Now that we have examined limits and continuity of functions of two variables, we can proceed to study derivatives. Finding derivatives of functions of two variables is the key concept in this chapter, with as many applications in mathematics, science, and engineering as differentiation of single-variable functions. However, we have already seen that limits and continuity of multivariable functions have new issues and require new terminology and ideas to deal with them. This carries over into differentiation as well.
When studying derivatives of functions of one variable, we found that one interpretation of the derivative is an instantaneous rate of change of
as a function of
Leibniz notation for the derivative is
which implies that
is the dependent variable and
is the independent variable. For a function
of two variables,
and
are the independent variables and
is the dependent variable. This raises two questions right away: How do we adapt Leibniz notation for functions of two variables? Also, what is an interpretation of the derivative? The answer lies in partial derivatives.
Let
be a function of two variables. Then the partial derivative of
with respect to
written as
or
is defined as
The partial derivative of
with respect to
written as
or
is defined as
This definition shows two differences already. First, the notation changes, in the sense that we still use a version of Leibniz notation, but the
in the original notation is replaced with the symbol
(This rounded
is usually called “partial,” so
is spoken as the “partial of
with respect to
This is the first hint that we are dealing with partial derivatives. Second, we now have two different derivatives we can take, since there are two different independent variables. Depending on which variable we choose, we can come up with different partial derivatives altogether, and often do.
The idea to keep in mind when calculating partial derivatives is to treat all independent variables, other than the variable with respect to which we are differentiating, as constants. Then proceed to differentiate as with a function of a single variable. To see why this is true, first fix
and define
as a function of
Then
The same is true for calculating the partial derivative of
with respect to
This time, fix
and define
as a function of
Then
All differentiation rules from Introduction to Derivatives apply.
Calculate
and
for the following functions by holding the opposite variable constant then differentiating:
treat the variable
as a constant. Then differentiate
with respect to
using the sum, difference, and power rules:
The derivatives of the third, fifth, and sixth terms are all zero because they do not contain the variable
so they are treated as constant terms. The derivative of the second term is equal to the coefficient of
which is
Calculating
These are the same answers obtained in [link].
treat the variable y as a constant. Then differentiate
with respect to x using the chain rule and power rule:
To calculate
treat the variable
as a constant. Then differentiate
with respect to
using the chain rule and power rule:
How can we interpret these partial derivatives? Recall that the graph of a function of two variables is a surface in
If we remove the limit from the definition of the partial derivative with respect to
the difference quotient remains:
This resembles the difference quotient for the derivative of a function of one variable, except for the presence of the
variable. [link] illustrates a surface described by an arbitrary function
In [link], the value of
is positive. If we graph
and
for an arbitrary point
then the slope of the secant line passing through these two points is given by
This line is parallel to the
Therefore, the slope of the secant line represents an average rate of change of the function
as we travel parallel to the
As
approaches zero, the slope of the secant line approaches the slope of the tangent line.
If we choose to change
instead of
by the same incremental value
then the secant line is parallel to the
and so is the tangent line. Therefore,
represents the slope of the tangent line passing through the point
parallel to the
and
represents the slope of the tangent line passing through the point
parallel to the
If we wish to find the slope of a tangent line passing through the same point in any other direction, then we need what are called directional derivatives, which we discuss in Directional Derivatives and the Gradient.
We now return to the idea of contour maps, which we introduced in Functions of Several Variables. We can use a contour map to estimate partial derivatives of a function
Use a contour map to estimate
at the point
for the function
The following graph represents a contour map for the function
The inner circle on the contour map corresponds to
and the next circle out corresponds to
The first circle is given by the equation
the second circle is given by the equation
The first equation simplifies to
and the second equation simplifies to
The
of the first circle is
and the
of the second circle is
We can estimate the value of
evaluated at the point
using the slope formula:
To calculate the exact value of
evaluated at the point
we start by finding
using the chain rule. First, we rewrite the function as
and then differentiate with respect to
while holding
constant:
Next, we evaluate this expression using
and
The estimate for the partial derivative corresponds to the slope of the secant line passing through the points
and
It represents an approximation to the slope of the tangent line to the surface through the point
which is parallel to the
Use a contour map to estimate
at point
for the function
Compare this with the exact answer.
