We have learned in class that for the graph of function z = f(x, y), the tangent

Question

We have learned in class that for the graph of function z = f(x, y), the tangent plane at point (x_0, y_0, z_0) is z- z_0 = f_x(x_0, y_0){x - x_0) + f_y(x_0, y_0)(y - y_0). For example, for function 2 = f{x, y) where f(x, y) = 4 - x^2 - 2y^2, since f_x(1, 1) = -2 and f_y(1, 1) = -4, we conclude that its tangent plane at point (1, 1, 1) is z - 1 = -2(x - 1) - 4 (y - 1). In this homework, we study the tangent plane from one other perspective. Consider the function F(x, y, z) = x^2 + 2y^2 + z - 4. Compute a scalar equation of the plane that passes through (1, 1, 1) with normal vector nabla F(1, 1, 1). This plane is indeed the tangent plane of the surface F(x, y, 2) = 0 at (1, 1, 1). Compute a symmetric equation of the line that passes through (1, 1, 1) and is parallel to the vector nabla F(1, 1, 1). This line is called the normal line of the surface F(x, y, z) = 0 at (1, 1, 1).

Explanation / Answer

1)F(x,y,z)=x2+2y2+z-4

F=<2x,4y,1>

at (1,1,1)

F=<2,4,1>

F.<x-1,y-1,z-1>=0

<2,4,1>.<x-1,y-1,z-1>=0

2(x-1)+4(y-1)+1(z-1)=0

2x-2+4y-4+z-1=0

2x+4y+z=7

equation of tangent plane is 2x+4y+z=7

2)vector equation of normal line is

r(t)=(1,1,1)+t<2,4,1>

r(t)=<1+2t,1+4t,1+t>

x=1+2t , y=1+4t ,z=1+t

(x-1)/2 =(y-1)/4 =(z-1) is symmetric equation of normal line

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