Parabolic Integration

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Real Life Parabolas

Parabolas are very well-known and are seen frequently in the field of mathematics. Their applications are varied and are apparent in our every day lives. For example, the main image on the right is of the Golden Gate Bridge in San Francisco, California. It has main suspension cables in the shape of a parabola. Of course there are many more examples of parabolic architecture such as roller coasters, flight paths, and probably the most recognized, the Golden Arches of McDonald's. With all of these appearances in real life, have you ever wondered how to find the area under one?

Real Life Parabolas

Field: Algebra
Created By: Aaron Logan

1 Basic Description
- 1.1 Basic Definition
2 A More Mathematical Explanation
3 Real World Application
4 Teaching Materials
5 References
6 Future Directions for this Page

Basic Description

Two methods for finding parabolic area exist. One is very accurate and the other is more of an approximation method. This procedure for approximation is known as the Rectangle Method and is used by finding the area of rectangles that can fit in the parabola. The area of these rectangles are added together, giving you the approximate area under or above the parabola.

For a detailed overview of parabolas, see the page, Parabola. However, we will provide a brief summary and description of parabolas below before explaining how to find the area beneath or above one.

Basic Definition

You may recall first learning about parabolas and your teacher telling you that it is a curve in the shape of a "u" and can be oriented to open upwards, downwards, sideways, or diagonally. To be a little more mathematical, a parabola is a conic section formed by the intersection of a cone and a plane. Below is an image illustrating this.

When you were first introduced to parabolas, you learned that the quadratic equation, $y= a(x-h)^2+ k$ is its algebraic representation (where $h$ and $k$ are the coordinates of the vertex and $x$ and $y$ are the coordinates of an arbitrary point on the parabola.

As you progressed in mathematics, you learned how to find the area of the space enclosed by the parabola. This can be accomplished in two different ways:

Using Definite Integration
Using the Rectangle method (Also referred to as finding the Riemann Sum)

We will explain both of these approaches by posing a problem and then solving it step by step. But first we are going to familiarize you with some parabolic architecture and occurrences found in the real world.

A More Mathematical Explanation

Integral Approach to Determining Area

Typically, when attempting to find the area underneath a pa [...]

Integral Approach to Determining Area

Typically, when attempting to find the area underneath a parabola, we take its integral. Below is a proposed problem with a numbered procedure of individual steps for completion:

Find the area under the curve $y =4-x^2$ between $x = 0$ and $x = 2$ and the $x-axis$ .

1.Graphing the function first will help you to visualize the curve. Below I have graphed the function using the mathematical program Derive, but you can easily graph it either using your calculator or by hand.

2.Now we will algebraically evaluate the expression by taking its integral; doing so will give us an EXACT area. Integration is shown and calculated by:

$Area$	$=$	$\int_{a}^{b}f(x)dx$
$\int_{a}^{b}f(x)dx$	$=$	$\left[F(x)\right]_{a}^{b}$
$\left[F(x)\right]_{a}^{b}$	$=$	$F(b)-F(a)$

3.So for this particular problem, our bounds ( $a$ and $b$ ) will be $x = 0$ and $x = 2$ respectfully and our integral will look like:

$\int_{0}^{2}(4-x^2)dx$

4. Now we will integrate the function and then substitute the bounds for $x$ as follows

$\int_{0}^{2}(4-x^2)$	$=$	$\left[4x-\frac{x^3}{3}\right]_{0}^{2}$
$\left[4x-\frac{x^3}{3}\right]_{0}^{2}$	$=$	$\left[4(2)-\frac{2^3}{3}\right]-\left[4(0)-\frac{0^3}{3}\right]$
$\left[4(2)-\frac{2^3}{3}\right]-\left[4(0)-\frac{0^3}{3}\right]$	$=$	$\left[\frac{24}{3}-\frac{8}{3}\right]-\left[0-0\right]$
$\left[\frac{24}{3}-\frac{8}{3}\right]-\left[0-0\right]$	$=$	$\frac{16}{3}$
$\therefore\int_{0}^{2}(4-x^2)dx$	$=$	$\frac{16}{3}$

