How To Find Directional Derivative Of A Vector

Differentiation of Functions of Several Variables

26 Directional Derivatives and the Slope

Learning Objectives

Determine the directional derivative in a given direction for a function of ii variables.
Determine the slope vector of a given real-valued office.
Explicate the significance of the gradient vector with regard to management of change along a surface.
Use the gradient to find the tangent to a level bend of a given role.
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 ii fractional 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 equally instantaneous rates of change (that is, equally 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 past $z=f\left(x,y\right),$ assuming the tangent line is parallel to the x-centrality. 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 cull a direction to travel from that point. Nosotros mensurate the management using an angle $\theta ,$ which is measured counterclockwise in the ten, y-plane, starting at zero from the positive 10-centrality ((Figure)). The altitude 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 past $z=f\left(a+h\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta ,b+h\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right).$

Finding the directional derivative at a point on the graph of $z=f\left(x,y\right).$ The slope of the black arrow on the graph indicates the value of the directional derivative at that indicate.

A shape in xyz space with point (a, b, f(a, b)). From the point, there is an arrow that represents the directional derivative. On the xy plane, the point (a, b) is marked and an angle of size θ is made between the projection of the directional derivative onto the plane and a line parallel to the x axis.

We can calculate the slope of the secant line by dividing the deviation in $z\text{-values}$ by the length of the line segment connecting the ii 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 detect the slope of the tangent line in the same management, nosotros have the limit as $h$ approaches zero.

Definition

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

(Figure) provides a formal definition of the directional derivative that can be used in many cases to calculate a directional derivative.

First of all, since $\text{cos}\phantom{\rule{0.2em}{0ex}}\theta =3\text{/}5$ and $\theta$ is astute, 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 summate $f\left(x+h\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta ,y+h\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right)\text{:}$

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

Nosotros substitute this expression into (Effigy):

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

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

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

(Encounter the following figure.)

Finding the directional derivative in a given management $u$ at a given point on a surface. The aeroplane is tangent to the surface at the given bespeak $\left(-1,2,15\right).$

The shape f(x, y) = x2 – xy + 3y2 in xyz space with tangent plane at point (–1, 2, 15). There are two arrows from the point, one seemingly along the surface of the shape and the other in a direction on the plane. The one that corresponds to the plane is marked u = 3/5 i + 4/5 j.

Another arroyo to calculating a directional derivative involves fractional derivatives, equally outlined in the following theorem.

Proof

(Figure) states that the directional derivative of f in the direction of $u=\text{cos}\phantom{\rule{0.2em}{0ex}}\theta i+\text{sin}\phantom{\rule{0.2em}{0ex}}\theta j$ is given past

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

Let $x=a+t\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta$ and $y=b+t\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta ,$ and define $g\left(t\right)=f\left(x,y\right).$ Since ${f}_{x}$ and ${f}_{y}$ both exist, we can utilize the chain rule for functions of two variables to calculate ${g}^{\prime }\left(t\right)\text{:}$

$\begin{array}{cc}\hfill {g}^{\prime }\left(t\right)& =\frac{\partial \mathit{\text{f}}}{\partial \mathit{\text{x}}}\phantom{\rule{0.1em}{0ex}}\frac{dx}{dt}+\frac{\partial \mathit{\text{f}}}{\partial \mathit{\text{y}}}\phantom{\rule{0.1em}{0ex}}\frac{dy}{dt}\hfill \\ & ={f}_{x}\left(x,y\right)\text{cos}\phantom{\rule{0.2em}{0ex}}\theta +{f}_{y}\left(x,y\right)\text{sin}\phantom{\rule{0.2em}{0ex}}\theta .\hfill \end{array}$

If $t=0,$ then $x={x}_{0}$ and $y={y}_{0},$ so

${g}^{\prime }\left(0\right)={f}_{x}\left({x}_{0},{y}_{0}\right)\text{cos}\phantom{\rule{0.2em}{0ex}}\theta +{f}_{y}\left({x}_{0},{y}_{0}\right)\text{sin}\phantom{\rule{0.2em}{0ex}}\theta .$

Past the definition of ${g}^{\prime }\left(t\right),$ it is as well true that

$\begin{array}{cc}\hfill {g}^{\prime }\left(0\right)& =\underset{t\to 0}{\text{lim}}\frac{g\left(t\right)-g\left(0\right)}{t}\hfill \\ & =\underset{t\to 0}{\text{lim}}\frac{f\left({x}_{0}+t\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta ,{y}_{0}+t\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right)-f\left({x}_{0},{y}_{0}\right)}{t}.\hfill \end{array}$