Using the curves corresponding to
we obtain
The exact answer is
Create a contour map for
using values of
from
Which of these curves passes through point
Suppose we have a function of three variables, such as
We can calculate partial derivatives of
with respect to any of the independent variables, simply as extensions of the definitions for partial derivatives of functions of two variables.
Let
be a function of three variables. Then, the *partial derivative of
with respect to x,* written as
or
is defined to be
The partial derivative of
with respect to
written as
or
is defined to be
The partial derivative of
with respect to
written as
or
is defined to be
We can calculate a partial derivative of a function of three variables using the same idea we used for a function of two variables. For example, if we have a function
of
and we wish to calculate
then we treat the other two independent variables as if they are constants, then differentiate with respect to
Use the limit definition of partial derivatives to calculate
for the function
Then, find
and
by setting the other two variables constant and differentiating accordingly.
We first calculate
using [link], then we calculate the other two partial derivatives by holding the remaining variables constant. To use the equation to find
we first need to calculate
and recall that
Next, we substitute these two expressions into the equation:
Then we find
by holding
constant. Therefore, any term that does not include the variable
is constant, and its derivative is zero. We can apply the sum, difference, and power rules for functions of one variable:
To calculate
we hold x and y constant and apply the sum, difference, and power rules for functions of one variable:
Use the limit definition of partial derivatives to calculate
for the function
Then find
and
by setting the other two variables constant and differentiating accordingly.
Use the strategy in the preceding example.
Calculate the three partial derivatives of the following functions.
In each case, treat all variables as constants except the one whose partial derivative you are calculating.
Calculate
and
for the function
Use the strategy in the preceding example.
Consider the function
Its partial derivatives are
Each of these partial derivatives is a function of two variables, so we can calculate partial derivatives of these functions. Just as with derivatives of single-variable functions, we can call these second-order derivatives, third-order derivatives, and so on. In general, they are referred to as higher-order partial derivatives. There are four second-order partial derivatives for any function (provided they all exist):
An alternative notation for each is
and
respectively. Higher-order partial derivatives calculated with respect to different variables, such as
and
are commonly called mixed partial derivatives.
Calculate all four second partial derivatives for the function
To calculate
and
we first calculate
To calculate
differentiate
with respect to
To calculate
differentiate
with respect to
To calculate
and
first calculate
To calculate
differentiate
with respect to
To calculate
differentiate
with respect to
Calculate all four second partial derivatives for the function
Follow the same steps as in the previous example.
At this point we should notice that, in both [link] and the checkpoint, it was true that
Under certain conditions, this is always true. In fact, it is a direct consequence of the following theorem.
Suppose that
is defined on an open disk
that contains the point
If the functions
and
are continuous on
then
Clairaut’s theorem guarantees that as long as mixed second-order derivatives are continuous, the order in which we choose to differentiate the functions (i.e., which variable goes first, then second, and so on) does not matter. It can be extended to higher-order derivatives as well. The proof of Clairaut’s theorem can be found in most advanced calculus books.
Two other second-order partial derivatives can be calculated for any function
The partial derivative
is equal to the partial derivative of
with respect to
and
is equal to the partial derivative of
with respect to
In Introduction to Differential Equations, we studied differential equations in which the unknown function had one independent variable. A partial differential equation is an equation that involves an unknown function of more than one independent variable and one or more of its partial derivatives. Examples of partial differential equations are
(heat equation in two dimensions)
(wave equation in two dimensions)
(Laplace’s equation in two dimensions)
In the first two equations, the unknown function
has three independent variables—
—and
is an arbitrary constant. The independent variables
are considered to be spatial variables, and the variable
represents time. In Laplace’s equation, the unknown function
has two independent variables
Verify that
is a solution to the wave equation
First, we calculate
and
Next, we substitute each of these into the right-hand side of [link] and simplify:
This verifies the solution.
