UBC Wiki - User contributions [en]

Course:Math110/003/Teams/Ticino/The Richter Scale

2011-02-03T03:22:32Z

BenJeffery: It's a start. More info would be awesome!

==The Richter Scale==
===Background===

The Richter Scale is a logarithmic scale used to measure the intensity of earthquakes. The earthquake in Chile last February measured an 8.8 on this scale, while the devastating Haitian earthquake was measured at 7.0. There is clearly a difference between these two numbers, but what does it really mean?

===Logarithmic functions===
The Richter Scale measures how much the ground moves during an earthquake.
The formula for the Richter scale is as follows, where A is the amplitude of the readings recorded by the seismographs:

<math>M_L = \log_{10} A - \log_{10}A_0(\delta)</math><ref>http://en.wikipedia.org/wiki/Richter_magnitude_scale</ref>

The important thing to note here is that this formula uses a log of base 10. This means that for every increase of 1.0 in <math>M_L</math>, the amplitude of the ground's motion has increased by a factor of 10. Examining the previously-mentioned earthquakes in Chile and Haiti, the amplitude of the ground motion was 63 times greater in Chile than in Haiti. <ref>http://www.backbenchmedia.com/richter-scale-equation-earthquakes/</ref>

===Why Logarithmic?===

Looking at this scale, one might wonder what the benefits of using a logarithmic scale. An increase of 1.0 does not intuitively suggest a tenfold increase in magnitude. However, if we look at the numbers involved, it makes a lot more sense.

Given that the amplitudes measured can vary so drastically, it makes sense to use a scale that deals in small numbers. The lowest rating on the scale that is typically felt by people is around 3.0. The "Big One" earthquake that is predicted for BC is expected to have a rating of approximately 9.0, or even higher. If we weren't using a logarithmic scale, we would be looking at something like the following:

3.0 on the scale: in the vicinity of <math>10^3</math>, or 1,000.

9.0 on the scale: somewhere around <math>10^9</math>, or 1,000,000,000.

It would be much more unwieldy talking about an earthquake that measured 532,054,942 on the amplitude scale. Being able to express this in logarithmic makes the information much more manageable and comprehensible to the average person.

<references/>

Course:Math110/003/Teams/Ticino/The Richter Scale

2011-02-03T03:10:50Z

BenJeffery:

Course:Math110/003/Teams/Ticino/The Richter Scale

2011-02-03T03:10:14Z

BenJeffery:

Course:Math110/003/Teams/Ticino/The Richter Scale

2011-02-03T03:09:22Z

BenJeffery:

Course:Math110/003/Teams/Ticino/The Richter Scale

2011-02-03T02:57:24Z

BenJeffery:

Course:Math110/003/Teams/Ticino/The Richter Scale

2011-02-03T02:02:08Z

BenJeffery:

Course:Math110/003/Teams/Ticino/The Richter Scale

2011-02-03T00:11:42Z

BenJeffery: Created page with "Filler"

Filler

Course:MATH110/Archive/2010-2011/003/Teams/Ticino

2011-02-02T21:39:35Z

BenJeffery:

{{
Infobox MATH110 Teams
| team name = Ticino
| member 1 = [[User:BenJeffery|BenJeffery]]
| member 2 = [[User:BernadetteHii|BernadetteHii]]
| member 3 = Caitlin Lastiwka-Farquharson
| member 4 = Micha Gutmanis
}}
In workshop M.

===Keywords===

Caitlin - Watermelon

Micha - Music

Bernadette - Purple

Ben - Coffee

===Group Work===

[[Course:MATH110/003/Teams/Ticino/Homework_11|Homework 11]]

[[Course:MATH110/003/Teams/Ticino/Homework_12|Homework 12]]

[[Course:Math110/003/Teams/Ticino/The_Richter_Scale|The Richter Scale]]

Course:MATH110/Archive/2010-2011/003/Teams/Ticino/The Richter Scale

2011-02-02T21:38:59Z

BenJeffery: Created page with "Filler"

Filler

User:BenJeffery

2011-01-28T15:59:31Z

BenJeffery:

Hi, I'm Ben Jeffery.
I'm in the BCS program with the department of Computer Science.
I haven't done algebra in about 9 years, so I may be rusty...

----

==Homework 12==

My area of information is computer science. Though there are many applications of calculus (mostly lambda calculus or predicate calculus), an "interesting page of text" would take far more time to write, in a higher level of expertise than I have.

== René Descartes ==

The French polymath René Descartes was one of the most influential people of the last 1000 years. He pioneered areas of mathematics, philosophy, physics, and even had an influence on the then-irrelevant field of psychology.
Descartes is most popularly known for his system of coordinates, the Cartesian coordinate system. Legend has it that Descartes was watching a fly walking on his ceiling when he had an epiphany. He realized that the location of that fly could be shown at any point as having a certain distance from the corner of the ceiling. <ref>[http://library.thinkquest.org/27694/Rene%20Descartes.htm]</ref> This story may not be true, but it serves as an interesting tale, and an example of how brilliant insights can come from the most innocuous moments.

Descartes wasn't solely a mathematician. In philosophy and anatomy, one of his most endearing, though not enduring, contributions was the concept that the soul (a very common belief at the time) was in fact connected to the body. As the nexus of this connection, he chose the rather peculiar Pineal gland, in the brain. René argued that this gland was the physical location that the soul, a non-corporeal concept, anchored itself to the body.

Despite his occasionally quirky ideas (to be fair, these were mostly a product of the times he lived in), Descartes has had an amazing influence on modern science, and from the Cartesian system that is used today in myriad applications, to his most famous quotation, "Cogito Ergo Sum.", he has shown that creativity, study, and a fresh approach to the world can create a resounding impact.

<references/>

Course:MATH110/Archive/2010-2011/003/Teams/Ticino/Homework 12

2011-01-26T00:11:34Z

BenJeffery:

==Altering a Logistical graph==

<math>p(t)=\dfrac{1}{1+e^{-t}}</math>

[[File:Ticino graph 1.png|300px]]

_____________________

To change the height of the horizontal asymptote to ''k'', we can add ''k'' to the formula as follows:

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{-t}}</math>

The graph shows what happens when ''k = 8'':
<math>p(t)=\dfrac{1}{\tfrac {1}{8}+e^{-t}}</math>

[[File:Ticino2.JPEG|300px]]

_____________________

In order to change the y-intercept of our function, we can add any number, ''q'', where ''q'' is any value between negative infinity and infinity, as shown:

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{-t+q}}</math>

The graph shows what happens when ''k = 8'' and ''q = 3'':
<math>p(t)=\dfrac{1}{\tfrac {1}{8}+e^{-t+3}}</math>

[[File:Ticino3.jpg|300px]]

_______________________

BONUS: to change the slope of the function, multiply ''t'' by ''s'', where ''s'' is any number.

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{-st + q}}</math>

Here's what happens when ''k = 8'', ''q = 3'' and ''s = 5'':
<math>p(t)=\dfrac{1}{\tfrac {1}{8}+e^{-5t + 3}}</math>

[[File:Ticino4.jpg|300px]]

==Applications of the data==

Our group has chosen market share as a real life example of our model. Specifically we have focused on the Lexus brand of luxury cars. We have denoted the X axis as time (10’s of years), and our Y axis as number of Lexus’ sold. As the Lexus brand comes out with a new car, initially there will be great increase in consumer interest thus increasing the number of cars sold. However, this example reflects our model as there is a fixed constraint, creating a vertical asymptote, at which the number of Lexus’ will never exceed. The fixed constraint in this case is the number of consumers with the means of purchasing the brand, as well as market share. At a certain point, denoted on our graphs at 8 (on the y-axis) marketing stops being affective and regardless of effort to sell more of the product, because of substitute goods as well as income levels the graph will level off at 8. The number of product sold is therefore dependent on the market.
An increasing slope indicates more sales of the car which suggests an effective marketing campaign. Likewise, a decreasing slope suggests marketing campaign is ineffective and the company should do something to increase sales.
The graph represents a logistic growth.
The y-intercept represents the current number of cars sold, the beginning stages of a growing company.
Our final altered graph, where our x-axis is years and our y-axis is millions of cars sold, our model predicts that with a very successful advertising campaign, in 18 months we should expect to have sold approximately 6.5 million cars.

=====References=====
http://billharlan.com/pub/papers/logistic/logistic.html

Course:MATH110/Archive/2010-2011/003/Teams/Ticino/Homework 12

2011-01-26T00:08:18Z

BenJeffery:

Course:MATH110/Archive/2010-2011/003/Teams/Ticino/Homework 12

2011-01-25T23:47:49Z

BenJeffery:

<math>p(t)=\dfrac{1}{1+e^{-t}}</math>

[[File:Ticino graph 1.png|300px]]

_____________________

To change the height of the horizontal asymptote to ''k'', we can add ''k'' to the formula as follows:

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{-t}}</math>

The graph shows what happens when ''k = 8'':
<math>p(t)=\dfrac{1}{\tfrac {1}{8}+e^{-t}}</math>

[[File:Ticino2.JPEG|300px]]

_____________________

In order to change the y-intercept of our function, we can add any number, ''q'', where ''q'' is any value between negative infinity and infinity, as shown:

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{-t+q}}</math>

The graph shows what happens when ''k = 8'' and ''q = 3'':
<math>p(t)=\dfrac{1}{\tfrac {1}{8}+e^{-t+3}}</math>

[[File:Ticino3.jpg|300px]]

_______________________

BONUS: to change the slope of the function, multiply ''t'' by ''s'', where ''s'' is any number.

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{-st + q}}</math>

Here's what happens when ''k = 8'', ''q = 3'' and ''s = 5'':
<math>p(t)=\dfrac{1}{\tfrac {1}{8}+e^{-5t + 3}}</math>

[[File:Ticino4.jpg|300px]]

Course:MATH110/Archive/2010-2011/003/Teams/Ticino/Homework 12

2011-01-25T23:36:00Z

BenJeffery:

<math>p(t)=\dfrac{1}{1+e^{-t}}</math>

[[File:Ticino graph 1.png|300px]]

_____________________

To change the height of the horizontal asymptote to ''k'', we can add ''k'' to the formula as follows:

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{-t}}</math>

The graph shows what happens when ''k = 8'':

[[File:Ticino2.JPEG|300px]]

_____________________

In order to change the y-intercept of our function, we can add any number, ''q'', where ''q'' is any value between negative infinity and infinity, as shown:

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{-t+q}}</math>

The graph shows what happens when ''k = 8'' and ''q = 3'':

[[File:Ticino3.jpg|300px]]

_______________________

BONUS: to change the slope of the function, multiply ''t'' by ''s'', where ''s'' is any number.