Therefore, ${D}_{u}f\left({x}_{0},{y}_{0}\right)={f}_{x}\left(x,y\right)\text{cos}\phantom{\rule{0.2em}{0ex}}\theta +{f}_{y}\left(x,y\right)\text{sin}\phantom{\rule{0.2em}{0ex}}\theta .$

□

First, nosotros must calculate the partial derivatives of $f\text{:}$

$\begin{array}{c}{f}_{x}=2x-y\hfill \\ {f}_{y}=\text{−}x+6y,\hfill \end{array}$

Then we utilise (Figure) with $\theta =\text{arccos}\left(3\text{/}5\right)\text{:}$

$\begin{array}{cc}\hfill {D}_{u}f\left(x,y\right)& ={f}_{x}\left(x,y\right)\text{cos}\phantom{\rule{0.2em}{0ex}}\theta +{f}_{y}\left(x,y\right)\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \hfill \\ & =\left(2x-y\right)\frac{3}{5}+\left(\text{−}x+6y\right)\frac{4}{5}\hfill \\ & =\frac{6x}{5}-\frac{3y}{5}-\frac{4x}{5}+\frac{24y}{5}\hfill \\ & =\frac{2x+21y}{5}.\hfill \end{array}$

To summate ${D}_{u}f\left(-1,2\right),$ let $x=-1$ and $y=2\text{:}$

${D}_{u}f\left(-1,2\right)=\frac{2\left(-1\right)+21\left(2\right)}{5}=\frac{-2+42}{5}=8.$

This is the same answer obtained in (Figure).

Find the directional derivative ${D}_{u}f\left(x,y\right)$ of $f\left(x,y\right)=3{x}^{2}y-4x{y}^{3}+3{y}^{2}-4x$ in the direction of $u=\left(\text{cos}\phantom{\rule{0.2em}{0ex}}\frac{\pi }{3}\right)i+\left(\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{\pi }{3}\right)j$ using (Figure). What is ${D}_{u}f\left(3,4\right)?$

$\begin{array}{}\\ \\ {D}_{u}f\left(x,y\right)=\frac{\left(6xy-4{y}^{3}-4\right)\left(1\right)}{2}+\frac{\left(3{x}^{2}-12x{y}^{2}+6y\right)\sqrt{3}}{2}\hfill \\ {D}_{u}f\left(3,4\right)=\frac{72-256-4}{2}+\frac{\left(27-576+24\right)\sqrt{3}}{2}=-94-\frac{525\sqrt{3}}{2}\hfill \end{array}$

Hint

Calculate the fractional derivatives and decide the value of $\theta .$

If the vector that is given for the management of the derivative is not a unit vector, then it is only necessary to divide past the norm of the vector. For example, if we wished to observe the directional derivative of the function in (Figure) in the management of the vector $〈-5,12〉,$ nosotros would first divide by its magnitude to get $u.$ This gives us $u=〈\text{−}\left(5\text{/}13\right),12\text{/}13〉.$ And then

$\begin{array}{cc}\hfill {D}_{u}f\left(x,y\right)& =\nabla f\left(x,y\right)·u\hfill \\ & =-\frac{5}{13}\left(2x-y\right)+\frac{12}{13}\left(\text{−}x+6y\right)\hfill \\ & =-\frac{22}{13}x+\frac{17}{13}y.\hfill \end{array}$

Gradient

The correct-paw side of (Effigy) is equal to ${f}_{x}\left(x,y\right)\text{cos}\phantom{\rule{0.2em}{0ex}}\theta +{f}_{y}\left(x,y\right)\text{sin}\phantom{\rule{0.2em}{0ex}}\theta ,$ which can be written as the dot product of two vectors. Define the first vector as $\nabla f\left(x,y\right)={f}_{x}\left(x,y\right)\text{i}+{f}_{y}\left(x,y\right)\text{j}$ and the second vector as $u=\left(\text{cos}\phantom{\rule{0.2em}{0ex}}\theta \right)i+\left(\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right)j.$ Then the correct-mitt side of the equation can exist written equally the dot product of these two vectors:

${D}_{u}f\left(x,y\right)=\nabla f\left(x,y\right)·u.$

The showtime vector in (Figure) has a special proper noun: the gradient of the function $f.$ The symbol $\nabla$ is called nabla and the vector $\nabla f$ is read $\text{``del}\phantom{\rule{0.2em}{0ex}}f\text{.''}$

