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Bangalore University 2008 B.Sc Mathematics III Maths VI - - Question Paper

Saturday, 23 March 2013 03:50Web



Ill Year B.A./B.Sc. Examination, November/December 2008 (1999-2000 & Onwards Scheme)

Paper - VI: MATHEMATICS

Max. Marks : '100

Time : 3 Hours


Instructions : 1) Answer all the questions.

2) Answers should be completely in Kannada or in English.

I. Answer any twenty questions of the following :

(20x2=40)

JL)lflr(t) = e x i + log(t2 + l)j - tant k, find the acceleration at t 0.

A f \

/TF A = t2i - tj-(2t+ l)k and B = (2t -1) i + j - tk, find (Ax Bj.

dt

3) Find the spherical polar coordinates of the point whose Cartesian coordinates are (1,1,1).    |

Find the direction cosines of the tangent at the point t =j of the curve r = 4 cost

! 2

i + cos 2tj + cos21 k.

5) If <|> = xyz, find V2<j) *

= x3 + cosy + z, find grad <j> at f 0, ~, 1

7)JRrn3 the directional derivative of (j) = xy + yz + zx at (1,2,0) in the direction of i+j+k.

the vector F = (ax + 3y + 4z) j + (x - 2y + 3z) j + (3x + 2y - z)k is solenoidal, find the value ofa.

)valuate : I xdx + ydy + zdz; where C is the curve X = t2 +13; y = t2 + 1,

c

z = et\ where -l<t<l.

P.T.O.


1

AN 1319    -2-

lQyEvafuate : J J (x2 + y2).dydx

o x

ill

jJEvaluate : J J J (x + y + z) dxdydz.

ooo

jifA = ti -12j + (t - l)k and B = 2t2i + 6tk, find jA.B dt.

13) Using Gauss divergence theorem, show that JJr .nds = 3V, where V is the volume

s

of any closed surface S.

JylJsing Greens theorem, show that the area bounded by a simple closed curve

C is given by j xdy - ydx. .

15) State Greens theorem.

sin t i + e 1 cost j, find |


17) Evaluate: Km


dfA

'dty


dt


'V + l

v 1 - zi y


18)    Show that f(z) = z is not analytic.

19)    Examine C -Requations for f(z) ~z~ z.

(J20)"<Verify whether u = x3 - 3xy2 + 3x2 - 3y2 + 1 is harrhonic.

21)    State fundamental theorem of Algebra.

22)    State Cauchys integral formula.

23)    Evaluate : j 3z2dz> where C is the line segment frJm 0 to 2i.

c    '    !

2z-l



-3-

II. Answer any two questions :

AN - 1319 (2x5=10)


r = ti +1 j + t k; find the unit tangent and the principal normal vector at t = 1.

2) Derive the formula for the torsion x of a space curve r = r(u) in the form [rrf]

[r x ?]

)HEfT= 3t2i -(t + 4)j + (t2 - 2t)k and b - sin t i +3e-ti j - 3 cost k , (axb),

find


dt

4) Show that the surface 5x2 - 2yz = 9x is orthogonal to the surface 4x2y + z3 = 4 at the point (1, -1, 2).

III. Answer any two questions :

(2x5=10)

1) Prove that div (curl F) = 0 and curl (grad <|>) 0 *

xi + yj + zk . . . i, ?s lrrotational.

2)Sfiow that the vector f =


\jx2 + y2 +z2'

A

+ sin x)k, then prove that


F - (sin y + z cosx)i + (x cosy + sin z) j + (y cosz

curl F = 0. Find <j> such that F = V<j>.

4) Find the equation of the tangent plane to the surface z = x2 - y2 at (2, -1, 3).

(2x5=10)


IV. Answer any two questions :

Devaluate : jF.dr, where F = x2i + (2xz + y)j + zk along the line C from (0, 0, 0) '    c

to (2, 3, 4).

2) Evaluate jj ydx dy; where R is the region bounded

by the parabolae y2 = 4ax

R

a 4 a*-x2 /a2-x2-y2

and x2 = 4ay.


dzdy dx


Evaluate : f f    f    7--,.

J    / 2 2 2 2

oo o    ya-x-y-zl


AN 1319    .4.

V. Answer any two questions :

(2x5=10)


1) Evaluate JJ [yzi + zyj + xykj.nds, where S is the surface of the sphere s

x2 + y2 + z2 - 1 in the first Octant.