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{-ts + q}}</math>

Here's what happens when ''k = 8'', ''q = 3'' and ''s = 5'':

[[File:Ticino4.jpg|300px]]

Course:MATH110/Archive/2010-2011/003/Teams/Ticino/Homework 12

2011-01-25T23:31:33Z

BenJeffery:

<math>p(t)=\dfrac{1}{1+e^{-t}}</math>

[[File:Ticino graph 1.png|300px]]

_____________________

To change the height of the horizontal asymptote to ''k'', we can add ''k'' to the formula as follows:

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{-t}}</math>

The graph shows what happens when ''k = 8'':

[[File:Ticino2.JPEG|300px]]

_____________________

In order to change the y-intercept of our function, we can add any number, ''q'', where ''q'' is any value between negative infinity and infinity, as shown:

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{-t+q}}</math>

The graph shows what happens when ''k = 8'' and ''q = 3''

[[File:Ticino3.jpg|300px]]

_______________________

BONUS: to change the slope of the function, multiply any number, ''q'', to ''t''

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{(s)-t}}</math>

Course:MATH110/Archive/2010-2011/003/Teams/Ticino/Homework 12

2011-01-25T23:26:12Z

BenJeffery:

<math>p(t)=\dfrac{1}{1+e^{-t}}</math>

[[File:Ticino graph 1.png|300px]]

To change the height of the horizontal asymptote to ''k'', we can add ''k'' to the formula as follows:

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{-t}}</math>

[[File:Ticino2.JPEG|300px]]

In order to change the y-intercept of our function, we can add any number, ''q'', where ''q'' is any value between negative infinity and infinity, as shown:

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{-t+q}}</math>

BONUS: to change the slope of the function, multiply any number, ''q'', to ''t''

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{-8t}}</math>

Course:MATH110/Archive/2010-2011/003/Teams/Ticino/Homework 12

2011-01-25T23:25:20Z

BenJeffery:

<math>p(t)=\dfrac{1}{1+e^{-t}}</math>

[[File:Ticino graph 1.png]]

To change the height of the horizontal asymptote to ''k'', we can add ''k'' to the formula as follows:

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{-t}}</math>

[[File:Ticino2.png]]

In order to change the y-intercept of our function, we can add any number, ''q'', where ''q'' is any value between negative infinity and infinity, as shown:

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{-t+q}}</math>

BONUS: to change the slope of the function, multiply any number, ''q'', to ''t''

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{-8t}}</math>

Course:MATH110/Archive/2010-2011/003/Teams/Ticino/Homework 12

2011-01-25T23:24:57Z

BenJeffery:

<math>p(t)=\dfrac{1}{1+e^{-t}}</math>

[[File:Ticino graph 1.png]

To change the height of the horizontal asymptote to ''k'', we can add ''k'' to the formula as follows:

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{-t}}</math>

[[File:Ticino2.png]]

In order to change the y-intercept of our function, we can add any number, ''q'', where ''q'' is any value between negative infinity and infinity, as shown:

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{-t+q}}</math>

BONUS: to change the slope of the function, multiply any number, ''q'', to ''t''

<math>p(t)=\dfrac{1}{\tfrac {1}{k}+e^{-8t}}</math>

Course:MATH110/Archive/2010-2011/003/Teams/Ticino/Homework 12

2011-01-25T23:19:14Z

BenJeffery:

Course:MATH110/Archive/2010-2011/003/Teams/Ticino/Homework 12

2011-01-25T22:17:49Z

BenJeffery: Blanked the page

Course:MATH110/Archive/2010-2011/003/Teams/Ticino/Homework 12

2011-01-25T22:16:38Z

BenJeffery: Created page with "5x^2-2"

5x^2-2

Course:MATH110/Archive/2010-2011/003/Teams/Ticino

2011-01-24T16:12:14Z

BenJeffery: /* Keywords */

{{
Infobox MATH110 Teams
| team name = Ticino
| member 1 = [[User:BenJeffery|BenJeffery]]
| member 2 = [[User:BerndetteHii|BernadetteHii]]
| member 3 = Caitlin Lastiwka-Farquharson
| member 4 = Micha Gutmanis
}}
In workshop M.

===Keywords===

Caitlin - Watermelon

Micha - Music

Bernadette - Purple

Ben - Coffee

===Group Work===

[http://wiki.ubc.ca/Course:MATH110/003/Teams/Ticino/Homework_11 Homework 11]

Course:MATH110/Archive/2010-2011/003/Teams/Ticino

2011-01-24T16:11:40Z

BenJeffery: /* Group Work */

{{
Infobox MATH110 Teams
| team name = Ticino
| member 1 = [[User:BenJeffery|BenJeffery]]
| member 2 = [[User:BerndetteHii|BernadetteHii]]
| member 3 = Caitlin Lastiwka-Farquharson
| member 4 = Micha Gutmanis
}}
In workshop M.

===Keywords===

Caitlin - Watermelon
Micha - Music
Bernadette - Purple
Ben - Coffee

===Group Work===

[http://wiki.ubc.ca/Course:MATH110/003/Teams/Ticino/Homework_11 Homework 11]

Course:MATH110/Archive/2010-2011/003/Teams/Ticino

2011-01-19T03:52:58Z

BenJeffery: /* Group Work */

{{
Infobox MATH110 Teams
| team name = Ticino
| member 1 = [[User:BenJeffery|BenJeffery]]
| member 2 = [[User:BerndetteHii|BernadetteHii]]
| member 3 = Caitlin Lastiwka-Farquharson
| member 4 = Micha Gutmanis
}}
In workshop M.

===Group Work===

[http://wiki.ubc.ca/Course:MATH110/003/Teams/Ticino/Homework_11 Homework 11]

Course:MATH110/Archive/2010-2011/003/Teams/Ticino

2011-01-19T03:52:29Z

BenJeffery: /* Group Work */

{{
Infobox MATH110 Teams
| team name = Ticino
| member 1 = [[User:BenJeffery|BenJeffery]]
| member 2 = [[User:BerndetteHii|BernadetteHii]]
| member 3 = Caitlin Lastiwka-Farquharson
| member 4 = Micha Gutmanis
}}
In workshop M.

===Group Work===

[[http://wiki.ubc.ca/Course:MATH110/003/Teams/Ticino/Homework_11 |Homework 11]]

Course:MATH110/Archive/2010-2011/003/Teams/Ticino

2011-01-19T03:52:08Z

BenJeffery: /* Group Work */

{{
Infobox MATH110 Teams
| team name = Ticino
| member 1 = [[User:BenJeffery|BenJeffery]]
| member 2 = [[User:BerndetteHii|BernadetteHii]]
| member 3 = Caitlin Lastiwka-Farquharson
| member 4 = Micha Gutmanis
}}
In workshop M.

===Group Work===

[[http://wiki.ubc.ca/Course:MATH110/003/Teams/Ticino/Homework_11|Homework_11]]

Course:MATH110/Archive/2010-2011/003/Teams/Ticino

2011-01-19T03:51:46Z

BenJeffery: /* Some Other Interesting Functions */

{{
Infobox MATH110 Teams
| team name = Ticino
| member 1 = [[User:BenJeffery|BenJeffery]]
| member 2 = [[User:BerndetteHii|BernadetteHii]]
| member 3 = Caitlin Lastiwka-Farquharson
| member 4 = Micha Gutmanis
}}
In workshop M.

===Group Work===

[[http://wiki.ubc.ca/Course:MATH110/003/Teams/Ticino/Homework_11|Homework 11]]

Course:MATH110/Archive/2010-2011/003/Teams/Ticino/Homework 11

2011-01-19T03:51:38Z

BenJeffery:

====The Flag of Ticino====

To produce flags of Ticino, we begin with the information that it costs $100 to make 20 flags.

<math>$100/20=5</math>

shows that to make one flag out of this group of 20, it costs $5.

To make the twenty-first flag, and onwards, it will now cost $7 per flag.

If it costs $7 to produce x units, we can write this as <math>7x</math>
making the model

<math>P(x)=7x</math>

But we need to consider that the first 20 flags produced are purchased at the cheaper price of $100, or $5 per flag. We will have to subtract something in our model to accommodate for this. To find out what this is, we calculate how much it would cost if there was no cheaper price of $5 per flag and the price was set at $7 per unit.
To produce those same 20 flags we would have

<math>($7)(20)=$140.</math>

Compared to:

<math>($5)(20)=$100</math>

<math>$140-$100=$40</math>, we have a difference of $40 saved.

The model now looks like:

<math>C = 7x-40 </math>

Where:

*x = number of items

*7 = $7 marginal cost

*C = Total production cost

*40 = the difference between buying 20 flags for $5 each or 20 flags for $7 each.

To produce 150 flags, our model predicts the following:

<math> $7(150) - $40 = $1010 </math>

So it will cost us $1010 to produce 150 flags.

The average cost of each flag if we bought 150 would be given as:

<math>$1010/150 = $6.73</math>

So each flag costs on average $6.73.

Our model shows that as we produce more and more flags, the average cost will approach $7. We should see a horizontal asymptote at $7 as <math> x -> \infty</math>.

==Some Other Interesting Functions===

In order to have a model where the average cost stays the same as production increases, you could have a very simple model. If, say, we are producing toy banks (banking being a huge part of Ticino's economy) we could have a model where each bank produced cost $10, with no scaling. Our model would be

<math> C = 10b </math>

where C is our cost and b is the number of banks created. Here our average cost would always stay the same.

___________________________________________

In order to have the average cost decrease, we could have a factory producing our toy banks. At such a situation, the cost is quite a bit higher to start production. Let's say that we have a flat rate per unit, $3, but a cost of $10,000 to set up the factory. In this case, our model would be as follows:

<math>C = 10000 + 3b</math>

Our average cost/unit would decrease quite quickly, to the point where it would greatly encourage mass production. This model is an example of an economy of scale, as our average cost always decreases after we start production.

____________________________________________

For a more interesting model, let's look at a hypothetical model where we're buying land. Say we are looking at the costs in millions of dollars per <math>km^2 </math>.

We may see the average cost diminish as we start buying larger sections of land. However, once we reach a certain point, the cost of each square kilometer will start increasing. Purely hypothetically, this could be described as follows:

When <math>L < 1000</math> then <math>C = \frac {2L(L-1)} {L+1}</math>

When <math> L > 1000 </math> then <math> C = L^2 </math>

where L is the amount of land in km^2, and C is the cost in millions of dollars.
We may see behaviour like this because Ticino is only <math> 2812 km^2 </math>. If you attempt to buy too much of the ''canton'', the cost may well start increasing very quickly. Obviously this is not an exact representation of real estate in Ticino, but it's an interesting model regardless.

Course:MATH110/Archive/2010-2011/003/Teams/Ticino

2011-01-19T03:51:01Z

BenJeffery: /* Group Work */

{{
Infobox MATH110 Teams
| team name = Ticino
| member 1 = [[User:BenJeffery|BenJeffery]]
| member 2 = [[User:BerndetteHii|BernadetteHii]]
| member 3 = Caitlin Lastiwka-Farquharson
| member 4 = Micha Gutmanis
}}
In workshop M.

===Group Work===

[[http://wiki.ubc.ca/Course:MATH110/003/Teams/Ticino/Homework_11|Homework 11]]

===Some Other Interesting Functions===

In order to have a model where the average cost stays the same as production increases, you could have a very simple model. If, say, we are producing toy banks (banking being a huge part of Ticino's economy) we could have a model where each bank produced cost $10, with no scaling. Our model would be

<math> C = 10b </math>

where C is our cost and b is the number of banks created. Here our average cost would always stay the same.

___________________________________________

In order to have the average cost decrease, we could have a factory producing our toy banks. At such a situation, the cost is quite a bit higher to start production. Let's say that we have a flat rate per unit, $3, but a cost of $10,000 to set up the factory. In this case, our model would be as follows:

<math>C = 10000 + 3b</math>

Our average cost/unit would decrease quite quickly, to the point where it would greatly encourage mass production. This model is an example of an economy of scale, as our average cost always decreases after we start production.

____________________________________________

For a more interesting model, let's look at a hypothetical model where we're buying land. Say we are looking at the costs in millions of dollars per <math>km^2 </math>.

We may see the average cost diminish as we start buying larger sections of land. However, once we reach a certain point, the cost of each square kilometer will start increasing. Purely hypothetically, this could be described as follows:

When <math>L < 1000</math> then <math>C = \frac {2L(L-1)} {L+1}</math>

When <math> L > 1000 </math> then <math> C = L^2 </math>

where L is the amount of land in km^2, and C is the cost in millions of dollars.
We may see behaviour like this because Ticino is only <math> 2812 km^2 </math>. If you attempt to buy too much of the ''canton'', the cost may well start increasing very quickly. Obviously this is not an exact representation of real estate in Ticino, but it's an interesting model regardless.

Course:MATH110/Archive/2010-2011/003/Teams/Ticino/Homework 11

2011-01-19T03:49:59Z

BenJeffery: Created page with "====The Flag of Ticino==== To produce flags of Ticino, we begin with the information that it costs $100 to make 20 flags. <math>$100/20=5</math> shows that to make one flag ou..."

