Differentiation of Functions of Several Variables
26 Directional Derivatives and the Gradient
Learning Objectives
- 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 has two partial derivatives:
and
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,
represents the slope of a tangent line passing through a given point on the surface defined by
assuming the tangent line is parallel to the x-axis. Similarly,
represents the slope of the tangent line parallel to the
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 Given a point
in the domain of
we choose a direction to travel from that point. We measure the direction using an angle
which is measured counterclockwise in the x, y-plane, starting at zero from the positive x-axis ((Figure)). The distance we travel is
and the direction we travel is given by the unit vector
Therefore, the z-coordinate of the second point on the graph is given by

We can calculate the slope of the secant line by dividing the difference in by the length of the line segment connecting the two points in the domain. The length of the line segment is
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 approaches zero.
Suppose is a function of two variables with a domain of
Let
and define
Then the directional derivative of
in the direction of
is given by

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.
Let Find the directional derivative
of
in the direction of
What is
First of all, since and
is acute, this implies

Using we first calculate

We substitute this expression into (Figure):

To calculate we substitute
and
into this answer:

(See the following figure.)


Another approach to calculating a directional derivative involves partial derivatives, as outlined in the following theorem.
Let be a function of two variables
and assume that
and
exist. Then the directional derivative of
in the direction of
is given by

Proof
(Figure) states that the directional derivative of f in the direction of is given by

Let and
and define
Since
and
both exist, we can use the chain rule for functions of two variables to calculate

If then
and
so

By the definition of it is also true that

Therefore,
□
Find the directional derivative of
in the direction of
using (Figure). What is
Calculate the partial derivatives and determine the value of
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 (Figure) in the direction of the vector we would first divide by its magnitude to get
This gives us
Then

Gradient
The right-hand side of (Figure) is equal to which can be written as the dot product of two vectors. Define the first vector as
and the second vector as
Then the right-hand side of the equation can be written as the dot product of these two vectors:

The first vector in (Figure) has a special name: the gradient of the function The symbol
is called nabla and the vector
is read
Let be a function of
such that
and
exist. The vector
is called the gradient of
and is defined as

The vector is also written as
Find the gradient of each of the following functions:
For both parts a. and b., we first calculate the partial derivatives and
then use (Figure).
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 and
is
then
Therefore, if the angle between
and
is
we have

The disappears because
is a unit vector. Therefore, the directional derivative is equal to the magnitude of the gradient evaluated at
multiplied by
Recall that
ranges from
to
If
then
and
and
both point in the same direction. If
then
and
and
point in opposite directions. In the first case, the value of
is maximized; in the second case, the value of
is minimized. If
then
for any vector
These three cases are outlined in the following theorem.
Suppose the function is differentiable at
((Figure)).
- If
then
for any unit vector
- If
then
is maximized when
points in the same direction as
The maximum value of
is
- If
then
is minimized when
points in the opposite direction from
The minimum value of
is
Find the direction for which the directional derivative of at
is a maximum. What is the maximum value?
The maximum value of the directional derivative occurs when and the unit vector point in the same direction. Therefore, we start by calculating

Next, we evaluate the gradient at

We need to find a unit vector that points in the same direction as so the next step is to divide
by its magnitude, which is
Therefore,

This is the unit vector that points in the same direction as To find the angle corresponding to this unit vector, we solve the equations

for Since cosine is negative and sine is positive, the angle must be in the second quadrant. Therefore,
The maximum value of the directional derivative at is
(see the following figure).

Find the direction for which the directional derivative of at
is a maximum. What is the maximum value?
The gradient of at
is
The unit vector that points in the same direction as
is
which gives an angle of
The maximum value of the directional derivative is
Evaluate the gradient of at point
(Figure) shows a portion of the graph of the function Given a point
in the domain of
the maximum value of the gradient at that point is given by
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 (Figure)). 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 then the vector
is tangent to the curve for every value of
in the domain. Now let’s assume
is a differentiable function of
and
is in its domain. Let’s suppose further that
and
for some value of
and consider the level curve
Define
and calculate
on the level curve. By the chain Rule,

But because
for all
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.
Suppose the function has continuous first-order partial derivatives in an open disk centered at a point
If
then
is normal to the level curve of
at
We can use this theorem to find tangent and normal vectors to level curves of a function.
For the function find a tangent vector to the level curve at point
Graph the level curve corresponding to
and draw in
and a tangent vector.
First, we must calculate

Next, we evaluate at

This vector is orthogonal to the curve at point We can obtain a tangent vector by reversing the components and multiplying either one by
Thus, for example,
is a tangent vector (see the following graph).