Finding Gradients

Find the gradient $\nabla f\left(x,y\right)$ of each of the following functions:

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

Notice the gradient $\nabla f\left(x,y\right)$ of $f\left(x,y\right)=\left({x}^{2}-3{y}^{2}\right)\text{/}\left(2x+y\right).$

$\nabla f\left(x,y\right)=\frac{2{x}^{2}+2xy+6{y}^{2}}{{\left(2x+y\right)}^{2}}i-\frac{{x}^{2}+12xy+3{y}^{2}}{{\left(2x+y\right)}^{2}}j$

Hint

Calculate the partial derivatives, then use (Figure).

The gradient has some of import properties. Nosotros take already seen one formula that uses the gradient: the formula for the directional derivative. Call up from The Dot Product that if the bending betwixt two vectors $a$ and $b$ is $\phi ,$ then $a·\text{b}=‖a‖‖\text{b}‖\text{cos}\phantom{\rule{0.2em}{0ex}}\phi .$ Therefore, if the angle between $\nabla f\left({x}_{0},{y}_{0}\right)$ and $u=\left(\text{cos}\phantom{\rule{0.2em}{0ex}}\theta \right)i+\left(\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right)j$ is $\phi ,$ we have

${D}_{u}f\left({x}_{0},{y}_{0}\right)=\nabla f\left({x}_{0},{y}_{0}\right)·u=‖\nabla f\left({x}_{0},{y}_{0}\right)‖‖u‖\text{cos}\phantom{\rule{0.2em}{0ex}}\phi =‖\nabla f\left({x}_{0},{y}_{0}\right)‖\text{cos}\phantom{\rule{0.2em}{0ex}}\phi .$

The $‖u‖$ disappears because $u$ is a unit vector. Therefore, the directional derivative is equal to the magnitude of the gradient evaluated at $\left({x}_{0},{y}_{0}\right)$ multiplied by $\text{cos}\phantom{\rule{0.2em}{0ex}}\phi .$ Retrieve that $\text{cos}\phantom{\rule{0.2em}{0ex}}\phi$ ranges from $-1$ to $1.$ If $\phi =0,$ then $\text{cos}\phantom{\rule{0.2em}{0ex}}\phi =1$ and $\nabla f\left({x}_{0},{y}_{0}\right)$ and $u$ both indicate in the same direction. If $\phi =\pi ,$ and then $\text{cos}\phantom{\rule{0.2em}{0ex}}\phi =-1$ and $\nabla f\left({x}_{0},{y}_{0}\right)$ and $u$ point in contrary directions. In the first case, the value of ${D}_{\text{u}}f\left({x}_{0},{y}_{0}\right)$ is maximized; in the second instance, the value of ${D}_{\text{u}}f\left({x}_{0},{y}_{0}\right)$ is minimized. If $\nabla f\left({x}_{0},{y}_{0}\right)=0,$ and so ${D}_{u}f\left({x}_{0},{y}_{0}\right)=\nabla f\left({x}_{0},{y}_{0}\right)·u=0$ for any vector $u.$ These iii cases are outlined in the post-obit theorem.

The gradient indicates the maximum and minimum values of the directional derivative at a point.

Finding a Maximum Directional Derivative

Find the management for which the directional derivative of $f\left(x,y\right)=3{x}^{2}-4xy+2{y}^{2}$ at $\left(-2,3\right)$ is a maximum. What is the maximum value?

The maximum value of the directional derivative occurs when $\nabla f$ and the unit vector point in the same direction. Therefore, we first past computing $\nabla f\left(x,y\right)\text{:}$

$\begin{array}{ccc}\hfill {f}_{x}\left(x,y\right)& =\hfill & 6x-4y\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{f}_{y}\left(x,y\right)=-4x+4y,\phantom{\rule{0.2em}{0ex}}\text{so}\hfill \\ \hfill \nabla f\left(x,y\right)& =\hfill & {f}_{x}\left(x,y\right)i+{f}_{y}\left(x,y\right)j=\left(6x-4y\right)i+\left(-4x+4y\right)j.\hfill \end{array}$

Next, nosotros evaluate the slope at $\left(-2,3\right)\text{:}$

$\nabla f\left(-2,3\right)=\left(6\left(-2\right)-4\left(3\right)\right)i+\left(-4\left(-2\right)+4\left(3\right)\right)j=-24i+20j.$

