Paper: Kenya National Examinations Council (KNEC), Diploma in Electrical and Electronic Engineering, Module III — Engineering Mathematics III, June/July 2020. Paper codes: 2521/303, 2601/303, 2602/303 and 2603/303.

The supplied paper contains eight questions; candidates are instructed to answer any five. The lesson retains the original numbering and printed marks. These independently prepared worked answers are revision notes, not an official KNEC marking scheme.

Question 1: Complex variables and harmonic functions

1(a) Cauchy–Riemann equations (6 marks)

Question. Determine the values of the constants a and b such that w = ax3 − 3axy2 + j(bx2y − y3) is analytic.

Worked answer.

Let u = ax3 − 3axy2 and v = bx2y − y3.
The Cauchy–Riemann equations require ux = vy and uy = −vx.
From ux = 3ax2 − 3ay2 and vy = bx2 − 3y2, equating coefficients gives a = 1 and b = 3.

The second equation gives the same relation.

Therefore, a = 1, b = 3.

1(b) Harmonic function (4 marks)

Question. Given u = e3xcos 3y + 4x + 3, show that u is harmonic.

Worked answer.

uxx = 9e3xcos 3y and uyy = −9e3xcos 3y.
Hence ∇2u = uxx + uyy = 0.

The added term 4x + 3 has zero second derivatives, so u is harmonic.

1(c) Image of a circle under a bilinear transformation (10 marks)

Question. The circle |z| = 3 is mapped onto the w-plane by w = (z + 2j)/(z − j). Determine the centre and radius of the image circle.

Worked answer.

Solving for z gives z = j(w + 2)/(w − 1).
Thus |z| = 3 becomes |w + 2| = 3|w − 1|.
Put w = u + jv and square: (u + 2)2 + v2 = 9[(u − 1)2 + v2].
Simplifying and completing the square gives (u − 11/8)2 + v2 = (9/8)2.

The image circle has centre (11/8, 0) and radius 9/8.

Question 2: Matrices and state transition matrices

2(a) Determining a matrix parameter (10 marks)

Question. Given that λ = −2 is an eigenvalue of A = [[x, 3], [4, x + 1]], determine the possible values of x.

Worked answer.

Since −2 is an eigenvalue, det(A + 2I) = 0.
Hence (x + 2)(x + 3) − 12 = 0, so x2 + 5x − 6 = (x + 6)(x − 1) = 0.
Therefore, x = −6 or x = 1.

2(b) State transition matrix (10 marks)

Question. A system is characterized by dx/dt = Ax, where A = [[0, 1], [−1, −2]]. Determine the state transition matrix Φ(t).

Worked answer.

Write A = −I + N, where N = A + I = [[1, 1], [−1, −1]] and N2 = 0.
Thus eAt = e−teNt = e−t(I + tN).
Therefore,

Φ(t) = e−t[[1 + t, t], [−t, 1 − t]].

At t = 0, Φ(0) = I.

Question 3: Fourier series

3(a) Half-range Fourier cosine series (11 marks)

Question. Determine the half-range Fourier cosine series of f(t) = t − t2, 0 < t < 1, and sketch the corresponding graph in −1 < t < 1.

Worked answer.

The even extension is f(t) = |t| − t2 on −1 <

t < 1.

It is zero at t = −1, 0, 1 and has equal maxima 1/4 at t = ±1/2.
For L = 1, a0/2 = ∫01(t − t2)dt = 1/6.
The cosine coefficients are an = 2∫01(t − t2)cos(nπt)dt = −2[1 + (−1)n]/(n2π2), so odd coefficients vanish.
Therefore,

f(t) = 1/6 − (1/π2)Σm=1∞cos(2mπt)/m2,   0 < t < 1.

