Light Bulb Project

Ski Movie Final Project

The purpose of this final project is to help us understand related rates through teaching them ourselves. We are also hoping that we can help to teach others and challenge them to a new Related Rates problem. About all everyday scenarios, such as skiing, contain some bit of calculus. We are proposing a scenario, in which two skiers are skiing away from each other. We are going to calculate the rate at which the distance between them is changing. This problem was created by skirdude, secret and Flying Slug.

Two skiers are having an extreme day on Snowmass Mountain. They have been skiing together all day, dropping fatty cliffs and schralping mad gnar. Its the end of the day, and they are getting tired, so they now decide they both want to go down different runs. They are total math buffs and decide to turn this into an awesome Related Rates problem!

You can find our solution here!

And a big shout out to the crew at Aspen for being very patient with us and giving us tons of help!

Math Geeks Will Rule the World

In the cover story in Business Week, Steven Baker addressed how math is becoming necessary in today's world. Because our world is now filled with oceans of data, math is being used more and more to 'mine', interpret, and use the data. The beginning of the article talked about how mathematical models can be used by companies to increase revenue:

"These studies crunched consumer data to measure the effectiveness of advertising in a host of media. The results came back in hard numbers. They indicated, for example, that Ford Motor Co. () could have sold an additional $625 million worth of trucks if it had lifted its online ad budget from 2.5% to 6% of the total."(3).

This kind of data is hugely valuable and could be potentially used on a federal level to calculate spending in the government. A lot of the article, however, talked about how math can be used with people, "The power of mathematicians to make sense of personal data and to model the behavior of individuals will inevitably continue to erode privacy." (2). Companies can use math to quantify and analyze their consumers and use that to create effective advertising.

I feel that, if taken too far, this could lead to a bad future. If we live in a world where every person is one of several million that fit a certain profile, that will be used for advertising, our society will continue to be controlled by media and consumerism. I once read a book call Fahrenheit 451, and this article reminded me of that book. In the book, books were banned, and nobody ever had independent thoughts or opinions. Everyone was stuffed full of information and data, but never taught to think about it or draw conclusions from it. Of course, this is a huge exaggeration of our world, but I feel that if the world moves more towards machines, and numbers, and data that generalize society into numerical patterns and equations, the opportunity for free thought and imagination will become less.

As for the part about how what we're leaning now is actually going to help us in the future, I think the importance of knowing high level math is overestimated. Really, we only need a few geniuses to write the computer programs and then the computers do all the math. I guess though, when math geeks and computers rule the world, those with calculus level math education have a better chance of survival.

Math Pictures!

Just click on the picture and it'll take you to flickr so you can see the tags and stuff.

Choices

1. For any point, (a,b), on an invertible function, the derivative of the inverse of the function evaluated at b is equal to the reciprocal of the derivative of the function evaluated at a.
2. In symbols this is:

3. The volume of a sphere is a function of its radius:

This is an invertible function because it passes the horizontal line test.

4. The inverse of this function is:


5. Here is a graph. The Black line is the original function, f(x) for the volume of a sphere. The Red line is a line tangent to f(x) at 1. The Blue line is the inverse of the original function, f^-1(x). The Green line is a line tangent to f^-1(x) at 4.189.


6. The slope of the tangent of f(x) is 12.566 and the slope of the tangent of f^-1(x) is 0.079 because 0.079=1/12.566

7. I came to class on virtual Wednesday because I really value a classroom environment for learning. I find i rely heavily on Bru and other classmates to be able to fully understand important parts of a lesson. I felt that in a virtual section I would get very frustrated because when I had questions I would not be able to get them answered to my satisfaction.

Anything, Anyone, Anywhere, Anytime

1. In the introduction to our weblog it was suggested that because of Web 2.0, “Anyone can learn anything, from anyone, anywhere, at anytime”. Do you agree or disagree with this statement? If you agree, then what role do schools and formal education play in your learning? If you disagree, then what gets in the way of learning?

Yes, I do agree that anyone can learn anything from anyone, anywhere, at anytime. However, I think that physical schools, teachers, and classmates are vital to learning. For me, there is something about learning from a real life person that betters my learning. Explanations are easier understood when given by a real person for a number of reasons. First, they are spoken rather than read. Secondly, visuals that go along with the explanation are happening right in front of you. And finally there are hands on activities (like the scary clown one) that you can do that really give you a good idea of how whatever you're studying pertains to the real world.

Some might say that all of this can be given through the web, but not in the same way. You can get verbal explanations along with visual aids through videos. There are plenty of learning activities through the web, and a social network of people that act as teachers and classmates, however, you cannot stop a video in the middle and ask what's going on, and commenting back and forth on blog posts to clear up something you don't understand could take weeks. Though the internet is definitely a huge aid in learning and teaching, I don't think the web could ever replace real human interaction. Not only this, but with the internet, there is always a question of validity. Though this exists to some extent with people too, it is much harder to tell if the information you are getting off the internet is right. Though this is not math related, think about foreign languages: an online translator could never give you the same results as a real person. Yes, you can get a real human to translate a language for you through the internet, but they cost money. And why would anyone pay for that when we have Dan Pittz?

Product Rule Scribe Post

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Live from Carbondale, it's Tuesday night! Welcome to your first scribe post. Our first scribe post is about the product rule, and since I can't think of any cool pictures that would represent the product rule, I'm just going to jump right in.

