4.6 Directional derivatives and the gradient

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• Determine the directional derivative in a given direction for a function of two variables.
• Determine the gradient vector of a given real-valued function.
• Explain the significance of the gradient vector with regard to direction of change along a surface.
• Use the gradient to find the tangent to a level curve of a given function.
• Calculate directional derivatives and gradients in three dimensions.

In Partial Derivatives we introduced the partial derivative. A function $z=f\left(x,y\right)$ has two partial derivatives: $\partial z\text{/}\partial x$ and $\partial z\text{/}\partial y.$ These derivatives correspond to each of the independent variables and can be interpreted as instantaneous rates of change (that is, as slopes of a tangent line). For example, $\partial z\text{/}\partial x$ represents the slope of a tangent line passing through a given point on the surface defined by $z=f\left(x,y\right),$ assuming the tangent line is parallel to the x -axis. Similarly, $\partial z\text{/}\partial y$ represents the slope of the tangent line parallel to the $y\text{-axis.}$ Now we consider the possibility of a tangent line parallel to neither axis.

Directional derivatives

We start with the graph of a surface defined by the equation $z=f\left(x,y\right).$ Given a point $\left(a,b\right)$ in the domain of $f,$ we choose a direction to travel from that point. We measure the direction using an angle $\theta ,$ which is measured counterclockwise in the x , y -plane, starting at zero from the positive x -axis ( [link] ). The distance we travel is $h$ and the direction we travel is given by the unit vector $u=\left(\text{cos}\phantom{\rule{0.2em}{0ex}}\theta \right)i+\left(\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right)j.$ Therefore, the z -coordinate of the second point on the graph is given by $z=f\left(a+h\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta ,b+h\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right).$ Finding the directional derivative at a point on the graph of z = f ( x , y ) . The slope of the black arrow on the graph indicates the value of the directional derivative at that point.

We can calculate the slope of the secant line by dividing the difference in $z\text{-values}$ by the length of the line segment connecting the two points in the domain. The length of the line segment is $h.$ Therefore, the slope of the secant line is

${m}_{\text{sec}}=\frac{f\left(a+h\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta ,b+h\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right)-f\left(a,b\right)}{h}.$

To find the slope of the tangent line in the same direction, we take the limit as $h$ approaches zero.

Definition

Suppose $z=f\left(x,y\right)$ is a function of two variables with a domain of $D.$ Let $\left(a,b\right)\in D$ and define $\text{u}=\text{cos}\phantom{\rule{0.2em}{0ex}}\theta i+\text{sin}\phantom{\rule{0.2em}{0ex}}\theta j.$ Then the directional derivative    of $f$ in the direction of $u$ is given by

${D}_{u}f\left(a,b\right)=\underset{h\to 0}{\text{lim}}\frac{f\left(a+h\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta ,b+h\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right)-f\left(a,b\right)}{h},$

provided the limit exists.

[link] provides a formal definition of the directional derivative that can be used in many cases to calculate a directional derivative.

Finding a directional derivative from the definition

Let $\theta =\text{arccos}\left(3\text{/}5\right).$ Find the directional derivative ${D}_{u}f\left(x,y\right)$ of $f\left(x,y\right)={x}^{2}-xy+3{y}^{2}$ in the direction of $\text{u}=\left(\text{cos}\phantom{\rule{0.2em}{0ex}}\theta \right)i+\left(\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right)j.$ What is ${D}_{\text{u}}f\left(-1,2\right)?$

First of all, since $\text{cos}\phantom{\rule{0.2em}{0ex}}\theta =3\text{/}5$ and $\theta$ is acute, this implies

$\text{sin}\phantom{\rule{0.2em}{0ex}}\theta =\sqrt{1-{\left(\frac{3}{5}\right)}^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}.$

Using $f\left(x,y\right)={x}^{2}-xy+3{y}^{2},$ we first calculate $f\left(x+h\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta ,y+h\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right)\text{:}$

$\begin{array}{cc}\hfill f\left(x+h\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta ,y+h\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right)& ={\left(x+h\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta \right)}^{2}-\left(x+h\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta \right)\left(y+h\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right)+3{\left(y+h\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right)}^{2}\hfill \\ & ={x}^{2}+2xh\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta +{h}^{2}{\text{cos}}^{2}\theta -xy-xh\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta -yh\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta \hfill \\ & \phantom{\rule{0.5em}{0ex}}{\mathit{\text{−h}}}^{2}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta +3{y}^{2}+6yh\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta +3{h}^{2}{\text{sin}}^{2}\theta \hfill \\ & ={x}^{2}+2xh\left(\frac{3}{5}\right)+\frac{9{h}^{2}}{25}-xy-\frac{4xh}{5}-\frac{3yh}{5}-\frac{12{h}^{2}}{25}+3{y}^{2}\hfill \\ & \phantom{\rule{0.5em}{0ex}}+6yh\left(\frac{4}{5}\right)+3{h}^{2}\left(\frac{16}{25}\right)\hfill \\ & ={x}^{2}-xy+3{y}^{2}+\frac{2xh}{5}+\frac{9{h}^{2}}{5}+\frac{21yh}{5}.\hfill \end{array}$

We substitute this expression into [link] :

$\begin{array}{cc}\hfill {D}_{u}f\left(a,b\right)& =\underset{h\to 0}{\text{lim}}\frac{f\left(a+h\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta ,b+h\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right)-f\left(a,b\right)}{h}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{\left({x}^{2}-xy+3{y}^{2}+\frac{2xh}{5}+\frac{9{h}^{2}}{5}+\frac{21yh}{5}\right)-\left({x}^{2}-xy+3{y}^{2}\right)}{h}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{\frac{2xh}{5}+\frac{9{h}^{2}}{5}+\frac{21yh}{5}}{h}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{2x}{5}+\frac{9h}{5}+\frac{21y}{5}\hfill \\ & =\frac{2x+21y}{5}.\hfill \end{array}$

To calculate ${D}_{u}f\left(-1,2\right),$ we substitute $x=-1$ and $y=2$ into this answer:

$\begin{array}{cc}\hfill {D}_{\text{u}}f\left(-1,2\right)& =\frac{2\left(-1\right)+21\left(2\right)}{5}\hfill \\ & =\frac{-2+42}{5}\hfill \\ & =8.\hfill \end{array}$

(See the following figure.) Finding the directional derivative in a given direction u at a given point on a surface. The plane is tangent to the surface at the given point ( −1 , 2 , 15 ) .

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