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→∞ (√(x2 + 4x) − x).

f(x) = √(x2 + 4x) − xx → ∞
L =
2.LimitsHard
xyx = 0

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

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

Let f(x) = (3x + 5)5. Find f′(−1).

f(x) = (3x + 5)5x = −1
f′(−1) =
4.DerivativesHard
xy

Let f(x) = x3 − 10.5x2 + 36x − 3. Find the x-coordinate x of the local minimum.

f(x) = x3 − 10.5x2 + 36x − 3
x =
5.IntegralsMedium
xy3y = x

The region under y = x from x = 0 to x = 3 is rotated about the x-axis. Find the volume V of the solid (a cone).

y = x0 ≤ x ≤ 3
V =
6.IntegralsHard
xyy = 0.5x2​y = 3x

Find the area A of the region enclosed by y = 0.5x2 and y = 3x.

y = 0.5x2y = 3x
A =
7.Complex numbersMedium
ReIm55−5−5θz

Find the argument θ of z = 2 + j8 in degrees, as the principal value between −180° and 180°.

z = 2 + j8
θ = °
8.Complex numbersHard
ReIm22−2−244−4−4ABCD

Exactly one of the points A–D is a root of z⁴ = 81∠180°. Click it.

w = 81∠180°n = 4
  1. 1A (5.4∠−45°)
  2. 2B (3∠0°)
  3. 3C (3∠180°)
  4. 4D (3∠−45°)
9.Differential equationsMedium
tyy(0) = 4y(0.5) = ?

Solve y″ + 6y′ + 5y = 0 with y(0) = 4 and y′(0) = 0. Find y(0.5).

y″ + 6y′ + 5y = 0y(0) = 4y′(0) = 0
y(0.5) =
10.Differential equationsHard
tyK = 2ymax​ = ?

A system obeys y″ + 2y′ + 5y = 10 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.

y″ + 2y′ + 5y = 10y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
A =5−20−3−533−25

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

A = [5 −2 0; −3 −5 3; 3 −2 5]
det A =
12.Matrices & determinantsHard
A =−4−3−24b =10−9

Solve the system −4x − 3y = 10 and −2x + 4y = −9. Find x.

−4x − 3y = 10−2x + 4y = −9
x =
13.Vectors: dot & cross productMedium
xyzab

Find the angle θ between the vectors a = (−3, −4, 0) and b = (−4, −6, 4), in degrees.

a = (−3, −4, 0)b = (−4, −6, 4)
θ = °
14.Vectors: dot & cross productHard
xyzab

Find the area A of the parallelogram spanned by a = (−6, −1, 6) and b = (1, 3, −6).

a = (−6, −1, 6)b = (1, 3, −6)
A =
15.ProbabilityMedium
xp(x)10.320.330.1540.25μ = 2.35

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

x: 1, 2, 3, 4p(x): 0.3, 0.3, 0.15, 0.25
σ =
16.ProbabilityHard
xp(x)10.320.0530.3540.3μ

X has the distribution shown, and Y = −2X + 1. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.3, 0.05, 0.35, 0.3Y = −2X + 1
σY =
17.Descriptive statisticsMedium
1520253035

Find the sample standard deviation s of the data: 37, 14, 27, 24, 24, 37.

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

A quantity is normally distributed with mean 170 and standard deviation 15. What proportion P of values is below 140?

μ = 170σ = 15x = 140
P =
19.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.6772

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.6772. Find z to two decimals.

A = 0.6772
z =
20.Normal distributionHard
35455565758595pointstop 40%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points40%
x = points

3Solutions

1.Limits

Answer: L = 2

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) = 4x/(√(x2 + 4x) + x)

Divide by x

f(x) = 4/(√(1 + 4/x) + 1)

Let x → ∞

L = 4/(1 + 1) = 2

Answer

L = 2
2.Limits

Answer: L = 4

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)]/[cos x]

Substitute x = 0

L = 4/1 = 4

Answer

L = 4
3.Derivatives

Answer: f′(−1) = 240

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·(3x + 5)4·3

Substitute

f′(−1) = 5·24·3 = 240

Answer

f′(−1) = 240
4.Derivatives

Answer: x = 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.

Derivative

f′(x) = 3x2 − 21x + 36 = 3·(x2 − 7x + 12)

Solve f′(x) = 0

x = 3 or x = 4

Which one is which

f″(x) = 6x − 21: negative at a maximum, positive at a minimum.

f″(3) = −3, f″(4) = 3
x = 4

Answer

x = 4
5.Integrals

Answer: V = 28.27

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.

