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→∞ (1 − 1/x)x. Give a decimal value.

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

Find the limit L = limx→0 ln(1 + 4x)/sin(5x), with x in radians.

f(x) = ln(1 + 4x)/sin(5x)x → 0
L =
3.DerivativesMedium
xyx = 1

Let f(x) = (2x − 1)5. Find f′(1).

f(x) = (2x − 1)5x = 1
f′(1) =
4.DerivativesHard
xyx = −2

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

f(x) = −x2 + 3xx = −2
b =
5.IntegralsMedium
xy210

Evaluate I = ∫210 4/x dx.

f(x) = 4/x[2, 10]
I =
6.IntegralsHard
xyy = 0.5x2​y = 2x

Find the area A of the region enclosed by y = 0.5x2 and y = 2x.

y = 0.5x2y = 2x
A =
7.Complex numbersMedium
ReIm55−5−5θz

Write z = 7 − j4 in exponential form r·ejθ. Give θ in radians, between −π and π.

z = 7 − j4
θ = rad
8.Complex numbersHard
ReIm22−2−244−4−4z₁z₂

One root of z³ + 8z² + 37z + 50 = 0 is z₁ = −3 + j4. The coefficients are real. Find the real root r.

z₁ = −3 + j4
r =
9.Differential equationsMedium
t (min)T (°C)8Ta​50 °C38 °C

A cup of coffee at 81 °C is left in a room at 24 °C. After 8 min its temperature is 50 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 38 °C?

T0 = 81 °CTa = 24 °CT(8) = 50 °CT = 38 °C
t = min
10.Differential equationsHard
ty1y(1) = ?

Solve y′ + 3y = 6e−2t with y(0) = 3. Find y(1).

y′ + 3y = 6e−2ty(0) = 3
y(1) =
11.Matrices & determinantsMedium
A =−322−35−4240

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

A = [−3 2 2; −3 5 −4; 2 4 0]
det A =
12.Matrices & determinantsHard
A =−6−1−2−3b =25

Solve the system −6x − y = 2 and −2x − 3y = 5. Find x.

−6x − y = 2−2x − 3y = 5
x =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (1, −7, 7) onto b = (−5, 1, 5), that is, the component of a in the direction of b.

a = (1, −7, 7)b = (−5, 1, 5)
compb a =
14.Vectors: dot & cross productHard
xyzab

For a = (0, 1, 1) and b = (1, −3, 2), find the y-component (a × b)y of the cross product a × b.

a = (0, 1, 1)b = (1, −3, 2)
(a × b)y =
15.ProbabilityMedium
xp(x)00.320.140.360.3μ = 3.2

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

x: 0, 2, 4, 6p(x): 0.3, 0.1, 0.3, 0.3
σ =
16.ProbabilityHard
n = 10, p = 0.4: P(X = 3) = ?k012345678910

A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 3 correct answers?

n = 10p = 0.4k = 3
P =
17.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 57, 12, 17, 32, 46.

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

A quantity is normally distributed with mean 50 and standard deviation 12. What proportion P of values is below 35.6?

μ = 50σ = 12x = 35.6
P =
19.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8064

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8064. Find z to two decimals.

A = 0.8064
z =
20.Normal distributionHard
μ = 178x = ?20%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 20% of them lie (the 20th percentile).

μ = 178σ = 7P = 20%
x = cm

3Solutions

1.Limits

Answer: L = 0.3679

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 − 1/x)

Evaluate

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

ln L = 1·(−1) = −1

Undo the logarithm

L = e−1 = 0.3679

Answer

L = 0.3679
2.Limits

Answer: L = 0.8

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)]/[5·cos 5x]

Substitute x = 0

L = 4/5 = 0.8

Answer

L = 0.8
3.Derivatives

Answer: f′(1) = 10

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·(2x − 1)4·2

Substitute

f′(1) = 5·14·2 = 10

Answer

f′(1) = 10
4.Derivatives

Answer: b = 4

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) = −10
f′(x) = −2x + 3 → f′(−2) = 7

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = −10 − 7·(−2) = 4

Answer

b = 4
5.Integrals

Answer: I = 6.438

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 10 − ln 2) = 4·ln 5 = 6.438

Answer

I = 6.438
6.Integrals

Answer: A = 5.333

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.

