Chapter 4: Partial Differentiation
Section 4.5: Gradient Vector
Show both graphically and analytically that the level curves of u=x2−y2−3 x+2 are orthogonal to the level curves of v=2⁢x⁢y−3 y.
Figure 4.5.8(a) shows the level curves of u in black, and the level curves of v in red. From the figure, it appears that these sets of level curves intersect at right angles.
Analytically, the orthogonality of these sets of level curves is shown by the computation
∇u·∇v=2 x−3−2 y·2 y2 x−3=0
which shows that the respective gradient vectors are orthogonal. Since these gradient vectors are themselves orthogonal to vectors tangent to the level curves, the sets of level curves are then orthogonal to each other.
use plots in
Figure 4.5.8(a) Level curves of u and v
Maple Solution - Interactive
Tools≻Load Package: Student Multivariate Calculus
Control-drag expressions for each of u and v.
Context Panel: Student Multivariate Calculus≻Differentiate≻Gradient
Context Panel: Assign to a Name≻Gu or Gv, as appropriate
x2−y2−3 x+2→gradient2⁢x−3−2⁢y→assign to a nameGu
2⁢x⁢y−3 y→gradient2⁢y2⁢x−3→assign to a nameGv
Common Symbols palette: Dot product operator
Context Panel: Evaluate and Display Inline
Gu·Gv = 0
Use the Plot Builder to generate Figure 4.5.8(a)
Write the rule for u.
Context Panel: Plot Builder≻2-D contour plot
Basic Options≻color 1≻black
Write the rule for v.
Basic Options≻color 1≻red
Copy the graph on the left and paste it onto the one on the right.
Maple Solution - Coded
Install the Student VectorCalculus package.
Define u and v.
u,v≔x2−y2−3 x+2,2 x y−3 y:
Use the Gradient and DotProduct commands.
DotProductGradientu,x,y,Gradientv,x,y = 0
Obtain Figure 4.5.8(a)
The contourplot and display commands from the plots package will produce Figure 4.5.8(a), as illustrated by the following code. Simply remove the colon from the last line and execute.
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