We demand to detect a unit of measurement vector that points in the same direction as $\nabla f\left(-2,3\right),$ and then the adjacent step is to divide $\nabla f\left(-2,3\right)$ by its magnitude, which is $\sqrt{{\left(-24\right)}^{2}+{\left(20\right)}^{2}}=\sqrt{976}=4\sqrt{61}.$ Therefore,

$\frac{\nabla f\left(-2,3\right)}{‖\nabla f\left(-2,3\right)‖}=\frac{-24}{4\sqrt{61}}i+\frac{20}{4\sqrt{61}}j=\frac{-6\sqrt{61}}{61}i+\frac{5\sqrt{61}}{61}j.$

This is the unit of measurement vector that points in the aforementioned direction equally $\nabla f\left(-2,3\right).$ To find the bending corresponding to this unit vector, we solve the equations

$\text{cos}\phantom{\rule{0.2em}{0ex}}\theta =\frac{-6\sqrt{61}}{61}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta =\frac{5\sqrt{61}}{61}$

for $\theta .$ Since cosine is negative and sine is positive, the angle must be in the 2nd quadrant. Therefore, $\theta =\pi -\text{arcsin}\left(\left(5\sqrt{61}\right)\text{/}61\right)\approx 2.45\phantom{\rule{0.2em}{0ex}}\text{rad.}$

The maximum value of the directional derivative at $\left(-2,3\right)$ is $‖\nabla f\left(-2,3\right)‖=4\sqrt{61}$ (see the following effigy).

The maximum value of the directional derivative at $\left(-2,3\right)$ is in the direction of the gradient.

An upward facing paraboloid f(x, y) = 3x2 – 4xy + 2y2 with tangent plane at the point (–2, 3, 54). The tangent plane has equation z = –24x + 20y – 54.

(Figure) shows a portion of the graph of the function $f\left(x,y\right)=3+\text{sin}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}y.$ Given a point $\left(a,b\right)$ in the domain of $f,$ the maximum value of the gradient at that point is given by $‖\nabla f\left(a,b\right)‖.$ This would equal the rate of greatest rising if the surface represented a topographical map. If nosotros went in the opposite direction, it would be the charge per unit of greatest descent.

A typical surface in ${ℝ}^{3}.$ Given a point on the surface, the directional derivative can be calculated using the gradient.

An surface in xyz space with point at f(a, b). There is an arrow in the direction of greatest descent.

When using a topographical map, the steepest gradient is always in the management where the contour lines are closest together (encounter (Figure)). This is analogous to the contour map of a office, assuming the level curves are obtained for equally spaced values throughout the range of that function.

Profile map for the part $f\left(x,y\right)={x}^{2}-{y}^{2}$ using level values between $-5$ and $5.$

Two crossing dashed lines that pass through the origin and a series of curved lines approaching the crosses dashed lines as if they are asymptotes.

Gradients and Level Curves

Recall that if a curve is defined parametrically by the part pair $\left(x\left(t\right),y\left(t\right)\right),$ then the vector ${x}^{\prime }\left(t\right)i+{y}^{\prime }\left(t\right)j$ is tangent to the curve for every value of $t$ in the domain. Now let's assume $z=f\left(x,y\right)$ is a differentiable function of $x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y,$ and $\left({x}_{0},{y}_{0}\right)$ is in its domain. Let's suppose farther that ${x}_{0}=x\left({t}_{0}\right)$ and ${y}_{0}=y\left({t}_{0}\right)$ for some value of $t,$ and consider the level bend $f\left(x,y\right)=k.$ Define $g\left(t\right)=f\left(x\left(t\right),y\left(t\right)\right)$ and calculate ${g}^{\prime }\left(t\right)$ on the level bend. Past the chain Rule,

${g}^{\prime }\left(t\right)={f}_{x}\left(x\left(t\right),y\left(t\right)\right){x}^{\prime }\left(t\right)+{f}_{y}\left(x\left(t\right),y\left(t\right)\right){y}^{\prime }\left(t\right).$

But ${g}^{\prime }\left(t\right)=0$ because $g\left(t\right)=k$ for all $t.$ Therefore, on the one hand,

