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 + 0.5/x)3x. Give a decimal value.

f(x) = (1 + 0.5/x)3xx → ∞
L =
2.LimitsHard
xyx = 0

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

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

Let f(x) = (2x − 1)5. Find f′(−1).

f(x) = (2x − 1)5x = −1
f′(−1) =
4.DerivativesHard
xy

Let f(x) = x3 − 6x2 − 2. Find the x-coordinate x of the local minimum.

f(x) = x3 − 6x2 − 2
x =
5.IntegralsMedium
xy3y = √(4x)

The region under y = √(4x) from x = 0 to x = 3 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).

y = √(4x)0 ≤ x ≤ 3
V =
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
ReIm55−5−5θz

Write z = −7 + j8 in exponential form r·ejθ. Give θ in radians, between −π and π.

z = −7 + j8
θ = rad
8.Complex numbersHard
ReIm11−1−122−2−2z₀z₁z₂z₃

The roots of zⁿ = w with n = 4 and w = 16∠180° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2, 3. Find the imaginary part of z₀.

w = 16∠180°n = 4k = 0
Im(z₀) =
9.Differential equationsMedium
tyy(0) = 5y(2) = ?

Solve y″ + 5y′ + 4y = 0 with y(0) = 5 and y′(0) = 4. Find y(2).

y″ + 5y′ + 4y = 0y(0) = 5y′(0) = 4
y(2) =
10.Differential equationsHard
tyy(0) = 3y(0.75) = ?

Solve y″ + 4y′ + 13y = 0 with y(0) = 3 and y′(0) = 0 (released from rest). Find y(0.75); angles in radians.

y″ + 4y′ + 13y = 0y(0) = 3y′(0) = 0
y(0.75) =
11.Matrices & determinantsMedium
A =0−3−3−24−24−4−1

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

A = [0 −3 −3; −2 4 −2; 4 −4 −1]
det A =
12.Matrices & determinantsHard
A =1−23−4b =−80

Solve the system x − 2y = −8 and 3x − 4y = 0. Find x.

x − 2y = −83x − 4y = 0
x =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (1, 5, 4) onto b = (3, 6, −5), that is, the component of a in the direction of b.

a = (1, 5, 4)b = (3, 6, −5)
compb a =
14.Vectors: dot & cross productHard
xyzrFOA

A force F = (50, −60, −40) N acts at point A(2, −3, −3). Coordinates are in metres. Find the y-component My of the moment of the force about the origin O(0, 0, 0).

F = (50, −60, −40) NA(2, −3, −3) mO(0, 0, 0) m
My = N·m
15.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6?7?8P(six) = 1/6 on each roll

A fair die is rolled 8 times. What is the probability P of getting at least one six?

n = 8
P =
16.ProbabilityHard
n = 11, p = 0.1: P(X = 2) = ?k01234567891011

A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 2 correct answers?

n = 11p = 0.1k = 2
P =
17.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 29, 13, 18, 36, 15, 58.

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

A quantity is normally distributed with mean 500 and standard deviation 75. What proportion P of values is below 593.8?

μ = 500σ = 75x = 593.8
P =
19.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.4515

Drag z so that the shaded area between −z and z under the standard normal curve is 0.4515. Find z to two decimals.

A = 0.4515
z =
20.Normal distributionHard
μ = 32x = ?75%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 75% of them lie (the 75th percentile).

μ = 32σ = 6P = 75%
x = min

3Solutions

1.Limits

Answer: L = 4.482

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→∞ 3·x·ln(1 + 0.5/x)

Evaluate

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

ln L = 3·0.5 = 1.5

Undo the logarithm

L = e1.5 = 4.482

Answer

L = 4.482
2.Limits

Answer: L = 1.333

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 [4/(1 + 4x)]/[3·cos 3x]

Substitute x = 0

L = 4/3 = 1.333

Answer

L = 1.333
3.Derivatives

Answer: f′(−1) = 810

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) = 5·(2x − 1)4·2

Substitute

f′(−1) = 5·(−3)4·2 = 810

Answer

f′(−1) = 810
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 − 12x = 3·(x2 − 4x)

Solve f′(x) = 0

x = 0 or x = 4

Which one is which

f″(x) = 6x − 12: negative at a maximum, positive at a minimum.

f″(0) = −12, f″(4) = 12
x = 4

Answer

x = 4
5.Integrals

Answer: V = 56.55

Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.

