Chapter 4: Partial Differentiation
Section 4.6: Surface Normal and Tangent Plane
Example 4.6.1
At P:2,−3 on the surface defined by z=fx,y≡5−x2/3−y2/2, obtain and draw both the normal and tangent plane.
Solution
Mathematical Solution
Figure 4.6.1(a) shows the surface in green, the tangent plane at Q:2,−3,−5/6 in red, and the normal at this point in black.
According to Table 4.6.1, N is obtained by evaluating −fx i−fy j+k at x,y=2,−3, yielding
N=4/3−31
The tangent plane is then given vectorially by
R−Q·N=0
use plots, Student:-VectorCalculus in module() local P,N,p1,p2,p3,p4; N:=RootedVector(root=[2,-3,-5/6],<4/3,-3,1>); P:=65/6-4*x/3+3*y; p1:=PlotVector(N,color=black,width=.4); p2:=plot3d(P,x=0..4,y=-5..5,style=surface,color=red,transparency=.3); p3:=plot3d(5-x^2/3-y^2/2,x=-4..4,y=-5..5,style=surface,color=green); p4:=display(p1,p2,p3,scaling=constrained,axes=frame,view=-3..5,labels=[x,y,z],tickmarks=[4,8,4],orientation=[-120,75,0]); print(p4); end module: end use:
Figure 4.6.1(a) Surface, normal, and tangent plane
and then by
0
=xyz−2−3−5/6·4/3−31
=43x−2−3y+3+z+5/6
=4 x3−3 y+z−83−9+56
=4 x3−3 y+z−656
Maple Solution - Interactive
A complete solution is available with the Student MultivariateCalculus package.
Let Q be the point on the surface that corresponds to the point x,y=2,−3.
Initialize
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Context Panel: Assign Function
fx,y=5−x2/3−y2/2→assign as functionf
Context Panel: Assign Name
Q=2,−3,f2,−3→assign
Obtain a surface normal at point Q
Matrix palette: Template for a 3-component vector
Calculus palette: Partial differentiation operator
Context Panel: Evaluate and Display Inline
Context Panel: Evaluate at a Point≻x=2,y=−3 (See Figure 4.6.1(b).)
Context Panel: Assign to a Name≻N
Figure 4.6.1(b) Dialog: Evaluate at a Point
−∂∂ x f(x,y)−∂∂ y f(x,y)1 = →evaluate at point →assign to a nameN
Obtain an equation for the tangent plane
Write a sequence of names for the point and normal that define the tangent plane.
Context Panel: Student Multivariate Calculus≻Lines & Planes≻Plane
Context Panel: Student Multivariate Calculus≻Lines & Planes≻Representation
Q,N→make plane<< Plane 1 >>→representation43x−3y+z=656
Maple also supports a solution from first principles.
Convert Q to the position vector A
Write the name for point Q.
Context Panel: Conversions≻Column Vector
Context Panel: Assign to a Name≻A
Q = 2,−3,−56→to Vector →assign to a nameA
Define the generic position vector R and implement the vector equation of a plane
R=x,y,z→assign
Write the vector equation of the plane that has normal N and passes through point A.
Press the Enter key.
R−A·N=0
43x−656−3y+z=0
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define the function f.
f≔x,y→5−x2/3−y2/2:
Obtain a vector normal to the surface
Use the diff and eval commands to obtain partial derivatives evaluated on the surface.
N≔eval−difffx,y,x,−difffx,y,y,1,x=2,y=−3
Use the Plane command to generate the plane data-structure.
Use the GetRepresentation command to extract the equation of the tangent plane.
GetRepresentationPlane2,−3,f2,−3,N
43x−3y+z=656
The tangent plane can also be obtained via the TangentPlane command in the Student VectorCalculus package.
Student:-VectorCalculus:-TangentPlanefx,y,x=2,y=−3 =
The plane is given in the form of a position vector, where the third component is interpreted as the equation z=−43x+656+3y.
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