Paper: Kenya National Examinations Council (KNEC), Diploma in Electrical and Electronic Engineering, Module III — Engineering Mathematics III, July 2019. This lesson follows the supplied archive scan and preserves its question numbering and printed marks.

Attempt any five of the eight questions, as directed on the paper. These independently prepared worked answers are for revision and are not an official KNEC marking scheme.

Question 1: Matrices and systems

1(a) Eigenvalues and eigenvectors (10 marks)

Question. Determine the eigenvalues and corresponding eigenvectors of A = [[3, 1], [2, 4]].

Worked answer.

det(A − λI) = (3 − λ)(4 − λ) − 2 = λ2 − 7λ + 10 = (λ − 5)(λ − 2).

The eigenvalues are 5 and 2.

For λ = 5, (A − 5I)v = 0 gives y = 2x, so an eigenvector is (1, 2)T.
For λ = 2, (A − 2I)v = 0 gives y = −x, so an eigenvector is (1, −1)T.

1(b) State transition matrix (10 marks)

Question. A linear time-invariant system is characterized by dx/dt = Ax, where A = [[0, 1], [3, −2]] and x(t) is the system state vector. Determine the state transition matrix Φ(t) of the system.

Worked answer.

The eigenvalues of A are 1 and −3.

The associated projectors are (A + 3I)/4 = (1/4)[[3, 1], [3, 1]] and (I − A)/4 = (1/4)[[1, −1], [−3, 3]].
Therefore,

Φ(t) = (1/4){et[[3, 1], [3, 1]] + e−3t[[1, −1], [−3, 3]]}.

At t = 0 this reduces to I, as required for a state transition matrix.

Question 2: Numerical methods

2(a) Newton–Raphson method (9 marks)

Question. Given that xn is an approximation to the root of x4 − x2 − 1 = 0, use Newton–Raphson to show that a better approximation is xn+1 = (3xn4 − xn2 + 1)/(4xn3 − 2xn). Taking x0 = 1.3, determine the root correct to four decimal places.

Worked answer.

For f(x) = x4 − x2 − 1, f′(x) = 4x3 − 2x.
Newton’s rule xn+1 = xn − f(xn)/f′(xn) simplifies to the stated iteration.

n xn
0 1.300000
1 1.273158
2 1.272022
3 1.272020

The root correct to four decimal places is 1.2720.

2(b) Newton–Gregory interpolation (11 marks)

Question. Table 1 represents a polynomial f(x): x = −1, 0, 1, 2, 3, 4 and f(x) = −1, −3, −1, 17, 87, 269. Use the Newton–Gregory interpolation formula to determine (i) f(0.2); (ii) f(4.6).

Worked answer.

The forward differences are Δy: −2, 2, 18, 70, 182; Δ2y: 4, 16, 52, 112; Δ3y: 12, 36, 60; and Δ4y: 24, 24.

Thus the data fit the quartic f(x) = x4 + x2 − 3.

Using Newton–Gregory forward interpolation from x0 = −1, h = 1, p = 1.2 gives

f(0.2) = −1 + p(−2) + [p(p − 1)/2](4) + [p(p − 1)(p − 2)/6](12) + [p(p − 1)(p − 2)(p − 3)/24](24) = −2.9584.

Using the backward formula from xn = 4 with p = 0.6 gives

f(4.6) = 269 + p(182) + [p(p + 1)/2](112) + [p(p + 1)(p + 2)/6](60) + [p(p + 1)(p + 2)(p + 3)/24](24) = 465.9056.

Question 3: Multiple integrals

3(a) Evaluating double and triple integrals (11 marks)

Question. Evaluate: (i) ∫02∫0√(2x−x²) [x/(x² + y²)] dy dx; (ii) ∫01∫0√(1−y²)∫0√(1−x²−y²) z dz dx dy.

Worked answer (i). The region x2 + y2 ≤ 2x, y ≥ 0 is the upper half of the disk (x − 1)2 + y2 ≤ 1. In polar coordinates, 0 ≤ θ ≤ π/2 and 0 ≤ r ≤ 2cosθ. The integrand times the Jacobian is cosθ, so the integral is ∫0π/2∫02cosθcosθ dr dθ = π/2.

