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 + x − 5)/(x2 + 4x + 5).

f(x) = (4x2 + x − 5)/(x2 + 4x + 5)x → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 ln(1 + 3x)/sin(4x), with x in radians.

f(x) = ln(1 + 3x)/sin(4x)x → 0
L =
3.DerivativesMedium
xyx = 1

Let f(x) = x·sin(2x), with x in radians. Find f′(1).

f(x) = x·sin(2x)x = 1 rad
f′(1) =
4.DerivativesHard
xyx = 3

The tangent line to f(x) = 2x2 + 2x − 5 at x = 3 crosses the y-axis at y = b. Find b.

f(x) = 2x2 + 2x − 5x = 3
b =
5.IntegralsMedium
xyπ/2y = x·sin x

Evaluate I = ∫₀π/2 x·sin x dx (x in radians).

∫₀π/2 x·sin x dx
I =
6.IntegralsHard
xyy = x2​y = 3x

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

y = x2y = 3x
A =
7.Complex numbersMedium
ReIm22−2−244−4−4z₁z₂w

Given z₁ = 4∠150° and z₂ = 4∠110°, find the real part of w = z₁/z₂.

z₁ = 4∠150°z₂ = 4∠110°
Re(w) =
8.Complex numbersHard
ReIm22−2−244−4−4z₁z₂

One root of z³ + 3z² − 15z − 125 = 0 is z₁ = −4 + j3. The coefficients are real. Find the real root r.

z₁ = −4 + j3
r =
9.Differential equationsMedium
t (min)T (°C)11Ta​21 °C24 °C

A can of soda at 6 °C is taken out of the fridge into a room at 29 °C. After 11 min its temperature is 21 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 24 °C?

T0 = 6 °CTa = 29 °CT(11) = 21 °CT = 24 °C
t = min
10.Differential equationsHard
xy12(1, −3)y(2) = ?

Solve the initial value problem y′ + 3y/x = 4x2 for x > 0, with y(1) = −3. Find y(2).

y′ + 3y/x = 4x2y(1) = −3
y(2) =
11.Matrices & determinantsMedium
A =364−6B =−602−1

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

A = [3 6; 4 −6]B = [−6 0; 2 −1]
c22 =
12.Matrices & determinantsHard
A =6451b =6−12

Solve the system 6x + 4y = 6 and 5x + y = −12. Find x.

6x + 4y = 65x + y = −12
x =
13.Vectors: dot & cross productMedium
xyzab

Find the dot product a·b of a = (−6, −2, −5) and b = (−6, −3, 1).

a = (−6, −2, −5)b = (−6, −3, 1)
a·b =
14.Vectors: dot & cross productHard
xyzab

The unit vector n̂ is perpendicular to both a = (5, 4, 3) and b = (−1, −4, 2) and has a positive z-component. Find its y-component n̂y.

a = (5, 4, 3)b = (−1, −4, 2)n̂z > 0
n̂y =
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)00.1510.2520.2530.35μ

X has the distribution shown, and Y = 0.5X + 10. Find the standard deviation σY of Y.

x: 0, 1, 2, 3p(x): 0.15, 0.25, 0.25, 0.35Y = 0.5X + 10
σY =
17.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 33, 50, 21, 57, 21, 47.

n = 6
s =
18.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 70 and standard deviation 8. Find the z-score z of a score of 62.

μ = 70σ = 8x = 62
z =
19.Normal distributionMedium
442550P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 442 and 550 h? Give P as a decimal.

μ = 500σ = 40x1 = 442x2 = 550
P =
20.Normal distributionHard
μ = 100x = ?90%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 90% of them lie (the 90th percentile).

μ = 100σ = 15P = 90%
x = points

3Solutions

1.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).

Divide by the highest power

(4 + 1/x + (−5)/x2)/(1 + 4/x + 5/x2)

Let x → ∞

L = 4/1 = 4

Answer

L = 4

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

2.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).

