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→∞ (−2x2 − 8)/(4x2 + x + 7).

f(x) = (−2x2 − 8)/(4x2 + x + 7)x → ∞
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) = ln(x2 + 1). Find f′(1).

f(x) = ln(x2 + 1)x = 1
f′(1) =
4.DerivativesHard
xy

Let f(x) = x3 − 3x − 2. Find the x-coordinate x of the local maximum.

f(x) = x3 − 3x − 2
x =
5.IntegralsMedium
xy02

Evaluate I = ∫02 2x·(x² + 1)3 dx.

f(x) = 2x·(x² + 1)3[0, 2]
I =
6.IntegralsHard
xy3y = x²eˣ

Evaluate I = ∫₀3 x²·eˣ dx.

∫₀3 x²eˣ dx
I =
7.Complex numbersMedium
ReIm55−5−5θz

Find the argument θ of z = −3 − j7 in degrees, as the principal value between −180° and 180°.

z = −3 − j7
θ = °
8.Complex numbersHard
ReIm11−1−122−2−2z₀z₁z₂z₃z₄z₅

The roots of zⁿ = w with n = 6 and w = 64∠−120° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, …, 5. Find the real part of z₀.

w = 64∠−120°n = 6k = 0
Re(z₀) =
9.Differential equationsMedium
tyy(0) = 1y(2) = ?

Solve y″ + 8y′ + 15y = 0 with y(0) = 1 and y′(0) = 6. Find y(2).

y″ + 8y′ + 15y = 0y(0) = 1y′(0) = 6
y(2) =
10.Differential equationsHard
tyK = 1y(0.75) = ?

A system obeys y″ + y′ + 9.25y = 9.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.75); angles in radians.

y″ + y′ + 9.25y = 9.25y(0) = 0y′(0) = 0
y(0.75) =
11.Matrices & determinantsMedium
A =3−206B =1−344

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

A = [3 −2; 0 6]B = [1 −3; 4 4]
c22 =
12.Matrices & determinantsHard
A =142−1

Find the larger eigenvalue λ of the matrix A shown.

A = [1 4; 2 −1]
λ =
13.Vectors: dot & cross productMedium
xyzdFAB

A constant force F = (−30, 10, 30) N acts on an object while it moves in a straight line from point A(−2, 1, 4) to point B(−3, −1, 1). Coordinates are in metres. Find the work W done by the force.

F = (−30, 10, 30) NA(−2, 1, 4) mB(−3, −1, 1) m
W = J
14.Vectors: dot & cross productHard
xyzab

For a = (6, 3, −3) and b = (1, 3, 5), find the z-component (a × b)z of the cross product a × b.

a = (6, 3, −3)b = (1, 3, 5)
(a × b)z =
15.ProbabilityMedium
xp(x)00.310.120.2530.35μ = 1.65

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

x: 0, 1, 2, 3p(x): 0.3, 0.1, 0.25, 0.35
Var(X) =
16.ProbabilityHard
xp(x)00.3510.320.35μ

X has the distribution shown, and Y = 3X + 1. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.35, 0.3, 0.35Y = 3X + 1
σY =
17.Descriptive statisticsMedium
202530354045

Find the sample standard deviation s of the data: 49, 20, 35, 40, 31.

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

A quantity is normally distributed with mean 60 and standard deviation 12. What proportion P of values is above 50.4?

μ = 60σ = 12x = 50.4
P =
19.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.8770

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.8770. Find z to two decimals.

A = 0.8770
z =
20.Normal distributionHard
μ = 178x = ?90%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 90% of them lie (the 90th percentile).

μ = 178σ = 7P = 90%
x = cm

3Solutions

1.Limits

Answer: L = −0.5

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

(−2 + (−8)/x2)/(4 + 1/x + 7/x2)

Let x → ∞

L = −2/4 = −0.5

Answer

L = −0.5

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

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) = 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.

Chain rule for the logarithm

f′(x) = 2x/(x2 + 1)

Substitute

f′(1) = 2/2 = 1

Answer

f′(1) = 1
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 − 3 = 3·(x2 − 1)

Solve f′(x) = 0

x = −1 or x = 1

Which one is which

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

f″(−1) = −6, f″(1) = 6
x = −1

Answer

x = −1
5.Integrals

Answer: I = 156

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 u3 du

Integrate and evaluate

I = [u4/4]15 = (625 − 1)/4 = 156

Answer

I = 156
6.Integrals

Answer: I = 98.43

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.

