Chapter 4: Partial Differentiation
Section 4.5: Gradient Vector
The gradient of fx,y is the vector ∇f=fx i+fy j, whereas the gradient of fx,y,z is the vector ∇f=fx i+fy j+fz k. Table 5.4.1 lists the five most important properties of the gradient vector.
The gradients of fx,y are orthogonal to the level curves y=yx defined implicitly by fx,y=c, where c is a real constant.
The gradients of fx,y,z are orthogonal to the level surfaces z=zx,y defined implicitly by fx,y,z=c, where c is a real constant.
At any point where ∇f≠0, the gradient ∇f points in the direction of increasing values of f.
Where ∇f≠0, it necessarily points in the direction of maximal increase in f.
The maximal rate of change in f, as measured by the directional derivative, has the value ∇f.
Table 5.4.1 Properties of the gradient vector
Obtaining the Gradient in Maple
The Del (or Nabla) operator ∇ =i ∂∂x+j ∂∂y+k ∂∂z, applied to a scalar fx,y,z, results in the gradient vector ∇f=fx i+fy j+fz k. Maple has ∇, the Nabla symbol, in its Common Symbols palette, but it only works as an operator when one of the VectorCalculus packages is loaded.
With the Student VectorCalculus package loaded, the ∇-operator will correctly compute the gradient in a Cartesian frame for any expression in either two or three names of coordinate variables. Thus, ∇x+y would correctly return i+j, but applied to a+x, would treat both a and x as independent variables, and differentiate with respect to both. The Gradient command itself admits a list of names with respect to which differentiation is to take place, but this list cannot be made know to the ∇-operator without the use of the SetCoordinates command. Finally, note that the gradient is returned as a VectorField, a more complex data structure than a "free vector" such as x,y.
The Gradient command in the Student MultivariateCalculus package never links to the Nabla symbol, and always requires as an additional parameter, a list of the variables of differentiation. However, this list of names can be set equal to a list of coordinates, so that the gradient is computed at a specific point.
Of course, the gradient of, say, fx,y, could also be obtained by the construct ∂∂ x f,∂∂ y f.
With the Student MultivariateCalculus package loaded, the Context Panel, launched on an expression in two or three variables, provides interactive access to the Gradient command.
There is also a Gradient command in the Physics:-Vectors package, but it only acts on vectors defined within that package. No use is made of that package, or its structures, in this work.
Let fx,y=5−2 x2−3 y2 and let P be the point 1,2.
Obtain ∇f at P.
Graph the surface z=fx,y.
On the same set of axes, graph the level curve through P, and ∇f at P.
At P, show that ∇f is orthogonal to a vector tangent to the level curve through P.
At P, obtain ψ=∇f·u, the directional derivative of f in the direction u=cost i+sint j . Show that ψ is a maximum when u is along ∇fP and that this maximum is ∇fP2.
Let w=x2+2 y2+3 z2 and let P be the point 1,1,1.
Obtain ∇w at P.
On the same set of axes, graph the level surface w=6 and ∇w at P.
At P, show that ∇w is orthogonal to the level surface w=6. Hint: Show that this gradient is orthogonal to the x- and y-coordinate curves through P.
Prove Property 1 in Table 4.5.1.
Prove Property 2 in Table 4.5.1.
Prove Property 3 in Table 4.5.1.
Prove Property 4 in Table 4.5.1.
Prove Property 5 in Table 4.5.1.
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.
Show both graphically and analytically that the level curves of u=sin⁡x⁢cosh⁡y are orthogonal to the level curves of v=cos⁡x⁢sinh⁡y.
If u=x4+2⁢x2⁢y2+y4−1x2+y2+2⁢y+1⁢x2+y2−2⁢y+1 and v=4⁢x⁢yx2+y2+2⁢y+1⁢x2+y2−2⁢y+1, show both graphically and analytically that their level curves are mutually orthogonal.
At P:2,1, determine the maximal rate of change and its direction for fx,y=x yx2+y3.
At P:1,2,3, determine the maximal rate of change and its direction for fx,y,z=x2−2 y+5 zx+y z.
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