Question

b) Consider the surface in R3 described by f(x,y,z) = 2x²y3 + z + ye*2 = 9 (i) Find Vf(x,y,z). [3 marks] (ii) Verify that (2,1,0) is a point on this surface. Find the cartesian equation of the tangent plane to this surface at the point (2.1,0). [5 marks]

Answer 1

To find the gradient of the surface we first need to rearrange the equation and write everything in one side.

Now

let it is a new fuction f1(x,y,z) then find the gradient and do all asked calculations.

See below attached complete solution.

ato =0+1+ yne" solution or x2 Given surface t(ny, 2) = 2x²y3 +2+ yena=g * writing in bevel surface fo omate. n21 ti (2,5,2) = 2x?+ 3+ yen3 (i) of (1,9, 2) = gradient a normal to the surfeface. de ato v 6, 614,3) tato it a to ay an 22 We have ato = xy²toty z ek? от C 12 ate= 6x²y² tot he oy ena 2 Hence normal to surface = 5+(99,9) is na v to 10,53) = juny2+42 km2) i +ku+gº+exa); + 1 ++ x^^) (ii) if point (2,1,0) is on the surface it must Satisfy the equation for surface. putting (2, 1,0) in fx,y,z) Any 2) = ? x²y3+2+ yenų -9 4,12,1,1) = 2x22x1 totie-9-8-850 Hence satisfied a verified it lês on surface.

is sit 259+ 3 Now, finding squation of tangent plane, at 12, 1,0) first find ot, (x,y,z) at (2,1,0) which normal to the surface. V6 (1,1, 2)= (4 137+92 €47) f+ (5x+y?+212) 3+(1+4x242) R => 86, (2,1,0) = (8+0) i+(24+0) 3 +11+2) & => ot, 2,16) = now, because of 121,0) is normal to the tangent plane also it must be orthogonal to every Vector in the tangent plane hence v6,6,10){(x-2)+ (9-4)3+ta-o)#') = 0 (3 i +2 53+3ť). [(x-2) +(9-13 +(2-0) f g = 0 86-21+2514-1)+32=0 sn't254 +32 = 41 It is required we have a) equation of tangent plane.