Chapter 2: Space Curves
Section 2.5: Principal Normal
Example 2.5.1
At x=2 on the graph of C, the curve defined by y=x2, compute N. Graph C, along with N2 and T2. Does N point towards the center of curvature?
Solution
Mathematical Solution
Write the position vector as R=xx2 so that R′=12 x and ρ=1+4 x2. Then
T=11+4 x212 x, dTds=21+4 x22−2 x1, κ=21+4 x23/2, N=11+4 x2−2 x1
Evaluating at x=2 gives T2=11714 and N2=117−41.
Note that N can be obtained from T by interchanging components and negating the first component to place N to the left of T.
The center of curvature for the point 2,4 is given by R2+N2/κ2, that is, by
24+117−41 173/22 = −3217/2
Hence, the center of curvature is to the left of the point 2,4.
In Figure 2.5.1(a), T2 is represented by the black arrow; and N2, by the green.
use Student:-VectorCalculus in module() local R,p1; R:=PositionVector([x,x^2]); p1:=PlotPositionVector(R,x=0..3, points=[2],normal,tangent, curveoptions=[scaling= constrained,labels=[x,y], size=[300,300]], tangentoptions=[width=.1], normaloptions=[width=.1]); print(p1); end module: end use:
Figure 2.5.1(a) Graph of C, T2,N2
Maple Solution - Interactive
Initialize
Tools≻Load Package: Student Vector Calculus
Loading Student:-VectorCalculus
Execute the BasisFormat command at the right, or use the task template.
BasisFormatfalse:
Define C as the position vector R
Enter the vector notation for C as per Table 1.1.1. Context Panel: Evaluate and Display Inline
Context Panel: Student Vector Calculus≻Conversions≻To Position Vector
Context Panel: Assign to a Name≻R
x,x2 = xx2→to position Vectorxx2→assign to a nameR
Obtain T2 and N2
Write R and press the Enter key.
Context Panel: Student Vector Calculus≻ Frenet Formalism≻Tangent Vector≻x
Context Panel: Student Vector Calculus≻ Normalize≻Euclidean
Context Panel: Evaluate at a Point≻x=2
Context Panel: Student Vector Calculus≻ Frenet Formalism≻Principal Normal≻x
Context Panel: Simplify≻Assuming Real
R
xx2
→tangent vector
12x
→2-normalize
14x2+12x4x2+1
→evaluate at point
171741717
→principal normal
−4x4x2+13224x2+132
−2xcsgn14x2+14x2+11csgn14x2+14x2+1
→assuming real
−2x4x2+114x2+1
−417171717
Construct Figure 2.5.1(a)
Control drag T2 and N2
Context Panel: Plots≻Arrow from point≻x=2,y=4
Write R Context Panel: Evaluate and Display Inline
Context Panel: Student Vector Calculus≻Conversions≻To List
Context Panel: Plots≻Plot Builder Set x∈0,3 Options: Constrained Scaling
Copy and paste the arrows onto the graph of C
1171741717→plot arrow
−4171711717→plot arrow
R = xx2→to listx,x2→
Maple Solution - Coded
Install the Student Vector Calculus package.
Use the BasisFormat command to set the display of vectors.
withStudent:-VectorCalculus:
Use the PositionVector command to define C as the position vector R.
R≔PositionVectorx,x2:
Use the PrincipalNormal command with the normalized option to obtain the general principal normal vector.
Use the eval command to obtain the principal normal vector at x=2.
N≔evalPrincipalNormalR,normalized,x=2:
Use the TangentVector command with the normalized option to obtain the general tangent vector along the curve.
Use the eval command to obtain the tangent vector at x=2.
T≔evalTangentVectorR,normalized,x=2:
Use the PlotPositionVector command to graph the curve along with the tangent and principal normal vectors at the single point x=2.
PlotPositionVectorR,x=0..3,points=2,normal,tangent,curveoptions=scaling=constrained,labels=x,y,size=300,300,tangentoptions=width=.1,normaloptions=width=.1
The principal normal indeed points towards the center of curvature. The components of N could be obtained by interchanging the components of T and negating the first component so that N points to the left of T.
<< Chapter Overview Section 2.5 Next Example >>
© Maplesoft, a division of Waterloo Maple Inc., 2024. All rights reserved. This product is protected by copyright and distributed under licenses restricting its use, copying, distribution, and decompilation.
For more information on Maplesoft products and services, visit www.maplesoft.com
Download Help Document
