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Chapter 4: Partial Differentiation
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Section 4.4: Directional Derivative
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Essentials


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The directional derivative of $f\left(x\,y\right)$ or $f\left(x\,y\,z\right)$ is the rate of change of $f$ in a given direction specified by a unit vector u.

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A common notation for the directional derivative at point P and in the direction u is ${\mathrm{D}}_{\mathbf{u}}f\left(P\right)$.

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When the directional derivative is known to exist, it can be computed as the dot product of the vector u and the vector

$\nabla f\={f}_{x}\mathbf{i}plus;{f}_{y}\mathbf{j}plus;{f}_{z}\mathbf{k}$
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called the gradient vector, to be studied at length in Section 4.5.
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Conceptually, the directional derivative is obtained as the rate of change of the values of $f$ along a line that is through point P and that has direction u. Since the line is parametrized by a single parameter such as $t$, the function values along this line are similarly parametrized, so the rate of change of

$w\left(t\right)\=f\left(x\left(t\right)\,y\left(t\right)\right)$ or $w\left(t\right)\=f\left(x\left(t\right)\,y\left(t\right)\,z\left(t\right)\right)$
is taken as the directional derivative of $f$. In particular, if the line is parametrized so that point P corresponds to $t\=0$, then the directional derivative at P is simply $w\prime \left(0\right)$. See Example 4.4.7 for an implementation of this concept.
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A Caution


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A standard example or exercise for the directional derivative is to give a function, a point P and a direction vector u, and ask for ${\mathrm{D}}_{\mathbf{u}}f\left(P\right)$. In an attempt to add "interest," a common variant is to give a base point P and a second point Q towards which the rate of change is to be determined. If the vector from P to Q is called v, then the direction vector is $\mathbf{u}\=\mathbf{v}\/\u2225\mathbf{v}\u2225$. Being warned of this "wrinkle" allows the student to avoid the error of interpreting Q as the direction vector u itself.
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The Directional Derivative in Maple


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The Student MultivariateCalculus package contains the DirectionalDerivative command whose arguments can be the function $f$, a list of its variables, and a vector (given as a list of components) in the appropriate direction. In this form, the command will automatically normalize the vector and return the generic directional derivative for an arbitrary point.

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Alternatively, to obtain the directional derivative at a specific point P:$\left(a\,b\,c\right)$, modify the list of variables by equating the list of variables to the list $\left[a\,b\,c\right]$.

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For functions of two variables, this command can return a graph that purports to show the surface, a plane tangent to the surface, the direction vector, and the direction vector scaled and projected onto the tangent plane. This visualization is more readily obtained via the
tutor.

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There is a DirectionalDiff command in the Student VectorCalculus package, and a DirectionalDiff command in the Physics:Vectors package. The command in the VectorCalculus package will not automatically normalize the direction vector, and makes no provision for evaluation at a point. The command in the Physics:Vectors package does not require normalization of the direction vector, but requires the complete machinery of vectors within this package. Neither of these two commands will be explored further.

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With the Student MultivariateCalculus package loaded, the Context Panel, launched on an expression in two or three variables, provides interactive access to the DirectionalDerivative command.



Examples


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Example 4.4.1

At the point P:$\left(1\,1\,0\right)$, and in the direction $\mathbf{v}\=2\mathbf{i}3\mathbf{j}plus;6\mathbf{k}$, obtain the directional derivative of $f\left(x\,y\,z\right)\={x}^{3}x{y}^{2}z$.

Example 4.4.2

At the point P:$\left(1\,1\,0\right)$, and in the direction of the point Q:$\left(2\,3\,6\right)$, obtain the directional derivative of $f\left(x\,y\,z\right)\={x}^{3}x{y}^{2}z$.

Example 4.4.3

At the point P:$\left(2\,0\,3\right)$, and in the direction $\mathbf{v}\=4\mathbf{i}\mathbf{plus;}5\mathbf{j}2\mathbf{k}$, obtain the directional derivative of $f\left(x\,y\,z\right)\=x{e}^{y}plus;y{z}^{2}$.

Example 4.4.4

At the point P:$\left(2\,0\,3\right)$, and in the direction of the point Q:$\left(4\,5\,2\right)$, obtain the directional derivative of $f\left(x\,y\,z\right)\=x{e}^{y}plus;y{z}^{2}$.

Example 4.4.5

At the point P:$\left(1\,2\,3\right)$, and in the direction $\mathbf{v}\=3\mathbf{i}7\mathbf{j}plus;5\mathbf{k}$, obtain the directional derivative of $f\left(x\,y\,z\right)\={e}^{\mathrm{sin}\left(yz\right)}\mathrm{cosh}\left(\frac{xy}{zy}\right)$.

Example 4.4.6

At the point P:$\left(1\,2\,3\right)$, and in the direction of the point Q:$\left(3\,7\,5\right)$, obtain the directional derivative of $f\left(x\,y\,z\right)\={e}^{\mathrm{sin}\left(yz\right)}\mathrm{cosh}\left(\frac{xy}{zy}\right)$.

Example 4.4.7

From first principles, obtain the directional derivative of $f\left(x\,y\,z\right)$ at the generic point $\left(a\,b\,c\right)$ and in the arbitrary direction $\mathbf{u}\=p\mathbf{i}plus;q\mathbf{j}plus;r\mathbf{k}$, where u is a unit vector.

Example 4.4.8

At point P, the directional derivative of $g$ in the direction $\mathbf{u}\=2\mathbf{i}3\mathbf{j}$ is 8, but in the direction $\mathbf{v}\=5\mathbf{i}plus;4\mathbf{j}$, it's $6$. Find the directional derivative of $g$ in the direction $\mathbf{w}\=7\mathbf{i}2\mathbf{j}$.



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