Advanced Calculus using Mathematica: NoteBook Edition is a complete text on calculus of several variables written in Mathematica NoteBooks.

Implicit 3D: Procedure 4.5.1

Procedure  4.5.1: Finding the Implicit Tangent Plane

Suppose f[x,y,z] is smooth in a neighborhood of a particular fixed X=X(x,y,z). To find the equation of the plane tangent to the implicit contour surface at X=X(x,y,z), for constant, do the following steps:

1.  Compute the general symbolic partial derivatives , , and .  Verify that the formulas f[x,y,z], , , and are valid in a box about our particular point of tangency, X(x,y,z) and write the symbolic total differential of the equation

(the partials are functions)

The right hand side is zero since the partial derivatives of a constant are zero.

2.  Calculate the specific values of the partial derivatives at the particular X(x,y,z), , , and , verify that they are not all zero, and write the specific total differential at this point:

p·d x+q·d y+r·d z=0 or G dX =0 (the coefficients are numbers)

3.  If you want the equation of the tangent plane in (x,y,z)-coordinates, say is your particular point of tangency. The local  (d x,d y,d z)-coordinates are related to (x,y,z)-coordinates by , , , so replace  d x,  d y, d z  with , , obtaining the equation:

The result of Step 2 an implicit plane through the (d x, d y, d z)-origin with perpendicular f[x,y,z].  You can plot the tangent plane using this equation in local coordinates  (d x, d y, d z) centered at the particular (x,y,z).  The equation says, “An unknown vector d X =(d x, d y, d z) (with origin at the point of tangency X(x,y,z)) lies on the tangent line provided it is perpendicular to the particular gradient G(p,q,r).”

The result of Step 3 is an implicit plane through the point with perpendicular .  This equation says, “An unknown vector X (with origin at (x,y,z)=(0,0,0)) lies on the tangent line provided its displacement from is perpendicular to the particular gradient G(p,q,r).”  Notice that to correctly evaluate this final expression, you need to perform steps (1), (2), and (3) in that order.

When G=0, the implicit equation does not define a plane.  In this case we may not have a tangent.  We take this up in Chapter 5 on The Implicit Function Theorem.  The rigorous justification for the procedure is the Implicit Function Theorem for one equation in 3 unknowns.

Example

Find the equation tangent to at

Solution

First, the function is smooth at all (x,y,z) by the 3-variable extension of the Theorem on Smooth Formulas.

The general symbolic total differential of the equation is

At the particular point , the differential is

⇔

The plane through perpendicular to

Figure 4.5.3: The Plane Tangent to an Ellipsoid