# 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 $z=f\left(a+h\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta ,b+h\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right).$

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.)

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