Course:MATH110/Archive/2010-2011/003/Teams/Ticino

2011-01-19T03:40:16Z

BenJeffery: /* The Flag of Ticino */

{{
Infobox MATH110 Teams
| team name = Ticino
| member 1 = [[User:BenJeffery|BenJeffery]]
| member 2 = [[User:BerndetteHii|BernadetteHii]]
| member 3 = Caitlin Lastiwka-Farquharson
| member 4 = Micha Gutmanis
}}
In workshop M.

===Group Work===
====The Flag of Ticino====

To produce flags of Ticino, we begin with the information that it costs $100 to make 20 flags.

<math>$100/20=5</math>

shows that to make one flag out of this group of 20, it costs $5.

To make the twenty-first flag, and onwards, it will now cost $7 per flag.

If it costs $7 to produce x units, we can write this as <math>7x</math>
making the model

<math>P(x)=7x</math>

But we need to consider that the first 20 flags produced are purchased at the cheaper price of $100, or $5 per flag. We will have to subtract something in our model to accommodate for this. To find out what this is, we calculate how much it would cost if there was no cheaper price of $5 per flag and the price was set at $7 per unit.
To produce those same 20 flags we would have

<math>($7)(20)=$140.</math>

Compared to:

<math>($5)(20)=$100</math>

<math>$140-$100=$40</math>, we have a difference of $40 saved.

The model now looks like:

<math>C = 7x-40 </math>

Where:

*x = number of items

*7 = $7 marginal cost

*C = Total production cost

*40 = the difference between buying 20 flags for $5 each or 20 flags for $7 each.

To produce 150 flags, our model predicts the following:

<math> $7(150) - $40 = $1010 </math>

So it will cost us $1010 to produce 150 flags.

The average cost of each flag if we bought 150 would be given as:

<math>$1010/150 = $6.73</math>

So each flag costs on average $6.73.

Our model shows that as we produce more and more flags, the average cost will approach $7. We should see a horizontal asymptote at $7 as <math> x -> \infty</math>.

===Some Other Interesting Functions===

In order to have a model where the average cost stays the same as production increases, you could have a very simple model. If, say, we are producing toy banks (banking being a huge part of Ticino's economy) we could have a model where each bank produced cost $10, with no scaling. Our model would be

<math> C = 10b </math>

where C is our cost and b is the number of banks created. Here our average cost would always stay the same.

___________________________________________

In order to have the average cost decrease, we could have a factory producing our toy banks. At such a situation, the cost is quite a bit higher to start production. Let's say that we have a flat rate per unit, $3, but a cost of $10,000 to set up the factory. In this case, our model would be as follows:

<math>C = 10000 + 3b</math>

Our average cost/unit would decrease quite quickly, to the point where it would greatly encourage mass production. This model is an example of an economy of scale, as our average cost always decreases after we start production.

____________________________________________

For a more interesting model, let's look at a hypothetical model where we're buying land. Say we are looking at the costs in millions of dollars per <math>km^2 </math>.

We may see the average cost diminish as we start buying larger sections of land. However, once we reach a certain point, the cost of each square kilometer will start increasing. Purely hypothetically, this could be described as follows:

When <math>L < 1000</math> then <math>C = \frac {2L(L-1)} {L+1}</math>

When <math> L > 1000 </math> then <math> C = L^2 </math>

where L is the amount of land in km^2, and C is the cost in millions of dollars.
We may see behaviour like this because Ticino is only <math> 2812 km^2 </math>. If you attempt to buy too much of the ''canton'', the cost may well start increasing very quickly. Obviously this is not an exact representation of real estate in Ticino, but it's an interesting model regardless.

Course:MATH110/Archive/2010-2011/003/Teams/Ticino

2011-01-19T03:39:19Z

BenJeffery: /* Some Other Interesting Functions */

{{
Infobox MATH110 Teams
| team name = Ticino
| member 1 = [[User:BenJeffery|BenJeffery]]
| member 2 = [[User:BerndetteHii|BernadetteHii]]
| member 3 = Caitlin Lastiwka-Farquharson
| member 4 = Micha Gutmanis
}}
In workshop M.

===Group Work===
====The Flag of Ticino====

To produce flags of Ticino, we begin with the information that it costs $100 to make 20 flags.

<math>$100/20=5</math>

shows that to make one flag out of this group of 20, it costs $5.

To make the twenty-first flag, and onwards, it will now cost $7 per flag.

If it costs $7 to produce x units, we can write this as <math>7x</math>
making the model

<math>P(x)=7x</math>

But we need to consider that the first 20 flags produced are purchased at the cheaper price of $100, or $5 per flag. We will have to subtract something in our model to accommodate for this. To find out what this is, we calculate how much it would cost if there was no cheaper price of $5 per flag and the price was set at $7 per unit.
To produce those same 20 flags we would have

<math>($7)(20)=$140.</math>

Compared to:

<math>($5)(20)=$100</math>

<math>$140-$100=$40</math>, we have a difference of $40 saved.

The model now looks like:

<math>C = 7x-40 </math>

Where:

*x = number of items

*7 = $7 marginal cost

*C = Total production cost

*40 = the difference between buying 20 flags for $5 each or 20 flags for $7 each.

To produce 150 flags, our model predicts the following:

<math> $7(150) - $40 = $1010 </math>
So it will cost us $1010 to produce 150 flags.

The average cost of each flag if we bought 150 would be given as:

<math>$1010/150 = $6.73</math>

So each flag costs on average $6.73.

Our model shows that as we produce more and more flags, the average cost will approach $7. We should see a horizontal asymptote as <math> x -> \infty</math>.

===Some Other Interesting Functions===

In order to have a model where the average cost stays the same as production increases, you could have a very simple model. If, say, we are producing toy banks (banking being a huge part of Ticino's economy) we could have a model where each bank produced cost $10, with no scaling. Our model would be

<math> C = 10b </math>

where C is our cost and b is the number of banks created. Here our average cost would always stay the same.

___________________________________________

In order to have the average cost decrease, we could have a factory producing our toy banks. At such a situation, the cost is quite a bit higher to start production. Let's say that we have a flat rate per unit, $3, but a cost of $10,000 to set up the factory. In this case, our model would be as follows:

<math>C = 10000 + 3b</math>

Our average cost/unit would decrease quite quickly, to the point where it would greatly encourage mass production. This model is an example of an economy of scale, as our average cost always decreases after we start production.

____________________________________________

For a more interesting model, let's look at a hypothetical model where we're buying land. Say we are looking at the costs in millions of dollars per <math>km^2 </math>.

We may see the average cost diminish as we start buying larger sections of land. However, once we reach a certain point, the cost of each square kilometer will start increasing. Purely hypothetically, this could be described as follows:

When <math>L < 1000</math> then <math>C = \frac {2L(L-1)} {L+1}</math>

When <math> L > 1000 </math> then <math> C = L^2 </math>

where L is the amount of land in km^2, and C is the cost in millions of dollars.
We may see behaviour like this because Ticino is only <math> 2812 km^2 </math>. If you attempt to buy too much of the ''canton'', the cost may well start increasing very quickly. Obviously this is not an exact representation of real estate in Ticino, but it's an interesting model regardless.

Course:MATH110/Archive/2010-2011/003/Teams/Ticino

2011-01-19T03:34:29Z

BenJeffery: /* Some Other Interesting Functions */

{{
Infobox MATH110 Teams
| team name = Ticino
| member 1 = [[User:BenJeffery|BenJeffery]]
| member 2 = [[User:BerndetteHii|BernadetteHii]]
| member 3 = Caitlin Lastiwka-Farquharson
| member 4 = Micha Gutmanis
}}
In workshop M.

===Group Work===
====The Flag of Ticino====

To produce flags of Ticino, we begin with the information that it costs $100 to make 20 flags.

<math>$100/20=5</math>

shows that to make one flag out of this group of 20, it costs $5.

To make the twenty-first flag, and onwards, it will now cost $7 per flag.

If it costs $7 to produce x units, we can write this as <math>7x</math>
making the model

<math>P(x)=7x</math>

But we need to consider that the first 20 flags produced are purchased at the cheaper price of $100, or $5 per flag. We will have to subtract something in our model to accommodate for this. To find out what this is, we calculate how much it would cost if there was no cheaper price of $5 per flag and the price was set at $7 per unit.
To produce those same 20 flags we would have

<math>($7)(20)=$140.</math>

Compared to:

<math>($5)(20)=$100</math>

<math>$140-$100=$40</math>, we have a difference of $40 saved.

The model now looks like:

<math>C = 7x-40 </math>

Where:

*x = number of items

*7 = $7 marginal cost

*C = Total production cost

*40 = the difference between buying 20 flags for $5 each or 20 flags for $7 each.

To produce 150 flags, our model predicts the following:

<math> $7(150) - $40 = $1010 </math>
So it will cost us $1010 to produce 150 flags.

The average cost of each flag if we bought 150 would be given as:

<math>$1010/150 = $6.73</math>

So each flag costs on average $6.73.

Our model shows that as we produce more and more flags, the average cost will approach $7. We should see a horizontal asymptote as <math> x -> \infty</math>.

===Some Other Interesting Functions===

In order to have a model where the average cost stays the same as production increases, you could have a very simple model. If, say, we are producing toy banks (banking being a huge part of Ticino's economy) we could have a model where each bank produced cost $10, with no scaling. Our model would be

<math> C = 10b </math>

where C is our cost and b is the number of banks created. Here our average cost would always stay the same.

In order to have the average cost decrease, we could have a factory producing our toy banks. At such a situation, the cost is quite a bit higher to start production. Let's say that we have a flat rate per unit, $3, but a cost of $10,000 to set up the factory. In this case, our model would be as follows:

<math>C = 10000 + 3b</math>

Our average cost/unit would decrease quite quickly, to the point where it would greatly encourage mass production. This model is an example of an economy of scale, as our average cost always decreases after we start production.

For a more interesting model, let's look at a hypothetical model where we're buying land. Say we are looking at the costs in millions of dollars per <math>km^2 </math>.

We may see the average cost diminish as we start buying larger sections of land. However, once we reach a certain point, the cost of each square kilometer will start increasing. Purely hypothetically, this could be described as follows:

When <math>L < 1000</math> then <math>C = \frac {2L(L-1)} {L+1}</math>

When <math> L > 1000 </math> then <math> C = L^2 </math>

where L is the amount of land in km^2, and C is the cost in millions of dollars.
We may see behaviour like this because Ticino is only <math> 2812 km^2 </math>. If you attempt to buy too much of the ''canton'', the cost may well start increasing very quickly. Obviously this is not an exact representation of real estate in Ticino, but it's an interesting model regardless.

Course:MATH110/Archive/2010-2011/003/Teams/Ticino

2011-01-19T03:29:35Z

BenJeffery: /* Some Other Interesting Functions */

{{
Infobox MATH110 Teams
| team name = Ticino
| member 1 = [[User:BenJeffery|BenJeffery]]
| member 2 = [[User:BerndetteHii|BernadetteHii]]
| member 3 = Caitlin Lastiwka-Farquharson
| member 4 = Micha Gutmanis
}}
In workshop M.

===Group Work===
====The Flag of Ticino====

To produce flags of Ticino, we begin with the information that it costs $100 to make 20 flags.

<math>$100/20=5</math>

shows that to make one flag out of this group of 20, it costs $5.

To make the twenty-first flag, and onwards, it will now cost $7 per flag.

If it costs $7 to produce x units, we can write this as <math>7x</math>
making the model

<math>P(x)=7x</math>

But we need to consider that the first 20 flags produced are purchased at the cheaper price of $100, or $5 per flag. We will have to subtract something in our model to accommodate for this. To find out what this is, we calculate how much it would cost if there was no cheaper price of $5 per flag and the price was set at $7 per unit.
To produce those same 20 flags we would have

<math>($7)(20)=$140.</math>

Compared to:

<math>($5)(20)=$100</math>

<math>$140-$100=$40</math>, we have a difference of $40 saved.

