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→∞ (6x2 + 2x + 3)/(x2 + 2x + 6).

f(x) = (6x2 + 2x + 3)/(x2 + 2x + 6)x → ∞
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 = 2

Let f(x) = x·sin(x), with x in radians. Find f′(2).

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

Let f(x) = x3 − 3x2 − 9x − 4. Find the x-coordinate x of the local maximum.

f(x) = x3 − 3x2 − 9x − 4
x =
5.IntegralsMedium
xyπ/3y = x·cos x

Evaluate I = ∫₀π/3 x·cos x dx (x in radians).

∫₀π/3 x·cos x dx
I =
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
ReIm11−1−122−2−233−3−3θz

Write z = 4∠35° in rectangular form a + jb and give the imaginary part b.

r = 4θ = 35°
b =
8.Complex numbersHard
ReIm11−1−122−2−2z₁z₂

Solve z² + 2z + 5 = 0. The roots are a ± jb; find the real part a.

p = 2q = 5
a =
9.Differential equationsMedium
t (min)T (°C)60Ta​45 °C31 °C

A forged steel part at 500 °C is left to cool in a workshop at 21 °C. After 60 min its temperature is 45 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 31 °C?

T0 = 500 °CTa = 21 °CT(60) = 45 °CT = 31 °C
t = min
10.Differential equationsHard
t (min)m (kg)75 minm = ?

A tank holds 500 L of water in which 10 kg of salt is dissolved. 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. How much salt m is in the tank after 75 min?

V = 500 Lm0 = 10 kgc = 0.1 kg/Lr = 2 L/min
m = kg
11.Matrices & determinantsMedium
A =−44−41−44−35−1

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

A = [−4 4 −4; 1 −4 4; −3 5 −1]
det A =
12.Matrices & determinantsHard
A =−126−2b =−36

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

−x + 2y = −36x − 2y = 6
x =
13.Vectors: dot & cross productMedium
xyzdFAB

A constant force F = (−15, 5, 20) N acts on an object while it moves in a straight line from point A(−2, 0, 4) to point B(0, −3, 1). Coordinates are in metres. Find the work W done by the force.

F = (−15, 5, 20) NA(−2, 0, 4) mB(0, −3, 1) m
W = J
14.Vectors: dot & cross productHard
xyzabc

Find the volume V of the parallelepiped whose edges from one corner are a = (2, 3, −4), b = (5, −3, 0) and c = (−1, −5, −4).

a = (2, 3, −4)b = (5, −3, 0)c = (−1, −5, −4)
V =
15.ProbabilityMedium
xp(x)10.1520.2530.2540.35μ = 2.8

For the distribution shown, find the variance Var(X) of X.

x: 1, 2, 3, 4p(x): 0.15, 0.25, 0.25, 0.35
Var(X) =
16.ProbabilityHard
n = 7, p = 0.3: P(X = 1) = ?k01234567

A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 1 correct answers?

n = 7p = 0.3k = 1
P =
17.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 58, 44, 54, 57, 17.

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

A quantity is normally distributed with mean 100 and standard deviation 5. What proportion P of values is above 97.5?

μ = 100σ = 5x = 97.5
P =
19.Normal distributionMedium
442580P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 442 and 580 h? Give P as a decimal.

μ = 500σ = 40x1 = 442x2 = 580
P =
20.Normal distributionHard
μ = 178x = ?95%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 95% of them lie (the 95th percentile).

μ = 178σ = 7P = 95%
x = cm

3Solutions

1.Limits

Answer: L = 6

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

(6 + 2/x + 3/x2)/(1 + 2/x + 6/x2)

Let x → ∞

L = 6/1 = 6

Answer

L = 6

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

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′(2) = 0.077

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′(2) = 0.9093 + 2·(−0.4161) = 0.077

Answer

f′(2) = 0.077
4.Derivatives

Answer: x = −1

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 = −1 or x = 3

Which one is which

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

f″(−1) = −12, f″(3) = 12
x = −1

Answer

x = −1
5.Integrals

Answer: I = 0.4069

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 = x, du = dx; dv = cos x dx, v = sin x