5. So, the area under the curve $y=4-x^2$ between $x=0$ and $x=2$ is $\frac{16}{3}units^2$ .

Determining Area Using the Riemann Sum

Another procedure exists for finding the area under a curve, but it gives an approximation, not an exact value like the definite integral. This method is known as the Rectangle Method, more widely known as the Riemann Sum. To compute the Riemann sum, you divide the space enclosed by the parabola into rectangles and add together their areas. This method was used before integration was developed, mostly by the ancient Greeks. As we did above with definite integration, we will propose a problem accompanied by the procedure used for solving it.

Find the area under the curve $y = 4-x^2$ between $x = 0$ and $x = 2$ for $n=10$ using the rectangle method.

1.To find the width of each of the 10 rectangles ( $n$ is the number of rectangles used to find the area under the curve) we will use the the formula,

$w=\vartriangle(x)=\frac{b-a}{n}$ where $a$ and $b$ are the boundaries and $\vartriangle(x)$ is the change in $x$ values

2.So for our particular curve, the above formula will look like this:

$w=\vartriangle(x)=\frac{2-0}{10}$

3.Therefore, the width of each of the 10 rectangles will be 0.2 units

$w=0.2$ units

4.Now that we have determined the width of each of the rectangles, we need to find each rectangle's individual height. To do this, we will use the equation of the curve:

$Rectangle 1$	$=$	$4-(0.0)^2=4$
$Rectangle 2$	$=$	$4-(0.2)^2=3.96$
$Rectangle 3$	$=$	$4-(0.4)^2=3.84$
$Rectangle 4$	$=$	$4-(0.6)^2=3.64$
$Rectangle 5$	$=$	$4-(0.8)^2=3.36$
$Rectangle 6$	$=$	$4-(1.0)^2=3$
$Rectangle 7$	$=$	$4-(1.2)^2=2.56$
$Rectangle 8$	$=$	$4-(1.4)^2=2.04$
$Rectangle 9$	$=$	$4-(1.6)^2=1.44$
$Rectangle 10$	$=$	$4-(1.8)^2=0.76$

5. Using the values above which represent the heights of the 10 rectangles under the parabola, we next multiply each height by the width determined earlier, $0.2$ . By doing this we are finding the area of each individual rectangle, $(h)\cdot(w)$

$Rectangle 1$	$=$	$(4)\cdot(0.2)=0.8$
$Rectangle 2$	$=$	$(3.96)\cdot(0.2)=0.792$
$Rectangle 3$	$=$	$(3.84)\cdot(0.2)=0.768$
$Rectangle 4$	$=$	$(3.64)\cdot(0.2)=0.728$
$Rectangle 5$	$=$	$(3.36)\cdot(0.2)=0.672$
$Rectangle 6$	$=$	$(3.00)\cdot(0.2)=0.600$
$Rectangle 7$	$=$	$(2.56)\cdot(0.2)=0.512$
$Rectangle 8$	$=$	$(2.04)\cdot(0.2)=0.408$
$Rectangle 9$	$=$	$(1.44)\cdot(0.2)=0.288$
$Rectangle 10$	$=$	$(0.76)\cdot(0.2)=0.152$

6 To determine the area, add together the area of the ten rectangles. When this is done, we have an area of $5.72$ $units^2$ . (This is the approximate area under the parabola)

The area using this method is slightly greater than the area determined using definite integration (only by approximately $0.4units^2$ ). We will explain why this is so in the following section.

Integral Approach v. Rectangle Method

We have now used two different methods to determine the area enclosed by a parabola. The first, Integration, resulted in an EXACT area, whereas the second, The Rectangle Method, provided only an approximation, meaning that the value could be larger or smaller than the actual area found using definite integration.

