Chapter 4: Partial Differentiation
Section 4.5: Gradient Vector
Example 4.5.12
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.
Solution
Mathematical Solution
At P:1,2,3, the direction of the maximal rate of change of f is ∇fP/∥∇fP∥ = 2525105−22110511525105.
The maximal rate of change itself is 5x2+5−2x−x2+1.
Maple Solution - Interactive
Initialize
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Obtain ∇fP
Control-drag the expression defining f.
Context Panel: Student Multivariate Calculus≻ Differentiate≻Gradient Complete the Gradient dialog as shown in Figure 4.5.12(a).
Context Panel: Select Element≻1
Context Panel: Assign to a Name≻GfP
Figure 4.5.12(a) Gradient dialog
x2−2 y+5 zx+y z→gradient →select entry 1 →assign to a nameGfP
Obtain ∇fP/∥∇fP∥
Write the name GfP. Context Panel: Evaluate and Display Inline
Context Panel: Normalize≻Euclidean
GfP = →Euclidean-normalize
Context Panel: Norm≻Euclidean
GfP = →Euclidean-norm549105
Maple Solution - Coded
Install the Student MultivariateCalculus package.
Define f.
f≔x2−2 y+5 zx+y z:
Calculate ∇fP
Use the Gradient command.
GfP≔Gradientf,x,y,z=1,2,3 =
Calculate ∇fP/∥∇fP∥
Apply the Normalize command.
NormalizeGfP =
Apply the Norm command.
NormGfP = 549105
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