Calculus Online: Lab 3

Welcome to Lab 3 of Math 101, Sections 203, 204, 207 and 209

Instructions

Answer all the questions below.
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Arc Length and Numerical Techniques for Integration

Up to this point in our course, we have seen many applications of the definite integral. Generally, we have evaluated definite integrals using the Fundamental Theorem of Calculus. However, the Fundamental Theorem of Calculus cannot always be applied: for instance, some functions do not have antiderivatives which are expressible in terms of familiar functions. When we cannot find a simple way to integrate a function, we may need to resort to some numerical approximation. In this lab, we will see one more application of the definite integral and explore a few ways to approximate definite integrals.

Question 1: Arc Length

It is straightforward to find the length of a straight line segment using the Pythagorean Theorem. However, it is sometimes helpful to know the length of a curve such as the circumference of a circle. In this situation, we cannot rely on the Pythagorean Theorem since the length is not being computed along a straight line. However, we can use straight lines to obtain an approximation to the curve.
In the exercise below, you are shown a curve which represents the shape of a cable suspended between two points. How long is the cable? That's a bit difficult to answer precisely. However, you will see a straight line segment connecting the ends of the cable and the length of this segment displayed below. You may add line segments by clicking the mouse near a point on the cable. The length of all the line segments will then be displayed. Notice that when you add quite a few line segments, you have a pretty good approximation to the length of the cable. You may remove points by selecting the button "Remove" and clicking to remove a particular point. For technical reasons, you are only allowed to add 15 points.

2 marks: Estimate the length of the cable and enter your estimate in the box above. (As always, no units please.)

Does it feel like there's a definite integral lurking in this problem? Let's see.
The cable in this problem is described by the graph of
$f(x) = 2(e^{x/2} + e^{-x/2})$
between x = 2 and x = -2. Let's look at a very small piece of the cable. In fact, we will assume that the piece is so small that it is indistinguishable from a straight line.

If we call the horizontal distance represented by this piece of the cable dx and the vertical distance dy, then the length of this small piece of cable is
$dl = \sqrt{dx^2 + dy^2} = \sqrt{1 + (\frac{dy}{dx})^2}~ dx$
Notice that the notation $\frac{dy}{dx}$ is no accident: this is the precisely the slope of the hypotenuse of this little triangle or, in other words, the slope of the cable. Since the scale is so small, the cable is indistinguishable from its tangent line and this means that the derivative $f^\prime = \frac{dy}{dx}$ as we might expect.
In the usual way, we will now add together all the lengths of these very small pieces of the cable by integrating. This shows that the length of the cable is
$l = \int_a^b \sqrt{1 + (\frac{dy}{dx})^2} ~dx = \int_a^b \sqrt{1 + (f^\prime(x))^2} ~dx$
This is a pretty nice expression. However, let's try to apply it to find the length of our cable where the function is $f(x) = 2(e^{x/2} + e^{-x/2}).$ Then we have $f^\prime(x) = e^{x/2} - e^{-x/2}$ and so
$\begin{eqnarray*} l & = & \int_{-2}^2 \sqrt{1 + (e^x -2 + e^{-x})} ~dx \\ & = & \int_{-2}^2 \sqrt{e^x - 1 + e^{-x}} ~dx \end{eqnarray*}$
This is an integral which we could evaluate using the Fundamental Theorem of Calculus but it would be quite a bit of work to find an antiderivative. In addition, there are some other functions, such as $e^{-x^2},$ whose antiderivatives cannot be expressed in terms of familiar functions. In situations like this---when an antiderivative is either difficult to find or impossible---the Fundamental Theorem of Calculus cannot be applied and we need some other tool for evaluating definite integrals.

Question 2: Numerical Techniques

To approximate a definite integral, the simplest thing we can try is the idea we used at the beginning of the term: think of the definite integral as representing an area under a graph and then approximate this area by the area of several rectangles as in the picture below.

