Chapter 9: Vector Calculus
Section 9.7: Conservative and Solenoidal Fields
Example 9.7.6
For F=2⁢x⁢z+y2 i+2 x y−z3 j+x2−3⁢y⁢z2 k, show that the line integral along any member of the family of curves x=t,y=tk,z=t2 k,t∈0,1, k≥1, has the value 1.
Solution
Mathematical Solution
The slider in Figure 9.7.6(a) varies the value of k over the range 1≤k≤25, and the curve thereby shown is a member of the given family of curves.
The astute reader will realize that for k=0, the extended family would not pass through the origin. Indeed, the general position vector would become
t i+j+k
which means that this particular curve would pass through 0,1,1 at t=0.
Writing the members of the family via the position vector
r=ttkt2 k
Explore(plots[:-spacecurve]([t,t^k,t^k],t=0..1,labels=[t,"",""],color=red,thickness=2,labels=[x,y,z],orientation=[-35,80,0],tickmarks=[2,2,2]),':-parameters'=[k=1.0..25.],':-initialvalues'=[k=3.0]);
k
Figure 9.7.6(a) The family of curves
the integral of F·dr becomes, after a simplification,
∫01−7⁢k⁢t7⁢k−1+2⁢k⁢t2⁢k+1+2⁢t2⁢k⁢k+2⁢t2⁢k+1+t2⁢k dt
Under the assumption k≥1, this integral evaluates to 1.
Maple Solution - Interactive
Table 9.7.6(a) provides a solution in which the LineInt command is accessed via the Context Panel.
Initialize
Tools≻Load Package: Student Vector Calculus
Loading Student:-VectorCalculus
Tools≻Tasks≻Browse: Calculus - Vector≻ Vector Algebra and Settings≻ Display Format for Vectors
Press the Access Settings button and select "Display as Column Vector"
Display Format for Vectors
Define the vector field F
Enter the components of F in a free vector. Context Panel: Evaluate and Display Inline
Context Panel: Student Vector Calculus≻Conversions≻To Vector Field
Context Panel: Assign to a Name≻F
2 x z+y2,2 x y−z3,x2−3 y z2 = 2⁢x⁢z+y2−z3+2⁢x⁢y−3⁢y⁢z2+x2→to Vector Field2⁢x⁢z+y2−z3+2⁢x⁢y−3⁢y⁢z2+x2→assign to a nameF
Form and evaluate the line integral of F along curves of the family
Write the name F and press the Enter key. Context Panel: Student Vector Calculus≻Line Integral Complete the dialog as per Figure 9.7.6(b)
Context Panel: Simplify≻Simplify
Context Panel: Evaluate Integral
Context Panel: Simplify≻Assuming Real Range Complete the dialog as per Figure 9.7.6(c).
Figure 9.7.6(b) Line Integral Domain dialog
Figure 9.7.6(c) RealRange
F
→line integral
∫012⁢t⁢t2⁢k+tk2+−t2⁢k3+2⁢t⁢tk⁢tk−1⁢k+2⁢−3⁢tk⁢t2⁢k2+t2⁢t2⁢k−1⁢kⅆt
= simplify
∫01−7⁢k⁢t7⁢k−1+2⁢k⁢t2⁢k+1+2⁢t2⁢k⁢k+2⁢t2⁢k+1+t2⁢kⅆt
=
limt→0+t7⁢k−t2⁢k+2−t2⁢k+1+1
→assuming real range
1
Table 9.7.6(a) Solution via the LineInt command accessed from the Context Panel
Maple Solution - Coded
Table 9.7.6(b) provides a solution via the LineInt command.
Install the Student VectorCalculus package.
withStudent:-VectorCalculus:
Set display of vectors via BasisFormat.
BasisFormatfalse:
Define F via the VectorField command.
F≔VectorField2 x z+y2,2 x y−z3,x2−3 y z2:
Apply the simplify command to the integral returned by the LineInt command.
Evaluate the line integral under an appropriate assumption on k.
simplifyLineIntF,Patht,tk,t2 k,t=0..1,output=integral
LineIntF,Patht,tk,t2 k,t=0..1 assuming k≥1 = 1
Table 9.7.6(b) Solution via the LineInt command
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