The model now looks like:

<math>C = 7x-40 </math>

Where:

*x = number of items

*7 = $7 marginal cost

*C = Total production cost

*40 = the difference between buying 20 flags for $5 each or 20 flags for $7 each.

To produce 150 flags, our model predicts the following:

<math> $7(150) - $40 = $1010 </math>
So it will cost us $1010 to produce 150 flags.

The average cost of each flag if we bought 150 would be given as:

<math>$1010/150 = $6.73</math>

So each flag costs on average $6.73.

Our model shows that as we produce more and more flags, the average cost will approach $7. We should see a horizontal asymptote as <math> x -> \infty</math>.

===Some Other Interesting Functions===

In order to have a model where the average cost stays the same as production increases, you could have a very simple model. If, say, we are producing toy banks (banking being a huge part of Ticino's economy) we could have a model where each bank produced cost $10, with no scaling. Our model would be

<math> C = 10b </math>

where C is our cost and b is the number of banks created. Here our average cost would always stay the same.

In order to have the average cost decrease, we could have a factory producing our toy banks. At such a situation, the cost is quite a bit higher to start production. Let's say that we have a flat rate per unit, $3, but a cost of $10,000 to set up the factory. In this case, our model would be as follows:

C = 10000 + 3b

Our average cost/unit would decrease quite quickly, to the point where it would greatly encourage mass production.

For a more interesting model, let's look at a hypothetical model where we're buying land. Say we are looking at the costs in millions of dollars per <math>km^2 </math>.

We may see the average cost diminish as we start buying larger sections of land. However, once we reach a certain point, the cost of each square kilometer will start increasing. Purely hypothetically, this could be described as follows:

When <math>L < 1000</math> then <math>C = \frac {2L(L-1)} {L+1}</math>

When <math> L > 1000 </math> then <math> C = L^2 </math>

where L is the amount of land in km^2, and C is the cost in millions of dollars.
We may see behaviour like this because Ticino is only <math> 2812 km^2 </math>. If you attempt to buy too much of the ''canton'', the cost may well start increasing very quickly. Obviously this is not an exact representation of real estate in Ticino, but it's an interesting model regardless.

Course:MATH110/Archive/2010-2011/003/Teams/Ticino

2011-01-19T03:13:42Z

BenJeffery: /* Some Other Interesting Functions */

{{
Infobox MATH110 Teams
| team name = Ticino
| member 1 = [[User:BenJeffery|BenJeffery]]
| member 2 = [[User:BerndetteHii|BernadetteHii]]
| member 3 = Caitlin Lastiwka-Farquharson
| member 4 = Micha Gutmanis
}}
In workshop M.

===Group Work===
====The Flag of Ticino====

To produce flags of Ticino, we begin with the information that it costs $100 to make 20 flags.

<math>$100/20=5</math>

shows that to make one flag out of this group of 20, it costs $5.

To make the twenty-first flag, and onwards, it will now cost $7 per flag.

If it costs $7 to produce x units, we can write this as <math>7x</math>
making the model

<math>P(x)=7x</math>

But we need to consider that the first 20 flags produced are purchased at the cheaper price of $100, or $5 per flag. We will have to subtract something in our model to accommodate for this. To find out what this is, we calculate how much it would cost if there was no cheaper price of $5 per flag and the price was set at $7 per unit.
To produce those same 20 flags we would have

<math>($7)(20)=$140.</math>

Compared to:

<math>($5)(20)=$100</math>

<math>$140-$100=$40</math>, we have a difference of $40 saved.

The model now looks like:

<math>C = 7x-40 </math>

Where:

*x = number of items

*7 = $7 marginal cost

*C = Total production cost

*40 = the difference between buying 20 flags for $5 each or 20 flags for $7 each.

To produce 150 flags, our model predicts the following:

<math> $7(150) - $40 = $1010 </math>
So it will cost us $1010 to produce 150 flags.

The average cost of each flag if we bought 150 would be given as:

<math>$1010/150 = $6.73</math>

So each flag costs on average $6.73.

Our model shows that as we produce more and more flags, the average cost will approach $7. We should see a horizontal asymptote as <math> x -> \infty</math>.

===Some Other Interesting Functions===

In order to have a model where the average cost stays the same as production increases, you could have a very simple model. If, say, we are producing toy banks (banking being a huge part of Ticino's economy) we could have a model where each bank produced cost $10, with no scaling. Our model would be

<math> C = 10b </math>

where C is our cost and b is the number of banks created. Here our average cost would always stay the same.

For a more interesting model, let's look at a hypothetical model where we're buying land. Say we are looking at the costs in millions of dollars per <math>km^2 </math>.

We may see the average cost diminish as we start buying larger sections of land. However, once we reach a certain point, the cost of each square kilometer will start increasing. Purely hypothetically, this could be described as follows:

When <math>L < 1000</math> then <math>C = \frac {2L (L-1)} {L+1})</math>

Course:MATH110/Archive/2010-2011/003/Teams/Ticino

2011-01-19T03:11:56Z

BenJeffery: /* Some Other Interesting Functions */

{{
Infobox MATH110 Teams
| team name = Ticino
| member 1 = [[User:BenJeffery|BenJeffery]]
| member 2 = [[User:BerndetteHii|BernadetteHii]]
| member 3 = Caitlin Lastiwka-Farquharson
| member 4 = Micha Gutmanis
}}
In workshop M.

===Group Work===
====The Flag of Ticino====

To produce flags of Ticino, we begin with the information that it costs $100 to make 20 flags.

<math>$100/20=5</math>

shows that to make one flag out of this group of 20, it costs $5.

To make the twenty-first flag, and onwards, it will now cost $7 per flag.

If it costs $7 to produce x units, we can write this as <math>7x</math>
making the model

<math>P(x)=7x</math>

But we need to consider that the first 20 flags produced are purchased at the cheaper price of $100, or $5 per flag. We will have to subtract something in our model to accommodate for this. To find out what this is, we calculate how much it would cost if there was no cheaper price of $5 per flag and the price was set at $7 per unit.
To produce those same 20 flags we would have

<math>($7)(20)=$140.</math>

Compared to:

<math>($5)(20)=$100</math>

<math>$140-$100=$40</math>, we have a difference of $40 saved.

The model now looks like:

<math>C = 7x-40 </math>

Where:

*x = number of items

*7 = $7 marginal cost

*C = Total production cost

*40 = the difference between buying 20 flags for $5 each or 20 flags for $7 each.

To produce 150 flags, our model predicts the following:

<math> $7(150) - $40 = $1010 </math>
So it will cost us $1010 to produce 150 flags.

The average cost of each flag if we bought 150 would be given as:

<math>$1010/150 = $6.73</math>

So each flag costs on average $6.73.

Our model shows that as we produce more and more flags, the average cost will approach $7. We should see a horizontal asymptote as <math> x -> \infty</math>.

===Some Other Interesting Functions===

In order to have a model where the average cost stays the same as production increases, you could have a very simple model. If, say, we are producing toy banks (banking being a huge part of Ticino's economy) we could have a model where each bank produced cost $10, with no scaling. Our model would be

<math> C = 10b </math>

where C is our cost and b is the number of banks created. Here our average cost would always stay the same.

For a more interesting model, let's look at a hypothetical model where we're buying land. Say we are looking at the costs in millions of dollars per <math>km^2 </math>.

We may see the average cost diminish as we start buying larger sections of land. However, once we reach a certain point, the cost of each square kilometer will start increasing. Purely hypothetically, this could be described as follows:

When <math>L < 1000</math> then <math>C = (2L (L-1))/(L+1))</math>

Course:MATH110/Archive/2010-2011/003/Teams/Ticino

2011-01-19T03:09:00Z

BenJeffery: /* Some Other Interesting Functions */

{{
Infobox MATH110 Teams
| team name = Ticino
| member 1 = [[User:BenJeffery|BenJeffery]]
| member 2 = [[User:BerndetteHii|BernadetteHii]]
| member 3 = Caitlin Lastiwka-Farquharson
| member 4 = Micha Gutmanis
}}
In workshop M.

===Group Work===
====The Flag of Ticino====

To produce flags of Ticino, we begin with the information that it costs $100 to make 20 flags.

<math>$100/20=5</math>

shows that to make one flag out of this group of 20, it costs $5.

To make the twenty-first flag, and onwards, it will now cost $7 per flag.

If it costs $7 to produce x units, we can write this as <math>7x</math>
making the model

<math>P(x)=7x</math>

But we need to consider that the first 20 flags produced are purchased at the cheaper price of $100, or $5 per flag. We will have to subtract something in our model to accommodate for this. To find out what this is, we calculate how much it would cost if there was no cheaper price of $5 per flag and the price was set at $7 per unit.
To produce those same 20 flags we would have

<math>($7)(20)=$140.</math>

Compared to:

<math>($5)(20)=$100</math>

<math>$140-$100=$40</math>, we have a difference of $40 saved.

The model now looks like:

<math>C = 7x-40 </math>

Where:

*x = number of items

*7 = $7 marginal cost

*C = Total production cost

*40 = the difference between buying 20 flags for $5 each or 20 flags for $7 each.

To produce 150 flags, our model predicts the following:

<math> $7(150) - $40 = $1010 </math>
So it will cost us $1010 to produce 150 flags.

The average cost of each flag if we bought 150 would be given as:

<math>$1010/150 = $6.73</math>

So each flag costs on average $6.73.

Our model shows that as we produce more and more flags, the average cost will approach $7. We should see a horizontal asymptote as <math> x -> \infty</math>.

===Some Other Interesting Functions===

In order to have a model where the average cost stays the same as production increases, you could have a very simple model. If, say, we are producing toy banks (banking being a huge part of Ticino's economy) we could have a model where each bank produced cost $10, with no scaling. Our model would be

<math> C = 10b </math>

where C is our cost and b is the number of banks created. Here our average cost would always stay the same.

For a more interesting model, let's look at a hypothetical model where we're buying land. Say we are looking at the costs in millions of dollars per <math>km^2 </math>.

We may see the average cost diminish as we start buying larger sections of land. However, once we reach a certain point, the cost of each square kilometer will start increasing. Purely hypothetically, this could be described as follows:

<math>f(C) = \begin{cases} (L (L-1))/L, & \mbox{if }L\mbox{ is < 1000} \\ L^2, & \mbox{if }L\mbox{ is > 1000} \end{cases}</math>

Course:MATH110/Archive/2010-2011/003/Teams/Ticino

2011-01-19T03:08:28Z

BenJeffery: /* Homework 11 */

{{
Infobox MATH110 Teams
| team name = Ticino
| member 1 = [[User:BenJeffery|BenJeffery]]
| member 2 = [[User:BerndetteHii|BernadetteHii]]
| member 3 = Caitlin Lastiwka-Farquharson
| member 4 = Micha Gutmanis
}}
In workshop M.

===Group Work===
====The Flag of Ticino====

To produce flags of Ticino, we begin with the information that it costs $100 to make 20 flags.

<math>$100/20=5</math>

shows that to make one flag out of this group of 20, it costs $5.

To make the twenty-first flag, and onwards, it will now cost $7 per flag.

If it costs $7 to produce x units, we can write this as <math>7x</math>
making the model

<math>P(x)=7x</math>

But we need to consider that the first 20 flags produced are purchased at the cheaper price of $100, or $5 per flag. We will have to subtract something in our model to accommodate for this. To find out what this is, we calculate how much it would cost if there was no cheaper price of $5 per flag and the price was set at $7 per unit.
To produce those same 20 flags we would have

<math>($7)(20)=$140.</math>

Compared to:

<math>($5)(20)=$100</math>

<math>$140-$100=$40</math>, we have a difference of $40 saved.

The model now looks like:

<math>C = 7x-40 </math>

Where:

*x = number of items

*7 = $7 marginal cost

*C = Total production cost

*40 = the difference between buying 20 flags for $5 each or 20 flags for $7 each.

