MTH101A (2016), Tentative Marking Scheme - End sem. exam

Transcription

1 MTH11A (16), Tentative Marking Scheme - End sem. eam 1. (a) Let f(, y, z) = yz and S be + y + z = 6. Using Lagrange multipliers method, find the maimum and minimum values of f on S. [7] Lag. Eqns.: yz = λ, z = λy, y = λz, + y + z = 6 Case I: λ = the candidates (± 6,, ), (, ± 6, ), (,, ± 6) [] Case II: λ 3yz = 1λ Hence λ = 4λ which implies = ± Similarly y = ± and z = ±. The candidates are (±, ±, ± ). The ma value is and min value is (This one mark is not to be released if Case I is NOT considered) (b) Evaluate the volume of the solid which is common to the cylinders + y = 4 and y + z = 4 using the method of double integrals. [5] V = = = y 4 y 4 y ddy [3] 4(4 y )dy (c) Let f : [ 1, 3 ] R be differentiable. Show that there eist c 1, c [ 1, 3 ] such that f (c ) = c f (c 1 ). [5] By CMVT f( 3 ) f( 1 ) 3 1 = f (c 1 ) 1, c 1 [ 1, 3 ] [] and f( 3 ) f( 1 ) ( 3 ) ( 1 ) = f (c ) c, c [ 1, 3 ] [] Therefore f (c 1 ) = f (c ) c. (a) Let the surface S be part of +3y z = which lies inside the region bounded by = 1, =, y = and y =. Evaluate dσ. [6] S Let R be the projection of the given region and f(, y) = + 3y. Then dσ = 1+f +fy ddy = 14ddy [] S R R = π/4 sec θ sec θ 14 rdrdθ (See Figure (a)) [3] r = 14 log(1 + )

2 (b) Let X, U R where U = 1 and f : R R be differentiable at X. Prove that the directional derivative D X f(u) of f at X in the direction U eists and D X f(u) = f( ) U. [5] Since f is differentiable at X, f(x +H) f(x ) f(x ) H as H H For H = tu, t R, U = 1 and t, f(x +tu) f(x ) f(x ) tu t. As t, f(x +tu) f(x ) f(x ) tu. t As t, f(x +tu) f(x ) f(x t ) U. (c) Let a n and n=1 (n3 a n 1) converge. Verify whether n a n n=1 converges. [6] Observe that n 3 a n 1 [] LCT with 1 n : a n / n 1/n = n 3 a n 1 [] The series converges. 3. (a) Let the curve C be described by R(t) = ((sin 3t) cos t, (sin 3t) sin t), t π 3. Sketch C. Evaluate yd + dy and yd + dy where C is oriented C C counterclockwise. [8] Note that C is r = sin 3θ, θ π 3. For the curve (see Figure 3(a)). Observe that yd + dy = (y) dr = by FTC of line integrals [] C C Observe that yd + dy = (Area enclosed by C). [] C Area = 1 π 3 sin (3θ)dθ = π 1. (b) Let D be the region that lies below the surface + y + z = 4z and above z = 3( + y ). Using the spherical coordinates epress + y D + z dv as three iterated single integrals. [4] The sphere is ρ = 4ρ cos ϕ, i.e., ρ = 4 cos ϕ. The cone is ρ cos ϕ = 3r = 3ρ sin ϕ, i.e., ϕ = π 6. Therefore D + y + z dv = π 6 4 cos ϕ ρρ sin ϕdρdϕdθ. []