${f}_{x}\left(x\left(t\right),y\left(t\right)\right){x}^{\prime }\left(t\right)+{f}_{y}\left(x\left(t\right),y\left(t\right)\right){y}^{\prime }\left(t\right)=0;$

on the other hand,

${f}_{x}\left(x\left(t\right),y\left(t\right)\right){x}^{\prime }\left(t\right)+{f}_{y}\left(x\left(t\right),y\left(t\right)\right){y}^{\prime }\left(t\right)=\nabla f\left(x,y\right)·〈{x}^{\prime }\left(t\right),{y}^{\prime }\left(t\right)〉\text{.}$

Therefore,

$\nabla f\left(x,y\right)·〈{x}^{\prime }\left(t\right),{y}^{\prime }\left(t\right)〉=0.$

Thus, the dot production of these vectors is equal to zero, which implies they are orthogonal. Still, the 2nd vector is tangent to the level curve, which implies the gradient must exist normal to the level curve, which gives rise to the following theorem.

We tin can utilize this theorem to find tangent and normal vectors to level curves of a role.

Finding Tangents to Level Curves

For the function $f\left(x,y\right)=2{x}^{2}-3xy+8{y}^{2}+2x-4y+4,$ find a tangent vector to the level curve at point $\left(-2,1\right).$ Graph the level curve respective to $f\left(x,y\right)=18$ and depict in $\nabla f\left(-2,1\right)$ and a tangent vector.

First, nosotros must calculate $\nabla f\left(x,y\right)\text{:}$

${f}_{x}\left(x,y\right)=4x-3y+2\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{f}_{y}=-3x+16y-4\phantom{\rule{0.2em}{0ex}}\text{so}\phantom{\rule{0.2em}{0ex}}\nabla f\left(x,y\right)=\left(4x-3y+2\right)i+\left(-3x+16y-4\right)j.$

Side by side, we evaluate $\nabla f\left(x,y\right)$ at $\left(-2,1\right)\text{:}$

$\nabla f\left(-2,1\right)=\left(4\left(-2\right)-3\left(1\right)+2\right)i+\left(-3\left(-2\right)+16\left(1\right)-4\right)j=-9i+18j.$

This vector is orthogonal to the curve at betoken $\left(-2,1\right).$ We tin obtain a tangent vector by reversing the components and multiplying either one by $-1.$ Thus, for example, $-18i-9j$ is a tangent vector (encounter the post-obit graph).

Tangent and normal vectors to $2{x}^{2}-3xy+8{y}^{2}+2x-4y+4=18$ at signal $\left(-2,1\right).$

A rotated ellipse with equation f(x, y) = 10. At the point (–2, 1) on the ellipse, there are drawn two arrows, one tangent vector and one normal vector. The normal vector is marked ∇f(–2, 1) and is perpendicular to the tangent vector.

For the function $f\left(x,y\right)={x}^{2}-2xy+5{y}^{2}+3x-2y+4,$ observe the tangent to the level curve at point $\left(1,1\right).$ Draw the graph of the level curve respective to $f\left(x,y\right)=8$ and draw $\nabla f\left(1,1\right)$ and a tangent vector.

Hint

Calculate the gradient at point $\left(1,1\right).$

Three-Dimensional Gradients and Directional Derivatives

The definition of a gradient tin be extended to functions of more than ii variables.

Definition

Permit $w=f\left(x,y,z\right)$ be a function of 3 variables such that ${f}_{x},{f}_{y},\text{and}\phantom{\rule{0.2em}{0ex}}{f}_{z}$ exist. The vector $\nabla f\left(x,y,z\right)$ is called the gradient of $f$ and is defined as

$\nabla f\left(x,y,z\right)={f}_{x}\left(x,y,z\right)i+{f}_{y}\left(x,y,z\right)j+{f}_{z}\left(x,y,z\right)k.$

$\nabla f\left(x,y,z\right)$ tin can as well be written as $\text{grad}\phantom{\rule{0.2em}{0ex}}f\left(x,y,z\right).$

Calculating the gradient of a function in three variables is very like to calculating the slope of a function in ii variables. First, we calculate the partial derivatives ${f}_{x},{f}_{y},$ and ${f}_{z},$ and and so we use (Figure).

