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

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

Find the limit L = limx→0 ln(1 + 3x)/sin(5x), with x in radians.

f(x) = ln(1 + 3x)/sin(5x)x → 0
L =
3.DerivativesMedium
xyx = 1.5

Let f(x) = x2·ex. Find f′(1.5).

f(x) = x2·exx = 1.5
f′(1.5) =
4.DerivativesHard
xy

Let f(x) = x3 − 12x2 + 45x + 4. Find the x-coordinate x of the local maximum.

f(x) = x3 − 12x2 + 45x + 4
x =
5.IntegralsMedium
xy01

Evaluate I = ∫01 2x·(x² + 1)2 dx.

f(x) = 2x·(x² + 1)2[0, 1]
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
ReIm22−2−244−4−4θz

The number z = 5·ej0.6 has its angle in radians. Find its real part a.

r = 5θ = 0.6 rad
a =
8.Complex numbersHard
ReIm11−1−122−2−233−3−3z₀z₁z₂z₃

The roots of zⁿ = w with n = 4 and w = 81∠135° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2, 3. Find the argument of z₂ in degrees, between −180° and 180°.

w = 81∠135°n = 4k = 2
θ₂ = °
9.Differential equationsMedium
t (min)T (°C)21Ta​189 °C

A forged steel part at 400 °C is left to cool in a workshop at 29 °C. After 21 min its temperature is 189 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 400 °CTa = 29 °CT(21) = 189 °C
k = min⁻¹
10.Differential equationsHard
xy13(1, −1)y(3) = ?

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

y′ + y/x = 2xy(1) = −1
y(3) =
11.Matrices & determinantsMedium
A =0−166B =250−2

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

A = [0 −1; 6 6]B = [2 5; 0 −2]
c22 =
12.Matrices & determinantsHard
A =52−5−5b =1212

Solve the system 5x + 2y = 12 and −5x − 5y = 12. Find x.

5x + 2y = 12−5x − 5y = 12
x =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (1, 3, 5) onto b = (−2, −2, −5), that is, the component of a in the direction of b.

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

The unit vector n̂ is perpendicular to both a = (−4, 5, 3) and b = (−2, 3, 0) and has a positive z-component. Find its y-component n̂y.

a = (−4, 5, 3)b = (−2, 3, 0)n̂z > 0
n̂y =
15.ProbabilityMedium
red: 8blue: 3

A bag holds 8 red and 3 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

8 / 3
P =
16.ProbabilityHard
xp(x)00.2510.320.0530.240.2μ

X has the distribution shown, and Y = 0.5X − 3. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.25, 0.3, 0.05, 0.2, 0.2Y = 0.5X − 3
σY =
17.Descriptive statisticsMedium
15202530354045505560

Find the sample standard deviation s of the data: 11, 38, 50, 13, 60, 26.

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

A quantity is normally distributed with mean 60 and standard deviation 15. What proportion P of values is below 69?

μ = 60σ = 15x = 69
P =
19.Normal distributionMedium
96 kg92.6 kgT = X₁ + … + X₈?

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 8 of them exceeds 92.6 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 8> 92.6 kg
P =
20.Normal distributionHard
μ = 500x = ?95%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 95% of them lie (the 95th percentile).

μ = 500σ = 40P = 95%
x = h

3Solutions

1.Limits

Answer: L = 20.09

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→∞ x·ln(1 + 3/x)

Evaluate

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

ln L = 1·3 = 3

Undo the logarithm

L = e3 = 20.09

Answer

L = 20.09
2.Limits

Answer: L = 0.6

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)]/[5·cos 5x]

Substitute x = 0

L = 3/5 = 0.6

Answer

L = 0.6
3.Derivatives

Answer: f′(1.5) = 23.53

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 rule

f′(x) = u′v + uv′ = ex·(2x + 1·x2)

Substitute

f′(1.5) = 4.482·(3 + 2.25) = 23.53

Answer

f′(1.5) = 23.53
4.Derivatives

Answer: x = 3

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 − 24x + 45 = 3·(x2 − 8x + 15)

Solve f′(x) = 0

x = 3 or x = 5

Which one is which

f″(x) = 6x − 24: negative at a maximum, positive at a minimum.

