Detailed Solutions A-35/C-35/T-35 December 2004

1. (a) D

which is circle in (u,v) plane with centre (3,2).

(c) C Let

z=0 is an essential singularity.

(d) B

Let f = x³-xyz-1,

The normal to surface at point (1,1,1) is

(e) A Using Green’s Lemma.

(f) B The given differential equation is

It is elliptic if B²-4AC<0 i.e. x²<y²

(g) A V(3x+4)=E((3x+4)²) – (E(3x+4))²

= 9E(x²)+24E(x)+16-(3E(x)+4)² = 9E(x²)-9(E(x))²= 9V(x)=9

(h) B As P(X < 0) = P(X > 2). 0, 2 are symmetrically placed around .

= .

2. (a) It is given that 3u+2v=y²-x²+16xy, thus differentiating partially w.r.t.x and y

or

Solving, we get u_x=2x+4y and v_x=-4x+2y

Thus f ’(z)= u_x+ iv_x=2x+4y+i(2y-4x)

By Milne’s Thomson method, putting x=z and y=0, we get

f’(z)=2(1-2i)z Thus f(z) = (1-2i)z²+c.

(b) Let w=u+iv, and

Since w is an analytic function, thus =and

Thus . Hence cut orthogonally.

3. (a) Let w = 1/z, then z = 1/w. Thus

Since y ≤ 1/2, thus u²+ v²+2v ≥ 0 or u²+ (v+1)² ≥ 1 i.e.

The boundary of this region is the outside of the circle with centre at(0,-1) and

radius 1,The region y ≥ 1/4 is transformed to

,

The boundary of this region is the outside of the circle with centre at (0,-2i) and

radius 2. Hence, the infinite strip maps into inside of the circle and outside of the circle . See the shaded region in the figure.

3(b) Since, w₁=0, w₂=1, w₃=∞, and z₁ = i, z₂=-1, z₃=1.

The bilinear transformation is given by

. Solving for z gives

Thus interior of the circle , in z plane is mapped onto the Imw > 0.

4. (a) It is given that

\ Given vector is irrotational. Thus it can be expressed as where f is scalar function.

,,

Integrating w.r.t. x, y, z we get

Since these three must be equal

+c

(b)

A vector normal to the surface xln(z) - y² + 4 = 0 is given by

which at point (-1,2,1) becomes . The required directional derivative is the component of along ,

.

5. (a)

At C: x² + y² = a², z=0, thus.

Work =

where R is the region bounded by circle x² + y² = a². Let x=rcosθ, y=rsinθ, then

dxdy=rdrdθ , where r changes from 0 to a and θ changes from 0 to 2π.

Thus work = .

(b)

\ Given vector represents a conservative field. Thus it can be expressed as where f is scalar function.

,,

Integrating w.r.t. x, y, z we get

Since these three must be equal

Also,

6. (i) Let the equation for conduction of heat be

Prior to temperature change at end B, when t = 0, the heat flow was independent of time (steady state condition), when u depends only on x i.e.

Since u = 100 for x = 0 and x = L

\ b = 100 and a = 0.

Thus initial condition is expressed as

The boundary conditions are

(ii) Assuming product solution u(x,t) = X(x). T(t) and substituting in equation (1), we get

Case I : If (4) gives

Solving we have the solution

This solution is rejected as exponential term makes temperature u(x,t) increases without bounds as t → ∞.

Case II : If = 0. (4) gives

Integrating, we obtain X = (A X + B) and T = C.

Thus we can write u(x,t) = (A₁ X + B₁), where A₁ = AC and B₁ =BC are arbitrary constant. Using boundary conditions (3), we get is a solution of heat equation.

Case III : If then from (4) (as in case (I)) we conclude that

Since we already have a solution (5) satisfying boundary conditions (3) we can find A, B in (6) by satisfying the condition u(0,t) = 0 =u(l,t) which gives A = 0, B sinλl = 0 .

As B = 0 leads to trivial solution we must have sinλl = 0 or , n =1,2,….. Combining (5) and (6), we have

as a general solution of (1).

Applying initial condition (2) to the general solution we must have

implies that are the coefficient in half range sine series expansion of

Putting in (7) we get required solution.

7(a)

If then f(z) = is analytic within and on C. Thus =0, by Cauchy Theorem if .

If the singularity of f(z) = lies within C and by Cauchy integral formula

Since ,

Since is constant, then .

7(b) ,

z=1, is apole of order 1 and z=-2 is a pole of order 2.

Expanding about z=1,

let z-1=t, i.e. z=t+1,

Since there is only one term in negative powers of (z-1), therefore z = 1,

is a pole of order 1. Residue at z = 1 is the coefficient of 1/t, which is 1/9.

.

Expanding about z=-2,

let z+2=t, i.e. z=t-2,

Since there are only two terms in negative powers of (z+2), therefore z = -2, is a pole of order 2. Residue is the coefficient of 1/t, which is 8/9.

.

8(a) Let the position vector of a point on C, in terms of arc length s be

. Then the tangent vector to C is given by

and a normal vector is given by

. Thus

,

since is a directional derivative of u in the direction of . Now, using Green’s theorem, we obtain

Also, ,

where S₁, S₂, S₃, S₄, S₅ and S₆ are the six faces of the cuboid.

On S₁,

On S₂ ,

On S₃,

On S₄,

On S₅,

On S₆,

Thus

Hence Gauss Divergence theorem is verified.

9(a) Let X be the number of engines that do not fail and let S_k denote the successful flight with k engine plane. Let 1-p=q,

P(S₂) = P(X ≥ 1) = 1-P(X=0)= 1-q²,

P(S₄) = P(X≥2) = 1-P(X=0)-P(X=1)= 1-4q³+3q⁴.

For P(S₄) > P(S₂), we have

1-4q³+3q⁴ > 1- q² or q²(1-q)(1-3q) > 0.

If 0<q<1/3, i.e. 2/3<p<1, the four engine plane is preferred.

9(b) Given that

Expected number of wires in a sample of 1000 with this specification

= 1000(0.6826)=682.6=683 approximately.

Hence, expected number among 1000 wires that cross the upper specification = 1000(0.1587)=158.7=159 approximately.

10(i) P(400-C≤ L≤ 400+C)=11/16

Since C≠ as it is either >1 or <1. Thus C=1/2.

(ii) E(L) = E(400+X) = 400+E(X)

Thus E(L) = 400.

V(L) = V(400+X) = V(X) = E(X²), as E(X) = 0

Thus E(L) = 400, V(L) = 0.2. Therefore,

E(2L+5) = 2E(L) + 5=805

V(2L+5) = 4V(L) = 0.8.

Since integrand is an even function, thus

Consider the contour integral and C is the path from –R to R along the real axis and from R to –R along C_R . Now f(z) is analytic in upper half of the plane except at z=ai, which is pole of order 1.

Residue of f(z) at

by Residue theorem.

Now, . Therefore, by Jordan’s Lemma

Equating imaginary part, we get

=