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 − 10x) − 2x).

f(x) = √(4x2 − 10x) − 2xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (1 − cos 4x)/(2x²), with x in radians.

f(x) = (1 − cos 4x)/(2x²)x → 0
L =
3.DerivativesMedium
xyx = −1

Let f(x) = (4x + 1)5. Find f′(−1).

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

Let f(x) = x3 − 9x2 + 15x − 4. Find the x-coordinate x of the local maximum.

f(x) = x3 − 9x2 + 15x − 4
x =
5.IntegralsMedium
xy02

Evaluate I = ∫02 2x·(x² + 1)1 dx.

f(x) = 2x·(x² + 1)1[0, 2]
I =
6.IntegralsHard
xyy = 2x2​y = 3x

Find the area A of the region enclosed by y = 2x2 and y = 3x.

y = 2x2y = 3x
A =
7.Complex numbersMedium
ReIm22−2−244−4−466−6−6θz

Find the argument θ of z = −2 + j6 in degrees, as the principal value between −180° and 180°.

z = −2 + j6
θ = °
8.Complex numbersHard
ReIm11−1−122−2−233−3−3z₁z₂

The equation z² + 4z + 13 = 0 has two complex roots. Find their modulus |z|.

p = 4q = 13
|z| =
9.Differential equationsMedium
t (min)T (°C)41Ta​T = ?

A forged steel part at 440 °C is left to cool in a workshop at 24 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.03 min⁻¹, find its temperature T after 41 min.

T0 = 440 °CTa = 24 °Ck = 0.03 min⁻¹t = 41 min
T = °C
10.Differential equationsHard
tyK = 10y(1) = ?

A system obeys y″ + 4y′ + 29y = 290 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1); angles in radians.

y″ + 4y′ + 29y = 290y(0) = 0y′(0) = 0
y(1) =
11.Matrices & determinantsMedium
A =51520−453−1

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

A = [5 1 5; 2 0 −4; 5 3 −1]
det A =
12.Matrices & determinantsHard
A =−21−6−4b =−12−1

Solve the system −2x + y = −12 and −6x − 4y = −1. Find x.

−2x + y = −12−6x − 4y = −1
x =
13.Vectors: dot & cross productMedium
xyzab

Find the value of k for which the vectors a = (−4, −2, −3) and b = (k, 6, 3) are perpendicular.

a = (−4, −2, −3)b = (k, 6, 3)
k =
14.Vectors: dot & cross productHard
xyzrFOA

A force F = (0, 20, 30) N acts at point A(−1, 1, 3). Coordinates are in metres. Find the z-component Mz of the moment of the force about point O(−2, −2, 0).

F = (0, 20, 30) NA(−1, 1, 3) mO(−2, −2, 0) m
Mz = N·m
15.ProbabilityMedium
xp(x)00.220.2540.260.35μ = 3.4

For the distribution shown, find the standard deviation σ of X.

x: 0, 2, 4, 6p(x): 0.2, 0.25, 0.2, 0.35
σ =
16.ProbabilityHard
n = 12, p = 0.25: P(X ≤ 1) = ?k0123456789101112

12 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?

n = 12p = 0.25
P =
17.Descriptive statisticsMedium
15202530354045505560

Find the sample standard deviation s of the data: 15, 39, 60, 12, 37.

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

A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is below 125?

μ = 100σ = 20x = 125
P =
19.Normal distributionMedium
242.5256.25P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 242.5 and 256.25 g? Give P as a decimal.

μ = 250σ = 5x1 = 242.5x2 = 256.25
P =
20.Normal distributionHard
235240245250255260265gtop 80%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 80% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g80%
x = g

3Solutions

1.Limits

Answer: L = −2.5

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).

Multiply by the conjugate

f(x) = −10x/(√(4x2 − 10x) + 2x)

Divide by x

f(x) = −10/(√(4 + (−10)/x) + 2)

Let x → ∞

L = −10/(2 + 2) = −2.5

Answer

L = −2.5
2.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).

First application

4·sin 4x/(4x) → 0/0

Second application

16·cos 4x/4

Substitute x = 0

L = 16/4 = 4

Answer

L = 4
3.Derivatives

Answer: f′(−1) = 1620

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·(4x + 1)4·4

Substitute

f′(−1) = 5·(−3)4·4 = 1620

Answer

f′(−1) = 1620
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 = 12

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.