To produce 150 flags, our model predicts the following:

<math> $7(150) - $40 = $1010 </math>
So it will cost us $1010 to produce 150 flags.

The average cost of each flag if we bought 150 would be given as:

<math>$1010/150 = $6.73</math>

So each flag costs on average $6.73.

Our model shows that as we produce more and more flags, the average cost will approach $7. We should see a horizontal asymptote as <math> x -> \infty</math>.

===Some Other Interesting Functions===

In order to have a model where the average cost stays the same as production increases, you could have a very simple model. If, say, we are producing toy banks (banking being a huge part of Ticino's economy) we could have a model where each bank produced cost $10, with no scaling. Our model would be

<math> C = 10b </math>

where C is our cost and b is the number of banks created. Here our average cost would always stay the same.

For a more interesting model, let's look at a hypothetical model where we're buying land. Say we are looking at the costs in millions of dollars per <math>km^2 </math>.

We may see the average cost diminish as we start buying larger sections of land. However, once we reach a certain point, the cost of each square kilometer will start increasing. Purely hypothetically, this could be described as follows:

f(C) = \begin{cases} (L (L-1))/L, & \mbox{if }L\mbox{ is < 1000} \\ L^2, & \mbox{if }L\mbox{ is > 1000} \end{cases}

Course:MATH110/Archive/2010-2011/003/Teams/Ticino

2011-01-19T02:38:10Z

BenJeffery: /* Homework 11 */

Course:MATH110/Archive/2010-2011/003/Teams/Ticino

2011-01-19T02:37:05Z

BenJeffery: /* Homework 11 */

Course:MATH110/Archive/2010-2011/003/Teams/Ticino

2011-01-19T02:34:10Z

BenJeffery: /* Homework 11 */ Added some stuff, changed formatting.

Course:MATH110/Archive/2010-2011/003/Teams/Ticino

2011-01-10T22:26:10Z

BenJeffery:

{{
Infobox MATH110 Teams
| team name = Ticino
| member 1 = [[User:BenJeffery|BenJeffery]]
| member 2 = Bernadette Hii
| member 3 = Caitlin Lastiwka-Farquharson
| member 4 =
}}
In workshop M.

Course talk:MATH110/003/Groups/Group 17/Basic Skills Project

2010-12-05T08:21:02Z

BenJeffery: /* Changes to the project page */

=Changes to the project page=

Hey guys. I made a ton of changes, reorganized everything so it's more suitable, fixed the formatting, got rid of some redundant text and incorrect examples, and incorporated David's feedback. Feel free to make any changes you think are relevant.

Thanks for the feedback, David!

[[User:BenJeffery|BenJeffery]] 08:20, 5 December 2010 (UTC)

=Remarks on your contribution to the Basic Skills Project (update)=
Dear group 17,

Here are some remarks on your page:
* In your description of the Pythagorean Theorem, why is there a line on the area of a triangle? I don't see the point.
* The water demo video (which is fun) cannot constitute a mathematical proof of the theorem, but it does make for a nice application of the theorem in the real world. A little discussion about this would be appreciated, I don't want the other students to believe this should be called a proof.
* As for applications in the trigonometric circle, I was hoping for an explanation of the identity: <math>\cos^2(\alpha)+\sin^2(\alpha)=1</math> which is true for any value of the angle <math>\alpha</math>.
* In terms of picture management, try to replace the code:
<pre>[[File:500px-Pythagorean.svg.png]]</pre>
:by the code
<pre>[[File:500px-Pythagorean.svg.png|thumb|left|A geometric visualization of the Pythagorean Theorem|300px]]</pre>
:It will look much better. Using this kind of tricks can help improve the visual appeal of your page.
* Please, don't sign where you contributed, I don't really care, this is a group effort. You will fill an evaluation form to distinguish who contributed more or less and by this way, redistribute the grade among yourselves.

Feel free to contact me if necessary of course. Cheers, -- [[User:DavidKohler|DavidKohler]]

=Remarks on your contribution to the Basic Skills Project=
Dear group 17,

You have a good start here, but I would like to see more of the content you still want to post. It does look promising so far, keep up the good work, but quickly if you want more feedback. Cheers, --
[[User:DavidKohler|DavidKohler]]

=Group 17 discussion=

Guys - Any ideas for our basic skills project offer?
Seeing as we can apparently all do Distance and Lines well, how about we offer to do an overview with examples of that part?

It's pretty straight-forward, so there shouldn't be too many challenges with getting it together. If no one answers this by tonight, I'll just put it up as our offer.

Cheers,
[[User:BenJeffery|BenJeffery]]

looks good to me! i would be between that or the pythagorean theorem (just because it seems more interesting) but for me, either or work. [[User:MarcoGasparian|MarcoGasparian]]

I'm comfortable with doing either- leaning more to pythagorean theorem

[[User:ChristaBicego|ChristaBicego]]

Was wondering what part of the guidelines Ben set out you guys would want me to do, and also how to compute them. I am a bit confused as to the guidelines of this project.

[[User:AviHarry|AviHarry]]

Course talk:MATH110/003/Groups/Group 17/Basic Skills Project

2010-12-05T08:20:34Z

BenJeffery: /* Changes to the project page */

=Changes to the project page=

Hey guys. I made a ton of changes, reorganized everything so it's more suitable, fixed the formatting, got rid of some redundant text and incorrect examples, and incorporated David's feedback. Feel free to make any changes you think are relevant.

[[User:BenJeffery|BenJeffery]] 08:20, 5 December 2010 (UTC)

=Remarks on your contribution to the Basic Skills Project (update)=
Dear group 17,

Here are some remarks on your page:
* In your description of the Pythagorean Theorem, why is there a line on the area of a triangle? I don't see the point.
* The water demo video (which is fun) cannot constitute a mathematical proof of the theorem, but it does make for a nice application of the theorem in the real world. A little discussion about this would be appreciated, I don't want the other students to believe this should be called a proof.
* As for applications in the trigonometric circle, I was hoping for an explanation of the identity: <math>\cos^2(\alpha)+\sin^2(\alpha)=1</math> which is true for any value of the angle <math>\alpha</math>.
* In terms of picture management, try to replace the code:
<pre>[[File:500px-Pythagorean.svg.png]]</pre>
:by the code
<pre>[[File:500px-Pythagorean.svg.png|thumb|left|A geometric visualization of the Pythagorean Theorem|300px]]</pre>
:It will look much better. Using this kind of tricks can help improve the visual appeal of your page.
* Please, don't sign where you contributed, I don't really care, this is a group effort. You will fill an evaluation form to distinguish who contributed more or less and by this way, redistribute the grade among yourselves.

Feel free to contact me if necessary of course. Cheers, -- [[User:DavidKohler|DavidKohler]]

=Remarks on your contribution to the Basic Skills Project=
Dear group 17,

You have a good start here, but I would like to see more of the content you still want to post. It does look promising so far, keep up the good work, but quickly if you want more feedback. Cheers, --
[[User:DavidKohler|DavidKohler]]

=Group 17 discussion=

Guys - Any ideas for our basic skills project offer?
Seeing as we can apparently all do Distance and Lines well, how about we offer to do an overview with examples of that part?

It's pretty straight-forward, so there shouldn't be too many challenges with getting it together. If no one answers this by tonight, I'll just put it up as our offer.

Cheers,
[[User:BenJeffery|BenJeffery]]

looks good to me! i would be between that or the pythagorean theorem (just because it seems more interesting) but for me, either or work. [[User:MarcoGasparian|MarcoGasparian]]

I'm comfortable with doing either- leaning more to pythagorean theorem

[[User:ChristaBicego|ChristaBicego]]

Was wondering what part of the guidelines Ben set out you guys would want me to do, and also how to compute them. I am a bit confused as to the guidelines of this project.

[[User:AviHarry|AviHarry]]

Course talk:MATH110/003/Groups/Group 17/Basic Skills Project

2010-12-05T08:20:12Z

BenJeffery: /* Remarks on your contribution to the Basic Skills Project (update) */

=Changes to the project page=

Hey guys. I made a ton of changes, reorganized everything so it's more suitable, fixed the formatting, got rid of some redundant text and incorrect examples, and incorporated David's feedback. Feel free to make any changes you think are relevant.
[[User:BenJeffery|BenJeffery]] 08:20, 5 December 2010 (UTC)

=Remarks on your contribution to the Basic Skills Project (update)=
Dear group 17,

Here are some remarks on your page:
* In your description of the Pythagorean Theorem, why is there a line on the area of a triangle? I don't see the point.
* The water demo video (which is fun) cannot constitute a mathematical proof of the theorem, but it does make for a nice application of the theorem in the real world. A little discussion about this would be appreciated, I don't want the other students to believe this should be called a proof.
* As for applications in the trigonometric circle, I was hoping for an explanation of the identity: <math>\cos^2(\alpha)+\sin^2(\alpha)=1</math> which is true for any value of the angle <math>\alpha</math>.
* In terms of picture management, try to replace the code:
<pre>[[File:500px-Pythagorean.svg.png]]</pre>
:by the code
<pre>[[File:500px-Pythagorean.svg.png|thumb|left|A geometric visualization of the Pythagorean Theorem|300px]]</pre>
:It will look much better. Using this kind of tricks can help improve the visual appeal of your page.
* Please, don't sign where you contributed, I don't really care, this is a group effort. You will fill an evaluation form to distinguish who contributed more or less and by this way, redistribute the grade among yourselves.

Feel free to contact me if necessary of course. Cheers, -- [[User:DavidKohler|DavidKohler]]

=Remarks on your contribution to the Basic Skills Project=
Dear group 17,

You have a good start here, but I would like to see more of the content you still want to post. It does look promising so far, keep up the good work, but quickly if you want more feedback. Cheers, --
[[User:DavidKohler|DavidKohler]]

=Group 17 discussion=

Guys - Any ideas for our basic skills project offer?
Seeing as we can apparently all do Distance and Lines well, how about we offer to do an overview with examples of that part?

It's pretty straight-forward, so there shouldn't be too many challenges with getting it together. If no one answers this by tonight, I'll just put it up as our offer.

Cheers,
[[User:BenJeffery|BenJeffery]]

looks good to me! i would be between that or the pythagorean theorem (just because it seems more interesting) but for me, either or work. [[User:MarcoGasparian|MarcoGasparian]]

I'm comfortable with doing either- leaning more to pythagorean theorem

[[User:ChristaBicego|ChristaBicego]]

Was wondering what part of the guidelines Ben set out you guys would want me to do, and also how to compute them. I am a bit confused as to the guidelines of this project.

[[User:AviHarry|AviHarry]]

Course:MATH110/Archive/2010-2011/003/Groups/Group 17/Basic Skills Project

2010-12-05T08:17:28Z

BenJeffery: /* How Do We Know It's True? */

==The Pythagorean Theorem==

=== What Is It? ===

[[File:500px-Pythagorean.svg.png|thumb|right|A geometric visualization of the Pythagorean Theorem|300px]]

'''History:'''

The theory was originally developed by a man named Pythagoras, who was born in the late 6th century B.C. on the island of Samos, Greece. He proved that for any right angle triangle, the two shorter sides squared and added together exactly equal the square of the longest side. This can be shown mathematically as:

<math>a^2 + b^2 = c^2</math>

'''Pythagorean Basics:'''

The theorem only works for triangles, specifically those with a right (90 degree) angle.

This image gives a visual example of how the Pythagorean theorem looks. The larger square has each side equal to the length of the hypotenuse (longest side) of the triangle. The two smaller squares are formed from the other two lengths on the triangle.

The theory, at it's most basic, states that the area of the largest square is equal to the combined areas of the two smaller squares. Since the area of a square is given by multiplying the length of a side by itself, we can label the sides above as shown, and bring ourselves right back to the mathematical formula! <math>a^2+b^2=c^2</math>

The following video shows a basic use of the theorem.