We started exploration 4-2 in class on Monday with this objective: Given a function that is a product of two other functions, find in one step, an equation for the derivative function. In math terms: if f(x)=g(x)*h(x), what will f'(x) be? First, we figured out if differentiation distributes over addition through this test:

• Using a function that was a sum of two functions,:

  y=(x³)+(5x+1)

• we first differentiated by adding the derivative of each separate function:

  d/dx(x³)+d/dx(5x+1)

• this equalled:

  =(3x²)+(5)

• Then we differentiated the function as a whole:

  d/dx(x³+5x+1)

• this equalled:

  =(3x²+5)

As you can see, the derivative of a sum of two funtions is equal to the sum of the derivatives of the two functions. This is how we discovered that YES, differentiation does distribute over addition. The we wondered, does it distribute over multiplication? We used similar steps to answer this question:

• we started with the same composition of equations, but this time they were multiplied:

  y=(x³)*(5x+1)

• we differentiated each equation seperately:

  d/dx(x³)*d/dx(5x+1)

• this equalled

  (3x²)*(5)

• which simplifies to:

  15x²

• then we wanted to differentiate as a whole equation, so first, we simplified by distributing the x³ to the 5x+1 which equalled:

  5x⁴+x³

• then we differentiated the equation as a whole:

  d/dx(5x⁴+x³)

• This equals:

  20x³+3x²

As you can see from this example, the derivative of a product of two functions is not equal to the the product of the derivatives of the two functions. This is how we discovered NO differentiation does NOT distribute over multiplications. So then we knew that we needed to find a way to find the derivative of a product of two functions. We then used the definition of derivative to derive the formula for the derivative of a product of two functions. Below is a slide from Bru's presentation that has the answers to the proof on the second page of your exploration. Below is an explanation of each step.

I'm really sorry if you can't read this, the same thing can be found of the slide show if the product rule that Bru posted.

1.-2.: Since y=uv, Δy=ΔuΔv and

(y+Δy)= (u+Δu)(v+Δv)

2.-3.: FOIL

3.-4.: The positive uv at the beginning and the negative uv at the end add to 0.

3.-4.: explained

4.-5.: You want the Δu or Δv to be in the fraction and the u or v to next to it. Bru did most of this step for you.

5.-6.: Explained, take the limit of all three seperately.

6.-7.: Δ symbol changes to d to symbolize that the limit is a derivative. The Δu in the third becomes zero based on the graph at the top of the page on the far left. We are taking the limit as Δx approaches 0 and as you can see by this graph, as Δx approaches 0, Δu also approaches 0.

So, the product rule is:

If f(x)=g(x)h(x)

Then f'(x)=g'(x)h(x)+g(x)h'(x)

We ended class with a couple of practice problems and learned that the algebra part (simplifying) is actually the hardest part. Here is one example we did in class. My commentary and additions are in red.

Check Your Understanding:

2. If f(x)=[(3x-8)⁷][(4x+9)⁵], find f'(x)

   I put in the red brackets to help me figure out what the two different functions are.

f'(x)= (7(3x-8)⁶•3)((4x+9)⁵)+((3x-8)⁷)•(5(4x+9)⁴•4)

Take a look at the first term. In this term, you differentiated the function (which happened to be a composite functions and required the chain rule), and multiplied it by the second function. Then in the second term, you took the first function and multiplied it by the derivative of the second function (which was also composite). As long as you can keep all this straight in your head, this step is relatively easy. But now you have to simplify.

f'(x)= (3x-8)⁶(4x-9)⁴[21(4x+9)+20(3x-8)]

In this step you factored out (3x-8)⁶ and (4x-9)⁴. Then, you were only left with what is inside the brackets.

f'(x)= (3x-8)⁶(4x-9)⁴[84x+189+60x-160]

Distribute within the brackets.

f'(x)= (3x-8)⁶(4x-9)⁴[144x+29]

Combine like terms within the brackets, and now the equation is fully simplified.

My answers for numbers 3, 4, and 5. Let me know if you agree or disagree.

3. e^sinxcos²x-e^sinxsinx

4. x^-6.3(1/x+ln4x*-6.3x^-1.3)

5. (21x⁶-60x4)cos10x-10(3x⁷-12x⁵)sin10x

   I think this one can be simplified more. I tried to factor out 3x⁴ but it didn't work, or maybe I did it wrong.

Enjoy! Comment if you wish.

Next Scribe is Babar.

Product Rule Site Comparison

Today I just googled 'product rule' and looked at one of the options that came up. From just looking at one site, I understood the product rule. The site I looked at: first gave a verbal definition and also an algebraic definition. Then, it had a whole list of problems that you could look at a detailed solution to. It included all kinds of problems full of exponents, trig, ln, etc. Also, in the solutions, each step was clearly written out and it was easy to see what had been done. When I looked at the book, however, I was confused, even though I had just learned the product rule. The verbal descriptions it had were very complex and confusing and I could not follow them. One thing I didn't like about the web site I found, though, was that there was not a verbal description for each of the steps in the solutions, which the text book offers. It helps me to see example with real numbers instead of just variables, but it also helps me to read how each step was done.
Definition:
if h(x)=f(x)g(x)
then h'(x)=f'(x)g(x)+f(x)g'(x)

Example:
if: h(x)= (x^2+5)(3x)
then: h'(x)= (2x)(3x)+(x^2+5)(3)
(derivative of the first)(the second)+(the first)(derivative of the second)
=3x^2+6x^2+15
=9x^2+15

MInd Map of Semester 1