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→∞ (√(9x2 + 7x) − 3x).

f(x) = √(9x2 + 7x) − 3xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (e−0.5x − 1 + 0.5x)/x².

f(x) = (e−0.5x − 1 + 0.5x)/x²x → 0
L =
3.DerivativesMedium
xyx = 1

Let f(x) = x2·e2x. Find f′(1).

f(x) = x2·e2xx = 1
f′(1) =
4.DerivativesHard
xy

Let f(x) = x3 − 4.5x2 + 6x − 1. Find the x-coordinate x of the local maximum.

f(x) = x3 − 4.5x2 + 6x − 1
x =
5.IntegralsMedium
xy12

Evaluate I = ∫12 1/x dx.

f(x) = 1/x[1, 2]
I =
6.IntegralsHard
xy1.667y = 5xy = 3x²

The region between the line y = 5x and the parabola y = 3x² is rotated about the x-axis. Find the volume V of the solid.

y = 5xy = 3x²
V =
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
ReIm22−2−244−4−4z₀z₁

The roots of zⁿ = w with n = 2 and w = 16∠−60° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1. Find the argument of z₁ in degrees, between −180° and 180°.

w = 16∠−60°n = 2k = 1
θ₁ = °
9.Differential equationsMedium
t (min)T (°C)33Ta​T = ?

A forged steel part at 420 °C is left to cool in a workshop at 21 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.04 min⁻¹, find its temperature T after 33 min.

T0 = 420 °CTa = 21 °Ck = 0.04 min⁻¹t = 33 min
T = °C
10.Differential equationsHard
tyK = 1y(0.5) = ?

A system obeys y″ + 4y′ + 13y = 13 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.5); angles in radians.

y″ + 4y′ + 13y = 13y(0) = 0y′(0) = 0
y(0.5) =
11.Matrices & determinantsMedium
A =−3565B =−1540

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

A = [−3 5; 6 5]B = [−1 5; 4 0]
c11 =
12.Matrices & determinantsHard
A =−14−58

Find the larger eigenvalue λ of the matrix A shown.

A = [−1 4; −5 8]
λ =
13.Vectors: dot & cross productMedium
xyzdFAB

A constant force F = (0, −10, −20) N acts on an object while it moves in a straight line from point A(−5, −1, 2) to point B(2, 2, −4). Coordinates are in metres. Find the work W done by the force.

F = (0, −10, −20) NA(−5, −1, 2) mB(2, 2, −4) m
W = J
14.Vectors: dot & cross productHard
xyzab

Find the area A of the parallelogram spanned by a = (−6, 2, 0) and b = (1, −2, 6).

a = (−6, 2, 0)b = (1, −2, 6)
A =
15.ProbabilityMedium
xp(x)00.3520.2540.0560.35μ = 2.8

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

x: 0, 2, 4, 6p(x): 0.35, 0.25, 0.05, 0.35
Var(X) =
16.ProbabilityHard
xp(x)100.2200.3300.25400.25μ

X has the distribution shown, and Y = 0.5X + 5. Find the standard deviation σY of Y.

x: 10, 20, 30, 40p(x): 0.2, 0.3, 0.25, 0.25Y = 0.5X + 5
σY =
17.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 34, 49, 22, 56, 42.

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

Exam scores are normally distributed with mean 70 and standard deviation 12. Find the z-score z of a score of 82.

μ = 70σ = 12x = 82
z =
19.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8098

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8098. Find z to two decimals.

A = 0.8098
z =
20.Normal distributionHard
235240245250255260265gbottom 60%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 60% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g60%
x = g

3Solutions

1.Limits

Answer: L = 1.167

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).

Multiply by the conjugate

f(x) = 7x/(√(9x2 + 7x) + 3x)

Divide by x

f(x) = 7/(√(9 + 7/x) + 3)

Let x → ∞

L = 7/(3 + 3) = 1.167

Answer

L = 1.167
2.Limits

Answer: L = 0.125

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

(−0.5·e−0.5x + 0.5)/(2x) → 0/0

Second application

0.25·e−0.5x/2

Substitute x = 0

L = 0.25·1/2 = 0.125

Answer

L = 0.125
3.Derivatives

Answer: f′(1) = 29.56

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·(2x + 2·x2)

Substitute

f′(1) = 7.389·(2 + 2) = 29.56

Answer

f′(1) = 29.56
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 − 9x + 6 = 3·(x2 − 3x + 2)

Solve f′(x) = 0

x = 1 or x = 2

Which one is which

f″(x) = 6x − 9: negative at a maximum, positive at a minimum.

f″(1) = −3, f″(2) = 3
x = 1

Answer

x = 1
5.Integrals

Answer: I = 0.6931

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 2 − ln 1) = ln 2 = 0.6931

Answer

I = 0.6931
6.Integrals

Answer: V = 48.48

Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.