Finding Gradients in Three Dimensions

Discover the slope $\nabla f\left(x,y,z\right)$ of each of the following functions:

$f\left(x,y\right)=5{x}^{2}-2xy+{y}^{2}-4yz+{z}^{2}+3xz$
$f\left(x,y,z\right)={e}^{-2z}\text{sin}\phantom{\rule{0.2em}{0ex}}2x\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}2y$

The directional derivative can as well be generalized to functions of three variables. To make up one's mind a direction in three dimensions, a vector with three components is needed. This vector is a unit vector, and the components of the unit of measurement vector are called directional cosines . Given a three-dimensional unit vector $u$ in standard form (i.e., the initial bespeak is at the origin), this vector forms three different angles with the positive $x-,y-,$ and z-axes. Let's call these angles $\alpha ,\beta ,$ and $\gamma .$ And then the directional cosines are given by $\text{cos}\phantom{\rule{0.2em}{0ex}}\alpha ,\text{cos}\phantom{\rule{0.2em}{0ex}}\beta ,$ and $\text{cos}\phantom{\rule{0.2em}{0ex}}\gamma .$ These are the components of the unit vector $u;$ since $u$ is a unit vector, it is truthful that ${\text{cos}}^{2}\alpha +{\text{cos}}^{2}\beta +{\text{cos}}^{2}\gamma =1.$

Definition

Suppose $w=f\left(x,y,z\right)$ is a function of 3 variables with a domain of $D.$ Permit $\left({x}_{0},{y}_{0},{z}_{0}\right)\in D$ and let $\text{u}=\text{cos}\phantom{\rule{0.2em}{0ex}}\alpha i+\text{cos}\phantom{\rule{0.2em}{0ex}}\beta j+\text{cos}\phantom{\rule{0.2em}{0ex}}\gamma k$ be a unit vector. And then, the directional derivative of $f$ in the direction of $u$ is given by

${D}_{u}f\left({x}_{0},{y}_{0},{z}_{0}\right)=\underset{t\to 0}{\text{lim}}\frac{f\left({x}_{0}+t\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\alpha ,{y}_{0}+t\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\beta ,{z}_{0}+t\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\gamma \right)-f\left({x}_{0},{y}_{0},{z}_{0}\right)}{t},$

provided the limit exists.

Nosotros tin can calculate the directional derivative of a function of three variables past using the gradient, leading to a formula that is analogous to (Figure).

Directional Derivative of a Function of Three Variables

Let $f\left(x,y,z\right)$ be a differentiable role of three variables and let $u=\text{cos}\phantom{\rule{0.2em}{0ex}}\alpha i+\text{cos}\phantom{\rule{0.2em}{0ex}}\beta j+\text{cos}\phantom{\rule{0.2em}{0ex}}\gamma k$ be a unit vector. Then, the directional derivative of $f$ in the direction of $u$ is given by

$\begin{array}{cc}\hfill {D}_{u}f\left(x,y,z\right)& =\nabla f\left(x,y,z\right)·u\hfill \\ & ={f}_{x}\left(x,y,z\right)\text{cos}\phantom{\rule{0.2em}{0ex}}\alpha +{f}_{y}\left(x,y,z\right)\text{cos}\phantom{\rule{0.2em}{0ex}}\beta +{f}_{z}\left(x,y,z\right)\text{cos}\phantom{\rule{0.2em}{0ex}}\gamma .\hfill \end{array}$

The three angles $\alpha ,\beta ,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\gamma$ determine the unit of measurement vector $u.$ In practice, we can use an capricious (nonunit) vector, then divide by its magnitude to obtain a unit of measurement vector in the desired direction.

Finding a Directional Derivative in Iii Dimensions

Calculate ${D}_{u}f\left(1,-2,3\right)$ in the management of $\text{v}=\text{−}i+2j+2k$ for the role

$f\left(x,y,z\right)=5{x}^{2}-2xy+{y}^{2}-4yz+{z}^{2}+3xz.$

Outset, we find the magnitude of $v\text{:}$

$‖v‖=\sqrt{{\left(-1\right)}^{2}+{\left(2\right)}^{2}}=3.$

Therefore, $\frac{v}{‖v‖}=\frac{\text{−}i+2j+2k}{3}=-\frac{1}{3}i+\frac{2}{3}j+\frac{2}{3}k$ is a unit vector in the direction of $v,$ and so $\text{cos}\phantom{\rule{0.2em}{0ex}}\alpha =-\frac{1}{3},\text{cos}\phantom{\rule{0.2em}{0ex}}\beta =\frac{2}{3},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\gamma =\frac{2}{3}.$ Adjacent, nosotros summate the partial derivatives of $f\text{:}$

$\begin{array}{ccc}\hfill {f}_{x}\left(x,y,z\right)& =\hfill & 10x-2y+3z\hfill \\ \hfill {f}_{y}\left(x,y,z\right)& =\hfill & -2x+2y-4z\hfill \\ \hfill {f}_{z}\left(x,y,z\right)& =\hfill & -4y+2z+3x,\hfill \end{array}$

then substitute them into (Figure):

