Question

Given f(x, y): 10-2x2-y, find a) The equation of the tangent plane to the surface at the point (2.2,6) b) The parametric equations of the normal line at the point (2, 2, 6) c) The outward unit normal vector to the surface at the point (2, 2,6) d) Sketch the surface and the outward unit normal vector at the point (2, 2,6). 1.

Answer 1

Consider the following equation f (x,y)-10-2x - point (2,2,6 (x.ozo) y, point- The tangent equation of the plane to the surface z 'f(x y) at the point (x0. У0,20 is, The partial derivatives of the function are, A(x,y)--(10-2x2-y?) =-2(2x) Treat all the other terms are as constants except x =-4x f (2,2)-4(2) 4(x,y)- (10-2x2-3,2) ,(2.2)-20) су v 2y Treat all the other terms are as constants except y So, the equation of the tangent plane at (2, 2,6)is, 6-8x+16-4y+8 --8x-4y+30 8x+4y +z 30

Let z=f(x,y)-10-2x2-y2 Take f (x,y,z)-10-2x -y -z The gradient of the function f (x,y,z)10-2x y-zis, 으( 10 = = (-4x-2y,-1) Vf(2,2,6)-(-4(2).-2(2),-1 The parametric equation of the normal line at the point (2, 2,6 is, = 2+(-8t) x(t) = 2-8t 2+(-4)t = 64(-1)t Therefore, the parametric equation is, (x(t).(t).z()) (2-8t,2-4t,6-t)

The outward unit normal vector is, Vf Here, Vf (-8,-4,-1) That is Vf n- 64+16+1 81

The sketch of the outward unit normal and the surface is shown below z= 8.25 z= 4.5 z= 0.75 y=-1 y=0 y=1