Homework Soursefind the tangent vector at 214 that is parallel to the yz pl
find the tangent vector at (2,1,4) that is parallel to the yz plane . let z=xy^2+xy^3
Solution
since z = xy2 + xy3
=> F(x,y,z) = xy2 + xy3 - z
then Grad F = < y2+y3, 2xy+3xy2, -1> ...[<derivative wrt x, wrt y, wrt z> ]
At the point (2,1,4), the normal vector is
Grad F(2,1,4) = <2, 10, -1>
Now use the point normal formula
<2, 10, -1> . <x-2, y-1, z-4> = 0
=> 2(x-2) +10(y-1) -1(z-4) = 0
=> 2x+10y-z = 10
| Let F(x,y,z) define a surface that is differentiable at a point (x0,y0,z0), thenthe tangent plane to F ( x, y, z ) at ( x0 , y0 , z0 ) is the plane with normal vector Grad F(x0,y0,z0) that passes through the point (x0,y0,z0). In Particular the equation of the tangent plane is Grad F(x0,y0,z0) . < x - x0 , y - y0 , z - z0 > = 0 |