Antiderivative

∫x cos x dx = x sin x − ∫sin x dx = x sin x + cos x

Limits

(π/3·sin π/3 + cos π/3) − (0 + 1) = 0.4069

Answer

I = 0.4069
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: b = 2.294

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 = 4·cos(35°) = 3.277
b = 4·sin(35°) = 2.294

Rectangular form

z = 3.277 + j2.294
b = 2.294

Answer

b = 2.294
8.Complex numbers

Answer: a = −1

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 = 4 − 20 = −16 < 0

Quadratic formula

z = (−2 ± j√16)/2 = −1 + j2, −1 − j2

Answer

a = −1

Answer

a = −1
9.Differential equations

Answer: t = 77.55 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(479/24)/60 = 0.04989 min⁻¹

Time to reach the target

t = ln(479/10)/0.04989 = 77.55 min

Answer

t = 77.55 min
10.Differential equations

Answer: m = 20.37 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 = 2·0.1 − (2/500)·m = 0.2 − 0.004·m

Solve

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

m(t) = 50 + (−40)·e−t/250

Substitute t

m(75) = 50 + (−40)·0.7408 = 20.37 kg

Answer

m = 20.37 kg
11.Matrices & determinants

Answer: det A = 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.

2×2 minors of the first row

M₁₁ = −16, M₁₂ = 11, M₁₃ = −7

Cofactor expansion

det A = (−4)·(−16) − 4·11 + (−4)·(−7) = 48

Answer

det A = 48
12.Matrices & determinants

Answer: x = 0.6

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)·(−2) − 2·6 = −10

Replace the x column

det Aₓ = (−3)·(−2) − 2·6 = −6
x = −6/(−10) = 0.6

Answer

x = 0.6
y = −1.2
13.Vectors: dot & cross product

Answer: W = −105 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 = (0, −3, 1) − (−2, 0, 4) = (2, −3, −3) m

Work

W = F·d = (−15)·2 + 5·(−3) + 20·(−3) = −105 J

Answer

W = −105 J

Negative work: the force acts mostly against the motion.

14.Vectors: dot & cross product

Answer: V = 196

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 b × c

(b × c)x = bycz − bzcy = (−3)·(−4) − 0·(−5) = 12
(b × c)y = bzcx − bxcz = 0·(−1) − 5·(−4) = 20
(b × c)z = bxcy − bycx = 5·(−5) − (−3)·(−1) = −28

Dot product with a

a·(b × c) = 2·12 + 3·20 + (−4)·(−28) = 196
V = |196| = 196

Answer

V = 196
15.Probability

Answer: Var(X) = 1.16

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 = 1·0.15 + 2·0.25 + 3·0.25 + 4·0.35 = 2.8

Mean of the squares

E(X²) = 1·0.15 + 4·0.25 + 9·0.25 + 16·0.35 = 9

Variance

Var(X) = 9 − 2.8² = 1.16

Answer

Var(X) = 1.16
16.Probability

Answer: P = 0.2471

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.

Binomial formula

P(X = 1) = 7·0.31·0.76 = 7·0.3·0.1176 = 0.2471

Answer

P = 0.2471
17.Descriptive statistics

Answer: s = 17.13

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 = 230/5 = 46

Squared deviations

Σ(x − x̄)² = 12² + (−2)² + 8² + 11² + (−29)² = 1174

Standard deviation

s = √(1174/4) = 17.13

Answer

s = 17.13
18.Descriptive statistics

Answer: P = 0.6915

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 = −0.5

Standard normal table

Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915

Answer

P = 0.6915
19.Normal distribution

Answer: P = 0.9037

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 = (442 − 500)/40 = −1.45
z₂ = (580 − 500)/40 = 2

Area between

P = Φ(2) − Φ(−1.45) = 0.9772 − 0.07353 = 0.9037

Answer

P = 0.9037
20.Normal distribution

Answer: x = 189.5 cm

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) = 0.95 → z = 1.645

Back to x

x = 178 + 1.645·7 = 189.51 cm

Answer

x = 189.51 cm

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