Verify that
is a solution to the heat equation
Calculate the partial derivatives and substitute into the right-hand side.
Since the solution to the two-dimensional heat equation is a function of three variables, it is not easy to create a visual representation of the solution. We can graph the solution for fixed values of t, which amounts to snapshots of the heat distributions at fixed times. These snapshots show how the heat is distributed over a two-dimensional surface as time progresses. The graph of the preceding solution at time
appears in the following figure. As time progresses, the extremes level out, approaching zero as t approaches infinity.
If we consider the heat equation in one dimension, then it is possible to graph the solution over time. The heat equation in one dimension becomes
where
represents the thermal diffusivity of the material in question. A solution of this differential equation can be written in the form
where
is any positive integer. A graph of this solution using
appears in [link], where the initial temperature distribution over a wire of length
is given by
Notice that as time progresses, the wire cools off. This is seen because, from left to right, the highest temperature (which occurs in the middle of the wire) decreases and changes color from red to blue.
During the late 1800s, the scientists of the new field of geology were coming to the conclusion that Earth must be “millions and millions” of years old. At about the same time, Charles Darwin had published his treatise on evolution. Darwin’s view was that evolution needed many millions of years to take place, and he made a bold claim that the Weald chalk fields, where important fossils were found, were the result of
million years of erosion.
At that time, eminent physicist William Thomson (Lord Kelvin) used an important partial differential equation, known as the heat diffusion equation, to estimate the age of Earth by determining how long it would take Earth to cool from molten rock to what we had at that time. His conclusion was a range of
million years, but most likely about
million years. For many decades, the proclamations of this irrefutable icon of science did not sit well with geologists or with Darwin.
Read Kelvin’s paper on estimating the age of the Earth.
Kelvin made reasonable assumptions based on what was known in his time, but he also made several assumptions that turned out to be wrong. One incorrect assumption was that Earth is solid and that the cooling was therefore via conduction only, hence justifying the use of the diffusion equation. But the most serious error was a forgivable one—omission of the fact that Earth contains radioactive elements that continually supply heat beneath Earth’s mantle. The discovery of radioactivity came near the end of Kelvin’s life and he acknowledged that his calculation would have to be modified.
Kelvin used the simple one-dimensional model applied only to Earth’s outer shell, and derived the age from graphs and the roughly known temperature gradient near Earth’s surface. Let’s take a look at a more appropriate version of the diffusion equation in radial coordinates, which has the form
Here,
is temperature as a function of
(measured from the center of Earth) and time
is the heat conductivity—for molten rock, in this case. The standard method of solving such a partial differential equation is by separation of variables, where we express the solution as the product of functions containing each variable separately. In this case, we would write the temperature as
is constant with respect to distance
and
is constant with respect to time
show that
and the right-hand side is only a function of
and they must be equal for all values of
Therefore, they both must be equal to a constant. Let’s call that constant
(The convenience of this choice is seen on substitution.) So, we have
Now, we can verify through direct substitution for each equation that the solutions are
and
where
Note that
is also a valid solution, so we could have chosen
for our constant. Can you see why it would not be valid for this case as time increases?
Which of the two constants,
or
must therefore be zero to keep
finite at
(Recall that
as
but
behaves very differently.)
For simplicity, let’s set
and find
such that this is the temperature there for all time
(Kelvin took the value to be
We can add this
constant to our solution later.) For this to be true, the sine argument must be zero at
Note that
has an infinite series of values that satisfies this condition. Each value of
represents a valid solution (each with its own value for
The total or general solution is the sum of all these solutions.
