Chapter 4: Partial Differentiation
Section 4.6: Surface Normal and Tangent Plane
Example 4.6.5
At P:a,b,c on the surface defined implicitly by fx,y,z=0, obtain an equation for the tangent plane in the form z=….
Solution
Mathematical Solution
If A is the position-vector representation of the point of contact of the plane on the surface; R, the generic position vector to the arbitrary point x,y,z; and N, a surface normal at the point of contact, then the vector equation for the tangent plane is R−A·N=0. This leads to
0
=xyz−abc·fx(a,b,c)fy(a,b,c)fz(a,b,c)
= fxa,b,c⋅x−a+fya,b,c⋅y−b+fza,b,c⋅z−c
an implicit form of the plane, more compactly written as
fxP⋅x−a+fyP⋅y−b+fzP⋅z−c=0
Maple Solution - Interactive
Guided by the Mathematical Solution, enter the appropriate equation, press the Enter key, and use the Context Panel's Solve option to isolate z. However, before Maple can isolate z, the symbol fz must be set as an Atomic Variable.
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
x,y,z−a,b,c·fxa,b,c,fya,b,c,f__za,b,c=0
x−a⁢fx⁡a,b,c+y−b⁢fy⁡a,b,c+z−c⁢f__z⁡a,b,c=0
→isolate for z
z=−x−a⁢fx⁡a,b,c−y−b⁢fy⁡a,b,cf__z⁡a,b,c+c
Tangent plane as first-degree Taylor polynomial
Alternatively, recognize the expression for tangent plane as the first-degree Taylor polynomial. Obtain this interactively via
Context Panel: Series≻Multivariate Taylor Polynomial. (Complete the resulting pop-up dialog as per Figure 4.6.5(a).)
Figure 4.6.5(a) Taylor Polynomial dialog
fx,y,z→Taylor polynomialf⁡a,b,c+D1⁡f⁡a,b,c⁢x−a+D2⁡f⁡a,b,c⁢y−b+D3⁡f⁡a,b,c⁢z−c
Maple Solution - Coded
An equation for a plane tangent to the surface fx,y,z=0 at the point a,b,c is given by the TangentPlane command in the Student VectorCalculus package.
Student:-VectorCalculus:-TangentPlanefx,y,z=0,x=a,y=b,z=c
∂∂a⁢f⁡a,b,c⁢x−a+∂∂b⁢f⁡a,b,c⁢y−b+∂∂c⁢f⁡a,b,c⁢z−c=0
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