2) Using Stokes theorem, evaluate

f [(x + y)dx + (2x - z)dy + (y + z)dz], where C is the boundary of the triangle

whose vertices are (2, 0, 0), (0, 3, 0) and (0, 0, 6).

3) Verify Green's theorem for <j(x2 + y2)dx - xy2dy

taken around the boundary

1

of the square whose vertices are (0, 0); (1, 0); (1, 1) and (0, 1). 4) State and prove Stokes theorem.

(4x5=20)


VI. Answer any four questions :

z-ll 71

1) Show that arg


- = represents a circle.

z + 1 J 3

frove that the necessary condition for a function if(z) = u(x, y) + iv(x, y) to be

, . . du dv du -dv

analytic is = and ---

dx dy dy dx

,    * x-y

he imaginary part is =-=-

x + y


3) Find the analytic function f(z) = u + iv, given that

4) Prove that the real and imaginary parts of an analytic function are harmonic. Is the converse true ? Justify.

iJvaluate J(3xy + iy2 )dz along the straight line C joining z = i and z = 2 i.

z -1

dz, where C is |z i| =

2.


c (z+1)2(z-2)

7)    Discuss the transformation W = Z2.

8)    State and prove Liouvilles theorem.


-5-    !    AN - 1319

i

I. cdtoradcta20    ;    (20x2=40)

1)    r(t) ~ e_t i + log(t2 + l)j - tant k, e?ddt = 0ste)wdd derfoedrdodb&ccoO.

I

2)    A = t2i - tj - (2t + l)k dbD B = (2t -1) i + j - tk 'sdd, (a x b)oc&&8o3oO.

3)    od> odb<D sstaeF&odW KcSers&rteb (1,1,1) dd, dd rfaed tpeccb derd3b odb&oso.    I

4)    r - 4 cos tin- cos 2tj + cos21 k eodb dde<d wdd, edd xfefcd c|d5ffs&?r

t - odbSoosoO.    i

2

5)    (j) = xyz, esdd, V2(|) odbSo&ctfoo.    i

?    * f 71 :    '

6)    (j) = x + cos y + z e?dd, grad (j) [ 0, , 1 d<odb&>coo.

J

7)    (1, 2,0) octa i + j + k odbcdD(|) = xy + jyz + zxrfQaia3

wdootfdsb tfodbSoccoo.    -    ' j    <

i. "

i

8)    F = (ax + 3y + 4z)j + (x - 2y + 3z) j + (3x + 2y - z)k*j.ot>*wto)oDrfor

Qcisdd a ddoi>;> iocso2cQoDDO.

=<. ,

9)    Coio x = t2 + t3;yt2+1, z = et! d~l <t <1 des3pd I xdx + ydy + zdz

j c

odbSoo2oo.    i

I

i

i Vx >    I

10)    odDSoacODO: j j(x2+y2)dydx.    j

X    !

iii ;

11)    odbMoOoO: J J J (x + y + z) dxdydz.    !

ooo    '    :

2__;

12)    A = ti - t2j + (t - l)k s)d B = 2t2i + 6tk ydd, JA.Bjdt odb2oSo3oO.

o


AN - 1319    -6

13)    rf    erodalfteF&, JJr.rids = 3V >odbatoeo&. <ao2oc>db

s

odtode dboi|d d3?d sjfc%>o.

14)    fyc? doeo&dab erodoirDeft, <j*xdy-ydx <aowodo C ou do>2|d

c

ddesdccood too;fe)ftdbd codb ;ft>eo.

15)

df v dtJ


16)    f(t) = e-t sin t i + e_t cost j, wdd j

dt c&ab odoSoox>o.


( 3 Z+1

v1 - zi y


lim

z-m


F&OtoQoM.


17)


18)    f (z) = z    estuaorib eOA.

19)    fz) = z-zmfC-R

oDodo dOea&.

eA

_dz cdb odbSoQosoo.


20)    u = x3 - 3xy2 + 3x2 - 3y2 + 1 arosftaeFDaf t?Adcdb

21)

22)    l>octo    pddab >oto&A>.

23)    C <aoioodb0 <5*od2i* ddrtodbddedd, J3:

c

2z 1

24) W =-atosroodd Ad ttoribrfcfcdb odoSoQcft}0.

t

(2x5=10)


A     A     A

1) r = ti +1 j + t k &oocb ddesj esdd t = 1    a&f db& scd 0020 Dddb

odb&CCDQ.

r r jr]

PF


eodb nJ3i


'ri.


2) r - f (u) toodb tssarf ddea? add, Scbzb x








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