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 (\nabla 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

\nabla f = ({\partial f}/{\partial x}, {\partial f}/{\partial y}, {\partial f}/{\partial z}) = (3 x^{2} y + z^{3}, x^{3} + 3 y^{2} z, y^{3} + 3 x z^{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 + 8 y + 6 z + 8 = 0 .

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