We call this the left endpoint approximation and we form it by dividing the interval [a,b] into n pieces of width h. (The picture above has n = 2. ) This gives us points x₀, x₁, ... , x_n and we call the value of the function at these points y₀,y₁, ... , y_n. Now the area of the first rectangle is y₀ h, the area of the second rectangle y₁ h and so on.
This gives the following formula for the left endpoint approximation
$\int_a^b f(x)~dx \approx L_n = h(y_0 + y_1 + \ldots + y_{n-1})$
In the picture above, you can see that there is a lot of error in this approximation because we are pretending that the function is constant along each little interval. In fact, we can do better if we assume that the function's graph is a straight line along each interval like this:

This is called the trapezoidal approximation: instead of using rectangles to approximate the area, we are using trapezoids. The area of the first trapezoid is $\frac h2(y_0 + y_1)$ while the area of the second is $\frac h2(y_1 + y_2)$ and so on. Putting all this together, we have
$\begin{eqnarray*} \int_a^b f(x)~dx \approx L_n & = & \frac h2(y_0 + y_1) + \frac h2(y_1 + y_2) + \ldots + \frac h2(y_{n-1} + y_n) \\ & = & \frac h2(y_0 + 2y_1 + 2y_2 + \ldots + 2y_{n-1} + y_n) \end{eqnarray*}$
Of course, we can do better still if we approximate the area of each little piece by the area under a parabola.

This is called Simpson's Rule and you can see that, at least in this picture, it gives quite a good approximation. It takes a little time to work this out and you will probably see it in class, but the area under the arc defined by the first three points works out to be $\frac h3(y_0 + 4y_1 + y_2).$ We can continue in this way if we have more steps.
$\begin{eqnarray*} \int_a^b f(x)~dx \approx S_{2n} & = & \frac h3(y_0 + 4y_1 + y_2) + \frac h3(y_2 + 4y_3 + y_4) + \ldots + \frac h3(y_{2n-2} + 4y_{2n-1} + y_{2n}) \\ & = & \frac h3(y_0 + 4y_1 + 2y_2 + \ldots + 2y_{2n-2} + 4y_{2n-1} + y_{2n}) \end{eqnarray*}$

Part a (2 marks)

Select "Part a" in the diagram above. You may move the boxes with the mouse. Notice that h = 1 and modify the points so that the left endpoint approximation
$L_2 = h(y_0 +y_1) = 5.$ There are many ways to accomplish this task.

Part b (2 marks)

Select "Part b" in the diagram. Modify the points so that the trapezoidal approximation
$T_2 = \frac h2(y_0 + 2y_1 + y_2) = 5.$ Once again, this can be done in many ways.

Part c (2 marks)

Select "Part c" in the diagram. Modify the points so that the approximation given by
$S_2 = \frac h3(y_0 + 4y_1 + y_2) = 3.$

Question 3

You can imagine that in all three of these approximations, the approximation becomes more accurate when we put in more steps. Still, the trapezoidal approximation will generally be more accurate than the left endpoint approximation since it is based on a better approximation of the function. In the same way, the approximation from Simpson's Rule should be better than the other two.
In this question, we'll look at the definite integral
$\int_0^1 \frac{4}{1+x^2}~dx = 4\tan^{-1}x ~|_0^1 = 4\tan^{-1}(1) = \pi.$
In other words, we know the value of this integral so we can use it to see how accurate our three approximations are.

Part a (2 marks)

Select "Part a" above and approximate the definite integral above using the left endpoint approximation with n = 2. Enter your result in the window above. ( Note: the width of the intervals h is no longer 1. )

Part b (2 marks)

Select "Part b" above and approximate the definite integral using the trapezoidal approximation with n = 2. Enter your result in the window above.

Part c (2 marks)

Select "Part c" above and approximate the definite integral using Simpson's Rule with 3 points. Enter your result in the window.

Of course, we know that the value of the integral is $\pi \approx 3.1415\ldots$ so you can see the relative accuracy of the approximations. In fact, by using smaller intervals, Simpson's Rule can be used to quickly generate quite accurate approximations to $\pi.$

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