Substitution

u = x² + 1, du = 2x dx
I = ∫15 u1 du

Integrate and evaluate

I = [u2/2]15 = (25 − 1)/2 = 12

Answer

I = 12
6.Integrals

Answer: A = 1.125

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

2x2 = 3x → x = 0, x = 1.5

Integrate upper minus lower

A = [1.5x2 − 0.6667x3]01.5 = 1.125

Answer

A = 1.125
7.Complex numbers

Answer: θ = 108.4 °

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.

Quadrant

a = −2, b = 6 → II

Reference angle

α = atan(6/2) = 71.57°

Argument

θ = 180° − α = 108.4°

Answer

θ = 108.4°

In polar form z = 6.325∠108.4°.

8.Complex numbers

Answer: |z| = 3.606

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 − 52 = −36 < 0

Quadratic formula

z = (−4 ± j√36)/2 = −2 + j3, −2 − j3

Answer

|z| = √((−2)² + 3²) = √13 = 3.606

Answer

|z| = 3.606
9.Differential equations

Answer: T = 145.6 °C

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.

Solution of the cooling law

T(t) = 24 + 416·e−0.03t

Substitute t

T(41) = 24 + 416·0.2923 = 145.6 °C

Answer

T = 145.6 °C
10.Differential equations

Answer: y(1) = 10.14

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.

Steady value and roots

yss = 10, r = −2 ± 5j

Constants from the initial values

y(0) = 10 + A = 0 → A = −10
y′(0) = −2A + 5B = 0 → B = −4

Evaluate

y(1) = 10·(1 − e−2(cos 5 + 0.4·sin 5)) = 10.14

Answer

y(1) = 10.14
11.Matrices & determinants

Answer: det A = 72

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 = 18, M₁₃ = 6

Cofactor expansion

det A = 5·12 − 1·18 + 5·6 = 72

Answer

det A = 72
12.Matrices & determinants

Answer: x = 3.5

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 = (−2)·(−4) − 1·(−6) = 14

Replace the x column

det Aₓ = (−12)·(−4) − 1·(−1) = 49
x = 49/14 = 3.5

Answer

x = 3.5
y = −5
13.Vectors: dot & cross product

Answer: k = −5.25

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 = (−4)·k + (−2)·6 + (−3)·3 = 0
−21 + (−4)·k = 0

Solve for k

k = −(−21)/(−4) = −5.25

Answer

k = −5.25
14.Vectors: dot & cross product

Answer: Mz = 20 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 = (−1, 1, 3) − (−2, −2, 0) = (1, 3, 3) m

Moment M = r × F

(r × F)x = ryFz − rzFy = 3·30 − 3·20 = 30 N·m
(r × F)y = rzFx − rxFz = 3·0 − 1·30 = −30 N·m
(r × F)z = rxFy − ryFx = 1·20 − 3·0 = 20 N·m

Requested component

Mz = 20 N·m

Answer

Mz = 20 N·m
15.Probability

Answer: σ = 2.289

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 = 0·0.2 + 2·0.25 + 4·0.2 + 6·0.35 = 3.4

Mean of the squares

E(X²) = 0·0.2 + 4·0.25 + 16·0.2 + 36·0.35 = 16.8

Variance

Var(X) = 16.8 − 3.4² = 5.24
σ = √5.24 = 2.289

Answer

σ = 2.289
16.Probability

Answer: P = 0.1584

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) = 0.7512 = 0.03168
P(X = 1) = 12·0.25·0.7511 = 0.1267

Add them

P = 0.03168 + 0.1267 = 0.1584

Answer

P = 0.1584
17.Descriptive statistics

Answer: s = 19.65

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 = 163/5 = 32.6

Squared deviations

Σ(x − x̄)² = (−17.6)² + 6.4² + 27.4² + (−20.6)² + 4.4² = 1545

Standard deviation

s = √(1545/4) = 19.65

Answer

s = 19.65
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: P = 0.8275

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 = (242.5 − 250)/5 = −1.5
z₂ = (256.25 − 250)/5 = 1.25

Area between

P = Φ(1.25) − Φ(−1.5) = 0.8944 − 0.06681 = 0.8275

Answer

P = 0.8275
20.Normal distribution

Answer: x = 245.8 g

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) = 0.2

z from the table

z ≈ −0.84

Back to the original units

x = 250 − 0.84·5 = 245.8 g

Answer

x = 245.8 g

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