Worked answer (ii). The region is the first-octant part of the unit sphere. Integrating first with respect to z gives 1/2(1 − x2 − y2). In polar coordinates, 0 ≤ r ≤ 1 and 0 ≤ θ ≤ π/2, so the integral is (1/2)∫0π/2∫01(1 − r2)r dr dθ = π/16.

3(b) Area between a curve and a line (9 marks)

Question. Use a double integral to determine the area of the region bounded by the curve x = y2 − y and the line y = x.

Worked answer.

Intersections satisfy y = y2 − y, so y = 0 or 2;

the points are (0, 0) and (2, 2).

For 0 ≤ y ≤ 2, the line x = y lies to the right of x = y2 − y.
Therefore, A = ∫02[y − (y2 − y)]dy = ∫02(2y − y2)dy = 4/3 square units.

Question 4: Line integrals and Green’s theorem

4(a) Line integral along a circular arc (4 marks)

Question. Evaluate ∫C(x2dx + xy dy), where C is the arc of the circle x2 + y2 = 1 from (0, 1) to (1, 0).

Worked answer.

Along the unit circle, x2 + y2 = 1, so differentiating gives x dx + y dy = 0.
Hence x2dx + xy dy = x(x dx + y dy) = 0 at every point of the arc.

The integral is 0.

4(b) Path independence and potential function (7 marks)

Question. Show that the line integral ∫(0,0)(1,π/2)(x − cos y)dx + (x sin y + 2y)dy is independent of path, and determine its value using a potential function.

Worked answer.

Let P = x − cos y and Q = x sin y + 2y.
Since ∂P/∂y = sin y = ∂Q/∂x, the field is conservative on the region.
A potential is φ(x,y) = x2/2 − x cos y + y2.
Thus the integral is φ(1, π/2) − φ(0, 0) = 1/2 + π2/4.

4(c) Green’s theorem on a triangle (9 marks)

Question. Use Green’s theorem in the plane to evaluate ∮C(x2 − y2)dx + x2dy, where C is the boundary of the triangle with vertices (0, 0), (1, 0) and (1/2, 1).

Worked answer.

For the positive (counter-clockwise) orientation, set P = x2 − y2 and Q = x2.
Then ∂Q/∂x − ∂P/∂y = 2x + 2y.
The triangular region is 0 ≤ y ≤ 1, y/2 ≤ x ≤ 1 − y/2.
Hence

∮C(P dx + Q dy) = ∫01∫y/21−y/22(x + y) dx dy = 5/6.

Question 5: Fourier series

5(a) Odd extension and half-range sine series (7 marks)

Question. Sketch the odd extension of f(t) = 1 − t2, 0 < t < 1, on the interval −2 < t < 2, and determine its half-range Fourier sine series.

Worked answer.

On −1 < t < 0 the odd extension is t2 − 1; on 0 < t < 1 it is 1 − t2. Repeat this shape with period 2. Over −2 < t < 2 the parabolic arcs are 1 − (t + 2)2 on (−2, −1), t2 − 1 on (−1, 0), 1 − t2 on (0, 1), and (t − 2)2 − 1 on (1, 2); there are jumps at even integers.

For 0 < t <

1, bn = 2∫01(1 − t2)sin(nπt)dt = 2/(nπ) + 4[1 − (−1)n]/(n3π3).
Therefore,

1 − t2 = Σn=1∞{2/(nπ) + 4[1 − (−1)n]/(n3π3)}sin(nπt),   0 < t < 1.

5(b) Fourier series of a periodic quadratic (13 marks)

Question. Given f(t) = t2, 0 ≤ t ≤ 2π, (i) sketch f(t) in the interval −2π < t < 2π; (ii) determine the Fourier series representation of f(t).

Worked answer.

Use the periodic extension with period 2π. On −2π < t < 0 the graph is (t + 2π)2;

on 0 ≤ t ≤ 2π it is t2.
Each period rises parabolically from 0 to 4π2 and then jumps back to 0.

For the 2π-periodic Fourier series, a0/2 = (1/2π)∫02πt2dt = 4π2/3, an = (1/π)∫02πt2cos(nt)dt = 4/n2, and bn = (1/π)∫02πt2sin(nt)dt = −4π/n.

Thus

f(t) = 4π2/3 + 4Σn=1∞[cos(nt)/n2 − π sin(nt)/n].

