Directional Derivatives and the Gradient

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.

Another approach to calculating a directional derivative involves partial derivatives, as outlined in the following theorem.

Proof

[link] states that the directional derivative of f in the direction of $u=\text{cos}\theta i+\text{sin}\theta j$

is given by

Let $x=a+t\text{cos}\theta$

and $y=b+t\text{sin}\theta ,$

and define $g\left(t\right)=f\left(x,y\right).$

Since $f_x$

and $f_y$

both exist, we can use the chain rule for functions of two variables to calculate $g^{\prime }\left(t\right)\text{:}$

If $t=0,$

then $x=x_0$

and $y=y_0,$

so

By the definition of $g^{\prime }\left(t\right),$

it is also true that

Therefore, $D_{u}f\left(x_0,y_0\right)=f_x\left(x,y\right)\text{cos}\theta +f_y\left(x,y\right)\text{sin}\theta .$

□

If the vector that is given for the direction of the derivative is not a unit vector, then it is only necessary to divide by the norm of the vector. For example, if we wished to find the directional derivative of the function in [link] in the direction of the vector $\langle \mathrm{-5},12\rangle ,$

we would first divide by its magnitude to get $u.$

This gives us $u=\langle \text{−}\left(5\text{/}13\right),12\text{/}13\rangle .$

Then

Gradient

The right-hand side of [link] is equal to $f_x\left(x,y\right)\text{cos}\theta +f_y\left(x,y\right)\text{sin}\theta ,$

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 $u=\left(\text{cos}\theta \right)i+\left(\text{sin}\theta \right)j.$

Then the right-hand side of the equation can be written as the dot product of these two vectors:

The first vector in [link] has a special name: the gradient of the function $f.$

The symbol $\nabla$

is called nabla and the vector $\nabla f$

is read $\text{“del}f\text{.”}$

The gradient has some important properties. We have already seen one formula that uses the gradient: the formula for the directional derivative. Recall from The Dot Product that if the angle between two vectors $a$

and $b$

is $\phi ,$

then $a·\text{b}=\Vert a\Vert \Vert \text{b}\Vert \text{cos}\phi .$

Therefore, if the angle between $\nabla f\left(x_0,y_0\right)$

and $u=\left(\text{cos}\theta \right)i+\left(\text{sin}\theta \right)j$

is $\phi ,$

we have

The $\Vert u\Vert$

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 $\text{cos}\phi .$

Recall that $\text{cos}\phi$

ranges from $\mathrm{-1}$

to $1.$

If $\phi =0,$

then $\text{cos}\phi =1$

and $\nabla f\left(x_0,y_0\right)$

and $u$

both point in the same direction. If $\phi =\pi ,$

then $\text{cos}\phi =\mathrm{-1}$

and $\nabla f\left(x_0,y_0\right)$

and $u$

point in opposite directions. In the first case, the value of $D_{\text{u}}f\left(x_0,y_0\right)$

is maximized; in the second case, the value of $D_{\text{u}}f\left(x_0,y_0\right)$

is minimized. If $\nabla f\left(x_0,y_0\right)=0,$

then $D_{u}f\left(x_0,y_0\right)=\nabla f\left(x_0,y_0\right)·u=0$

for any vector $u.$

These three cases are outlined in the following theorem.

[link] shows a portion of the graph of the function $f\left(x,y\right)=3+\text{sin}x\text{sin}y.$

Given a point $\left(a,b\right)$

in the domain of $f,$

the maximum value of the gradient at that point is given by $\Vert \nabla f\left(a,b\right)\Vert .$

This would equal the rate of greatest ascent if the surface represented a topographical map. If we went in the opposite direction, it would be the rate of greatest descent.

When using a topographical map, the steepest slope is always in the direction where the contour lines are closest together (see [link]). This is analogous to the contour map of a function, assuming the level curves are obtained for equally spaced values throughout the range of that function.

Gradients and Level Curves

Recall that if a curve is defined parametrically by the function 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 $x\text{and}y,$

and $\left(x_0,y_0\right)$

is in its domain. Let’s suppose further 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 curve $f\left(x,y\right)=k.$

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 curve. By the chain Rule,

But $g^{\prime }\left(t\right)=0$

because $g\left(t\right)=k$

for all $t.$

Therefore, on the one hand,

on the other hand,

Therefore,

Thus, the dot product of these vectors is equal to zero, which implies they are orthogonal. However, the second vector is tangent to the level curve, which implies the gradient must be normal to the level curve, which gives rise to the following theorem.

We can use this theorem to find tangent and normal vectors to level curves of a function.

Three-Dimensional Gradients and Directional Derivatives

The definition of a gradient can be extended to functions of more than two variables.

Calculating the gradient of a function in three variables is very similar to calculating the gradient of a function in two variables. First, we calculate the partial derivatives $f_x,f_y,$

and $f_z,$

and then we use [link].

The directional derivative can also be generalized to functions of three variables. To determine a direction in three dimensions, a vector with three components is needed. This vector is a unit vector, and the components of the unit vector are called directional cosines. Given a three-dimensional unit vector $u$

in standard form (i.e., the initial point 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 $\gamma .$

Then the directional cosines are given by $\text{cos}\alpha ,\text{cos}\beta ,$

and $\text{cos}\gamma .$

These are the components of the unit vector $u;$

since $u$

is a unit vector, it is true that ${\text{cos}}^2\alpha +{\text{cos}}^2\beta +{\text{cos}}^2\gamma =1.$

We can calculate the directional derivative of a function of three variables by using the gradient, leading to a formula that is analogous to [link].

The three angles $\alpha ,\beta ,\text{and}\gamma$

determine the unit vector $u.$

In practice, we can use an arbitrary (nonunit) vector, then divide by its magnitude to obtain a unit vector in the desired direction.

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 one variable.

Key Equations

• directional derivative (two dimensions)

  ---

  $D_{u}f\left(a,b\right)=\underset{h\to 0}{\text{lim}}\frac{f\left(a+h\text{cos}\theta ,b+h\text{sin}\theta \right)-f\left(a,b\right)}{h}$

or

• gradient (two dimensions)

  ---

  $\nabla f\left(x,y\right)=f_x\left(x,y\right)i+f_y\left(x,y\right)j$
• gradient (three dimensions)

  ---

  $\nabla f\left(x,y,z\right)=f_x\left(x,y,z\right)i+f_y\left(x,y,z\right)j+f_z\left(x,y,z\right)k$
• directional derivative (three dimensions)

  ---

  $\begin{array}{cc}\hfill D_{u}f\left(x,y,z\right)& =\nabla f\left(x,y,z\right)·u\hfill \\ & =f_x\left(x,y,z\right)\text{cos}\alpha +f_y\left(x,y,z\right)\text{cos}\beta +f_x\left(x,y,z\right)\text{cos}\gamma \hfill \end{array}$