

−6x + y = −106x − 2y = 3
x =
13.Vectors: dot & cross productMedium
xyzab

Find the dot product a·b of a = (4, 4, 6) and b = (−5, −6, 1).

a = (4, 4, 6)b = (−5, −6, 1)
a·b =
14.Vectors: dot & cross productHard
xyzab

Find the area A of the parallelogram spanned by a = (5, −3, 6) and b = (−5, 1, −5).

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

A bag holds 3 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

3 / 5
P =
16.ProbabilityHard
xp(x)100.3200.2300.35400.15μ

X has the distribution shown, and Y = 2X + 1. Find the standard deviation σY of Y.

x: 10, 20, 30, 40p(x): 0.3, 0.2, 0.35, 0.15Y = 2X + 1
σY =
17.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 52, 47, 40, 12, 36.

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

Exam scores are normally distributed with mean 170 and standard deviation 10. Find the z-score z of a score of 165.

μ = 170σ = 10x = 165
z =
19.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9312

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9312. Find z to two decimals.

A = 0.9312
z =
20.Normal distributionHard
157164171178185192199cmbottom 40%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 40% is to the left of it. Find the cut-off x.

μ = 178 cmσ = 7 cm40%
x = cm

3Solutions

1.Limits

Answer: L = 0.1353

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).

Take the logarithm

ln L = limx→∞ 2·x·ln(1 − 1/x)

Evaluate

ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.

ln L = 2·(−1) = −2

Undo the logarithm

L = e−2 = 0.1353

Answer

L = 0.1353
2.Limits

Answer: L = 8

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/(2x) → 0/0

Second application

16·cos 4x/2

Substitute x = 0

L = 16/2 = 8

Answer

L = 8
3.Derivatives

Answer: f′(1.5) = 150.6

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′ = e2x·(2x + 2·x2)

Substitute

f′(1.5) = 20.09·(3 + 4.5) = 150.6

Answer

f′(1.5) = 150.6
4.Derivatives

Answer: x = 4

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 − 15x + 12 = 3·(x2 − 5x + 4)

Solve f′(x) = 0

x = 1 or x = 4

Which one is which

f″(x) = 6x − 15: negative at a maximum, positive at a minimum.

f″(1) = −9, f″(4) = 9
x = 4

Answer

x = 4
5.Integrals

Answer: I = 0.6321

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) = e−x/(−1)

Evaluate at the limits

I = (e−1 − 1)/(−1) = (0.3679 − 1)/(−1) = 0.6321

Answer

I = 0.6321
6.Integrals

Answer: A = 0.6667

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 = x → x = 0, x = 2

Integrate upper minus lower

A = [0.5x2 − 0.1667x3]02 = 0.6667

Answer

A = 0.6667
7.Complex numbers

Answer: a = 5.438

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.

Components

a = 6·cos(25°) = 5.438
b = 6·sin(25°) = 2.536

Rectangular form

z = 5.438 + j2.536
a = 5.438

Answer

a = 5.438
8.Complex numbers

Answer: |z| = 3.202

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.

Discriminant

p² − 4q = 16 − 41 = −25 < 0

Quadratic formula

z = (4 ± j√25)/2 = 2 + j2.5, 2 − j2.5

Answer

|z| = √(2² + 2.5²) = √10.25 = 3.202

Answer

|z| = 3.202
9.Differential equations

Answer: t = 9.363 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(-20/-2)/27 = 0.08528 min⁻¹

Time to reach the target

t = ln(-20/(−9))/0.08528 = 9.363 min

Answer

t = 9.363 min
10.Differential equations

Answer: y(0.5) = 1.337

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 = e3t
(e3ty)′ = 2e2.5t

General solution

y = 2/2.5·e−0.5t + C·e−3t = 0.8·e−0.5t + C·e−3t

Initial condition

4 = 0.8 + C → C = 3.2

Evaluate

y(0.5) = 0.8·0.7788 + 3.2·0.2231 = 1.337

Answer

y(0.5) = 1.337
11.Matrices & determinants

Answer: c22 = 48

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 = (−6)·(−3) + (−5)·(−6) = 48

Answer

c22 = 48
12.Matrices & determinants

Answer: x = 2.833

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 = (−6)·(−2) − 1·6 = 6

Replace the x column

det Aₓ = (−10)·(−2) − 1·3 = 17
x = 17/6 = 2.833

Answer

x = 2.833
y = 7
13.Vectors: dot & cross product

Answer: a·b = −38

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.

Multiply matching components and add

a·b = 4·(−5) + 4·(−6) + 6·1 = (−20) + (−24) + 6 = −38

Answer

a·b = −38

Negative: the angle between the vectors is more than 90°.

14.Vectors: dot & cross product

Answer: A = 14.35

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) − 6·1 = 9
(a × b)y = azbx − axbz = 6·(−5) − 5·(−5) = −5
(a × b)z = axby − aybx = 5·1 − (−3)·(−5) = −10

Area

A = |a × b| = √(9² + (−5)² + (−10)²) = 14.35

Answer

A = 14.35
15.Probability

Answer: P = 0.1071

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 = 3/8 · 2/7 = 6/56 = 0.1071

Answer

P = 0.1071
16.Probability

Answer: σY = 21.24

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 = 23.5, Var(X) = 112.8, σX = 10.62

Linear change

σY = |2|·10.62 = 21.24

Answer

σY = 21.24

The mean does move: E(Y) = 2·23.5 + 1 = 48.

17.Descriptive statistics

Answer: s = 15.49

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 = 187/5 = 37.4

Squared deviations

Σ(x − x̄)² = 14.6² + 9.6² + 2.6² + (−25.4)² + (−1.4)² = 959.2

Standard deviation

s = √(959.2/4) = 15.49

Answer

s = 15.49
18.Descriptive statistics

Answer: z = −0.5

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 = (165 − 170)/10 = −0.5

Answer

z = −0.5
19.Normal distribution

Answer: z = 1.82

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 + 0.9312)/2 = 0.9656

Read the z-table backwards

Find 0.9656 inside the table and read off z.

z = 1.82

Answer

z = 1.82
20.Normal distribution

Answer: x = 176.3 cm

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

Φ(z) = 0.4

z from the table

z ≈ −0.25

Back to the original units

x = 178 − 0.25·7 = 176.3 cm

Answer

x = 176.3 cm

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