4.6 Directional derivatives and the gradient  (Page 13/8)

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 $u=\left(\text{cos}\theta \right)i+\left(\text{sin}\theta \right)j.$ Therefore, the z -coordinate of the second point on the graph is given by $z=f\left(a+h\text{cos}\theta ,b+h\text{sin}\theta \right).$

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

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

[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 $f\left(x,y\right)=x^2-xy+3y^2$ in the direction of $\text{u}=\left(\text{cos}\theta \right)i+\left(\text{sin}\theta \right)j.$ What is $D_{\text{u}}f\left(\mathrm{-1},2\right)?$

First of all, since $\text{cos}\theta =3\text{/}5$ and $\theta$ is acute, this implies

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

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

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

We substitute this expression into [link] :

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

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

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

(See the following figure.)

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