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 + 4x − 2)/(4x2 + 6x + 9).

f(x) = (−6x2 + 4x − 2)/(4x2 + 6x + 9)x → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (1 − cos 3x)/(x²), with x in radians.

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

Let f(x) = ln(3x2 + 1). Find f′(2).

f(x) = ln(3x2 + 1)x = 2
f′(2) =
4.DerivativesHard
xyx = −2

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

f(x) = 3x2 + 5x + 3x = −2
b =
5.IntegralsMedium
xy26

Evaluate I = ∫26 4/x dx.

f(x) = 4/x[2, 6]
I =
6.IntegralsHard
xyy = 2x2​y = 4x

Find the area A of the region enclosed by y = 2x2 and y = 4x.

y = 2x2y = 4x
A =
7.Complex numbersMedium
ReIm11−1−122−2−233−3−344−4−455−5−566−6−677−7−788−8−8zABCD

The diagram shows z = −3 − j2. With w = 2∠−90°, click the point that represents z·w.

z = −3 − j2w = 2∠−90°
  1. 1A (−6 − j4)
  2. 2B (−4 + j6)
  3. 3C (4 − j6)
  4. 4D (−2 + j3)
8.Complex numbersHard
ReIm11−1−122−2−233−3−3ABCD

Exactly one of the points A–D is a root of z⁶ = 64∠120°. Click it.

w = 64∠120°n = 6
  1. 1A (3.6∠−160°)
  2. 2B (2∠160°)
  3. 3C (2∠120°)
  4. 4D (2∠−160°)
9.Differential equationsMedium
t (min)T (°C)38Ta​T = ?

A forged steel part at 500 °C is left to cool in a workshop at 29 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.02 min⁻¹, find its temperature T after 38 min.

T0 = 500 °CTa = 29 °Ck = 0.02 min⁻¹t = 38 min
T = °C
10.Differential equationsHard
tyy(0) = 2y(1.5) = ?

Solve y″ + 4y′ + 8y = 0 with y(0) = 2 and y′(0) = 0 (released from rest). Find y(1.5); angles in radians.

y″ + 4y′ + 8y = 0y(0) = 2y′(0) = 0
y(1.5) =
11.Matrices & determinantsMedium
A =−4542B =5−6−31

Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).

A = [−4 5; 4 2]B = [5 −6; −3 1]
c21 =
12.Matrices & determinantsHard
A =4−61−4b =104

Solve the system 4x − 6y = 10 and x − 4y = 4. Find x.

4x − 6y = 10x − 4y = 4
x =
13.Vectors: dot & cross productMedium
xyzab

Find the angle θ between the vectors a = (−3, 4, −6) and b = (−6, 5, 1), in degrees.

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

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

a = (4, −5, −4)b = (−1, 1, −4)c = (−2, 3, 0)
V =
15.ProbabilityMedium
xp(x)10.2520.330.3540.1μ = 2.3

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

x: 1, 2, 3, 4p(x): 0.25, 0.3, 0.35, 0.1
σ =
16.ProbabilityHard
xp(x)00.3510.3520.3μ

X has the distribution shown, and Y = −2X − 3. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.35, 0.35, 0.3Y = −2X − 3
σY =
17.Descriptive statisticsMedium
121416182022242628

Find the sample standard deviation s of the data: 22, 29, 11, 18, 21.

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

A quantity is normally distributed with mean 70 and standard deviation 12. What proportion P of values is above 52?

μ = 70σ = 12x = 52
P =
19.Normal distributionMedium
120 kg118.7 kgT = X₁ + … + X₁₀?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 10 of them is below 118.7 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 10< 118.7 kg
P =
20.Normal distributionHard
235240245250255260265gbottom 40%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 40% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g40%
x = g

3Solutions

1.Limits

Answer: L = −1.5

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).

Divide by the highest power

(−6 + 4/x + (−2)/x2)/(4 + 6/x + 9/x2)

Let x → ∞

L = −6/4 = −1.5

Answer

L = −1.5

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

2.Limits

Answer: L = 4.5

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

3·sin 3x/(2x) → 0/0

Second application

9·cos 3x/2

Substitute x = 0

L = 9/2 = 4.5

Answer

L = 4.5
3.Derivatives

Answer: f′(2) = 0.9231

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 for the logarithm

f′(x) = 6x/(3x2 + 1)

Substitute

f′(2) = 12/13 = 0.9231

Answer

f′(2) = 0.9231
4.Derivatives

Answer: b = −9

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(−2) = 5
f′(x) = 6x + 5 → f′(−2) = −7

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 5 − (−7)·(−2) = −9

Answer

b = −9
5.Integrals

Answer: I = 4.394

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.