$\begin{array}{cc}\hfill {D}_{u}f\left(x,y,z\right)& ={f}_{x}\left(x,y,z\right)\text{cos}\phantom{\rule{0.2em}{0ex}}\alpha +{f}_{y}\left(x,y,z\right)\text{cos}\phantom{\rule{0.2em}{0ex}}\beta +{f}_{z}\left(x,y,z\right)\text{cos}\phantom{\rule{0.2em}{0ex}}\gamma \hfill \\ & =\left(10x-2y+3z\right)\left(-\frac{1}{3}\right)+\left(-2x+2y-4z\right)\left(\frac{2}{3}\right)+\left(-4y+2z+3x\right)\left(\frac{2}{3}\right)\hfill \\ & =-\frac{10x}{3}+\frac{2y}{3}-\frac{3z}{3}-\frac{4x}{3}+\frac{4y}{3}-\frac{8z}{3}-\frac{8y}{3}+\frac{4z}{3}+\frac{6x}{3}\hfill \\ & =-\frac{8x}{3}-\frac{2y}{3}-\frac{7z}{3}.\hfill \end{array}$

Final, to observe ${D}_{u}f\left(1,-2,3\right),$ we substitute $x=1,y=-2,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}z=3\text{:}$

$\begin{array}{cc}\hfill {D}_{u}f\left(1,-2,3\right)& =-\frac{8\left(1\right)}{3}-\frac{2\left(-2\right)}{3}-\frac{7\left(3\right)}{3}\hfill \\ & =-\frac{8}{3}+\frac{4}{3}-\frac{21}{3}\hfill \\ & =-\frac{25}{3}.\hfill \end{array}$

Calculate ${D}_{u}f\left(x,y,z\right)$ and ${D}_{u}f\left(0,-2,5\right)$ in the direction of $\text{v}=-3i+12j-4k$ for the function $f\left(x,y,z\right)=3{x}^{2}+xy-2{y}^{2}+4yz-{z}^{2}+2xz.$

$\begin{array}{ccc}\hfill {D}_{u}f\left(x,y,z\right)& =\hfill & -\frac{3}{13}\left(6x+y+2z\right)+\frac{12}{13}\left(x-4y+4z\right)-\frac{4}{13}\left(2x+4y-2z\right)\hfill \\ \hfill {D}_{u}f\left(0,-2,5\right)& =\hfill & \frac{384}{13}\hfill \end{array}$

Hint

First, split up $v$ past its magnitude, calculate the partial derivatives of $f,$ and so utilise (Figure).

Key Concepts

A directional derivative represents a rate of change of a function in any given direction.
The gradient can be used in a formula to calculate the directional derivative.
The gradient indicates the direction of greatest change of a function of more than ane variable.

Cardinal Equations

For the following exercises, observe the directional derivative using the limit definition only.

$f\left(x,y\right)={y}^{2}\text{cos}\left(2x\right)$ at point $P\left(\frac{\pi }{3},2\right)$ in the management of $\text{u}=\left(\text{cos}\phantom{\rule{0.2em}{0ex}}\frac{\pi }{4}\right)i+\left(\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{\pi }{4}\right)j$

$-3\sqrt{3}$

For the following exercises, discover the directional derivative of the function at point $P$ in the management of $v.$

$h\left(x,y\right)={e}^{x}\text{sin}\phantom{\rule{0.2em}{0ex}}y,P\left(1,\frac{\pi }{2}\right),v=\text{−}i$

$h\left(x,y,z\right)=xyz,P\left(2,1,1\right),\text{v}=2i+j-k$

$\frac{2}{\sqrt{6}}$

$f\left(x,y\right)=xy,P\left(1,1\right),u=〈\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}〉$

$f\left(x,y\right)={x}^{2}-{y}^{2},\begin{array}{cc}u=〈\frac{\sqrt{3}}{2},\frac{1}{2}〉,\hfill & P\left(1,0\right)\hfill \end{array}$

$\sqrt{3}$

$f\left(x,y\right)=3x+4y+7,\begin{array}{cc}u=〈\frac{3}{5},\frac{4}{5}〉,\hfill & P\left(0,\frac{\pi }{2}\right)\hfill \end{array}$