With this specific problem, we got two very numerically close values for the area of the parabola. Using definite integration, we found an area of $5.33$ $units^2$ , but with the rectangle method, an area of $5.72$ $units^2$ . If you look back to the image of the parabola with the ten green rectangles, you will notice that the tops of the rectangles extend over the top and sides of the parabola. This accounts for the larger area value and is known as the left Riemann Sum. If the rectangles would have been too short, we would call this taking the right Riemann Sum.

Real World Application

Within this section we are going to briefly introduce you to the magnificent Golden Gate Bridge (see the main image of this page) and then derive an equation that will represent the main parabolic suspension cable. Lastly, we will find the area under this suspension cable using integration.

The Golden Gate Bridge, located in San Francisco, California is a very well known parabolic suspension bridge, being one of the longest suspension bridge in America as well as a modern wonder of the world. (The longest suspension bridge in America as well as in the entire Western hemisphere is the Mackinac Bridge located in Michigan. It spans 26,372 ft, almost 5 miles! ) The Golden Gate spans the San Francisco Bay into the Pacific Ocean and was completed in 1937. Below we have included dimensions of the bridge that will be useful in determining the parabolic equation of its main suspension cables:

The length of the main span is 4,200 ft (1,280 m)
The height of the towers from the roadway to the cable is 500 ft (152 m)

Using these measurements, we are able to find that the parabolic equation for the parabolic main suspension cable is:

$y=(1.1337\cdot10^{-4})x^2$

This may seem a little overwhelming in the sense that I haven't provided a step by step procedure for this equation, but we will do so in order to ensure you understand!

Okay, there are two possible forms for writing a parabolic equation. The first is standard form and the second is vertex form. We are going to choose to use the vertex form and then later convert it into standard form. For this form we are going to need two sets of points, a vertex and another point on the parabola. So, we know the vertex form of a parabola is represented by $y=a(x-h)^2 +k$ , where $h$ and $k$ are the coordinates of the vertex and the points $x$ and $y$ are the coordinates of the other point.

So now you are probably wondering what our vertex here could be, right? Well, we are making the vertex $(0,0)$ and the other known point $(2100, 500)$ . We want to first direct your attention to the image pictured below: it shows the Golden Gate Bridge and the measurements we are using in our vertex form equation. As you can see, the length of the main suspension is $4,200 ft$ and the height of the tower is $500 ft$ .

For easier calculations it is best to divide the parabola in half symmetrically so that the bounds for taking the integral will be $0$ and $2100$ . This way we can simple multiply the result by $2$ in order to account for the other portion. Doing this, we can make a vertex at $(0,0)$ and chose our point of intersection to be the top of the tower represented by $(2100, 500)$ . Now we have the two coordinate sets needed to plug into our vertex form equation. Shown below is the equation with the vertex $(0,0)$ substituted for $h$ and $k$ :

$y$	$=$	$a(x-0)^2+0$
$y$	$=$	$ax^2$

Now we are going to substitute the other point $(2100, 500)$ for $x$ and $y$ in the equation above, giving us:

$500$	$=$	$2100^2a$
$500$	$=$	$4,410,000$
$\frac{500}{4,410,000}$	$=$	$a$
$a$	$=$	$1.1337\cdot10^{-4}$

Now that we have found the value of $a$ we can substitute it into the equation $y=ax^2$ . Doing this will get us our desired parabolic equation:

$y=(1.1337\cdot10^{-4})x^2$

The next step is to find the area under this curve and then multiply it by $2$ so to account for the other side of the parabola. To do this we are going to take the definite integral with a lower bound $0$ and an upper bound $2100$ . As you will notice, we have changed $1.1337\cdot10{-4}$ into standard form. This explains why we have used the sign for approximation.

