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Subsections
The purpose of this lab is to use Maple to help you visualize and
understand the mean value theorem (MVT) and its applications.
The mean value theorem is one of the most important and useful
theorems from calculus. Before stating it, here is a (rather silly)
application of it.
Just after you get on the Pike by Auburn, you pass a car pulled over
by the highway patrol. Feeling glad that it isn't you, and
relatively safe because you have a radar detector, you blast on by. One
hour and eighty
miles down the road, however, another patrolman pulls you over. You're
not worried, because your detector had warned you to slow down.
Imagine your shock, however, when he proceeds to write you a ticket
for going 80 miles per hour! When you protest, he tells you that the
patrolman at Auburn had radioed your position ahead. He then claims that
since your average speed was eighty, the MVT says you must have been
going eighty at least once in your journey. Then you see the
``MathNet'' patch on his uniform, and you really start to get that sinking
feeling.
Mathematically, the MVT can be stated as follows.
Theorem 1 (MVT)
Suppose that
is continuous on the closed
interval
and differentiable on the open interval
. Then
for some
between
and
,
Geometrically, the MVT says that for at least one point strictly
between and , the slope of the tangent line to the graph of
is equal to the slope of the straight line between the two
points and . (Note: This line is often called the
secant line between and .) Obtaining a formula for the secant
line isn't hard, but to cut down on the amount of Maple drudgery, a
procedure, called secantline has been written that does this for you.
The secantline command takes three arguments. The first is a
function or expression, the next one is the base point, and the third
is the increment, .
Before you can use this procedure, you must first load it using the
with(CalcP) command, as shown in the following example session.
>
with(CalcP):
>
f := x -> x^3;
>
g := secantline(f(x),x=-1,2);
>
plot(f(x),g,x=-1..1);
Looking at the graph displayed in the previous example, you should be
able to see that there are two values of between and
where the slope of is the same as the slope of the secant
line between the two points and .
To find these values, we need to solve for the values of in the
interval where the derivative, , is equal to the slope
of the secant line, which is in this example. Maple
commands to do this, and plot the two tangent lines are shown
below. In this case the solve command finds both solutions. Note
the use of the label sol and the notation sol[1] and sol[2] to access the two roots.
>
sol := solve(diff(f(x),x)=1,x);
>
plot(f(x),g,tangentline(f(x),x=sol[1]),
tangentline(f(x),x=sol[2]),x=-1..1);
The solve command can't always do the job if the function
is complicated. When the solve command fails, there is no output
from the command, as shown in the example below. In this case, use the
fsolve command instead, as shown below.
>
f := x -> sin(x^2);
>
g := secantline(f(x),x=4,1);
>
plot(f(x),g,x=4..5);
>
solve(diff(f(x),x)=diff(g,x),x);
>
fsolve(diff(f(x),x)=diff(g,x),x=4..5);
It is important to restrict the interval in the fsolve command to in this case, because the equation
has an infinite number of solutions. In fact, there are three
solutions to this equation in the interval . The fsolve
command given above found one of them. The other two can be found by
restricting the range in the fsolve command to an interval that
includes only the desired root. To do this, you need to determine
approximately where the roots are. Looking at the graph of the function
and the secant line is probably the best way to do this - just look
for points on the
curve where the slope of the tangent line is the same as the slope of
the secant line.
If you do this for the previous example,
, you
should be able to recognize that one of the roots is in the interval
. Using this interval in the fsolve command produces
the following results.
>
fsolve(diff(f(x),x)=diff(g,x),x=4..4.3);
There is nothing special about the interval ; any other
interval that included the root at
, but didn't
include any other roots would have worked. That is, the interval
would have worked equally as well, but the interval
is not a good choice because it includes two roots.
- For each of the following functions, verify graphically that the
MVT holds and find all values of , , that satisfy the
condition
Your solution should include a graph of the function and the secant
line on the same plot.
-
, , .
-
, , .
-
, , .
- Explain how the MVT applies to support the patrolman's claim.
- In some cases, the slope of the secant line can be interpreted
as an average value. For example consider the falling body problem
from the text. If is the position of a body, then the quantity
is change in position over the change in time, or the average velocity
of the body over the time interval. Because the derivative of
is the velocity , the mean value theorem says that
there is a time between and at which the instantaneous
velocity is equal to the average velocity.
Suppose that a body starts at a
height of 140 feet and that air resistance can be neglected. At time
zero, the body is released and subsequently falls to the ground. The
height is given by
.
- Find the time
at which the body hits the
ground.
- Find the average velocity of the body from when it was released
to when it hits the ground.
- Find the time at which the instantaneous velocity is equal to
the average velocity.
- In the previous problem, you should have found that the time at
which the instantaneous velocity is equal to the average velocity is
exactly half the time it takes for the body to hit the ground. Is this
always true? Go through the following steps to show that it is. Your
answers to parts a and b should be in terms of .
- Use
as the position function, where
is the initial height of
the body, and find the
time at which the body hits the ground.
- Using the value of you just computed, find the average
velocity of the body.
- Use to compute the value of at which the
instantaneous velocity is equal to the average velocity. Then show
that this value is half the time it takes for the body to hit the
ground.
Next: About this document ...
Up: lab_template
Previous: lab_template
Jane E Bouchard
2000-10-31