Chapter 4: Partial Differentiation
Section 4.5: Gradient Vector
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.
At P:1,2,3, the direction of the maximal rate of change of f is ∇fP/∥∇fP∥ = 2525⁢105−221⁢10511525⁢105.
The maximal rate of change itself is ∇fP=549⁢105.
Maple Solution - Interactive
Tools≻Load Package: Student Multivariate Calculus
Control-drag the expression defining f.
Context Panel: Student Multivariate Calculus≻
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→gradient249−50491149→select entry 1249−50491149→assign to a nameGfP
Write the name GfP.
Context Panel: Evaluate and Display Inline
Context Panel: Normalize≻Euclidean
GfP = 249−50491149→Euclidean-normalize2525⁢105−221⁢10511525⁢105
Context Panel: Norm≻Euclidean
GfP = 249−50491149→Euclidean-norm549⁢105
Maple Solution - Coded
Install the Student MultivariateCalculus package.
f≔x2−2 y+5 zx+y z:
Use the Gradient command.
GfP≔Gradientf,x,y,z=1,2,3 = 249−50491149
Apply the Normalize command.
NormalizeGfP = 2525⁢105−221⁢10511525⁢105
Apply the Norm command.
NormGfP = 549⁢105
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