Engineering Mathematics: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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→∞ (3x2 − x + 2)/(4x2 + 2x + 8).

f(x) = (3x2 − x + 2)/(4x2 + 2x + 8)x → ∞
L =
2.LimitsHard
xyx = 0

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

f(x) = ln(1 + x)/sin(4x)x → 0
L =
3.DerivativesMedium
xyx = 0.5

Let f(x) = x3·e2x. Find f′(0.5).

f(x) = x3·e2xx = 0.5
f′(0.5) =
4.DerivativesHard
xyx = −1

The tangent line to f(x) = x2 + 2x + 3 at x = −1 crosses the y-axis at y = b. Find b.

f(x) = x2 + 2x + 3x = −1
b =
5.IntegralsMedium
xy28

Evaluate I = ∫28 1/x dx.

f(x) = 1/x[2, 8]
I =
6.IntegralsHard
xy2y = x²eˣ

Evaluate I = ∫₀2 x²·eˣ dx.

∫₀2 x²eˣ dx
I =
7.Complex numbersMedium
ReIm22−2−244−4−466−6−6θz

Write z = 8∠−130° in rectangular form a + jb and give the real part a.

r = 8θ = −130°
a =
8.Complex numbersHard
ReIm11−1−122−2−233−3−3z₁z₂

The equation z² + 2z + 10 = 0 has two complex roots. Find their modulus |z|.

p = 2q = 10
|z| =
9.Differential equationsMedium
tyy(0) = 3y(0.5) = ?

Solve y″ + 8y′ + 15y = 0 with y(0) = 3 and y′(0) = 4. Find y(0.5).

y″ + 8y′ + 15y = 0y(0) = 3y′(0) = 4
y(0.5) =
10.Differential equationsHard
t (min)m (kg)5 minm = ?

A tank holds 100 L of pure water. Brine with 0.1 kg of salt per litre flows in at 20 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 5 min?

V = 100 Lm0 = 0 kgc = 0.1 kg/Lr = 20 L/min
m = kg
11.Matrices & determinantsMedium
A =0−6−5−4B =−6−5−3−5

Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).

A = [0 −6; −5 −4]B = [−6 −5; −3 −5]
c11 =
12.Matrices & determinantsHard
A =2−66−1b =−11−5

Solve the system 2x − 6y = −11 and 6x − y = −5. Find x.

2x − 6y = −116x − y = −5
x =
13.Vectors: dot & cross productMedium
xyzab

Find the value of k for which the vectors a = (0, 5, 1) and b = (5, k, 1) are perpendicular.

a = (0, 5, 1)b = (5, k, 1)
k =
14.Vectors: dot & cross productHard
xyzab

For a = (−1, 5, 5) and b = (−1, −3, 2), find the z-component (a × b)z of the cross product a × b.

a = (−1, 5, 5)b = (−1, −3, 2)
(a × b)z =
15.ProbabilityMedium
red: 7blue: 6

A bag holds 7 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

7 / 6
P =
16.ProbabilityHard
xp(x)00.3510.220.1530.3μ

X has the distribution shown, and Y = 5X + 4. Find the standard deviation σY of Y.

x: 0, 1, 2, 3p(x): 0.35, 0.2, 0.15, 0.3Y = 5X + 4
σY =
17.Descriptive statisticsMedium
152025303540

Find the sample standard deviation s of the data: 12, 44, 22, 11, 25, 25, 15.

n = 7
s =
18.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is below 74.8?

μ = 70σ = 8x = 74.8
P =
19.Normal distributionMedium
4975.5P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 49 and 75.5 points? Give P as a decimal.

μ = 65σ = 10x1 = 49x2 = 75.5
P =
20.Normal distributionHard
557085100115130145pointsbottom 95%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 95% is to the left of it. Find the cut-off x.