Disc method

V = π∫₀3 1x² dx

Integrate

V = π·1·3³/3 = 28.27

Answer

V = 28.27

Check: a cone of radius 3 and height 3 has (1/3)πr²h = 28.27.

6.Integrals

Answer: A = 18

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 = 3x → x = 0, x = 6

Integrate upper minus lower

A = [1.5x2 − 0.1667x3]06 = 18

Answer

A = 18
7.Complex numbers

Answer: θ = 75.96 °

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 = 2, b = 8 → I

Reference angle

α = atan(8/2) = 75.96°

Argument

θ = α = 75.96°

Answer

θ = 75.96°

In polar form z = 8.246∠75.96°.

8.Complex numbers

Answer: D (3∠−45°)

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 and arguments of the roots

|z| = 3
θ = 45°, 135°, −135°, −45°

The point

D = 3∠−45°

Answer

D (3∠−45°)
9.Differential equations

Answer: y(0.5) = 2.951

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 + 1)(r + 5) = 0 → r = −1, −5

Constants from the initial values

A + B = 4, −1A − 5B = 0
A = 5, B = −1

Evaluate

y(0.5) = 5·e−0.5 + (−1)·e−2.5 = 2.951

Answer

y(0.5) = 2.951
10.Differential equations

Answer: ymax = 2.416

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 = 2, r = −1 ± 2j

Constants from the initial values

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

Peak

t = π/2 = 1.571
ymax = 2·(1 + e−1·π/2) = 2.416

Answer

ymax = 2.416
11.Matrices & determinants

Answer: det A = −143

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 = −19, M₁₂ = −24, M₁₃ = 21

Cofactor expansion

det A = 5·(−19) − (−2)·(−24) + 0·21 = −143

Answer

det A = −143
12.Matrices & determinants

Answer: x = −0.5909

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 = (−4)·4 − (−3)·(−2) = −22

Replace the x column

det Aₓ = 10·4 − (−3)·(−9) = 13
x = 13/(−22) = −0.5909

Answer

x = −0.5909
y = −2.545
13.Vectors: dot & cross product

Answer: θ = 29.18 °

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 = (−3)·(−4) + (−4)·(−6) + 0·4 = 36

Magnitudes

|a| = √((−3)² + (−4)² + 0²) = 5
|b| = √((−4)² + (−6)² + 4²) = 8.246

Angle

cos θ = 36/(5·8.246) = 0.8731
θ = arccos(0.8731) = 29.18°

Answer

θ = 29.18°
14.Vectors: dot & cross product

Answer: A = 36.51

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 = (−1)·(−6) − 6·3 = −12
(a × b)y = azbx − axbz = 6·1 − (−6)·(−6) = −30
(a × b)z = axby − aybx = (−6)·3 − (−1)·1 = −17

Area

A = |a × b| = √((−12)² + (−30)² + (−17)²) = 36.51

Answer

A = 36.51
15.Probability

Answer: σ = 1.152

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 = 1·0.3 + 2·0.3 + 3·0.15 + 4·0.25 = 2.35

Mean of the squares

E(X²) = 1·0.3 + 4·0.3 + 9·0.15 + 16·0.25 = 6.85

Variance

Var(X) = 6.85 − 2.35² = 1.328
σ = √1.328 = 1.152

Answer

σ = 1.152
16.Probability

Answer: σY = 2.39

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 = 2.65, Var(X) = 1.428, σX = 1.195

Linear change

σY = |−2|·1.195 = 2.39

Answer

σY = 2.39

The mean does move: E(Y) = −2·2.65 + 1 = −4.3.

17.Descriptive statistics

Answer: s = 8.796

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 = 163/6 = 27.17

Squared deviations

Σ(x − x̄)² = 9.833² + (−13.17)² + (−0.1667)² + (−3.167)² + (−3.167)² + 9.833² = 386.8

Standard deviation

s = √(386.8/5) = 8.796

Answer

s = 8.796
18.Descriptive statistics

Answer: P = 0.02275

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 = −2

Standard normal table

Φ(−2) = 0.02275
P = 0.02275

Answer

P = 0.02275
19.Normal distribution

Answer: z = −0.46

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 − 0.6772 = 0.3228

Read the z-table backwards

Find 0.3228 inside the table and read off z.

z = −0.46

Answer

z = −0.46
20.Normal distribution

Answer: x = 67.5 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) = 0.6

z from the table

z ≈ 0.25

Back to the original units

x = 65 + 0.25·10 = 67.5 points

Answer

x = 67.5 points

ExamProblems · examproblems.com/courses/engineering-mathematics