3(b) Fourier series of a 2π-periodic function (9 marks)

Question. A function f(t), of period 2π, is symmetrical about the origin and is defined for 0 < t < π by f(t) = 4t/π for 0 < t < π/2 and f(t) = 4(1 − t/π) for π/2 < t < π. (i) Sketch f(t) in −π < t < π; (ii) determine its Fourier series.

Worked answer.

Origin symmetry means f(−t) = −f(t).

The graph is an odd triangular pulse: it is zero at −π, 0 and π;

it reaches −2 at t = −π/2 and 2 at t = π/2, with straight-line segments between these points.

The odd symmetry gives a0 = an = 0.

For bn = (2/π)∫0πf(t)sin(nt)dt, integration by parts on the two linear segments gives bn = 16sin(nπ/2)/(π2n2).
Hence

f(t) = (16/π2)[sin t − sin 3t/32 + sin 5t/52 − ···].

Question 4: Numerical methods

4(a) Newton–Raphson method (10 marks)

Question. Show that one root of x3 − 6x2 + 4 = 0 lies between x = 0 and x = 1, and use Newton–Raphson to determine the root correct to four decimal places.

Worked answer.

Let f(x) = x3 − 6x2 + 4.
Since f(0) = 4 and f(1) = −1, a root lies in (0, 1).
With f′(x) = 3x2 − 12x, the iteration is xn+1 = xn − (xn3 − 6xn2 + 4)/(3xn2 − 12xn).
Taking the midpoint x0 = 0.5:

n xn
0 0.500000
1 1.000000
2 0.888889
3 0.884259
4 0.884251

The root is 0.8843 correct to four decimal places.

4(b) Newton–Gregory interpolation from steam-table data (10 marks)

Question. The data in Table 1 are extracted from a steam table. Temperature (°C): 140, 150, 160, 170, 180. Pressure (kgf/cm2): 3.685, 4.854, 6.302, 8.076, 10.225. Use the Newton–Gregory forward difference interpolation formula to determine the pressure when the temperature is 145°C.

Worked answer.

The temperature interval is h = 10°C and p = (145 − 140)/10 = 0.5.

The forward differences are ΔP = 1.169, 1.448, 1.774, 2.149;

Δ2P = 0.279, 0.326, 0.375;
Δ3P = 0.047, 0.049;
Δ4P = 0.002.

P(145) = 3.685 + p(1.169) + [p(p − 1)/2](0.279) + [p(p − 1)(p − 2)/6](0.047) + [p(p − 1)(p − 2)(p − 3)/24](0.002) = 4.2375 kgf/cm2 (approximately).

Question 5: Double integrals and divergence theorem

5(a) Double integral over a circular region (9 marks)

Question. Evaluate ∬R[x/√(x2 + y2)]dxdy, where R is the region bounded by x2 + y2 = x.

Worked answer.

In polar coordinates, the boundary is r = cosθ, with −π/2 ≤ θ ≤ π/2 and 0 ≤ r ≤ cosθ.

The integrand is cosθ, and dA = r dr dθ.

Therefore, ∬R…dA = ∫−π/2π/2∫0cosθr cosθ dr dθ = (1/2)∫−π/2π/2cos3θ dθ = 2/3.

5(b) Verifying the divergence theorem (11 marks)

Question. Verify the divergence theorem for F = (x − z)i + (2y − x)j + (y − z)k, where S is the surface of the tetrahedron in the first octant bounded by the coordinate planes and x + y + z = 1, oriented by outward unit normals.

Worked answer.

The divergence is ∇·F = 1 + 2 − 1 = 2.
The tetrahedron has volume 1/6, so the volume integral is ∭V2dV = 1/3.

For the four faces, the outward fluxes are: x = 0 gives 1/6;

y = 0 gives 1/6;
z = 0 gives −1/6;
and x + y + z = 1 gives 1/6.

Their sum is 1/3, equal to the volume integral. Thus the divergence theorem is verified.

Question 6: Surface integrals and Stokes’ theorem

6(a) Flux across a paraboloid (9 marks)

Question. Evaluate ∬SF · n dS for F = xi + yj + zk, where S is the surface of the paraboloid z = 9 − x2 − y2 above the xy-plane.

