Multivariate Calculus - Bruce E. Shapiro

96 LECTURE 13. TANGENT PLANES
Solution. In our discussion of the equation of a plane in section 14.2 we found that
when a plane is written in the form
ax + by + cz = d
then its normal vector is n = (a, b, c). Hence we want to find a point on the surface
of z = 2x 2 + 3y 2 where the normal vector is parallel to
n = (8, −3, −1)
Rewriting equation 13.3 in the form F (x, y, z) = 0,
F (x, y, z) = 2x 2 + 3y 2 − z = 0
A general form of the normal vector is given by the gradient vector
∇F (x, y, z) = (4x, 6y, −1)
Since the two gradient vectors are parallel
(4x, 6y, −1) = k(8, −3, −1)
for some number k. The solution is k = 1 (which we find from the z-component)
and hence x = 2 and y = −1/2. Thus the point on the surface where the normal
vector has the right direction is
(x, y, z) = (2, −1/2, f(2, −1/2)) = (2, −1/2, 8.75)
Revised December 6, 2006. Math 250, Fall 2006