Antiderivative

F(x) = 4·ln x

Evaluate at the limits

I = 4·(ln 6 − ln 2) = 4·ln 3 = 4.394

Answer

I = 4.394
6.Integrals

Answer: A = 2.667

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 = 4x → x = 0, x = 2

Integrate upper minus lower

A = [2x2 − 0.6667x3]02 = 2.667

Answer

A = 2.667
7.Complex numbers

Answer: B (−4 + j6)

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.

Effect of w

|z·w| = 2·|z|, arg(z·w) = arg z + (−90)°

The point

z·w = −4 + j6 → B

Answer

B (−4 + j6)
8.Complex numbers

Answer: D (2∠−160°)

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 and arguments of the roots

|z| = 2
θ = 20°, 80°, 140°, −160°, −100°, −40°

The point

D = 2∠−160°

Answer

D (2∠−160°)
9.Differential equations

Answer: T = 249.3 °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) = 29 + 471·e−0.02t

Substitute t

T(38) = 29 + 471·0.4677 = 249.3 °C

Answer

T = 249.3 °C
10.Differential equations

Answer: y(1.5) = −0.08453

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 ± 2j

Constants

A = 2, B = 2·2/2 = 2

Evaluate

y(1.5) = e−3(2·cos 3 + 2·sin 3) = −0.08453

Answer

y(1.5) = −0.08453
11.Matrices & determinants

Answer: c21 = 14

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

c21 = 4·5 + 2·(−3) = 14

Answer

c21 = 14
12.Matrices & determinants

Answer: x = 1.6

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 = 4·(−4) − (−6)·1 = −10

Replace the x column

det Aₓ = 10·(−4) − (−6)·4 = −16
x = −16/(−10) = 1.6

Answer

x = 1.6
y = −0.6
13.Vectors: dot & cross product

Answer: θ = 58.64 °

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 = (−3)·(−6) + 4·5 + (−6)·1 = 32

Magnitudes

|a| = √((−3)² + 4² + (−6)²) = 7.81
|b| = √((−6)² + 5² + 1²) = 7.874

Angle

cos θ = 32/(7.81·7.874) = 0.5203
θ = arccos(0.5203) = 58.64°

Answer

θ = 58.64°
14.Vectors: dot & cross product

Answer: V = 12

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 = 1·0 − (−4)·3 = 12
(b × c)y = bzcx − bxcz = (−4)·(−2) − (−1)·0 = 8
(b × c)z = bxcy − bycx = (−1)·3 − 1·(−2) = −1

Dot product with a

a·(b × c) = 4·12 + (−5)·8 + (−4)·(−1) = 12
V = |12| = 12

Answer

V = 12
15.Probability

Answer: σ = 0.9539

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 = 1·0.25 + 2·0.3 + 3·0.35 + 4·0.1 = 2.3

Mean of the squares

E(X²) = 1·0.25 + 4·0.3 + 9·0.35 + 16·0.1 = 6.2

Variance

Var(X) = 6.2 − 2.3² = 0.91
σ = √0.91 = 0.9539

Answer

σ = 0.9539
16.Probability

Answer: σY = 1.609

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 = 0.95, Var(X) = 0.6475, σX = 0.8047

Linear change

σY = |−2|·0.8047 = 1.609

Answer

σY = 1.609

The mean does move: E(Y) = −2·0.95 − 3 = −4.9.

17.Descriptive statistics

Answer: s = 6.535

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 = 101/5 = 20.2

Squared deviations

Σ(x − x̄)² = 1.8² + 8.8² + (−9.2)² + (−2.2)² + 0.8² = 170.8

Standard deviation

s = √(170.8/4) = 6.535

Answer

s = 6.535
18.Descriptive statistics

Answer: P = 0.9332

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 = −1.5

Standard normal table

Φ(−1.5) = 0.06681
P = 1 − 0.06681 = 0.9332

Answer

P = 0.9332
19.Normal distribution

Answer: P = 0.305

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 total

μT = 120 kg, σT = 0.8·√10 = 2.53 kg

Standardise

z = (118.7 − 120)/2.53 = −0.51

Probability

P = Φ(−0.51) = 0.305

Answer

P = 0.305
20.Normal distribution

Answer: x = 248.8 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) = 0.4

z from the table

z ≈ −0.25

Back to the original units

x = 250 − 0.25·5 = 248.8 g

Answer

x = 248.8 g

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