{{#ev:youtube | ku4rEwRxZOc | 400}}

For an interesting visual example, watch the following video. It shows that the water from the large square, formed by the hypotenuse, fills the smaller squares formed by the other two sides. Note that though this is an interesting visualization of the theorem, it is not actually a concrete proof, simply an interesting visual demonstration.

{{#ev:youtube | CAkMUdeB06o | 400}}

=== How Do We Know It's True? ===

[[File:Proof_2.png|thumb|right|An example of a proof of the Pythagorean Theorem|300px]]

There are many proofs of the theorem. One of the most straightforward is the following:

Take four right-angled triangles and connect them into a square as shown, so that they hypotenuses form an inner square as follows:

We can see that the length of each side of the outer square is <math>a+b</math>, and that the area must be <math>(a+b)^2</math>.

The pieces of the square are the inner square and our four triangles.

The area of each triangle can be given as <math>\tfrac{1}{2}ab</math>.

Meanwhile, the inner square has side length of <math>c</math>, and area of <math>c^2</math>.

From this, we can see that:

<math>(a+b)^2 = 2ab + c^2</math>

Expanding, we see that:

<math> \ a^2+2ab+b^2 = 2ab +c^2</math>

Balance the equation by subtracting <math>2ab</math> to get:

<math>a^2+b^2 = c^2</math>

This is, of course, the formula for the theorem!

=== What Can We Do With It?===

Once you know the fundamentals of Pythagorean Theorem you can broaden the range of questions that you do and you can solve much more difficult questions with multiple steps involved.

One example of a real-life application of the Pythagorean theorem would be figuring out the necessary height of an object. Lets suppose you have a 15 meter high wall. You want to find out how long a ladder has to be if it is 5 meters from the base of the wall. Implementing the Pythagorean theorem one is able to solve this issue with little to no issues. By simply substituting 5 and 15 for a and b (in the formula) we can obtain the result which will give us the necessary height of the ladder.

Therefore:

<math>5^2 + 15^2 = 250</math>

<math>\sqrt{250} = 15.8m </math>

15.8 meters is the exact height the ladder must be to fit the given parameters of the wall.

As another example, let’s assume two friends Fred and Matt want to meet at a specific point, say a shopping mall. Fred is 8 km North of the mall and Matt is 7 km East of the mall (forming a right angle from the mall.) How do we find the distance between the two of them? By using the Pythagorean theorem we are able to determine the distance between them.

Let's call <math>a</math> the distance between the mall and Fred and <math>b</math> the distance between Matt and the mall. This means that the distance between the two friends can be labelled <math>c</math>. We can input these values into our formula:

<math>5^2 + 7^2 = c^2 </math>

So, <math>c = \sqrt{25 + 49}</math>

<math>\sqrt {74} = 8.6 km </math>

So we now know that Fred and Matt are 8.6 km from each other.

Other common questions could be based on discovering how much time you save by cutting across a field, calculating the length of a shadow cast by a building, or even calculating the height of a building based on the shadow. It can be used to find the length between the corners of a ceiling or floor while laying tiles.

A couple examples of other things you could use the theorem for would be computing Euclidean distance in various coordinate systems, Pythagorean trigonometric identities, along with complex arithmetic and various other questions. The following pdf attachment contains many problems of increasing complexity.

[http://bodmas.org/bnd/docs/y08_maths_worksheet_pythagoras.pdf Follow this link for further examples.]

They Pythagorean Theorem can even be applied in video games! As awkward as this game may be, it shows the integration of mathematics into the gaming world.
{{#ev:youtube | jQ2QDnpImcg | 400}}

=== What Does This Have To Do With the Trigonometric Circle?===

Given that the radius of a trigonometric circle is one, we can easily find out the sine length given the cosine length, or the reverse, using the Pythagorean theorem.

It is important to remember that when you draw a radius line at any angle from the origin, these rules apply. The sine value <math>sin(\theta)</math> will always be the distance on the y-axis between the point on the circle and the x-axis (labeled ''opposite'' on the image below). The cosine value <math>cos(\theta)</math> is therefore the distance between the point on the circle and the y-axis (labeled ''adjacent'' on the image below). The triangle formed by these distances will always have a hypotenuse length equal to 1 (the radius of the circle). Therefore, using the Pythagorean theorem, we can see that:

<math>\sin^2(\theta) + \cos^2(\theta) = 1</math>

[[File:Defining-Trigonometric-functions-from-a-right-angle-triangle-inside-a-circle.jpg|thumb|left|Application of the Pythagorean Theorem to a trigonometric circle|300px]]

Let's try an example with the cosine length equal to <math>\tfrac{1}{2}</math> We now have enough information to able to convert the Pythagoras Theorem to get the value for the opposite sine length.

<math>c^2 = a^2 + b^2</math>

<math>1^2 = a^2 + \tfrac{1}{2}^2</math>

Since we don't have one of the legs of the theorem we must arrange the formula so that we can isolate the piece to the puzzle we don't know.

<math>c^2 - b^2 = a^2</math>

<math>1^2 - \tfrac{1}{2}^2 = a^2</math>

<math>1 - \tfrac{1}{2} = a^2</math>

<math>0.75 = a^2</math>

<math>a = \tfrac{\sqrt{3}}{2}</math>

This method is quite simple and can also be used to solve the Cosine length, as long as we have one of the legs in the triangle.

==Proposition for Basic Skills Project==

We propose to put up information about the Pythagorean Theorem. We could put up some proofs, various applications of the theorem, how to use it, and other such things. This could certainly tie in to distance and lines, as well, as getting the distance between two points on a graph is essentially the same thing.

[[User:BenJeffery|BenJeffery]]

----

Using information from the group members who responded (Christa, Marco, Ben):

==Things we all can do well==

10. Distance and Lines

12. Construction of Graphs

13a. Pythagorean Theorem

14. Areas and Volumes

15. Mathematical Writing

==Things some of us can do well==

3. Equations

4. Inequalities

5. Composition of Functions

8. Intersections of Functions

9. Reading Graphs of Functions

11. Operations on Graphs of Functions

13b. Trigonometry

== Things none of us can do well==

1. Basic Functions

2. Properties of Functions

6. Polynomial Long Division

7. Graphs of Functions

Course:MATH110/Archive/2010-2011/003/Groups/Group 17/Basic Skills Project

2010-12-05T08:16:37Z

BenJeffery: /* The Pythagorean Theorem */

==The Pythagorean Theorem==

=== What Is It? ===

[[File:500px-Pythagorean.svg.png|thumb|right|A geometric visualization of the Pythagorean Theorem|300px]]

'''History:'''

The theory was originally developed by a man named Pythagoras, who was born in the late 6th century B.C. on the island of Samos, Greece. He proved that for any right angle triangle, the two shorter sides squared and added together exactly equal the square of the longest side. This can be shown mathematically as:

<math>a^2 + b^2 = c^2</math>

'''Pythagorean Basics:'''

The theorem only works for triangles, specifically those with a right (90 degree) angle.

This image gives a visual example of how the Pythagorean theorem looks. The larger square has each side equal to the length of the hypotenuse (longest side) of the triangle. The two smaller squares are formed from the other two lengths on the triangle.

The theory, at it's most basic, states that the area of the largest square is equal to the combined areas of the two smaller squares. Since the area of a square is given by multiplying the length of a side by itself, we can label the sides above as shown, and bring ourselves right back to the mathematical formula! <math>a^2+b^2=c^2</math>

The following video shows a basic use of the theorem.

{{#ev:youtube | ku4rEwRxZOc | 400}}

For an interesting visual example, watch the following video. It shows that the water from the large square, formed by the hypotenuse, fills the smaller squares formed by the other two sides. Note that though this is an interesting visualization of the theorem, it is not actually a concrete proof, simply an interesting visual demonstration.

{{#ev:youtube | CAkMUdeB06o | 400}}

=== How Do We Know It's True? ===

[[File:Proof_2.png|thumb|right|An example of a proof of the Pythagorean Theorem|300px]]

There are many proofs of the theorem. One of the most straightforward is the following:

Take four right-angled triangles and connect them into a square as shown, so that they hypotenuses form an inner square as follows:

We can see that the length of each side of the outer square is <math>a+b</math>, and that the area must be <math>(a+b)^2</math>.

The pieces of the square are the inner square and our four triangles.

The area of each triangle can be given as <math>(1/2)ab</math>.

Meanwhile, the inner square has side length of <math>c</math>, and area of <math>c^2</math>.

From this, we can see that:

<math>(a+b)^2 = 2ab + c^2</math>

Expanding, we see that:

<math> \ a^2+2ab+b^2 = 2ab +c^2</math>

Balance the equation by subtracting <math>2ab</math> to get:

<math>a^2+b^2 = c^2</math>

This is, of course, the formula for the theorem!

=== What Can We Do With It?===

Once you know the fundamentals of Pythagorean Theorem you can broaden the range of questions that you do and you can solve much more difficult questions with multiple steps involved.

One example of a real-life application of the Pythagorean theorem would be figuring out the necessary height of an object. Lets suppose you have a 15 meter high wall. You want to find out how long a ladder has to be if it is 5 meters from the base of the wall. Implementing the Pythagorean theorem one is able to solve this issue with little to no issues. By simply substituting 5 and 15 for a and b (in the formula) we can obtain the result which will give us the necessary height of the ladder.

Therefore:

<math>5^2 + 15^2 = 250</math>

<math>\sqrt{250} = 15.8m </math>

15.8 meters is the exact height the ladder must be to fit the given parameters of the wall.

As another example, let’s assume two friends Fred and Matt want to meet at a specific point, say a shopping mall. Fred is 8 km North of the mall and Matt is 7 km East of the mall (forming a right angle from the mall.) How do we find the distance between the two of them? By using the Pythagorean theorem we are able to determine the distance between them.

Let's call <math>a</math> the distance between the mall and Fred and <math>b</math> the distance between Matt and the mall. This means that the distance between the two friends can be labelled <math>c</math>. We can input these values into our formula:

<math>5^2 + 7^2 = c^2 </math>

So, <math>c = \sqrt{25 + 49}</math>

<math>\sqrt {74} = 8.6 km </math>

So we now know that Fred and Matt are 8.6 km from each other.

Other common questions could be based on discovering how much time you save by cutting across a field, calculating the length of a shadow cast by a building, or even calculating the height of a building based on the shadow. It can be used to find the length between the corners of a ceiling or floor while laying tiles.

A couple examples of other things you could use the theorem for would be computing Euclidean distance in various coordinate systems, Pythagorean trigonometric identities, along with complex arithmetic and various other questions. The following pdf attachment contains many problems of increasing complexity.

[http://bodmas.org/bnd/docs/y08_maths_worksheet_pythagoras.pdf Follow this link for further examples.]

They Pythagorean Theorem can even be applied in video games! As awkward as this game may be, it shows the integration of mathematics into the gaming world.
{{#ev:youtube | jQ2QDnpImcg | 400}}

=== What Does This Have To Do With the Trigonometric Circle?===

Given that the radius of a trigonometric circle is one, we can easily find out the sine length given the cosine length, or the reverse, using the Pythagorean theorem.

It is important to remember that when you draw a radius line at any angle from the origin, these rules apply. The sine value <math>sin(\theta)</math> will always be the distance on the y-axis between the point on the circle and the x-axis (labeled ''opposite'' on the image below). The cosine value <math>cos(\theta)</math> is therefore the distance between the point on the circle and the y-axis (labeled ''adjacent'' on the image below). The triangle formed by these distances will always have a hypotenuse length equal to 1 (the radius of the circle). Therefore, using the Pythagorean theorem, we can see that:

<math>\sin^2(\theta) + \cos^2(\theta) = 1</math>

[[File:Defining-Trigonometric-functions-from-a-right-angle-triangle-inside-a-circle.jpg|thumb|left|Application of the Pythagorean Theorem to a trigonometric circle|300px]]

Let's try an example with the cosine length equal to <math>\tfrac{1}{2}</math> We now have enough information to able to convert the Pythagoras Theorem to get the value for the opposite sine length.

