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Chapter 4: Partial Differentiation
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Section 4.5: Gradient Vector
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Essentials


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The gradient of $f\left(x\,y\right)$ is the vector $\nabla f\={f}_{x}\mathbf{i}plus;{f}_{y}\mathbf{j}$, whereas the gradient of $f\left(x\,y\,z\right)$ is the vector $\nabla f\={f}_{x}\mathbf{i}plus;{f}_{y}\mathbf{j}plus;{f}_{z}\mathbf{k}$. Table 5.4.1 lists the five most important properties of the gradient vector.
Property

Description

1

The gradients of $f\left(x\,y\right)$ are orthogonal to the level curves $y\=y\left(x\right)$ defined implicitly by $f\left(x\,y\right)\=c$, where $c$ is a real constant.

2

The gradients of $f\left(x\,y\,z\right)$ are orthogonal to the level surfaces $z\=z\left(x\,y\right)$ defined implicitly by $f\left(x\,y\,z\right)\=c$, where $c$ is a real constant.

3

At any point where $\nabla f\ne \mathbf{0}$, the gradient $\nabla f$ points in the direction of increasing values of $f$.

4

Where $\nabla f\ne \mathbf{0}$, it necessarily points in the direction of maximal increase in $f$.

5

The maximal rate of change in $f$, as measured by the directional derivative, has the value $\u2225\nabla f\u2225$.

Table 5.4.1 Properties of the gradient vector



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Obtaining the Gradient in Maple


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•

The Del (or Nabla) operator $\nabla equals;\mathbf{i}\frac{\partial}{\partial x}plus;\mathbf{j}\frac{\partial}{\partial y}plus;\mathbf{k}\frac{\partial}{\partial z}$, applied to a scalar $f\left(x\,y\,z\right)$, results in the gradient vector $\nabla f\={f}_{x}\mathbf{i}plus;{f}_{y}\mathbf{j}plus;{f}_{z}\mathbf{k}$. 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.

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•

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, $\nabla \left(x\+y\right)$ would correctly return $\mathbf{i}\+\mathbf{j}$, 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 $\u27e8x\,y\u27e9$.

•

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, $f\left(x\,y\right)$, could also be obtained by the construct $\u27e8\frac{\partial}{\partial x}fcomma;\frac{\partial}{\partial y}f\u27e9$.

•

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.

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•

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.

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Examples


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Example 4.5.1

Let $f\left(x\,y\right)\=52{x}^{2}3{y}^{2}$ and let P be the point $\left(1\,2\right)$.
a)

Obtain $\nabla f$ at P.

b)

Graph the surface $z\=f\left(x\,y\right)$.

c)

On the same set of axes, graph the level curve through P, and $\nabla f$ at P.

d)

At P, show that $\nabla f$ is orthogonal to a vector tangent to the level curve through P.

e)

At P, obtain $\mathrm{\ψ}\=\left(\nabla f\right)\xb7\mathbf{u}$, the directional derivative of $f$ in the direction $\mathbf{u}\=\mathrm{cos}\left(t\right)\mathbf{i}plus;\mathrm{sin}\left(t\right)\mathbf{j}$ . Show that $\mathrm{\ψ}$ is a maximum when u is along $\nabla f\left(\mathrm{P}\right)$ and that this maximum is ${\u2225\nabla f\left(\mathrm{P}\right)\u2225}^{2}$.


Example 4.5.2

Let $w\={x}^{2}\+2{y}^{2}plus;3{z}^{2}$ and let P be the point $\left(1\,1\,1\right)$.
a)

Obtain $\nabla w$ at P.

b)

On the same set of axes, graph the level surface $w\=6$ and $\nabla w$ at P.

c)

At P, show that $\nabla w$ is orthogonal to the level surface $w\=6$. Hint: Show that this gradient is orthogonal to the $x$ and $y$coordinate curves through P.


Example 4.5.3

Prove Property 1 in Table 4.5.1.

Example 4.5.4

Prove Property 2 in Table 4.5.1.

Example 4.5.5

Prove Property 3 in Table 4.5.1.

Example 4.5.6

Prove Property 4 in Table 4.5.1.

Example 4.5.7

Prove Property 5 in Table 4.5.1.

Example 4.5.8

Show both graphically and analytically that the level curves of $u\={x}^{2}{y}^{2}3xplus;2$ are orthogonal to the level curves of $v\=2xy3y$.

Example 4.5.9

Show both graphically and analytically that the level curves of $u\=\mathrm{sin}\left(x\right)\mathrm{cosh}\left(y\right)$ are orthogonal to the level curves of $v\=\mathrm{cos}\left(x\right)\mathrm{sinh}\left(y\right)$.

Example 4.5.10

If $u\=\frac{{x}^{4}\+2{x}^{2}{y}^{2}\+{y}^{4}1}{\left({x}^{2}\+{y}^{2}\+2y\+1\right)\left({x}^{2}\+{y}^{2}2y\+1\right)}$ and $v\=\frac{4xy}{\left({x}^{2}\+{y}^{2}\+2y\+1\right)\left({x}^{2}\+{y}^{2}2y\+1\right)}$, show both graphically and analytically that their level curves are mutually orthogonal.

Example 4.5.11

At P:$\left(2\,1\right)$, determine the maximal rate of change and its direction for $f\left(x\,y\right)\=\frac{xy}{{x}^{2}plus;{y}^{3}}$.

Example 4.5.12

At P:$\left(1\,2\,3\right)$, determine the maximal rate of change and its direction for $f\left(x\,y\,z\right)\=\frac{{x}^{2}2yplus;5z}{xplus;yz}$.



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