$Area$ $\approx2\int_{0}^{2100}0.000113x^2dx$

Next, rewrite using the upper and lower bounds after taking the integral:

$\int_{0}^{2100}0.000113x^2dx=\left[\frac{0.000113x^3}{3}\right]_{0}^{2100}$

Substitute the bounds into the already taken integral:

$\left[\frac{0.000113x^3}{3}\right]_{0}^{2100}=\left(\frac{0.000113(2100)^3}{3}\right)-\left(\frac{0.000113(0)^3}{3}\right)$

Simplify:

$\left(\frac{0.000113(2100)^3}{3}\right)-\left(\frac{0.000113(0)^3}{3}\right)=\left(\frac{1046493}{3}\right)-\left(\frac{0}{3}\right)$

Simplify:

$\left(\frac{1046493}{3}\right)-\left(\frac{0}{3}\right)=348,831$

Multiply $348,831$ by $2$ to account for the other side of the parabola:

$\therefore2\int_{0}^{2100}0.000113x^2dx=697,662$

Now we know that the area under the main parabolic suspension cable of the Golden Gate bridge is $697,662 Ft^2$ . Remember, this is the area from the road, not from the water.

Teaching Materials

There are currently no teaching materials for this page. Add teaching materials.

References

[1] About the Bridge. Retrieved from http://www.mackinacbridge.org/about-the-bridge-8/

[2] Books, P. "What Is the Difference Between a Parabola and a Catenary? (2011, April 27). Retrieved from http://journaleducation.net/classroom/mathematics/difference-parabola-catenary-3065.html

[3] Bridge Design and Construction Statistics. Retrieved from http://www.goldengatebridge.org/research/factsGGBDesign.php

[4] Bourne, M. The Area under a Curve. Retrieved from http://www.intmath.com/integration/3-area-under-curve.php

[5] Calvert, J.B. Parabola. (2002, May 3). Retrieved from http://mysite.du.edu/~jcalvert/math/parabola.htm

[6] Glydon, N. Suspension Bridges. Retrieved from http://mathcentral.uregina.ca/beyond/articles/Architecture/Bridges.html

[7] Golden Gate Bridge. Retrieved from http://en.wikipedia.org/wiki/Golden_Gate_Bridge

[8] Hague, C.H. "What shapes do roller coaster hills/dips follow? (parabolas? circles? etc.). (2003, July 23). Retrieved from http://www.madsci.org/posts/archives/2003-07/1059246836.Eg.r.html

[9] Ian. Applications of Parabolas. (2000, October 24). Retrieved from http://mathforum.org/library/drmath/view/53224.html

[10] Kukel, P. "Hanging With Galileo. Retrieved from http://whistleralley.com/hanging/hanging.htm

[11] Parabolic Reflectors and the Ideal Flashlight. Retrieved from http://www.maplesoft.com/applications/view.aspx?SID=5523

[12] Sandborg, D. Physics of Roller Coasters. Retrieved from http://cec.chebucto.org/Co-Phys.html

[13] Suspension Bridges. Retrieved from http://carondelet.net/Family/Math/03210/page4.htm

[14] Williams, S. Types of Parabolic Bridges. (2010, December 31) Retrieved from http://www.ehow.com/list_7712164_types-parabolic-bridges.html

Future Directions for this Page

The possibilities for further exploration and addition to this page are bountiful. Developing a section explaining the difference between a parabola and a catenary would be fantastic, as so often people confuse the two curves. Also, finding the area between the suspension cable and the road by use of the Riemann Sum would be very beneficial!

If you are able, please consider adding to or editing this page!
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Categories: Higher Education | Math Images In Progress | Algebra | Aaron Logan

Parabolic Integration

From Math Images

Contents

Basic Description

Basic Definition

A More Mathematical Explanation

Integral Approach to Determining Area

Integral Approach to Determining Area

Determining Area Using the Riemann Sum

Integral Approach v. Rectangle Method

Real World Application

Teaching Materials

References

Future Directions for this Page

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