<math>c^2 = a^2 + b^2</math>

<math>1^2 = a^2 + \tfrac{1}{2}^2</math>

Since we don't have one of the legs of the theorem we must arrange the formula so that we can isolate the piece to the puzzle we don't know.

<math>c^2 - b^2 = a^2</math>

<math>1^2 - \tfrac{1}{2}^2 = a^2</math>

<math>1 - \tfrac{1}{2} = a^2</math>

<math>0.75 = a^2</math>

<math>a = \tfrac{\sqrt{3}}{2}</math>

This method is quite simple and can also be used to solve the Cosine length, as long as we have one of the legs in the triangle.

==Proposition for Basic Skills Project==

We propose to put up information about the Pythagorean Theorem. We could put up some proofs, various applications of the theorem, how to use it, and other such things. This could certainly tie in to distance and lines, as well, as getting the distance between two points on a graph is essentially the same thing.

[[User:BenJeffery|BenJeffery]]

----

Using information from the group members who responded (Christa, Marco, Ben):

==Things we all can do well==

10. Distance and Lines

12. Construction of Graphs

13a. Pythagorean Theorem

14. Areas and Volumes

15. Mathematical Writing

==Things some of us can do well==

3. Equations

4. Inequalities

5. Composition of Functions

8. Intersections of Functions

9. Reading Graphs of Functions

11. Operations on Graphs of Functions

13b. Trigonometry

== Things none of us can do well==

1. Basic Functions

2. Properties of Functions

6. Polynomial Long Division

7. Graphs of Functions

Course:MATH110/Archive/2010-2011/003/Groups/Group 17/Basic Skills Project

2010-12-05T08:16:01Z

BenJeffery: /* What Is It? */

==The Pythagorean Theorem==

=== What Is It? ===

[[File:500px-Pythagorean.svg.png|thumb|right|A geometric visualization of the Pythagorean Theorem|300px]]

'''History:'''

The theory was originally developed by a man named Pythagoras, who was born in the late 6th century B.C. on the island of Samos, Greece. He proved that for any right angle triangle, the two shorter sides squared and added together exactly equal the square of the longest side. This can be shown mathematically as:

<math>a^2 + b^2 = c^2</math>

'''Pythagorean Basics:'''

The theorem only works for triangles, specifically those with a right (90 degree) angle.

This image gives a visual example of how the Pythagorean theorem looks. The larger square has each side equal to the length of the hypotenuse (longest side) of the triangle. The two smaller squares are formed from the other two lengths on the triangle.

The theory, at it's most basic, states that the area of the largest square is equal to the combined areas of the two smaller squares. Since the area of a square is given by multiplying the length of a side by itself, we can label the sides above as shown, and bring ourselves right back to the mathematical formula! <math>a^2+b^2=c^2</math>

The following video shows a basic use of the theorem.

{{#ev:youtube | ku4rEwRxZOc | 400}}

For an interesting visual example, watch the following video. It shows that the water from the large square, formed by the hypotenuse, fills the smaller squares formed by the other two sides. Note that though this is an interesting visualization of the theorem, it is not actually a concrete proof, simply an interesting visual demonstration.

{{#ev:youtube | CAkMUdeB06o | 400}}

=== How Do We Know It's True? ===

[[File:Proof_2.png|thumb|right|An example of a proof of the Pythagorean Theorem|300px]]

There are many proofs of the theorem. One of the most straightforward is the following:

Take four right-angled triangles and connect them into a square as shown, so that they hypotenuses form an inner square as follows:

We can see that the length of each side of the outer square is <math>a+b</math>, and that the area must be <math>(a+b)^2</math>.

The pieces of the square are the inner square and our four triangles.

The area of each triangle can be given as <math>(1/2)ab</math>.

Meanwhile, the inner square has side length of <math>c</math>, and area of <math>c^2</math>.

From this, we can see that:

<math>(a+b)^2 = 2ab + c^2</math>

Expanding, we see that:

<math> \ a^2+2ab+b^2 = 2ab +c^2</math>

Balance the equation by subtracting <math>2ab</math> to get:

<math>a^2+b^2 = c^2</math>

This is, of course, the formula for the theorem!

=== What Can We Do With It?===

Once you know the fundamentals of Pythagorean Theorem you can broaden the range of questions that you do and you can solve much more difficult questions with multiple steps involved.

One example of a real-life application of the Pythagorean theorem would be figuring out the necessary height of an object. Lets suppose you have a 15 meter high wall. You want to find out how long a ladder has to be if it is 5 meters from the base of the wall. Implementing the Pythagorean theorem one is able to solve this issue with little to no issues. By simply substituting 5 and 15 for a and b (in the formula) we can obtain the result which will give us the necessary height of the ladder.

Therefore:

<math>5^2 + 15^2 = 250</math>

<math>\sqrt{250} = 15.8m </math>

15.8 meters is the exact height the ladder must be to fit the given parameters of the wall.

As another example, let’s assume two friends Fred and Matt want to meet at a specific point, say a shopping mall. Fred is 8 km North of the mall and Matt is 7 km East of the mall (forming a right angle from the mall.) How do we find the distance between the two of them? By using the Pythagorean theorem we are able to determine the distance between them.

Let's call <math>a</math> the distance between the mall and Fred and <math>b</math> the distance between Matt and the mall. This means that the distance between the two friends can be labelled <math>c</math>. We can input these values into our formula:

<math>5^2 + 7^2 = c^2 </math>

So, <math>c = \sqrt{25 + 49}</math>

<math>\sqrt {74} = 8.6 km </math>

So we now know that Fred and Matt are 8.6 km from each other.

Other common questions could be based on discovering how much time you save by cutting across a field, calculating the length of a shadow cast by a building, or even calculating the height of a building based on the shadow. It can be used to find the length between the corners of a ceiling or floor while laying tiles.

A couple examples of other things you could use the theorem for would be computing Euclidean distance in various coordinate systems, Pythagorean trigonometric identities, along with complex arithmetic and various other questions. The following pdf attachment contains many problems of increasing complexity.

[http://bodmas.org/bnd/docs/y08_maths_worksheet_pythagoras.pdf Follow this link for further examples.]

They Pythagorean Theorem can even be applied in video games! As awkward as this game may be, it shows the integration of mathematics into the gaming world.
{{#ev:youtube | jQ2QDnpImcg | 400}}

=== What Does This Have To Do With the Trigonometric Circle?===

Given that the radius of a trigonometric circle is one, we can easily find out the sine length given the cosine length, or the reverse, using the Pythagorean theorem.

It is important to remember that when you draw a radius line at any angle from the origin, these rules apply. The sine value <math>sin(\theta)</math> will always be the distance on the y-axis between the point on the circle and the x-axis (labeled ''opposite'' on the image below). The cosine value <math>cos(\theta)</math> is therefore the distance between the point on the circle and the y-axis (labeled ''adjacent'' on the image below). The triangle formed by these distances will always have a hypotenuse length equal to 1 (the radius of the circle). Therefore, using the Pythagorean theorem, we can see that:

<math>\sin^2(\theta) + \cos^2(\theta) = 1</math>

[[File:Defining-Trigonometric-functions-from-a-right-angle-triangle-inside-a-circle.jpg|thumb|left|Application of the Pythagorean Theorem to a trigonometric circle|300px]]

Let's try an example with the cosine length equal to <math>\tfrac{1}{2}</math> We now have enough information to able to convert the Pythagoras Theorem to get the value for the opposite sine length.

<math>c^2 = a^2 + b^2</math>

<math>1^2 = a^2 + \tfrac{1}{2}^2</math>

Since we don't have one of the legs of the theorem we must arrange the formula so that we can isolate the piece to the puzzle we don't know.

<math>c^2 - b^2 = a^2</math>

<math>1^2 - \tfrac{1}{2}^2 = a^2</math>

<math>1 - \tfrac{1}{2} = a^2</math>

<math>0.75 = a^2</math>

<math>a = \tfrac{\sqrt{3}}{2}</math>

This method is quite simple and can also be used to solve the Cosine length, as long as we have one of the legs in the triangle.

==Proposition for Basic Skills Project==

We propose to put up information about the Pythagorean Theorem. We could put up some proofs, various applications of the theorem, how to use it, and other such things. This could certainly tie in to distance and lines, as well, as getting the distance between two points on a graph is essentially the same thing.

[[User:BenJeffery|BenJeffery]]

----

Using information from the group members who responded (Christa, Marco, Ben):

==Things we all can do well==

10. Distance and Lines

12. Construction of Graphs

13a. Pythagorean Theorem

14. Areas and Volumes

15. Mathematical Writing

==Things some of us can do well==

3. Equations

4. Inequalities

5. Composition of Functions

8. Intersections of Functions

9. Reading Graphs of Functions

11. Operations on Graphs of Functions

13b. Trigonometry

== Things none of us can do well==

1. Basic Functions

2. Properties of Functions

6. Polynomial Long Division

7. Graphs of Functions

Course:MATH110/Archive/2010-2011/003/Groups/Group 17/Basic Skills Project

2010-12-05T08:15:30Z

BenJeffery: /* What Can We Do With It? */

==The Pythagorean Theorem==

=== What Is It? ===

[[File:500px-Pythagorean.svg.png|thumb|right|A geometric visualization of the Pythagorean Theorem|300px]]

'''History:'''

The theory was originally developed by a man named Pythagoras, who was born in the late 6th century B.C. on the island of Samos, Greece. He proved that for any right angle triangle, the two shorter sides squared and added together exactly equal the square of the longest side. This can be shown mathematically as:

<math>a^2 + b^2 = c^2</math>

'''Pythagorean Basics:'''

The theorem only works for triangles, specifically those with a right (90 degree) angle.

This image gives a visual example of how the Pythagorean theorem looks. The larger square has each side equal to the length of the hypotenuse (longest side) of the triangle. The two smaller squares are formed from the other two lengths on the triangle.

The theory, at it's most basic, states that the area of the largest square is equal to the combined areas of the two smaller squares. Since the area of a square is given by multiplying the length of a side by itself, we can label the sides above as shown, and bring ourselves right back to the mathematical formula! <math>a^2+b^2=c^2</math>

The following video shows a basic use of the theorem.

{{#ev:youtube | ku4rEwRxZOc | 400}}

For an interesting visual example, watch the following video. It shows that the water from the large square, formed by the hypotenuse, fills the smaller squares formed by the other two sides. Note that though this is an interesting visualization of the theorem, it is not actually a concrete proof, simply an interesting visual demonstration.

{{#ev:youtube | CAkMUdeB06o | 400}}

=== How Do We Know It's True? ===

[[File:Proof_2.png|thumb|right|An example of a proof of the Pythagorean Theorem|300px]]

There are many proofs of the theorem. One of the most straightforward is the following:

Take four right-angled triangles and connect them into a square as shown, so that they hypotenuses form an inner square as follows:

We can see that the length of each side of the outer square is <math>a+b</math>, and that the area must be <math>(a+b)^2</math>.

The pieces of the square are the inner square and our four triangles.

The area of each triangle can be given as <math>(1/2)ab</math>.

Meanwhile, the inner square has side length of <math>c</math>, and area of <math>c^2</math>.

From this, we can see that:

<math>(a+b)^2 = 2ab + c^2</math>

Expanding, we see that:

<math> \ a^2+2ab+b^2 = 2ab +c^2</math>

Balance the equation by subtracting <math>2ab</math> to get:

<math>a^2+b^2 = c^2</math>

This is, of course, the formula for the theorem!

=== What Can We Do With It?===

Once you know the fundamentals of Pythagorean Theorem you can broaden the range of questions that you do and you can solve much more difficult questions with multiple steps involved.

