Question

(2) Let S be the surface parametrized by r(u, v) = (u? – 12)i + (u + v)j + (u? + 3v)k. (a) Find a normal vector to S at the point (3,1,1). (b) Find an equation of the tangent plane to S at (3, 1, 1).

Answer 1

Solution:- @ Unit Noemcel vectce to s at point (3,1,1) given t(UV) = (u=v2) i +(+ v)j + (u?+3v) ñ - žux Er will give us not mul vecice : Za = 2u1 + i +248 - ²x = - zvů + i +3² . 124 , zu 2u 3 1-2V 124 24 II 24 în t sal 34 - 34 . - V 4. ñ = (3-24) i-(60+40v) † + (24 +2V) to determine value of UAV * Scanned with CamScanner SCENE

we de goint to set the individual components of the parametric tepresentation equal to the points in question » u-v2-3, U+V=1, u?+3V= 1 î → = (U-v) (u+v)= 3 + +0= 1 4-V= 3 p uP+3v=) 34-3V = 9 + u² +3v=1 Butu?=10 † u (3+4)=10 - U+34-10=0 (4+5) (4-2)=0 = u=5 oc 71 u=2 (U+=17 =1f U= 5 > y=-4 ť 14 U=2 =) V=-1 satisfy but U=5 &v=-4 does not i-y? 3 or 4P3v = || CS Camseanl=27€ V=-1 Scanned with

:. put u=? Vet in ean we get . Ñ = (3-4)7'- (12-8) 1 + 2X .. Phoemal vectore= ñ = --41+2 I Tangent plane : We know that ean of tangent plane to sueface s cet (H0, 90, 80) is a (x-40 ) + 6(y-yo) +(3-30) = 0 where nemuel vectve û = (a,b,c) : .. ñ-i-4j+2k =) a=-1, b= -4, C=2 € Ho=3, Yo=1, 30=1 eq? of tangent plane is -1 (2-3) – 4 (7-1) +2 (3-1) = 0 - Je-44 +23+ 3 +4-2=0 -3e-44 +23+5= 0 bethy -23=55 tangent plane =) - Scanned with CamScanner