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 − 6x) − x).

f(x) = √(x2 − 6x) − xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 ln(1 + 5x)/sin(4x), with x in radians.

f(x) = ln(1 + 5x)/sin(4x)x → 0
L =
3.DerivativesMedium
xyx = 2

Let f(x) = (3x − 1)4. Find f′(2).

f(x) = (3x − 1)4x = 2
f′(2) =
4.DerivativesHard
xy

Let f(x) = x3 − 9x2 + 15x − 2. Find the x-coordinate x of the local maximum.

f(x) = x3 − 9x2 + 15x − 2
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
xyy = x2​y = 4x

Find the area A of the region enclosed by y = x2 and y = 4x.

y = x2y = 4x
A =
7.Complex numbersMedium
ReIm0.50.5−0.5−0.511−1−11.51.5−1.5−1.5θz

The number z = 1.5·ej2 has its angle in radians. Find its imaginary part b.

r = 1.5θ = 2 rad
b =
8.Complex numbersHard
ReIm11−1−122−2−2z₀z₁z₂

The roots of zⁿ = w with n = 3 and w = 8∠−90° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2. Find the real part of z₀.

w = 8∠−90°n = 3k = 0
Re(z₀) =
9.Differential equationsMedium
t (min)T (°C)11Ta​11 °C7 °C

A can of soda at 5 °C is taken out of the fridge into a room at 22 °C. After 11 min its temperature is 11 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 7 °C?

T0 = 5 °CTa = 22 °CT(11) = 11 °CT = 7 °C
t = min
10.Differential equationsHard
ty0.5y(0.5) = ?

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

y′ + 3y = 4e−0.5ty(0) = 0
y(0.5) =
11.Matrices & determinantsMedium
A =3−40−3−30−325

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

A = [3 −4 0; −3 −3 0; −3 2 5]
det A =
12.Matrices & determinantsHard
A =−443−6b =−9−11

Solve the system −4x + 4y = −9 and 3x − 6y = −11. Find x.

−4x + 4y = −93x − 6y = −11
x =
13.Vectors: dot & cross productMedium
xyzab

Find the value of k for which the vectors a = (−2, −5, −2) and b = (−1, k, −1) are perpendicular.

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

For a = (−3, 3, 3) and b = (−4, −4, −6), find the x-component (a × b)x of the cross product a × b.

a = (−3, 3, 3)b = (−4, −4, −6)
(a × b)x =
15.ProbabilityMedium
xp(x)00.310.3520.230.15μ = 1.2

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

x: 0, 1, 2, 3p(x): 0.3, 0.35, 0.2, 0.15
Var(X) =
16.ProbabilityHard
n = 11, p = 0.7: P(X ≤ 1) = ?k01234567891011

11 components are tested, and each one fails with probability 0.7, independently. What is the probability P that at most one fails?

n = 11p = 0.7
P =
17.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 52, 54, 14, 57, 51.

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

Exam scores are normally distributed with mean 170 and standard deviation 8. Find the z-score z of a score of 158.

μ = 170σ = 8x = 158
z =
19.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9761

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9761. Find z to two decimals.

A = 0.9761
z =
20.Normal distributionHard
0.05 mm0D − d?

Hole diameters D are normal with mean 25.05 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.05 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.015 mm
P =

3Solutions

1.Limits

Answer: L = −3

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) = −6x/(√(x2 − 6x) + x)

Divide by x

f(x) = −6/(√(1 + (−6)/x) + 1)

Let x → ∞

L = −6/(1 + 1) = −3

Answer

L = −3
2.Limits

Answer: L = 1.25

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 [5/(1 + 5x)]/[4·cos 4x]

Substitute x = 0

L = 5/4 = 1.25

Answer

L = 1.25
3.Derivatives

Answer: f′(2) = 1500

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) = 4·(3x − 1)3·3

Substitute

f′(2) = 4·53·3 = 1500

Answer

f′(2) = 1500
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 − 18x + 15 = 3·(x2 − 6x + 5)

Solve f′(x) = 0

x = 1 or x = 5

Which one is which

f″(x) = 6x − 18: negative at a maximum, positive at a minimum.

f″(1) = −12, f″(5) = 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 = 10.67

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 = 4x → x = 0, x = 4

Integrate upper minus lower

A = [2x2 − 0.3333x3]04 = 10.67

Answer

A = 10.67
7.Complex numbers

Answer: b = 1.364

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.

