Hey guys and gals! Hope your all enjoying the new snow. Here is the scribe post for our class on Jan. 26, 2010.
Hope fully by the end of this post you will be feeling like one of those :)
Hope fully by the end of this post you will be feeling like one of those :)
To start of the post with an absolute value attitude here is a math joke for all of you fella's who like those nerdy pick up lines.
"I don't know if you're in my range, but I'd sure like to take you home to my domain."
Now back to business...Please take a look at the slides from class today, they are important for understanding the post. It has pictures of the worksheet we did in class, which is debreifed in this post. If you loose your homework, you can either get organized or just look at the slides from today.
Today in class we discussed differentiability and continuity. We practiced our understanding of these two topics by creating graphs under certain guidelines. The following is a list of the guidelines we had to follow for creating our first 5 graphs:
Draw a graph that is...
1. Differentiable and continuous at point c.2. Has a limit as x approaches c, but fails to be continuous there due to f(c) being undefined.
3. Has a limit as x approaches c, and has a value for f(c), but fails to be continuous there
4. Has a value for f(c), but has no limit as x approaches c
5. Is continuous at the point (c,y), but is not differentiable at this point.
I added these problems from our worksheet for anybody who wants to be an ogre achiever (ha, get it, ogre) and continue to use them as practice for the chapter 4 test. Seeing as I have ESP I predict that test is looming very near, sorry guys. I think these are a great way to do something similar to gymnastics for your brain (as Sara used to say for those of you who were in her class last year).
So whats the whole point of this?
The whole point of this is to come up with some rules that dictate when a function is non-differentiable at a point x=c. Here are the rules I came up with.
1) If there is no value for the function at point x=c, it is not differentiable there
2) If there is no value for the derivative function when f'(c), then it is not differentiable at that point
3) If f'(c) is not a real number, then it is not differentiable at that point.
4) Examples of non-differentiable characteristics that can be found at c: cusp, asymptote, hole, discontinuity.
If all you people of Calculopolis out there have any additions or modification to this leave a comment and I will edit this post accordingly.
Another important topic to discuss from class today was problem number 6. Problem number 6 gave us two functions, the first function
was to be within the domain
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and the second function
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(listed below)was to be within the domain of
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Yet this piecewise function also had continuous and differentiable. Our Quest was to find the values of a and b Sorry for the visible formatting issues here.
The first step to find the value of these two variables is to derive the first function (x^2). The derivative of the first function is 2x. When we substitute in the last number in the domain of the first function's derivative we end up with 2(6) which equals 12. Next, substitute 12 in for the variable a. This is to ensure that the slope of the linear function matches up with the slope of the first function, x^2 at the point x=6. This is to prevent a hole and keep the instantaneous rate of change (the derivative) the same when x=6 for both functions. Because b controls the vertical translation of the linear function, we have to make it match up with the original function x^2. To find where the end of x^2 is, you substitute in 6, which gives you 36. We now know how high the linear portion of the function has to be in order to match up with the first function x^2. The variable b we now know equals 36. Ta Da! We have now successfully solved the problem.
This essentially conclueds our lessons of the day.
Hopefully the world of Calculoplis is a little easier to understand now.
Thanks,
mc Casper
p.s. almost forgot, I dub J-tron foteen as our next scribe
The first step to find the value of these two variables is to derive the first function (x^2). The derivative of the first function is 2x. When we substitute in the last number in the domain of the first function's derivative we end up with 2(6) which equals 12. Next, substitute 12 in for the variable a. This is to ensure that the slope of the linear function matches up with the slope of the first function, x^2 at the point x=6. This is to prevent a hole and keep the instantaneous rate of change (the derivative) the same when x=6 for both functions. Because b controls the vertical translation of the linear function, we have to make it match up with the original function x^2. To find where the end of x^2 is, you substitute in 6, which gives you 36. We now know how high the linear portion of the function has to be in order to match up with the first function x^2. The variable b we now know equals 36. Ta Da! We have now successfully solved the problem.
This essentially conclueds our lessons of the day.
Hopefully the world of Calculoplis is a little easier to understand now.
Thanks,
mc Casper
p.s. almost forgot, I dub J-tron foteen as our next scribe
Posts
Hello mc Casper. Nice job with the post. One thing I might suggest is that when you explained the graphs we drew, you might have given and example of an appropriate graph and equation for those criteria. You also could've pasted in that slide from class which had graphs and equations. This would probably help people who may have missed class figure out what exactly the exploration was asking.
ReplyDeleteWho ever you are I want you to know that I couldn't help from smiling and laughing while reading your post. Great job with the humor and thanks for the heads up on the upcoming test! Also I appreciate your "the point of this is" statement, I sometimes wonder about the point of things. The only problem I had with your post was the last paragraph which was a little long without any color breaks or anything like that. Other than that though I really like your post!
ReplyDeleteThanks
Beston
I agree with Beston and Secret. I thought this was a fantastic post, very clear and very understandable (although you did go a bit color-happy in the beginning). I think when you explained the graphs you could have posted a slide to explain them. Same with the last paragraph. It very clearly described in words what was necessary to solve the two equations, however, a visual aid could have really been a huge benefit (which is in fact the last slide for the day, so not a serious problem).
ReplyDeleteOther than that I really enjoyed reading this post. It was by no means dull and provided me a very large umbrella to protect against the confusing differential rains of calculopolis.
THANKS!
Agreed. Great Post. Clever, colors, math jokes, cities devoted to math... what else to you need? Great job. I will say that I sometimes like having outside sources to find extra help. Maybe finding a good video or site that could provide outside help could be beneficial...
ReplyDeleteAnd don't forget to credit your picture!
Well done!
mc Casper,
ReplyDeleteThis scribe post is really great. I like the light and energetic mood you create with the jokes and the picture in the beginning. I also really liked how you brought back a thing or two from Sarah's class in order to help us prep for this future test. Although that last paragraph was a little long, I like how you took the time to explain the process in words... Sometimes just equations can get a little bit heavy. My only suggestion would be to maybe insert a visual aid or one of the slides from class. Other than that, excellent job!
This post is really great. I like the usage of the picture saying King of the math. Usage of different colors were interesting; however, it wasn't a good idea to use yellow or brown. It is hard to notice the difference for brown, and yellow is hard color to recognize from the post. Also in the middle of the post, it would been better if you use the lined format for the equations.
ReplyDelete