First round

∫x²eˣ dx = x²eˣ − ∫2x eˣ dx

Second round

∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)

Limits

20.09·(5) − 1·2 = 98.43

Answer

I = 98.43
7.Complex numbers

Answer: θ = −113.2 °

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 = −3, b = −7 → III

Reference angle

α = atan(7/3) = 66.8°

Argument

θ = −(180° − α) = −113.2°

Answer

θ = −113.2°

In polar form z = 7.616∠−113.2°.

8.Complex numbers

Answer: Re(z₀) = 1.879

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

641/6 = 2

Argument of the root

θ₀ = (−120° + 360°·0)/6 = −20°

Rectangular form

z₀ = 2·(cos −20° + j sin −20°) = 1.879 − j0.684
Re(z₀) = 1.879

Answer

Re(z₀) = 1.879

The 6 roots lie on the circle of radius 2, 60° apart.

9.Differential equations

Answer: y(2) = 0.01343

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 + 3)(r + 5) = 0 → r = −3, −5

Constants from the initial values

A + B = 1, −3A − 5B = 6
A = 5.5, B = −4.5

Evaluate

y(2) = 5.5·e−6 + (−4.5)·e−10 = 0.01343

Answer

y(2) = 0.01343
10.Differential equations

Answer: y(0.75) = 1.343

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 = 1, r = −0.5 ± 3j

Constants from the initial values

y(0) = 1 + A = 0 → A = −1
y′(0) = −0.5A + 3B = 0 → B = −0.1667

Evaluate

y(0.75) = 1·(1 − e−0.375(cos 2.25 + 0.1667·sin 2.25)) = 1.343

Answer

y(0.75) = 1.343
11.Matrices & determinants

Answer: c22 = 24

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 = 0·(−3) + 6·4 = 24

Answer

c22 = 24
12.Matrices & determinants

Answer: λ = 3

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.

Characteristic equation

λ² − (0)λ + (−9) = 0

Solve

λ = (0 ± √(36))/2 → λ = −3, 3
λ = 3

Answer

λ = 3
13.Vectors: dot & cross product

Answer: W = −80 J

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.

Displacement

d = B − A = (−3, −1, 1) − (−2, 1, 4) = (−1, −2, −3) m

Work

W = F·d = (−30)·(−1) + 10·(−2) + 30·(−3) = −80 J

Answer

W = −80 J

Negative work: the force acts mostly against the motion.

14.Vectors: dot & cross product

Answer: (a × b)z = 15

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 = 3·5 − (−3)·3 = 24
(a × b)y = azbx − axbz = (−3)·1 − 6·5 = −33
(a × b)z = axby − aybx = 6·3 − 3·1 = 15

Requested component

(a × b)z = 15

Answer

(a × b)z = 15

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

15.Probability

Answer: Var(X) = 1.528

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 + 1·0.1 + 2·0.25 + 3·0.35 = 1.65

Mean of the squares

E(X²) = 0·0.3 + 1·0.1 + 4·0.25 + 9·0.35 = 4.25

Variance

Var(X) = 4.25 − 1.65² = 1.528

Answer

Var(X) = 1.528
16.Probability

Answer: σY = 2.51

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, Var(X) = 0.7, σX = 0.8367

Linear change

σY = |3|·0.8367 = 2.51

Answer

σY = 2.51

The mean does move: E(Y) = 3·1 + 1 = 4.

17.Descriptive statistics

Answer: s = 10.75

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 = 175/5 = 35

Squared deviations

Σ(x − x̄)² = 14² + (−15)² + 0² + 5² + (−4)² = 462

Standard deviation

s = √(462/4) = 10.75

Answer

s = 10.75
18.Descriptive statistics

Answer: P = 0.7881

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 = −0.8

Standard normal table

Φ(−0.8) = 0.2119
P = 1 − 0.2119 = 0.7881

Answer

P = 0.7881
19.Normal distribution

Answer: z = −1.16

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 − 0.8770 = 0.1230

Read the z-table backwards

Find 0.1230 inside the table and read off z.

z = −1.16

Answer

z = −1.16
20.Normal distribution

Answer: x = 187 cm

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) = 0.9 → z = 1.282

Back to x

x = 178 + 1.282·7 = 186.97 cm

Answer

x = 186.97 cm

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