Question

7. Find an equation of the tangent plane to the given parametric surface r(u, v) = uvi+u sin(n)j + v cos(u)k, at u = 0, v = . 8. Find the area of the part of the surface 2 = 2 + 5x + 2y that lies above the triangle with vertices (0.0), (0,1), and (2,1).

Answer 1

90 12- First we wrt. u & v. 7. Find the equation of the tangent plane to the given parametric surface 2 (0,0) = u vi + usine h e coon r at 4.0, van compute the tangent rectors Tu se î+ since ĝ + x[-sinu) & v ît sine 9 - (sin ure du auît uit u copre f t cosuh. find the normal rector to the tangent plane so od Then we -) ĥ ܠܢܬ sing - usinu U u cosu cosu 2 we î(sinvecasu + uv sinu coop) -h recosu+ uusinu) + û luusin cose-usin o) Now putting noo, va get, (sinn caso + +0)-1(76000+0) ta (oro) A ño (Tuxto), Uzo, Vza *g on Now the position rector of a point the plane is given by (uv, usinne, u cosu, 2 (0.0,-) leo V za the two where is given by (x-o, yo (-a). 5-0 then , Then the egn of plane comporized tangent rectors ru & ro & passing through 2(0,0,7) re(x, y, z). 0,2-a). (0, -a, 0) => 0(2-0) - *(y-0)+ 0 (2-1) - 0 -x y zo 720 so, the required tangent plane is the xz-plane. as a to

2 + 5x + 2y² 12 lies 8. Find the area of the part of the surface 2-2 + 5x+ above the triangle with vertices (0,0), (0, 1) & (2, 1) 2010 - for the surface area of the form surface of the form zof(x,y) is given by Ild3 = $fety da, where Dis the projection of the surface the xy-plane The line joining (0,0) 2 (2, 1) is on the 160.0 2-0 (0,oo y = = x22y Then Di={(x, y) / Osasay; O sys1} f(x, y) = 2 + 5x + 2y² in our problem so, Ex=15 by = 47 in surface sl √ 1 +52 + (4972 Now , 27 area a de) dy. 0 0 1 2y o 27 dy (1126 116ye de) dy x 26 +16y? ] 27 126 11 by²dy Let, 26+16 y = te D 3) 32 ydy - Belt 42 1 sve. VE. I dt. 16 t 26 42 26 26 to 1347 (62)-303 42 140 - 26126 21/42 - 13126 12 24