thank you Find the tangent planes to the following surfaces at the given points: x2 + 2y2 + 3xz = 10, at point (1, 2. 1/3 ) y2 – x2 = 3. at point (1, 2, 8) xyz = 1, at point (1, 1. 1)
Answer 1
Let ,f(x,y,z)=x” + 2y’ +3xz-10 The surface gradient is, är =2x+0+32(1 )- 0 s , -(x+2y+30) ä 0 2 (2y)+0-0 左-(x2 +2y2 +3xz-10) Oz =0+0+ 3x(1)-0 =3x (EN ) = 3( 1)
The vector n normal to the given surface is (3,8,3 The equation of the plane is 9(-1)+86-2)+3 3x+8y +3z 3-16-1 0 3x+8y + 3z 20 a surface Consider the following equation of the Given point: (1,2.8) Let f(x,y,z) = y2-x2-3 The gradient of the surface is, ox L2.8) = 23-0-0 ( 4)( 1.23) = 2(2)
小匙2-/-3) oz 3 Oz oz oz = 0-0-0 0 2 12.8 The vector n normal to the given surface is (12.8) = (- The equation of the plane is, ) 2, 4, 0
-2(x-1) +4(y-2)-0(z-8)=0 -2x+ 2+4y-8 0 2x-4y +6 = 0 Consider the following surface equation nz = 1 Given point : ( 1.1,1) Let f(x,y,z) =nz-1 The surface gradient is, (nz-1 左 =yz (x)–(1) ox ax = yz (A)(LU) – 1 Oz Oz=xy-1-0 (ALLI)-1-1
The vector n is normal to the given surface is, n (Vfan (LLl) 1,1,1 The plane equation is, x1-1+1(z-1) 0 xtytz-30