Disc method

V = π∫₀3 4x dx

Integrate

V = π·4·3²/2 = 56.55

Answer

V = 56.55
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: θ = 2.29 rad

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

r = √(49 + 64) = 10.63

Argument in radians

θ = 131.2° = 131.2·π/180 = 2.29 rad

Exponential form

z = 10.63·ej2.29

Answer

θ = 2.29 rad
8.Complex numbers

Answer: Im(z₀) = 1.414

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

161/4 = 2

Argument of the root

θ₀ = (180° + 360°·0)/4 = 45°

Rectangular form

z₀ = 2·(cos 45° + j sin 45°) = 1.414 + j1.414
Im(z₀) = 1.414

Answer

Im(z₀) = 1.414

The 4 roots lie on the circle of radius 2, 90° apart.

9.Differential equations

Answer: y(2) = 1.082

Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.

Characteristic roots

(r + 1)(r + 4) = 0 → r = −1, −4

Constants from the initial values

A + B = 5, −1A − 4B = 4
A = 8, B = −3

Evaluate

y(2) = 8·e−2 + (−3)·e−8 = 1.082

Answer

y(2) = 1.082
10.Differential equations

Answer: y(0.75) = −0.07327

Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.

Complex roots

r = −2 ± 3j

Constants

A = 3, B = 2·3/3 = 2

Evaluate

y(0.75) = e−1.5(3·cos 2.25 + 2·sin 2.25) = −0.07327

Answer

y(0.75) = −0.07327
11.Matrices & determinants

Answer: det A = 54

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 = −12, M₁₂ = 10, M₁₃ = −8

Cofactor expansion

det A = 0·(−12) − (−3)·10 + (−3)·(−8) = 54

Answer

det A = 54
12.Matrices & determinants

Answer: x = 16

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·(−4) − (−2)·3 = 2

Replace the x column

det Aₓ = (−8)·(−4) − (−2)·0 = 32
x = 32/2 = 16

Answer

x = 16
y = 12
13.Vectors: dot & cross product

Answer: compb a = 1.554

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

a·b = 1·3 + 5·6 + 4·(−5) = 13

Length of b

|b| = √(3² + 6² + (−5)²) = 8.367

Scalar projection

compb a = a·b/|b| = 13/8.367 = 1.554

Answer

compb a = 1.554
14.Vectors: dot & cross product

Answer: My = −70 N·m

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.

Position vector from O to A

r = A − O = (2, −3, −3) − (0, 0, 0) = (2, −3, −3) m

Moment M = r × F

(r × F)x = ryFz − rzFy = (−3)·(−40) − (−3)·(−60) = −60 N·m
(r × F)y = rzFx − rxFz = (−3)·50 − 2·(−40) = −70 N·m
(r × F)z = rxFy − ryFx = 2·(−60) − (−3)·50 = 30 N·m

Requested component

My = −70 N·m

Answer

My = −70 N·m
15.Probability

Answer: P = 0.7674

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)8 = 0.2326
P = 1 − 0.2326 = 0.7674

Answer

P = 0.7674
16.Probability

Answer: P = 0.2131

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 = 2) = 55·0.12·0.99 = 55·0.01·0.3874 = 0.2131

Answer

P = 0.2131
17.Descriptive statistics

Answer: s = 17.08

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 = 169/6 = 28.17

Squared deviations

Σ(x − x̄)² = 0.8333² + (−15.17)² + (−10.17)² + 7.833² + (−13.17)² + 29.83² = 1459

Standard deviation

s = √(1459/5) = 17.08

Answer

s = 17.08
18.Descriptive statistics

Answer: P = 0.8944

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 = 1.25

Standard normal table

Φ(1.25) = 0.8944
P = 0.8944

Answer

P = 0.8944
19.Normal distribution

Answer: z = 0.6

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 + 0.4515)/2 = 0.7257

Read the z-table backwards

Find 0.7257 inside the table and read off z.

z = 0.60

Answer

z = 0.60
20.Normal distribution

Answer: x = 36.05 min

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) = 0.75 → z = 0.6745

Back to x

x = 32 + 0.6745·6 = 36.05 min

Answer

x = 36.05 min

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