L'Hôpital's rule

L = limx→0 [3/(1 + 3x)]/[4·cos 4x]

Substitute x = 0

L = 3/4 = 0.75

Answer

L = 0.75
3.Derivatives

Answer: f′(1) = 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(2x) + 2x·cos(2x)

Substitute

f′(1) = 0.9093 + 2·(−0.4161) = 0.077

Answer

f′(1) = 0.077
4.Derivatives

Answer: b = −23

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) = 19
f′(x) = 4x + 2 → f′(3) = 14

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 19 − 14·3 = −23

Answer

b = −23
5.Integrals

Answer: I = 1

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 = sin x dx, v = −cos x

Antiderivative

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

Limits

(−π/2·cos π/2 + sin π/2) − (0 + 0) = 1

Answer

I = 1
6.Integrals

Answer: A = 4.5

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

x2 = 3x → x = 0, x = 3

Integrate upper minus lower

A = [1.5x2 − 0.3333x3]03 = 4.5

Answer

A = 4.5
7.Complex numbers

Answer: Re(w) = 0.766

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.

Modulus and argument of w

|w| = 4 / 4 = 1
θ = 150° − 110° = 40°

Rectangular form

Re(w) = 1·cos(40°) = 0.766
Im(w) = 1·sin(40°) = 0.6428
Re(w) = 0.766

Answer

Re(w) = 0.766
8.Complex numbers

Answer: r = 5

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 = −4 − j3
(z − z₁)(z − z₂) = z² − 2·(−4)·z + ((−4)² + 3²) = z² + 8z + 25

Factor out the quadratic

z³ + 3z² − 15z − 125 = (z² + 8z + 25)(z − 5)

Real root

r = 5

Answer

r = 5

Check with the sum of the roots: z₁ + z₂ + r = −8 + 5 = −3 = −(3).

9.Differential equations

Answer: t = 15.9 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(-23/-8)/11 = 0.096 min⁻¹

Time to reach the target

t = ln(-23/(−5))/0.096 = 15.9 min

Answer

t = 15.9 min
10.Differential equations

Answer: y(2) = 4.875

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 = e∫3/x dx = x3

Integrate

(x3y)′ = 4x5 → x3y = 4x6/6 + C

Initial condition

−3 = 4/6 + C → C = −3.667

Evaluate

y(2) = (4·26/6 + (−3.667))/23 = 4.875

Answer

y(2) = 4.875
11.Matrices & determinants

Answer: c22 = 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.

Row times column

c22 = 4·0 + (−6)·(−1) = 6

Answer

c22 = 6
12.Matrices & determinants

Answer: x = −3.857

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 = 6·1 − 4·5 = −14

Replace the x column

det Aₓ = 6·1 − 4·(−12) = 54
x = 54/(−14) = −3.857

Answer

x = −3.857
y = 7.286
13.Vectors: dot & cross product

Answer: a·b = 37

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 = (−6)·(−6) + (−2)·(−3) + (−5)·1 = 36 + 6 + (−5) = 37

Answer

a·b = 37

Positive: the angle between the vectors is less than 90°.

14.Vectors: dot & cross product

Answer: n̂y = 0.4526

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 = 4·2 − 3·(−4) = 20
(a × b)y = azbx − axbz = 3·(−1) − 5·2 = −13
(a × b)z = axby − aybx = 5·(−4) − 4·(−1) = −16

Normalise

The z-component of a × b is negative, so use −(a × b).

|a × b| = √(20² + (−13)² + (−16)²) = 28.72
n̂ = −(20, −13, −16)/28.72 = (−0.6963, 0.4526, 0.557)
n̂y = 0.4526

Answer

n̂y = 0.4526
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 = 0.5385

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 = 1.8, Var(X) = 1.16, σX = 1.077

Linear change

σY = |0.5|·1.077 = 0.5385

Answer

σY = 0.5385

The mean does move: E(Y) = 0.5·1.8 + 10 = 10.9.

17.Descriptive statistics

Answer: s = 15.42

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 = 229/6 = 38.17

Squared deviations

Σ(x − x̄)² = (−5.167)² + 11.83² + (−17.17)² + 18.83² + (−17.17)² + 8.833² = 1189

Standard deviation

s = √(1189/5) = 15.42

Answer

s = 15.42
18.Descriptive statistics

Answer: z = −1

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 = (62 − 70)/8 = −1

Answer

z = −1
19.Normal distribution

Answer: P = 0.8208

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 = (550 − 500)/40 = 1.25

Area between

P = Φ(1.25) − Φ(−1.45) = 0.8944 − 0.07353 = 0.8208

Answer

P = 0.8208
20.Normal distribution

Answer: x = 119.2 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) = 0.9 → z = 1.282

Back to x

x = 100 + 1.282·15 = 119.22 points

Answer

x = 119.22 points

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