Mini Test: The Calculus Strikes Back

 

I really need to stop with these late night scribe posts.

 

For the record, Joseph is scribe, and the list REALLY needs updating.

 

Question (1)

 

In this question, the answer should be straight forward. If you approach X from the left (low), then the largest integer value is 1. This is demonstrated in the first line. If you approach from the right (high), the largest integer value is 2. These are also known as roof and floor values. Because the right approach (floor) is not equal to the low approach (roof) then the value cannot exist, meaning the answer is (E).

 

 

Question (2)

 

For this question, we need to absolutely remember what the definition of a derivative is. Because this is it! Written right there. 1 is the value of f(x), which is also equal to sin(π/2). You will never actually see this, but you need to recognize it. Once you see it, this question becomes easy. The derivative of sin(π/2) is 0, because at sin(π/2) you are at the top of the curve and the tangent line (derivative) will be perfectly horizontal. Thus, the value must be zero, or (C).

 

 

Question (3)

 

Just keep deriving. You’ll get there eventually. (C)

 

Question (4)

A painfully simple solution to this question. To find a line, you need to have the slope and a point. You are given the point. Therefore, logically you should find the slope. Derive the equation of your curve, and you know that equation also equals the line with y = 1 and x = –1 (the equation for any line is y – y1 = m(x – x1). Solve for m with algebra. Once done that, do some rearranging with algebra, but the answer should become obvious as being (A).

 

Question (5)

Speed is the scalar version of velocity, which has vector (direction). When velocity is negative, you are moving backwards. Because speed is scalar and has no direction, it is the absolute value of velocity, so in terms of speed even if you are moving “backwards”, your speed will be positive because you are moving period. So if you re plot the absolute value of the graph, you will see the greatest value is at t = 8 which is answer (E).

 

Question (6)

If at t = 3 our object is at the origin, and its velocity is positive, it means the object is moving away from the origin in the positive direction for 3 units (the triangle formed by the graph from 3 to 6 is base 3 and height 2, so (1/2)2(3) = 3). So, we want to find where the object moves in the negative direction for 3 units, Since the value of the triangle created by the graph from 6 to 7 is base 1 and height 2, the area and total distance changed is 1 in the negative direction, meaning we havent gone backwards enough to be back at the origin. If we find the area from 6 to 8 then the base is and the height is 4, so by t = 8 you’ve moved 4 units negatively, which is past the origin, but we can logically deduce that at some point between t = 7 and t =8 we had moved 3 units negatively and thus returned to the origin. So, logically, the answer is (E).

 

Question (7) Free Response

For (a), you know S(0) = 6 so use that to solve for C. Quick math determines that C = 6 because e ^(0k) = 1 so 1C = 6. You also know that consumption doubles every 5 years, so S(5) = 2(6) and 2(6) = 6e^(5k) so e^5k = 2. Take the ln of both sides, so ln2 = 5k, and solve for k.

 

In (b), you want the integral from 3 to 13 (remember that your function counts years from 1980 so 1983 – 1980 = 3 as your starting value, and you want a 10 year period so 10+3 = 13 as your other limit) of the function. Calculator please. Really, you dont want to integrate by hand, but be my guest.

 

(d): “The integral of S(t)dt from 3 to 7 gives the total amount of cola consumed in the United States in billions of gallons per year during the time period from January 1, 1983 to January 1, 1987.”

 

 

Remember guys to start making your significant edits to the wiki solutions manual, and to at least try 3 exam questions a night. This is crunch time!

 

 

I dont really have a new funny/interesting youtube video, so here’s Intergalactic.

 

BOB of Differential Equations

I understood most of this unit, and most of it made sense. Looking at the pre-test, the thing I want to work on most is identifying what kind of solution I should use for a problem. But I liked how a lot of the stuff we did before sort of mixed together in this unit, and it was almost like a review of older stuff (particularly, logarithms...). I found I sometimes got held up on the pre-calculus, which means I might want to brush up on that. 

Also, Newton's Law of Cooling was a little confusing, but I think I get it. I only run into problems when solving for C or pinning down the original function when given a point on the original function.

I think the test should go well for me, and hopefully well for everyone else. Good luck guys!

The simple scribe post

 

Last Thursday, we were given this sheet.

 

 

(click to enlarge)

 

With the answers on the other side:

 

 

(click to enlarge)

 

Its due Monday (tomorrow).

 

I’ll try and post solutions to the best of my ability. Didn’t get far with this one…

 

I have questions 1 and 2… sad. I’ll explain how I solved these and I sort of grasp the concept to solving 4 and 5 so I’ll try and describe those.