$\begin{array}{ccc}f\left(x,y\right)={e}^{x}\text{cos}\phantom{\rule{0.2em}{0ex}}y,\hfill & u=〈0,1〉,\hfill & P=\left(0,\frac{\pi }{2}\right)\hfill \end{array}$

$-1.0$

$\begin{array}{ccc}f\left(x,y\right)={y}^{10},\hfill & u=〈0,-1〉,\hfill & P=\left(1,-1\right)\hfill \end{array}$

$f\left(x,y\right)=\text{ln}\left({x}^{2}+{y}^{2}\right),\begin{array}{cc}u=〈\frac{3}{5},\frac{4}{5}〉,\hfill & P\left(1,2\right)\hfill \end{array}$

$\frac{22}{25}$

$f\left(x,y\right)={x}^{2}y,\begin{array}{cc}P\left(-5,5\right),\hfill & v=3i-4j\hfill \end{array}$

$f\left(x,y\right)={y}^{2}+xz,\begin{array}{cc}P\left(1,2,2\right),\hfill & v=〈2,-1,2〉\hfill \end{array}$

$\frac{2}{3}$

For the post-obit exercises, detect the directional derivative of the role in the management of the unit vector $u=\text{cos}\phantom{\rule{0.2em}{0ex}}\theta i+\text{sin}\phantom{\rule{0.2em}{0ex}}\theta j.$

$f\left(x,y\right)={x}^{2}+2{y}^{2},\theta =\frac{\pi }{6}$

$f\left(x,y\right)=\frac{y}{x+2y},\theta =-\frac{\pi }{4}$

$\frac{\text{−}\sqrt{2}\left(x+y\right)}{2{\left(x+2y\right)}^{2}}$

$f\left(x,y\right)=\text{cos}\left(3x+y\right),\theta =\frac{\pi }{4}$

$w\left(x,y\right)=y{e}^{x},\theta =\frac{\pi }{3}$

$\frac{{e}^{x}\left(y+\sqrt{3}\right)}{2}$

$\begin{array}{cc}f\left(x,y\right)=x\phantom{\rule{0.2em}{0ex}}\text{arctan}\left(y\right),\hfill & \theta =\frac{\pi }{2}\hfill \end{array}$

$\begin{array}{cc}f\left(x,y\right)=\text{ln}\left(x+2y\right),\hfill & \theta =\frac{\pi }{3}\hfill \end{array}$

$\frac{1+2\sqrt{3}}{2\left(x+2y\right)}$

For the post-obit exercises, find the gradient.

Find the gradient of $f\left(x,y\right)=\frac{14-{x}^{2}-{y}^{2}}{3}.$ And then, find the gradient at bespeak $P\left(1,2\right).$

Observe the slope of $f\left(x,y,z\right)=xy+yz+xz$ at point $P\left(1,2,3\right).$

$〈5,4,3〉$

$f\left(x,y,z\right)=4{x}^{5}{y}^{2}{z}^{3},\begin{array}{cc}P\left(2,-1,1\right),\hfill & u=\frac{1}{3}i+\frac{2}{3}j-\frac{2}{3}k\hfill \end{array}$

$-320$

For the post-obit exercises, notice the directional derivative of the function at point $P$ in the direction of $Q.$

$f\left(x,y\right)={x}^{2}+3{y}^{2},\begin{array}{cc}P\left(1,1\right),\hfill & Q\left(4,5\right)\hfill \end{array}$

$f\left(x,y,z\right)=\frac{y}{x+z},\begin{array}{cc}P\left(2,1,-1\right),\hfill & Q\left(-1,2,0\right)\hfill \end{array}$

$\frac{3}{\sqrt{11}}$

For the following exercises, notice the derivative of the function at $P$ in the direction of $u.$

$f\left(x,y\right)=-7x+2y,\begin{array}{cc}P\left(2,-4\right),\hfill & u=4i-3j\hfill \end{array}$

$f\left(x,y\right)=\text{ln}\left(5x+4y\right),\begin{array}{cc}P\left(3,9\right),\hfill & u=6i+8j\hfill \end{array}$

$\frac{31}{255}$

[T] Use technology to sketch the level curve of $f\left(x,y\right)={x}^{2}+4{y}^{2}$ that passes through $P\left(-2,0\right)$ and describe the slope vector at $P.$

The top of half of an ellipse centered at the origin with major axis horizontal and of length 4 and minor axis 2. The point (–2, 0) is marked, and there is an arrow pointing out from it to the left marked –4i.