Multi-Variable Calculus Quiz Solution

Quiz of February 23, 2001

Wednesday, March 7, 2001

The Question

Find the equation of the plane in R^{3} that is tangent at the point (−2, 0, −1) to the surface

x^{3}y + y^{3}z + z^{3}x − 2 = 0 .

Solution

One obtains the equation of a plane in R^{3} either by finding three points in the plane or by finding (1) a point in the plane and (2) a vector normal to the plane. In this case the relevant fact is:

Theorem. If f is a differentiable function in R^{3} and a = (a_{1}, a_{2}, a_{3}) is a point in R^{3}, then the vector (∇f)(a) that is the gradient of f evaluated at the point a is a normal vector at a to the plane that is tangent there to the level surface of f, i.e., the surface given by the equation

f(x, y, z) = f(a_{1}, a_{2}, a_{3}) .

In the case at hand the given surface is the level surface at a = (−2, 0, −1) of the function

f(x, y, z) = x^{3}y + y^{3}z + z^{3}x .

Its gradient at an arbitrary point (x, y, z) is

∇f = (\frac{∂f}{∂x} , \frac{∂f}{∂y} , \frac{∂f}{∂z}) = (3x^{2}y + z^{3}, x^{3} + 3y^{2}z, y^{3} + 3xz^{2}) .

Its value at (−2, 0, −1) is the vector

(−1, −8, −6) ,

which is, therefore, a normal to the required tangent plane. Thus, the tangent plane is the plane normal to that vector passing through the given point a = (−2, 0, −1). Its equation is

x + 8y + 6z + 8 = 0 .

[Processed from GELLMU to L^{A}TEX on Tue 6 Mar 2001 at 23:07:00 EST]