we assume that all of Earth was at an initial hot temperature
(Kelvin took this to be about
The application of this boundary condition involves the more advanced application of Fourier coefficients. As noted in part b. each value of
represents a valid solution, and the general solution is a sum of all these solutions. This results in a series solution:
Note how the values of
come from the boundary condition applied in part b. The term
is the constant
for each term in the series, determined from applying the Fourier method. Letting
examine the first few terms of this solution shown here and note how
in the exponential causes the higher terms to decrease quickly as time progresses:
Near time
many terms of the solution are needed for accuracy. Inserting values for the conductivity
and
for time approaching merely thousands of years, only the first few terms make a significant contribution. Kelvin only needed to look at the solution near Earth’s surface ([link]) and, after a long time, determine what time best yielded the estimated temperature gradient known during his era
increase per
He simply chose a range of times with a gradient close to this value. In [link], the solutions are plotted and scaled, with the
surface temperature added. Note that the center of Earth would be relatively cool. At the time, it was thought Earth must be solid.
Epilog
On May
physicist Ernest Rutherford spoke at the Royal Institution to announce a revised calculation that included the contribution of radioactivity as a source of Earth’s heat. In Rutherford’s own words:
“I came into the room, which was half-dark, and presently spotted Lord Kelvin in the audience, and realised that I was in for trouble at the last part of my speech dealing with the age of the Earth, where my views conflicted with his. To my relief, Kelvin fell fast asleep, but as I came to the important point, I saw the old bird sit up, open an eye and cock a baleful glance at me.
Then a sudden inspiration came, and I said Lord Kelvin had limited the age of the Earth, provided no new source [of heat] was discovered. That prophetic utterance referred to what we are now considering tonight, radium! Behold! The old boy beamed upon me.”
Rutherford calculated an age for Earth of about
million years. Today’s accepted value of Earth’s age is about
billion years.
**Partial derivative of
with respect to
**
**Partial derivative of
with respect to
**
For the following exercises, calculate the partial derivative using the limit definitions only.
for
for
For the following exercises, calculate the sign of the partial derivative using the graph of the surface.
The sign is negative.
The partial derivative is zero at the origin.
For the following exercises, calculate the partial derivatives.
for
for
and
for
and
for
Find
for
Let
Find
and
Let
Find
and
Let
Find
and
Let
Find
and
Let
Evaluate
and
Let
Find
and
Evaluate the partial derivatives at point
Find
at
for
Given
find
and
Given
find
and
The area of a parallelogram with adjacent side lengths that are
and in which the angle between these two sides is
is given by the function
Find the rate of change of the area of the parallelogram with respect to the following:
Express the volume of a right circular cylinder as a function of two variables:
and its height
a.
b.
c.
Calculate
for
Find the indicated higher-order partial derivatives.
for
for
Let
Find
and
Given
find
and
Given
find
and
Given
show that
Show that
is a solution of the differential equation
Find
for
Let
Find
Let
Find
Given
find all points at which
simultaneously.
Given
find all points at which
and
simultaneously.
Given
find all points on
at which
simultaneously.
Given
find all points at which
simultaneously.
Show that
satisfies the equation
Show that
solves Laplace’s equation
Show that
satisfies the heat equation
Find
for
Find
for
Find
for
Find
for
The function
gives the pressure at a point in a gas as a function of temperature
and volume
The letters
are constants. Find
and
and explain what these quantities represent.
The equation for heat flow in the
is
Show that
is a solution.
The basic wave equation is
Verify that
and
are solutions.
The law of cosines can be thought of as a function of three variables. Let
and
be two sides of any triangle where the angle
is the included angle between the two sides. Then,
gives the square of the third side of the triangle. Find
and
when
and
Suppose the sides of a rectangle are changing with respect to time. The first side is changing at a rate of
in./sec whereas the second side is changing at the rate of
in/sec. How fast is the diagonal of the rectangle changing when the first side measures
in. and the second side measures
in.? (Round answer to three decimal places.)
A Cobb-Douglas production function is
where
represent the amount of labor and capital available. Let
and
Find
and
at these values, which represent the marginal productivity of labor and capital, respectively.
at
at
The apparent temperature index is a measure of how the temperature feels, and it is based on two variables:
which is relative humidity, and
which is the air temperature.
Find
and
when
and
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