Compute the partials:
Now, try to find those points where the tangent plane is horizontal.
The points where the derivatives equal 0.
Solve for x,y, and z.
hi
Question:
find the points on the surfaces xy + yz + zx - x- z^2 = 0
where the tangent plane is parallel to the xy-plane?
my solution:
tangent line is parallel to xy-plane = normal line to surface orthognal to xy-plane
since normal line is orthognal to xy-plane that mean its intersect xy-plane
and the normal is gradient f
all what i will do is finding the parametric equation of the normal line and putting z=0
equation of normal line:
x = x0 + (y+z-1) t
y = y0 + (x+z) t
z = z0 + (y+x-2z) t
putting z=0
-z0 = (y+x-2z)t
and then ??!!
Edit:
i saw question like this .. but it was given a point at which the normal line starts from
i.e. x0,y0 and z0 are known ..
y+z-1 = 0 ... (1)
x+z = 0 ... (2)
x + y - 2z = 0 ... (3)
from (1) ---> y = -z + 1
from (2) ---> x = -z
substitute this values in (3)
-z - z + 1 - 2z = 0
-4z + 1 = 0
z = 1/4
x = -1/4
y = 1 - (1/4) = 3/4
point is ( -1/4 , 3/4 , 1/4 )
is this right?