Euler's formula

z = 1.5·(cos 2 + j sin 2)

Components (radians)

a = 1.5·cos 2 = −0.6242
b = 1.5·sin 2 = 1.364
b = 1.364

Answer

b = 1.364
8.Complex numbers

Answer: Re(z₀) = 1.732

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.

Modulus of the roots

81/3 = 2

Argument of the root

θ₀ = (−90° + 360°·0)/3 = −30°

Rectangular form

z₀ = 2·(cos −30° + j sin −30°) = 1.732 − j1
Re(z₀) = 1.732

Answer

Re(z₀) = 1.732

The 3 roots lie on the circle of radius 2, 120° apart.

9.Differential equations

Answer: t = 3.163 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(-17/-11)/11 = 0.03957 min⁻¹

Time to reach the target

t = ln(-17/(−15))/0.03957 = 3.163 min

Answer

t = 3.163 min
10.Differential equations

Answer: y(0.5) = 0.8891

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)′ = 4e2.5t

General solution

y = 4/2.5·e−0.5t + C·e−3t = 1.6·e−0.5t + C·e−3t

Initial condition

0 = 1.6 + C → C = −1.6

Evaluate

y(0.5) = 1.6·0.7788 + (−1.6)·0.2231 = 0.8891

Answer

y(0.5) = 0.8891
11.Matrices & determinants

Answer: det A = −105

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 = −15, M₁₂ = −15, M₁₃ = −15

Cofactor expansion

det A = 3·(−15) − (−4)·(−15) + 0·(−15) = −105

Answer

det A = −105
12.Matrices & determinants

Answer: x = 8.167

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 = (−4)·(−6) − 4·3 = 12

Replace the x column

det Aₓ = (−9)·(−6) − 4·(−11) = 98
x = 98/12 = 8.167

Answer

x = 8.167
y = 5.917
13.Vectors: dot & cross product

Answer: k = 0.8

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 equals zero

a·b = (−2)·(−1) + (−5)·k + (−2)·(−1) = 0
4 + (−5)·k = 0

Solve for k

k = −4/(−5) = 0.8

Answer

k = 0.8
14.Vectors: dot & cross product

Answer: (a × b)x = −6

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, component by component

(a × b)x = aybz − azby = 3·(−6) − 3·(−4) = −6
(a × b)y = azbx − axbz = 3·(−4) − (−3)·(−6) = −30
(a × b)z = axby − aybx = (−3)·(−4) − 3·(−4) = 24

Requested component

(a × b)x = −6

Answer

(a × b)x = −6

Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.

15.Probability

Answer: Var(X) = 1.06

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 = 0·0.3 + 1·0.35 + 2·0.2 + 3·0.15 = 1.2

Mean of the squares

E(X²) = 0·0.3 + 1·0.35 + 4·0.2 + 9·0.15 = 2.5

Variance

Var(X) = 2.5 − 1.2² = 1.06

Answer

Var(X) = 1.06
16.Probability

Answer: P = 4.724 × 10⁻⁵

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.

Zero and one failure

P(X = 0) = 0.311 = 1.771 × 10⁻⁶
P(X = 1) = 11·0.7·0.310 = 4.547 × 10⁻⁵

Add them

P = 1.771 × 10⁻⁶ + 4.547 × 10⁻⁵ = 4.724 × 10⁻⁵

Answer

P = 4.724 × 10⁻⁵
17.Descriptive statistics

Answer: s = 17.81

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 = 228/5 = 45.6

Squared deviations

Σ(x − x̄)² = 6.4² + 8.4² + (−31.6)² + 11.4² + 5.4² = 1269

Standard deviation

s = √(1269/4) = 17.81

Answer

s = 17.81
18.Descriptive statistics

Answer: z = −1.5

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 = (158 − 170)/8 = −1.5

Answer

z = −1.5
19.Normal distribution

Answer: z = −1.98

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 − 0.9761 = 0.0239

Read the z-table backwards

Find 0.0239 inside the table and read off z.

z = −1.98

Answer

z = −1.98
20.Normal distribution

Answer: P = 0.9909

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 difference

μ = 25.05 − 25 = 0.05 mm, σ = 0.02121 mm

Standardise

z = (0 − 0.05)/0.02121 = −2.36

Probability

P = 1 − Φ(−2.36) = 0.9909

Answer

P = 0.9909

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