For the function find the tangent to the level curve at point
Draw the graph of the level curve corresponding to
and draw
and a tangent vector.
Tangent vector: or
Calculate the gradient at point
Three-Dimensional Gradients and Directional Derivatives
The definition of a gradient can be extended to functions of more than two variables.
Let be a function of three variables such that
exist. The vector
is called the gradient of
and is defined as

can also be written as
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 and
and then we use (Figure).
Find the gradient of each of the following functions:
For both parts a. and b., we first calculate the partial derivatives and
then use (Figure).
Find the gradient of
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 in standard form (i.e., the initial point is at the origin), this vector forms three different angles with the positive
and z-axes. Let’s call these angles
and
Then the directional cosines are given by
and
These are the components of the unit vector
since
is a unit vector, it is true that
Suppose is a function of three variables with a domain of
Let
and let
be a unit vector. Then, the directional derivative of
in the direction of
is given by

provided the limit exists.
We can calculate the directional derivative of a function of three variables by using the gradient, leading to a formula that is analogous to (Figure).
Let be a differentiable function of three variables and let
be a unit vector. Then, the directional derivative of
in the direction of
is given by

The three angles determine the unit vector
In practice, we can use an arbitrary (nonunit) vector, then divide by its magnitude to obtain a unit vector in the desired direction.
Calculate in the direction of
for the function

First, we find the magnitude of

Therefore, is a unit vector in the direction of
so
Next, we calculate the partial derivatives of

then substitute them into (Figure):

Last, to find we substitute

Calculate and
in the direction of
for the function
First, divide by its magnitude, calculate the partial derivatives of
then use (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 one variable.
Key Equations
- directional derivative (two dimensions)
or
- gradient (two dimensions)
- gradient (three dimensions)
- directional derivative (three dimensions)
For the following exercises, find the directional derivative using the limit definition only.
at point
in the direction of
at point
in the direction of
Find the directional derivative of at point
in the direction of
For the following exercises, find the directional derivative of the function at point in the direction of
For the following exercises, find the directional derivative of the function in the direction of the unit vector
For the following exercises, find the gradient.
Find the gradient of Then, find the gradient at point
Find the gradient of at point
Find the gradient of at
and in the direction of
For the following exercises, find the directional derivative of the function at point in the direction of
For the following exercises, find the derivative of the function at in the direction of
[T] Use technology to sketch the level curve of that passes through
and draw the gradient vector at
[T] Use technology to sketch the level curve of that passes through
and draw the gradient vector at

For the following exercises, find the gradient vector at the indicated point.
For the following exercises, find the derivative of the function.
at point
in the direction the function increases most rapidly
at point
in the direction the function increases most rapidly
at point
in the direction the function increases most rapidly
at point
in the direction the function increases most rapidly
at point
in the direction the function increases most rapidly
For the following exercises, find the maximum rate of change of at the given point and the direction in which it occurs.
For the following exercises, find equations of
- the tangent plane and
- the normal line to the given surface at the given point.
The level curve for
at point
at point
a. b.
at point
at point
a. b.
For the following exercises, solve the problem.
The temperature in a metal sphere is inversely proportional to the distance from the center of the sphere (the origin:
The temperature at point
is
- Find the rate of change of the temperature at point
in the direction toward point
- Show that, at any point in the sphere, the direction of greatest increase in temperature is given by a vector that points toward the origin.
The electrical potential (voltage) in a certain region of space is given by the function
- Find the rate of change of the voltage at point
in the direction of the vector
- In which direction does the voltage change most rapidly at point
- What is the maximum rate of change of the voltage at point
a. b.
c.
If the electric potential at a point in the xy-plane is
then the electric intensity vector at
is
- Find the electric intensity vector at
- Show that, at each point in the plane, the electric potential decreases most rapidly in the direction of the vector
In two dimensions, the motion of an ideal fluid is governed by a velocity potential The velocity components of the fluid
in the x-direction and
in the y-direction, are given by
Find the velocity components associated with the velocity potential
Glossary
- directional derivative
- the derivative of a function in the direction of a given unit vector
- gradient
- the gradient of the function
is defined to be
which can be generalized to a function of any number of independent variables