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The derivative has been a useful tool for us so far. For
instance, it has helped us determine where a function is increasing
or decreasing and where it has extrema. In this lab, we'll look at
another application of derivatives: providing approximate solutions to
problems we cannot solve exactly.

Remember that we started the course by looking at the graph of a
function near a point ** x**_{0} on a very
small scale. When we did this, the graph essentially looked like a
straight line which we called the tangent line.

Another way to state this observation is like this:

*
Near the point *** x**_{0}, the tangent line at
** x**_{0} provides a good
approximation for the graph.

#### Question 1: How to solve equations with Newton's Method

This question will explain a technique called * Newton's Method. *
It's a common problem to encounter equations we cannot easily solve
and Newton's Method is a means of finding approximate solutions.

Suppose we would like to solve an equation ** f(x) =
0.** Geometrically, this means that we want the values of
** x ** at which the graph of ** f **
crosses the ** x ** axis. If the graph of ** f(x)
** were a straight line, this would be easy. However, in other
cases, we might not be able to solve the equation ** f(x) = 0
** directly so we need a technique for finding approximate
solutions.

Suppose that we know there is a solution near the point
** x**_{0 }. (We'll see a more concrete
example of this in a minute.)
If we cannot solve the
equation ** f(x) = 0, ** we will instead look at the
tangent line at the point ** x**_{0 }
which is shown in the picture below.

The point at which the tangent line crosses the ** x
** axis can be thought of as an approximation of the solution
to the equation ** f(x) = 0. **

Now we will tackle the specific problem of solving ** f(x) =
x**^{2} - 5 = 0. Of course, we know that a solution is
but let's pretend we don't know that.
From our experience with squares, we can guess that the square root of
five is
between ** 2 ** and ** 3 ** since **
2**^{2} = 4 and ** 3**^{2} = 9.
In fact, let's assume that our initial guess is not even that good: we
will just say that we know the solution is somewhere around **
1. **

** Part a (2 marks) **

If you select "Question 1a" above, you will see the
graph of ** f(x) = x**^{2} - 5. On this graph,
you will also see a point which can be moved together with the
tangent line at that point. Move the point to **
x**_{0} = 1. This is our original
guess for the solution.

** Part b (2 marks) **

Now this is where the fun begins: we make the initial guess that a
solution is near **x**_{0} = 1 and we produce
an approximate solution by considering where the tangent line
crosses the axis. We will call this point **
x**_{1}. You should notice that this value is about
** 3 ** and it is a better approximation to the solution
than **x**_{0}. Select "Question 1b" above and
you will see
another dot which can be moved together with its tangent line. Now
move the dot to the point ** x**_{1}.

** Part c (2 marks) **

In Part b, you have found another approximate
solution ** x**_{2} where the tangent line to
the graph at ** x**_{1} crosses the ** x
** axis. Notice that this is a better approximation because
it is closer to the location of the root that we are
trying to find.
Now select "Question 1c" and
move the new dot to ** x**_{2}. You may want to
zoom in to position the dot carefully. In the same way, this produces
a new approximate solution ** x**_{3} which is an
even better approximation. Of course, you could continue this
indefinitely, but the approximate solutions start to stabilize at a
solution to ** f(x) = 0.** Notice that with three simple
steps, we have gotten quite close to the actual solution.

To summarize, starting with a rather bad guess
for the value of
we have used a repeated
procedure to generate better and better approximations to the true
value. This is the crux of Newton's method.

#### Question 2: A Practical Example of Newton's Method

Question 1 hopefully convinces you that Newton's Method is a pretty
simple trick: you simply replace a problem you cannot
solve---find where ** f(x) = 0 **---by a problem you
can solve---find where the tangent line crosses the ** x
**axis. This produces an approximate solution and the process
can be repeated. When the approximate solutions stabilize, then you
have found a solution.

In class, we'll see how to compute using Newton's Method,
but for now, we'll take it for granted that this can be done. In this
question, we would like to find the roots of the degree 3 polynomial:

Below is a device which will perform Newton's Method for you.
Simply enter a value for ** x**_{0} and click
the ** Step ** button. You will see the next approximate
solution appear. If you click the button again, you will see the next
result and so on.