 

 

   (click to enlarge)

 

Solution 1, is pretty straightforward. Since the variables are already on their respective sides, just antidifferentiate. Since we’re solving for C, we’ll put that on the left and move the rest to the right. Simplify. As Benchmen pointed out, because C is a constant, even if you multiply it it will remain C, hence the part that says “let (2) C = C”

 

Solution 2 is a little longer, but also pretty straight forward. Organize your variables to their respective sides, integrate/antidifferentiate both sides. Take the ln of both sides, and solve for A which is e^c. The solving for A is a bit of simple algebra.

 

Solution 3 I tried twice but kept getting stuck. I figure the way to solve it is to follow the same steps as in 2, IE organize and integrate, and once you have an equation (this is where I got stuck) solve for t using the given values.

 

Solution 4 I have no clue, sorry guys. Feel free to post your solutions in the comments.

 

Solution 5 I know HOW to solve, but cant actually do it because I get stuck (as you can see) at the point after integrating. However, this is very much like question 3, where you are given inital values. In this question, you would organize, integrate, and then use the inital values to pinpoint the graph since when you integrate/antidifferentiate you are finding the family of functions and you want the specific function that passes through a given point.

 

Solution 6 - 10 use Newtons cooling law.

 

 

As usual, this has been a late scribe post, although from what I remember on Thursday you guys are probably ahead of me on this stuff anyway. See you all tomorrow.

 

Oh and the next scribe is Francis.

Not Paul After Work But Before Pi Day Scribe Post EXTRAVAGANZA (oh and we did some math or something)

Digital Ethics: Or How I Learned To Start Worrying And Think About My Online Identity

 

So today as usual we did our little “AP Exam Practice Quiz.” This one was, in my opinion, easier than the earlier ones because half the quiz was just about the 1st and 2nd derivative (I note though that I got these questions wrong too so hah).

 

The aforementioned questions are these:

 

 

So the first question is just asking you for local minimums. Just by looking at the graph, we can sort of rough out a sketch of the original function f (shown here in red). And that’s all we really need! Just by looking at our rough sketch we can say the original function f has local minimums at f = 5 and f = 0. Remember, we’re talking local not global minimums. In fact, unless your graph is always decreasing or something, your bound to have min/maxes at the beginning and end of a graph. Good thing to remember.

 

In this question, things get even easier. A local max or min on the first derivative is a critical number on the second derivative. 0 and 5 dont work, but 2 and 3 sure do. 2 is an inflection point, although not a local max because the second derivative keeps increasing. 3 is where the second derivative hits a local max and starts going downwards (decreasing).

 

 

If anything, this would be the trickiest question, because you have to remember the relationship between Velocity, Acceleration and Position. Velocity is the anti-derivative of Acceleration, and Position is the anti-derivative of Velocity.  Remember to use the given values to find C so your equation matches the ones given. Otherwise, a cakewalk. Remember, Physics and Calculus are almost the same thing, seeing as Calculus was developed to advance Physics.

 

 

This is all stuff we should know. Anti-deriving, use of the Fundamental Theorem, algebra. If you’re really stuck or lazy, your graphing calculator can even do the question for you. Math, 0 is the equation solver. Either input the equation into the solver directly or put it in the graphing variables then select Vars, Y-Vars, Y1. Let it do its magic. Seriously Mr. K, in the future we’d like to know this sooner.

 

 

Either before or after the Practice Quiz, Mr. K also had a little bit more of a talk with us about Digital Ethics and our online identity. He reminded us that nothing “disappears” on the internet, and that online, what goes up stays up. He reminded us to keep in mind how the content we create now will always be there and to be mindful of how that content will be perceived in the future.

 

Mr. K even suggested that a great graduation present would be our own domain name, which I totally agree with. Comon! http://paulsantos.com/ isnt even taken yet!

 

Anyway, after that we continued work on the Greater Boston question. We determined that this question was different in a few ways, namely:

 

• That we cannot use the Fundamental Theorem because delta r cannot be changed. Since we cannot make r infinitely small, we cannot integrate it, and we have to use a Riemann sum.
• That we should “integrate” from 1 to 8 instead of 0 to 8 because the value doesn’t change from 0 to 1. Instead, we'll just add that extra population.
• That unlike our previous cylindrical shells questions, we’re dealing with half a shell not a whole shell.

The class ran out of time, and so we were left to ponder the population of Greater Boston by the seashore another day.

 

The homework was the remaining questions for 8.5 as well as finishing the Greater Boston question.