At the jump points t = 2kπ, the Fourier series converges to the mean of the one-sided limits, 2π2.

Question 6: Vector calculus

6(a) Surface area of a cone (5 marks)

Question. Find the surface area of the part of the cone z2 = 16(x2 + y2) that lies between the planes z = 0 and z = 8.

Worked answer.

For the upper cone, z = 4r, so 0 ≤ r ≤ 2.
The surface element is dS = √(1 + (dz/dr)2)r dr dθ = √17 r dr dθ.
Therefore, S = ∫02π∫02√17 r dr dθ = 4π√17 square units.

6(b) Divergence theorem on a sphere (5 marks)

Question. Use the divergence theorem to evaluate the surface integral ∬SF · n dS for F = (x − z)i + (y − x)j + (z − y)k, where S is the sphere x2 + y2 + z2 = 9.

Worked answer.

∇·F = 1 + 1 + 1 = 3.
The sphere has radius 3 and volume 4π(33)/3 = 36π.
The outward flux is ∭V3dV = 3(36π) = 108π.

6(c) Stokes’ theorem on a triangular plane (10 marks)

Question. Use Stokes’ theorem to evaluate ∮CF · dr for F = zi + xj + yk, where C is the boundary of the plane x + y + z = 2 in the first octant.

Worked answer.

∇×F = (1, 1, 1).

Take the positive orientation corresponding to the upward/outward normal (1, 1, 1).

The plane is z = 2 − x − y and its projection is x ≥ 0, y ≥ 0, x + y ≤ 2.
Then (∇×F)·n dS = 3 dx dy, so the line integral is ∬D3 dx dy = 3(2) = 6.

Reversing the boundary orientation changes the sign.

Question 7: Harmonic functions and transformations

7(a) Harmonic conjugate (10 marks)

Question. Given u = e3xcos 3y, (i) show that u is a harmonic function; (ii) determine a conjugate harmonic function V(x,y) such that f(z) = u + jV is analytic.

Worked answer.

uxx = 9e3xcos 3y and uyy = −9e3xcos 3y, so ∇2u = 0.
From the Cauchy–Riemann equations, Vy = ux = 3e3xcos 3y;
integrating gives V = e3xsin 3y + g(x).
Using Vx = −uy gives g′(x) = 0.
Hence V(x,y) = e3xsin 3y + C.

7(b) Image of the unit circle under a bilinear transformation (10 marks)

Question. The circle |z| = 1 is mapped onto the w-plane by w = 1/(z + 2j). Determine the centre and radius of the image circle.

Worked answer.

Rearranging gives z = 1/w − 2j.
On |z| = 1, |1 − 2jw| = |w|.
Put w = u + jv and square: (1 + 2v)2 + 4u2 = u2 + v2.
Thus u2 + (v + 2/3)2 = 1/9.

The image circle has centre (0, −2/3) and radius 1/3.

Question 8: Numerical methods and Fourier series

8(a) Newton–Raphson method and root bracketing (8 marks)

Question. Show that one root of x4 + x2 − 10 = 0 lies between x = 1 and x = 2, and use Newton–Raphson to determine the root correct to four decimal places.

Worked answer.

For f(x) = x4 + x2 − 10, f(1) = −8 and f(2) = 10, so a root lies between 1 and 2.
Newton’s iteration is xn+1 = xn − (xn4 + xn2 − 10)/(4xn3 + 2xn).
Taking the midpoint x0 = 1.5:

n xn
0 1.500000
1 1.662879
2 1.643940
3 1.643643
4 1.643643

The root is 1.6436 correct to four decimal places.

8(b) Fourier series of a half-wave rectifier (12 marks)

Question. The emf produced by a half-wave rectifier is e(t) = 10 sin t for 0 < t < π, e(t) = 0 for π < t < 2π, and e(t + 2π) = e(t). Determine the Fourier series representation of e(t).

Worked answer.

The period is 2π.

The coefficients are a0/2 = 10/π, b1 = 5, bn = 0 for n ≠ 1, an = 0 for odd n, and a2m = −20/[π(4m2 − 1)].
Therefore,

e(t) = 10/π + 5 sin t − (20/π)Σm=1∞cos(2mt)/(4m2 − 1).

Revision note: these independently prepared explanations are for study and are not the official KNEC marking scheme. Check the supplied scan for the original notation and instructions.

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