One example of a real-life application of the Pythagorean theorem would be figuring out the necessary height of an object. Lets suppose you have a 15 meter high wall. You want to find out how long a ladder has to be if it is 5 meters from the base of the wall. Implementing the Pythagorean theorem one is able to solve this issue with little to no issues. By simply substituting 5 and 15 for a and b (in the formula) we can obtain the result which will give us the necessary height of the ladder.

Therefore:

<math>5^2 + 15^2 = 250</math>

<math>\sqrt{250} = 15.8m </math>

15.8 meters is the exact height the ladder must be to fit the given parameters of the wall.

As another example, let’s assume two friends Fred and Matt want to meet at a specific point, say a shopping mall. Fred is 8 km North of the mall and Matt is 7 km East of the mall (forming a right angle from the mall.) How do we find the distance between the two of them? By using the Pythagorean theorem we are able to determine the distance between them.

Let's call <math>a</math> the distance between the mall and Fred and <math>b</math> the distance between Matt and the mall. This means that the distance between the two friends can be labelled <math>c</math>. We can input these values into our formula:

<math>5^2 + 7^2 = c^2 </math>

So, <math>c = \sqrt{25 + 49}</math>

<math>\sqrt {74} = 8.6 km </math>

So we now know that Fred and Matt are 8.6 km from each other.

Other common questions could be based on discovering how much time you save by cutting across a field, calculating the length of a shadow cast by a building, or even calculating the height of a building based on the shadow. It can be used to find the length between the corners of a ceiling or floor while laying tiles.

A couple examples of other things you could use the theorem for would be computing Euclidean distance in various coordinate systems, Pythagorean trigonometric identities, along with complex arithmetic and various other questions. The following pdf attachment contains many problems of increasing complexity.

[http://bodmas.org/bnd/docs/y08_maths_worksheet_pythagoras.pdf Follow this link for further examples.]

They Pythagorean Theorem can even be applied in video games! As awkward as this game may be, it shows the integration of mathematics into the gaming world.
{{#ev:youtube | jQ2QDnpImcg | 400}}

=== What Does This Have To Do With the Trigonometric Circle?===

Given that the radius of a trigonometric circle is one, we can easily find out the sine length given the cosine length, or the reverse, using the Pythagorean theorem.

It is important to remember that when you draw a radius line at any angle from the origin, these rules apply. The sine value <math>sin(\theta)</math> will always be the distance on the y-axis between the point on the circle and the x-axis (labeled ''opposite'' on the image below). The cosine value <math>cos(\theta)</math> is therefore the distance between the point on the circle and the y-axis (labeled ''adjacent'' on the image below). The triangle formed by these distances will always have a hypotenuse length equal to 1 (the radius of the circle). Therefore, using the Pythagorean theorem, we can see that:

<math>\sin^2(\theta) + \cos^2(\theta) = 1</math>

[[File:Defining-Trigonometric-functions-from-a-right-angle-triangle-inside-a-circle.jpg|thumb|left|Application of the Pythagorean Theorem to a trigonometric circle|300px]]

Let's try an example with the cosine length equal to <math>\tfrac{1}{2}</math> We now have enough information to able to convert the Pythagoras Theorem to get the value for the opposite sine length.

<math>c^2 = a^2 + b^2</math>

<math>1^2 = a^2 + \tfrac{1}{2}^2</math>

Since we don't have one of the legs of the theorem we must arrange the formula so that we can isolate the piece to the puzzle we don't know.

<math>c^2 - b^2 = a^2</math>

<math>1^2 - \tfrac{1}{2}^2 = a^2</math>

<math>1 - \tfrac{1}{2} = a^2</math>

<math>0.75 = a^2</math>

<math>a = \tfrac{\sqrt{3}}{2}</math>

This method is quite simple and can also be used to solve the Cosine length, as long as we have one of the legs in the triangle.

==Proposition for Basic Skills Project==

We propose to put up information about the Pythagorean Theorem. We could put up some proofs, various applications of the theorem, how to use it, and other such things. This could certainly tie in to distance and lines, as well, as getting the distance between two points on a graph is essentially the same thing.

[[User:BenJeffery|BenJeffery]]

----

Using information from the group members who responded (Christa, Marco, Ben):

==Things we all can do well==

10. Distance and Lines

12. Construction of Graphs

13a. Pythagorean Theorem

14. Areas and Volumes

15. Mathematical Writing

==Things some of us can do well==

3. Equations

4. Inequalities

5. Composition of Functions

8. Intersections of Functions

9. Reading Graphs of Functions

11. Operations on Graphs of Functions

13b. Trigonometry

== Things none of us can do well==

1. Basic Functions

2. Properties of Functions

6. Polynomial Long Division

7. Graphs of Functions

Course:MATH110/Archive/2010-2011/003/Groups/Group 17/Basic Skills Project

2010-12-05T08:14:32Z

BenJeffery: /* What Does This Have To Do With the Trigonometric Circle? */

==The Pythagorean Theorem==

=== What Is It? ===

[[File:500px-Pythagorean.svg.png|thumb|right|A geometric visualization of the Pythagorean Theorem|300px]]

'''History:'''

The theory was originally developed by a man named Pythagoras, who was born in the late 6th century B.C. on the island of Samos, Greece. He proved that for any right angle triangle, the two shorter sides squared and added together exactly equal the square of the longest side. This can be shown mathematically as:

<math>a^2 + b^2 = c^2</math>

'''Pythagorean Basics:'''

The theorem only works for triangles, specifically those with a right (90 degree) angle.

This image gives a visual example of how the Pythagorean theorem looks. The larger square has each side equal to the length of the hypotenuse (longest side) of the triangle. The two smaller squares are formed from the other two lengths on the triangle.

The theory, at it's most basic, states that the area of the largest square is equal to the combined areas of the two smaller squares. Since the area of a square is given by multiplying the length of a side by itself, we can label the sides above as shown, and bring ourselves right back to the mathematical formula! <math>a^2+b^2=c^2</math>

The following video shows a basic use of the theorem.

{{#ev:youtube | ku4rEwRxZOc | 400}}

For an interesting visual example, watch the following video. It shows that the water from the large square, formed by the hypotenuse, fills the smaller squares formed by the other two sides. Note that though this is an interesting visualization of the theorem, it is not actually a concrete proof, simply an interesting visual demonstration.

{{#ev:youtube | CAkMUdeB06o | 400}}

=== How Do We Know It's True? ===

[[File:Proof_2.png|thumb|right|An example of a proof of the Pythagorean Theorem|300px]]

There are many proofs of the theorem. One of the most straightforward is the following:

Take four right-angled triangles and connect them into a square as shown, so that they hypotenuses form an inner square as follows:

We can see that the length of each side of the outer square is <math>a+b</math>, and that the area must be <math>(a+b)^2</math>.

The pieces of the square are the inner square and our four triangles.

The area of each triangle can be given as <math>(1/2)ab</math>.

Meanwhile, the inner square has side length of <math>c</math>, and area of <math>c^2</math>.

From this, we can see that:

<math>(a+b)^2 = 2ab + c^2</math>

Expanding, we see that:

<math> \ a^2+2ab+b^2 = 2ab +c^2</math>

Balance the equation by subtracting <math>2ab</math> to get:

<math>a^2+b^2 = c^2</math>

This is, of course, the formula for the theorem!

=== What Can We Do With It?===

Once you know the fundamentals of Pythagorean Theorem you can broaden the range of questions that you do and you can solve much more difficult questions with multiple steps involved.

One example of a real-life application of the Pythagorean theorem would be figuring out the necessary height of an object. Lets suppose you have a 15 meter high wall. You want to find out how long a ladder has to be if it is 5 meters from the base of the wall. Implementing the Pythagorean theorem one is able to solve this issue with little to no issues. By simply substituting 5 and 15 for a and b (in the formula) we can obtain the result which will give us the necessary height of the ladder.

Therefore:

<math>5^2 + 15^2 = 250</math>

<math>\sqrt{250} = 15.8m </math>

<math>15.8m</math> is the exact height the ladder must be to fit the given parameters of the wall.

As another example, let’s assume two friends Fred and Matt want to meet at a specific point, say a shopping mall. Fred is 8 km North of the mall and Matt is 7 km East of the mall (forming a right angle from the mall.) How do we find the distance between the two of them? By using the Pythagorean theorem we are able to determine the distance between them.

Let's call <math>a</math> the distance between the mall and Fred and <math>b</math> the distance between Matt and the mall. This means that the distance between the two friends can be labelled <math>c</math>. We can input these values into our formula:

<math>5^2 + 7^2 = c^2 </math>

So, <math>c = sqrt(25 + 49)</math>

<math>\sqrt {74} = 8.6 km </math>

So we now know that Fred and Matt are 8.6 km from each other.

Other common questions could be based on discovering how much time you save by cutting across a field, calculating the length of a shadow cast by a building, or even calculating the height of a building based on the shadow. It can be used to find the length between the corners of a ceiling or floor while laying tiles.

A couple examples of other things you could use the theorem for would be computing Euclidean distance in various coordinate systems, Pythagorean trigonometric identities, along with complex arithmetic and various other questions. The following pdf attachment contains many problems of increasing complexity.

[http://bodmas.org/bnd/docs/y08_maths_worksheet_pythagoras.pdf Follow this link for further examples.]

They Pythagorean Theorem can even be applied in video games! As awkward as this game may be, it shows the integration of mathematics into the gaming world.
{{#ev:youtube | jQ2QDnpImcg | 400}}

=== What Does This Have To Do With the Trigonometric Circle?===

Given that the radius of a trigonometric circle is one, we can easily find out the sine length given the cosine length, or the reverse, using the Pythagorean theorem.

It is important to remember that when you draw a radius line at any angle from the origin, these rules apply. The sine value <math>sin(\theta)</math> will always be the distance on the y-axis between the point on the circle and the x-axis (labeled ''opposite'' on the image below). The cosine value <math>cos(\theta)</math> is therefore the distance between the point on the circle and the y-axis (labeled ''adjacent'' on the image below). The triangle formed by these distances will always have a hypotenuse length equal to 1 (the radius of the circle). Therefore, using the Pythagorean theorem, we can see that:

<math>\sin^2(\theta) + \cos^2(\theta) = 1</math>

[[File:Defining-Trigonometric-functions-from-a-right-angle-triangle-inside-a-circle.jpg|thumb|left|Application of the Pythagorean Theorem to a trigonometric circle|300px]]

Let's try an example with the cosine length equal to <math>\tfrac{1}{2}</math> We now have enough information to able to convert the Pythagoras Theorem to get the value for the opposite sine length.

<math>c^2 = a^2 + b^2</math>

<math>1^2 = a^2 + \tfrac{1}{2}^2</math>

Since we don't have one of the legs of the theorem we must arrange the formula so that we can isolate the piece to the puzzle we don't know.

<math>c^2 - b^2 = a^2</math>

<math>1^2 - \tfrac{1}{2}^2 = a^2</math>

<math>1 - \tfrac{1}{2} = a^2</math>

<math>0.75 = a^2</math>

<math>a = \tfrac{\sqrt{3}}{2}</math>

This method is quite simple and can also be used to solve the Cosine length, as long as we have one of the legs in the triangle.

==Proposition for Basic Skills Project==

We propose to put up information about the Pythagorean Theorem. We could put up some proofs, various applications of the theorem, how to use it, and other such things. This could certainly tie in to distance and lines, as well, as getting the distance between two points on a graph is essentially the same thing.

[[User:BenJeffery|BenJeffery]]

----

Using information from the group members who responded (Christa, Marco, Ben):

==Things we all can do well==

10. Distance and Lines

12. Construction of Graphs

13a. Pythagorean Theorem

14. Areas and Volumes

15. Mathematical Writing

==Things some of us can do well==

3. Equations

4. Inequalities

5. Composition of Functions

8. Intersections of Functions

9. Reading Graphs of Functions

11. Operations on Graphs of Functions

13b. Trigonometry

== Things none of us can do well==

1. Basic Functions

2. Properties of Functions

6. Polynomial Long Division

7. Graphs of Functions