 

Happy Pi/Pre-Test day guys, see you tomorrow. Jamie, cant wait for your cheese cake. As a reward for making such delicious food, I pass the scribe torch onto you.

 

And some extra value added content to make this scribe post a delicious and nutritious part of your daily scribe serving:

 

Falcon Pawnch:

 

 

Good night guys. Hopefully I'll wake up in time to pick up some delicious blueberry pie.

The nearly forgotten blog post: Solids

Hey guise, its *squeak* Not Paul here, blogging about our class.

The nearly forgotten part is because its like 1AM and I was about to call it a night when I booted up my browser to check my email. What pops up? The blog. And I'm like, where's the scribe?

OH RIGHT!

So yeah, here we go again.

Today we worked on the stuff we sort of got introduced to yesterday, which was Solids. How do you find the area between two functions?! THATS CRAZY MAN.

In reality, it ain't so crazy, man. thinking about it logically, the integral can be found by integrating over the interval that ranges from the first intercept of the two functions to the second intercept (they NEED to have two intercepts, oth

erwise it'd go on forever).

I think this slide describes better what I'm tryin

g to say. Because me English so good not as much.

[Click pics for a larger version]

What's going on here is that we have two functions: f(x) and g(x). f(x) is equal to root x and g(x) is equal to x squared. Our mission, should we choose to accept it, is to find the area between the two functions. To do this, first we find the intercepts. The first one we know is zero, just by looking at the graph/sketch, or by simple reasoning. The second one can be found by using the equation f(x) = g(x) and solving for the variable (in this case x). This is because we want to find the value where f(x) and g(x) are equal. Doing this yields us the value 1 for the second intercept. Yay!

Now, we create an equation to find the area. Subtract one integral from the other, but both must be on the same interval (intercept 1 to intercept 2). Antidifferentiate, solve with fundamental theorem of calculus! Bada bing.

Now we move onto something a little trickier. Our first function, f(x), is e to the power x. Funny how that little guy seems to pop up everywhere. The second function, g(x), is 4 minus x squared.

Now, I said a little trickier, but really, is it any harder? No! The only difference is it might take a little longer due to anti differentiation, but longer is not equal to harder! You can even speed things up by graphing and using your graphing calculator to find the intercepts.  Do a little antidifferentiation... algebra, and you're done before you know it.

Eaaaaaaaaasy.

 

Bread: designed to make us hungry. *eats toast*

But, also a cool reminder that in math that it’s okay to approximate just a little. I mean, nothing is really truly exact. So what if that slice of bread really has that little bit of z axis breadth (pun intended)? Treat it like a 2 dimensional object, relatively, its no big deal.

 

 

 

 

Now here’s when we get into the good stuff. Like shapes. Here, we use the circle area equation to find the integral of this cone. However, the cone is mapped to a 2D plane, so you cant really see all of it, but rest assured, its a cone. Now, see, this cone has the property that its radius at a certain height can be found using the equation r = –(1/2)x + 4. Do we really need any more information? We have the interval to integrate from (0 to 8, because the height is 8), and the function to integrate (area). Solution in the next slide:

 

 

And finally, a general rule to guide us. This should come in handy…

 

And with that, Im am done and tired, and in desperate need of some sleeeeeep.

 

Next scribe is the one and only Kristina.

 

Night guys.

BOBbing for Calculuses: Antiderivatives Edition

So yeah, test tomorrow (or today if you wanna get technical). What's it about? Antidifferentiation! 

Woo. Hoo.

I thing is, when we were going through the course, I feel like I understood what was going on, but now I don't really think I understand it as much after all. The integrals stuff was definitely my least favourite stuff, and some of the antidifferentiation rules and techniques were a little hazy on me. Specifically, substitution. Arc trig stuff was fairly simple and straightforward though.

However, the recent stuff I feel good with. Perhaps all this can be clarified by a bit of note reading.

Found some cool stuff, Team PSY video

http://www.youtube.com/watch?v=NxJq60ZxFrw

Took so loooong

Anyway, guys, I also ran across some great stuff you may be interested in. MIT has an online site called OCW (Online Course Work), which uses Youtube to hold videos of course lectures. Im sure we could all use these for studies and to get a feel for University style study.

The main page is http://ocw.mit.edu, with the mathematics section being here:

http://ocw.mit.edu/OcwWeb/web/courses/av/index.htm#Mathematics

They keep links to pages with the videos, and they're even avaliable to download so you can peruse them in your free offline time, or on your iPod! (neat)

Here's one page that has video links to stuff we've already learned. Good for brushing up.

http://ocw.mit.edu/OcwWeb/Mathematics/18-01Fall-2006/VideoLectures/index.htm

I havent gone through much of it, but this site also seems to cover coursework and lectures for more advanced calculus, in case anybody is interested in seeing what we're doing next.