** Part a (2 marks) **

Select "Question 2a" above and enter an inital guess
as to where a root to the polynomial ** p(x) ** is. Use
the "Step" button to find a root by repeating the process until the
result stabilizes.

** Part b (2 marks) **

Select "Question 2b" above and find another root to
the polynomial ** p(x) ** by making a different initial
guess. If you cannot quickly find a different root, you may wish to
think about the polynomial for a minute (for instance, where are its
critical points?)

#### Question 3: Euler's Method Explained

In the last two problems, we used Newton's Method to find
approximate solutions to * algebraic equations. * In this
problem, we will describe * Euler's Method, * a means of
finding approximate solutions to differential equations.

We will consider the differential equation

together with the initial value ** y(0) = 1. ** Of
course, we know that a solution is ** y(t) =
e**^{-t} so there is no need to find an approximate
solution. However, we will pretend that we don't know this so that we
we might illustrate the method.

Shown below is the graph of the solution ** y(t) =
e**^{-t}. We will build an approximate solution piece
by piece. The
approximate solution will not be a formula or a smooth
function. Rather, it will consist of a collection of straight line
segments which approximately "track" the true solution.

** Part a (1 mark) **

Select "a", enter the initial value ** y(0)
** for the function in the
window above the graph and press the button "OK". You will see a dot
appear on the graph at the corresponing point.

** Part b (1 mark) **

Since the differential equation is
and ** y(0) = 1,** what is the slope of the tangent
line (i.e. the derivative) at ** t = 0?** Select "b" above,
enter the slope and click "OK". You should see the tangent line
appear.

** Part c (1 mark) **

Generally, we apply Euler's Method when we do not know the
solution of the differential equation. So we approximate the solution
by the tangent line whose graph we have just constructed.

Select "c" above and enter the approximate value of the solution at
** t = 0.5 ** obtained from the tangent line. When you
click "OK", you should see a dot appear on the * tangent line*
you constructed in Part b.

** Part d (1 mark) **

As in Newton's Method, we can repeat this idea.
Remember that the differential equation is
Using your value for the approximate solution in Part c, what is the
approximate slope of the tangent line at ** t = 0.5.**
(It is minus the value you entered in Part c. Why?) Select "d"
above, enter your result and click "OK". You will see the next piece
of the approximation.

** Part e (1 mark) **

What is the value of your approximation at ** t
= 1.** Select "e" above, enter your result and click "OK".

The idea you have just seen is called Euler's Method. The point
is that we often face a differential equation which
cannot be easily solved. However, the differential equation gives us
the slope of the tangent line and so we can approximate the solution
by the tangent line. We can stop every so often and take a new
bearing: the differential equation will give us a new slope at our
approximate position. You can see that our approximate solution
does indeed track the actual solution.

#### Question 4: A Practical Example of Euler's Method (2 marks)

In the last question, we saw how to build approximate
solutions to a differential equation by approximating the solution
with its tangent lines. Of course, the tangent line at some point
only looks like the graph near that point. You can see this above:
the further away from the original point we get, the worse the
approximation is. We can get a better approximation by taking more
and smaller steps: this guarantees that the tangent line is only
being used in a region where the approximation is relatively good.
In this question, we will look at the differential equation

You may recognize this as the
differential equation from the last question in Lab 4. We will build
an approximate solution to this differential equation with the initial
value ** y(0) = 0.8 **

If you look below, you will see a graph with a window above
it. Using this window, you can experiment with the number of
steps and see how an increase in the number of steps produces
different approximate solutions. Begin with a small number of steps
and steadily increase. You should see that the approximations start
to stabilize when you use a large number of steps.

Enter the approximate time for which
** y(t) = 0.2** in the window below the graph.

The important thing to notice is that we originally had no idea
about how the solution to this equation behaved *
quantitatively.* (We did see how it behaved *
qualitatively* in Lab 4.) However, Euler's Method gives us a means
to build approximate solutions to any degree of accuracy by increasing
the number of steps.

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