Intersection points

5x = 3x² → x = 0, x = 1.667

Washer method

V = π∫₀1.667 ((5x)² − (3x²)²) dx

Integrate

V = π(25·1.667³/3 − 9·1.667⁵/5) = π·15.43 = 48.48

Answer

V = 48.48
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: θ₁ = 150 °

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

161/2 = 4

Argument of the root

θ₁ = (−60° + 360°·1)/2 = 150°

Argument

θ₁ = 150°

Answer

θ₁ = 150°

The 2 roots lie on the circle of radius 4, 180° apart.

9.Differential equations

Answer: T = 127.6 °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) = 21 + 399·e−0.04t

Substitute t

T(33) = 21 + 399·0.2671 = 127.6 °C

Answer

T = 127.6 °C
10.Differential equations

Answer: y(0.5) = 0.7293

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 = −2 ± 3j

Constants from the initial values

y(0) = 1 + A = 0 → A = −1
y′(0) = −2A + 3B = 0 → B = −0.6667

Evaluate

y(0.5) = 1·(1 − e−1(cos 1.5 + 0.6667·sin 1.5)) = 0.7293

Answer

y(0.5) = 0.7293
11.Matrices & determinants

Answer: c11 = 23

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 = (−3)·(−1) + 5·4 = 23

Answer

c11 = 23
12.Matrices & determinants

Answer: λ = 4

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 − (7)λ + (12) = 0

Solve

λ = (7 ± √(1))/2 → λ = 3, 4
λ = 4

Answer

λ = 4
13.Vectors: dot & cross product

Answer: W = 90 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 = (2, 2, −4) − (−5, −1, 2) = (7, 3, −6) m

Work

W = F·d = 0·7 + (−10)·3 + (−20)·(−6) = 90 J

Answer

W = 90 J
14.Vectors: dot & cross product

Answer: A = 39.24

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 = 2·6 − 0·(−2) = 12
(a × b)y = azbx − axbz = 0·1 − (−6)·6 = 36
(a × b)z = axby − aybx = (−6)·(−2) − 2·1 = 10

Area

A = |a × b| = √(12² + 36² + 10²) = 39.24

Answer

A = 39.24
15.Probability

Answer: Var(X) = 6.56

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 = 0·0.35 + 2·0.25 + 4·0.05 + 6·0.35 = 2.8

Mean of the squares

E(X²) = 0·0.35 + 4·0.25 + 16·0.05 + 36·0.35 = 14.4

Variance

Var(X) = 14.4 − 2.8² = 6.56

Answer

Var(X) = 6.56
16.Probability

Answer: σY = 5.356

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 = 25.5, Var(X) = 114.8, σX = 10.71

Linear change

σY = |0.5|·10.71 = 5.356

Answer

σY = 5.356

The mean does move: E(Y) = 0.5·25.5 + 5 = 17.75.

17.Descriptive statistics

Answer: s = 13.22

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 = 203/5 = 40.6

Squared deviations

Σ(x − x̄)² = (−6.6)² + 8.4² + (−18.6)² + 15.4² + 1.4² = 699.2

Standard deviation

s = √(699.2/4) = 13.22

Answer

s = 13.22
18.Descriptive statistics

Answer: z = 1

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 = (82 − 70)/12 = 1

Answer

z = 1
19.Normal distribution

Answer: z = 1.31

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 + 0.8098)/2 = 0.9049

Read the z-table backwards

Find 0.9049 inside the table and read off z.

z = 1.31

Answer

z = 1.31
20.Normal distribution

Answer: x = 251.3 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) = 0.6

z from the table

z ≈ 0.25

Back to the original units

x = 250 + 0.25·5 = 251.3 g

Answer

x = 251.3 g

ExamProblems · examproblems.com/courses/engineering-mathematics