Worked answer.

The projection is the disk r ≤ 3. For the upward orientation, n dS = (2x, 2y, 1)dA.

With z = 9 − r2, F · n dS = (2x2 + 2y2 + z)dA = (9 + r2)dA.
Therefore, the flux is ∫02π∫03(9 + r2)r dr dθ = 243π/2.

6(b) Stokes’ theorem on a triangular plane (11 marks)

Question. Use Stokes’ theorem to evaluate ∮CF · dr, given F = 2zi + 3xj + yk and C is the boundary of the plane x + y + z = 2 in the first octant, with counter-clockwise orientation.

Worked answer.

∇×F = (1, 2, 3).

For counter-clockwise orientation viewed from above, take the upward normal;

for z = 2 − x − y, n dS = (1, 1, 1)dxdy.

Their dot product is 6.

The projected triangle x ≥ 0, y ≥ 0, x + y ≤ 2 has area 2.
Hence the line integral is ∬D6 dxdy = 12.

Question 7: Line integrals and Green’s theorem

7(a) Line integral along a circular arc (6 marks)

Question. Evaluate ∫C(x2dx − xy dy) along the arc of the circle x2 + y2 = 4 from (2, 0) to (0, 2).

Worked answer.

On the first-quadrant arc, x = 2cosθ, y = 2sinθ, 0 ≤ θ ≤ π/2.
Substitution gives the integral −16∫0π/2cos2θ sinθ dθ = −16/3.

7(b) Path independence and potential function (6 marks)

Question. Show that the line integral ∫(0,0)(1,π)2x cos y dx − x2sin y dy is path independent, and determine its value by using a potential function.

Worked answer.

Let P = 2x cos y and Q = −x2sin y.
Since Py = −2x sin y = Qx, the integral is path independent.
A potential is φ(x,y) = x2cos y.
Thus the value is φ(1,π) − φ(0,0) = −1.

7(c) Green’s theorem on a triangle (8 marks)

Question. Use Green’s theorem to evaluate ∮T(3x − 2y)dx + (4x + 3y)dy, where T is the positively oriented boundary of the triangle with vertices (0, 0), (2, 0) and (0, 2).

Worked answer.

With P = 3x − 2y and Q = 4x + 3y, ∂Q/∂x − ∂P/∂y = 4 − (−2) = 6.
The triangle has area 2, so the line integral is 6 × 2 = 12.

Question 8: Multiple integrals

8(a) Double integral in polar coordinates (8 marks)

Question. Determine ∬R√(x2 + y2)dxdy, where R is the xy-plane region bounded by x2 + y2 = y.

Worked answer.

In polar coordinates the boundary is r = sinθ, so 0 ≤ θ ≤ π and 0 ≤ r ≤ sinθ.
Since √(x2 + y2) = r and dA = r dr dθ, the integral is ∫0π∫0sinθr2dr dθ = (1/3)∫0πsin3θ dθ = 4/9.

8(b) Changing the order of integration (7 marks)

Question. Change the order of integration and evaluate ∫01∫y1ex²dxdy.

Worked answer.

The region is 0 ≤ y ≤ x ≤ 1.
Reversing the order gives ∫01∫0xex²dy dx = ∫01xex²dx = (e − 1)/2.

8(c) Volume using a triple integral (5 marks)

Question. Use a triple integral to determine the volume of the solid bounded by the parabolic cylinder z = 1 − y2 and the planes x + z = 1, x = 0 and z = 0.

Worked answer.

The bounds are −1 ≤ y ≤ 1, 0 ≤ z ≤ 1 − y2, and 0 ≤ x ≤ 1 − z.
Thus

V = ∫−11∫01−y²∫01−zdx dz dy = ∫−11(1/2 − y4/2)dy = 4/5 cubic units.

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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