μ = 100 pointsσ = 15 points95%
x = points

3Solutions

1.Limits

Answer: L = 0.75

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

(3 + (−1)/x + 2/x2)/(4 + 2/x + 8/x2)

Let x → ∞

L = 3/4 = 0.75

Answer

L = 0.75

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

2.Limits

Answer: L = 0.25

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 [1/(1 + x)]/[4·cos 4x]

Substitute x = 0

L = 1/4 = 0.25

Answer

L = 0.25
3.Derivatives

Answer: f′(0.5) = 2.718

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′ = e2x·(3x2 + 2·x3)

Substitute

f′(0.5) = 2.718·(0.75 + 0.25) = 2.718

Answer

f′(0.5) = 2.718
4.Derivatives

Answer: b = 2

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(−1) = 2
f′(x) = 2x + 2 → f′(−1) = 0

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 2 − 0·(−1) = 2

Answer

b = 2
5.Integrals

Answer: I = 1.386

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) = ln x

Evaluate at the limits

I = (ln 8 − ln 2) = ln 4 = 1.386

Answer

I = 1.386
6.Integrals

Answer: I = 12.78

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

7.389·(2) − 1·2 = 12.78

Answer

I = 12.78
7.Complex numbers

Answer: a = −5.142

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.

Components

a = 8·cos(−130°) = −5.142
b = 8·sin(−130°) = −6.128

Rectangular form

z = −5.142 − j6.128
a = −5.142

Answer

a = −5.142
8.Complex numbers

Answer: |z| = 3.162

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 = 4 − 40 = −36 < 0

Quadratic formula

z = (−2 ± j√36)/2 = −1 + j3, −1 − j3

Answer

|z| = √((−1)² + 3²) = √10 = 3.162

Answer

|z| = 3.162
9.Differential equations

Answer: y(0.5) = 1.586

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 = 3, −3A − 5B = 4
A = 9.5, B = −6.5

Evaluate

y(0.5) = 9.5·e−1.5 + (−6.5)·e−2.5 = 1.586

Answer

y(0.5) = 1.586
10.Differential equations

Answer: m = 6.321 kg

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.

Set up the equation

dm/dt = 20·0.1 − (20/100)·m = 2 − 0.2·m

Solve

The integrating factor is ert/V; the solution approaches the equilibrium c·V.

m(t) = 10 + (−10)·e−t/5

Substitute t

m(5) = 10 + (−10)·0.3679 = 6.321 kg

Answer

m = 6.321 kg
11.Matrices & determinants

Answer: c11 = 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

c11 = 0·(−6) + (−6)·(−3) = 18

Answer

c11 = 18
12.Matrices & determinants

Answer: x = −0.5588

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·(−1) − (−6)·6 = 34

Replace the x column

det Aₓ = (−11)·(−1) − (−6)·(−5) = −19
x = −19/34 = −0.5588

Answer

x = −0.5588
y = 1.647
13.Vectors: dot & cross product

Answer: k = −0.2

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 = 0·5 + 5·k + 1·1 = 0
1 + 5·k = 0

Solve for k

k = −1/5 = −0.2

Answer

k = −0.2
14.Vectors: dot & cross product

Answer: (a × b)z = 8

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 = 5·2 − 5·(−3) = 25
(a × b)y = azbx − axbz = 5·(−1) − (−1)·2 = −3
(a × b)z = axby − aybx = (−1)·(−3) − 5·(−1) = 8

Requested component

(a × b)z = 8

Answer

(a × b)z = 8

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

15.Probability

Answer: P = 0.2692

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 = 7/13 · 6/12 = 42/156 = 0.2692

Answer

P = 0.2692
16.Probability

Answer: σY = 6.205

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 = 1.4, Var(X) = 1.54, σX = 1.241

Linear change

σY = |5|·1.241 = 6.205

Answer

σY = 6.205

The mean does move: E(Y) = 5·1.4 + 4 = 11.

17.Descriptive statistics

Answer: s = 11.34

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 = 154/7 = 22

Squared deviations

Σ(x − x̄)² = (−10)² + 22² + 0² + (−11)² + 3² + 3² + (−7)² = 772

Standard deviation

s = √(772/6) = 11.34

Answer

s = 11.34
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 = 0.7983

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 = (49 − 65)/10 = −1.6
z₂ = (75.5 − 65)/10 = 1.05

Area between

P = Φ(1.05) − Φ(−1.6) = 0.8531 − 0.0548 = 0.7983

Answer

P = 0.7983
20.Normal distribution

Answer: x = 124.6 points

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) = 0.95

z from the table

z ≈ 1.64

Back to the original units

x = 100 + 1.64·15 = 124.6 points

Answer

x = 124.6 points

ExamProblems · examproblems.com/courses/engineering-mathematics