3 (c) Let (a n ) be in (, 1) and 4a n (1 a n+1 ) > 1 for all n 1. Discuss the convergence/divergence of the series n=1 (a n 1). [5] If a n a for some a, then 4a (1 a ) 1. [] Since (a 1), a = 1. [] Since a n 1, (a n 1) does NOT converge. 4. (a) Consider the arc ( ) + y = 4, y. Using a theorem of Pappus, find the surface area of the surface generated by revolving the arc about the line y + =. [6] Let the coordinate of the centroid of the arc be (, y ). By Pappus theorem, 4π = πy π [] Hence y = 4 π. Distance of the line from the centroid is ρ = +y 1+ By Pappus theorem, the required area is πρπ. (b) Find the equation of the surface generated by the normals to the surface y + z + yz = at all points on the z-ais. [5] Normal is (z + yz, 1 + z, + yz). Normal at (,, z ) is (z, 1, ). If (, y, z) lies on the surface then, z = y, z = z 1. [] The equation of the surfaces is = zy. (c) Let f : [, ) [, ) be such that f () > for every and f()d converges. Show that n f()d nf(n) and f(n ). [6] By Taylor s theorem, for [, n], f() f( n) + f ( n)( n ). [] n Hence f()d nf( n) + f ( n) n f ( n) n. [] Since f()d converges, there eists M > such that f( n) M n. [] n 5. (a) For p > 1, consider the curve C : p + y p = 1. Evaluate C ( y + e (sin))d + ( + y(siny))dy where C is oriented counter- clockwise. [6] Given integral is C y d + dy + C (e (sin))d + (y(siny))dy.

4 By Green s theorem, C (e (sin))d + (y(siny))dy =. Observe that ). By Green s theorem, y d + dy = y C C r d + dy, [] where C r is a circle of radius r and C r lies inside the region enclosed by C. r sin td(r cos t)+r cos td(r sin t) Hence the required value is = π. r ( ) = ( y y (b) Consider the surface S : + y + z = 8, 1 z. i. Find a vector field F such that curlf = (,, 8). ii. Find the unit (outward) normal to S. iii. If C 1 is + y = 4, z = then evaluate C 1 F dr. iv. Evaluate S zdσ. (i) F (, y, z) = ( y 8, 8, ). (ii) The normal ˆn = 1 8 (, y, z). (iii) Parametrization of C 1 is ( cos θ, sin θ, ), θ π The value of the line integral is 8 sin θd( cos θ)+ 8 cos θd( sin θ) = 8 8π [] (iv) Observe that zdσ = curlf ndσ. S S By Stoke s theorem, curlf ndσ = ( )F dr. [] S C C 1 where C : + y = 7, z = 1. F dr = 14 8π. C Hence S zdσ = 6 8π 6. (a) Sketch the graph of f() = 3 after finding the intervals of decreasing/increasing, intervals of concavity/conveity, points of local maimum 1 and asymptotes. [5] f() = f () = 1 = 1, = 1 and y = 3 are the asymptotes. f is on (, 1), ( 1, ) and on (, 1), (1, ). ( 1) f () = (3 +1) conve on (, 1), (1, ) and concave on ( 1, 1) ( 1) 3 = is a point of local maimum. For the graph (see Figure 6(a)).

5 (b) Consider the surfaces [11] S 1 = {(, y, + 1) : + y 1 9 } and S = {(, y, 1) : + y 1 9 }. Let the surface S 3 be the part of the cylinder + y = 1 9 that lies between the surfaces S 1 and S. Let D denote the region enclosed by S 1, S and S 3. Let F (, y, z) = ρ 3 (, y, z) for (, y, z) where ρ = + y + z. i. Find the unit normals to the surfaces S 1, S and S 3. ii. Find DivF. iii. Evaluate (z )dσ ( + zdσ + 3( )dσ. )ρ 3 ρ 3 ρ 3 S S 3 S 1 (i) Unit normal on S 1 : n 1 = 1 ( 1,, 1) or ( n 1 ). Unit normal on S : n = (,, 1) or ( n ). Unit normal on S 3 : n 3 = (3, 3y, ) or ( n 3 ). (ii) (ρ) = ρ ( ρ 3 ) = 1 ρ 3 3 ρ 5 [] divf = 3 ρ 3 3ρ ρ 5 = (iii) The given integral I = S 1 F n 1 dσ + S F n dσ + S 3 F n 3 dσ. [] By divergence theorem I = S F ndσ where S : + y + z = r, r < 1 and n = 1 (, y, z). 3 r I = 4π. []