f″(3) = −6, f″(5) = 6
x = 3

Answer

x = 3
5.Integrals

Answer: I = 2.333

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 = ∫12 u2 du

Integrate and evaluate

I = [u3/3]12 = (8 − 1)/3 = 2.333

Answer

I = 2.333
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: a = 4.127

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 = 5·(cos 0.6 + j sin 0.6)

Components (radians)

a = 5·cos 0.6 = 4.127
b = 5·sin 0.6 = 2.823
a = 4.127

Answer

a = 4.127
8.Complex numbers

Answer: θ₂ = −146.3 °

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

811/4 = 3

Argument of the root

θ₂ = (135° + 360°·2)/4 = 213.8° → −146.3°

Argument

θ₂ = −146.3°

Answer

θ₂ = −146.3°

The 4 roots lie on the circle of radius 3, 90° apart.

9.Differential equations

Answer: k = 0.04005 min⁻¹

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.

Ratio of temperature differences

e−21k = 160/371

Solve for k

k = ln(371/160)/21 = 0.04005 min⁻¹

Answer

k = 0.04005 min⁻¹
10.Differential equations

Answer: y(3) = 5.444

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∫1/x dx = x1

Integrate

(x1y)′ = 2x2 → x1y = 2x3/3 + C

Initial condition

−1 = 2/3 + C → C = −1.667

Evaluate

y(3) = (2·33/3 + (−1.667))/31 = 5.444

Answer

y(3) = 5.444
11.Matrices & determinants

Answer: c22 = 18

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 = 6·5 + 6·(−2) = 18

Answer

c22 = 18
12.Matrices & determinants

Answer: x = 5.6

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 = 5·(−5) − 2·(−5) = −15

Replace the x column

det Aₓ = 12·(−5) − 2·12 = −84
x = −84/(−15) = 5.6

Answer

x = 5.6
y = −8
13.Vectors: dot & cross product

Answer: compb a = −5.745

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·(−2) + 3·(−2) + 5·(−5) = −33

Length of b

|b| = √((−2)² + (−2)² + (−5)²) = 5.745

Scalar projection

compb a = a·b/|b| = −33/5.745 = −5.745

Answer

compb a = −5.745

Negative: a has a component pointing against b.

14.Vectors: dot & cross product

Answer: n̂y = 0.5455

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 = 5·0 − 3·3 = −9
(a × b)y = azbx − axbz = 3·(−2) − (−4)·0 = −6
(a × b)z = axby − aybx = (−4)·3 − 5·(−2) = −2

Normalise

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

|a × b| = √((−9)² + (−6)² + (−2)²) = 11
n̂ = −(−9, −6, −2)/11 = (0.8182, 0.5455, 0.1818)
n̂y = 0.5455

Answer

n̂y = 0.5455
15.Probability

Answer: P = 0.5091

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.

First and second draw

P = 8/11 · 7/10 = 56/110 = 0.5091

Answer

P = 0.5091
16.Probability

Answer: σY = 0.7517

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) = 2.26, σX = 1.503

Linear change

σY = |0.5|·1.503 = 0.7517

Answer

σY = 0.7517

The mean does move: E(Y) = 0.5·1.8 − 3 = −2.1.

17.Descriptive statistics

Answer: s = 19.88

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 = 198/6 = 33

Squared deviations

Σ(x − x̄)² = (−22)² + 5² + 17² + (−20)² + 27² + (−7)² = 1976

Standard deviation

s = √(1976/5) = 19.88

Answer

s = 19.88
18.Descriptive statistics

Answer: P = 0.7257

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 0.7257

Answer

P = 0.7257
19.Normal distribution

Answer: P = 0.9332

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 = 96 kg, σT = 0.8·√8 = 2.263 kg

Standardise

z = (92.6 − 96)/2.263 = −1.5

Probability

P = 1 − Φ(−1.5) = 0.9332

Answer

P = 0.9332
20.Normal distribution

Answer: x = 565.8 h

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) = 0.95 → z = 1.645

Back to x

x = 500 + 1.645·40 = 565.8 h

Answer

x = 565.8 h

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