Intersection points

0.5x2 = 2x → x = 0, x = 4

Integrate upper minus lower

A = [x2 − 0.1667x3]04 = 5.333

Answer

A = 5.333
7.Complex numbers

Answer: θ = −0.5191 rad

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.

Modulus

r = √(49 + 16) = 8.062

Argument in radians

θ = −29.74° = −29.74·π/180 = −0.5191 rad

Exponential form

z = 8.062·e−j0.5191

Answer

θ = −0.5191 rad
8.Complex numbers

Answer: r = −2

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.

Conjugate pair

z₂ = −3 − j4
(z − z₁)(z − z₂) = z² − 2·(−3)·z + ((−3)² + 4²) = z² + 6z + 25

Factor out the quadratic

z³ + 8z² + 37z + 50 = (z² + 6z + 25)(z + 2)

Real root

r = −2

Answer

r = −2

Check with the sum of the roots: z₁ + z₂ + r = −6 + (−2) = −8 = −(8).

9.Differential equations

Answer: t = 14.31 min

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.

Cooling constant

k = ln(57/26)/8 = 0.09812 min⁻¹

Time to reach the target

t = ln(57/14)/0.09812 = 14.31 min

Answer

t = 14.31 min
10.Differential equations

Answer: y(1) = 0.6627

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 = e3t
(e3ty)′ = 6et

General solution

y = 6/1·e−2t + C·e−3t = 6·e−2t + C·e−3t

Initial condition

3 = 6 + C → C = −3

Evaluate

y(1) = 6·0.1353 + (−3)·0.04979 = 0.6627

Answer

y(1) = 0.6627
11.Matrices & determinants

Answer: det A = −108

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 = 16, M₁₂ = 8, M₁₃ = −22

Cofactor expansion

det A = (−3)·16 − 2·8 + 2·(−22) = −108

Answer

det A = −108
12.Matrices & determinants

Answer: x = −0.0625

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 = (−6)·(−3) − (−1)·(−2) = 16

Replace the x column

det Aₓ = 2·(−3) − (−1)·5 = −1
x = −1/16 = −0.0625

Answer

x = −0.0625
y = −1.625
13.Vectors: dot & cross product

Answer: compb a = 3.221

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·(−5) + (−7)·1 + 7·5 = 23

Length of b

|b| = √((−5)² + 1² + 5²) = 7.141

Scalar projection

compb a = a·b/|b| = 23/7.141 = 3.221

Answer

compb a = 3.221
14.Vectors: dot & cross product

Answer: (a × b)y = 1

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 = 1·2 − 1·(−3) = 5
(a × b)y = azbx − axbz = 1·1 − 0·2 = 1
(a × b)z = axby − aybx = 0·(−3) − 1·1 = −1

Requested component

(a × b)y = 1

Answer

(a × b)y = 1

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

15.Probability

Answer: σ = 2.4

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 + 2·0.1 + 4·0.3 + 6·0.3 = 3.2

Mean of the squares

E(X²) = 0·0.3 + 4·0.1 + 16·0.3 + 36·0.3 = 16

Variance

Var(X) = 16 − 3.2² = 5.76
σ = √5.76 = 2.4

Answer

σ = 2.4
16.Probability

Answer: P = 0.215

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.

Binomial formula

P(X = 3) = 120·0.43·0.67 = 120·0.064·0.02799 = 0.215

Answer

P = 0.215
17.Descriptive statistics

Answer: s = 18.99

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 = 164/5 = 32.8

Squared deviations

Σ(x − x̄)² = 24.2² + (−20.8)² + (−15.8)² + (−0.8)² + 13.2² = 1443

Standard deviation

s = √(1443/4) = 18.99

Answer

s = 18.99
18.Descriptive statistics

Answer: P = 0.1151

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 0.1151

Answer

P = 0.1151
19.Normal distribution

Answer: z = 1.3

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 + 0.8064)/2 = 0.9032

Read the z-table backwards

Find 0.9032 inside the table and read off z.

z = 1.30

Answer

z = 1.30
20.Normal distribution

Answer: x = 172.1 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) = 0.2 → z = −0.8416

Back to x

x = 178 + −0.8416·7 = 172.11 cm

Answer

x = 172.11 cm

ExamProblems · examproblems.com/courses/engineering-mathematics