Also, NOT just Calc stuff there, MANY subjects covered. I personally will be looking into the computer science stuff they have, neat stuff.

The Curious Case of Substitution

I apologise for the lateness of this scribe post, work went well into the night.

So guys, im Not Paul, as opposed to Paul, who incidentally came into our class to tell us about scholarships. He directed us to this page:

http://www.danielmac.net/dmci/index.php?option=com_content&view=category&layout=blog&id=133&Itemid=166

Which gives you links to alot of good resources not just for university application but job hunting. Be sure to check it out and head to Mr. Ssembwere's office to pick up some scholarship applications!

Anyway, after that short interlude, we went back to working on understanding the substitution method.

We didnt go really in depth with it, mostly reviewing what we've already done and trying to understand the idea. The the figure on the second slide explains the process pretty clearly:

Take integral of original function and substitute a function within that integral with u. Make sure the integral is in terms of u, and now antidifferentiate the simplified function. Replace the substituted function into the new function and volia, you antidifferentiated an integral.

Slide three is the first example we did. This question introduced the idea that we would have to find x in terms of u in order to write the function in terms of u. However, once you encounter and overcome that small hurdle, the rest is rather easy.

For the fourth slide, we were tasked with solving the same question, except instead of being able to choose x + 7 = u, we were given that u = (x + 7)^(1/2). In this case, the solution is similar, only longer. But remember kids, longer does not nessecarily mean harder!

>__>

Anyway, the new thing we learned today was how to find the antiderivative of an integral at certain limits. The trick is quite simple, and there's two ways to do it:

Antidifferentiate using the substitution method, then input the limit values into that function. You will get two sets of values, and you can use the integral ?theorem/law/thingeh? that states integral = (f(b) - f(a)) to find your answer.

ORRRR, and this may be easier, harder or even the same depending on the situation, you can rearrange your function in terms of u where u is any function inside the integral. In this way, you can have u in terms of x, and you know x because of the limits. Solve for u, and input into newly simplified function. Use thingeh, find value, which will be the same as if you did it the other way! Cool amirite?

Now on the final slide, we are given a beast of an integral and tasked with antidifferentation. This is no easy task. The problem is the function doesnt really jive with what we're doing. Like when we find the derivative du, we cant seem to find a place to put it in. Thus we have to do some mathematical massage (its easy and you dont even need a license) to get it in there. The trick works like this, depicted since it's not clearly explained on the slides:

And urgh, need more practise with using my tablet. My writing is all wonky >:

Anyway, since we do that, we can put du in there, moving one step closer to our plans of integral domination. And by domination I mean simplification. Since we have u, we just rearrange to solve for x in terms of u, substitute u in so our integral is in terms of u, antidifferentiate and put the good stuff back in. Simple as pie. Even better that our simplified integral is a polynomial (because those radicals are nasty) since that makes it easier to antidifferentiate.

Anyway, that concludes todays lesson. Hope you guys did your homework.

As for the Team PSY video which is oddly missing, I decided to resync some audio and redo some parts I rushed through, so I hopefully will have it uploaded Friday - Saturday at the latest. Hopefully will not disappoint when all is said and done.

Anyway, Im going to bed, see you all tomorrow (today).

Oh and someone's the scribe. Duck, duck, might want to remove Shelly and cross out my name and J+ME's, duck, duck, Joyce. Also what is it with this whole

Accumulation Functions Cont'd

This scribe post will be short as I hope to fully complete it tomorrow (hopefully) or Saturday (likely). Its like 2am. I think my name is George.

Today we started by reviewing the idea that an accumulation function is a composite function, with one function (the function defining limits) taking place within another function (the function we are finding the integral/area under of).

After that review, we worked on some problems. A good thing to remember is that if the integral is going from a value to your function, you want to use that inverse thingy and make it negative so you're going from your function to a value. This is on Slide 3, the second example. The integral from 2 to x is equal to the negative of the integral from x to 2, which is simpler to solve out.

After that he gave us a question that involved finding the total area under a cubic function. However, we hit a snag because nobody quite remembered how to find the roots of a cubic function! After a long demonstration and reminder about synthetic division, Mr. K gave us the roots. I highly reccomend going over the Gr 11 notes he added to the slides (Slides 6 - 11) if you dont remember synthetic division and or want to brush up.

However, after finding the roots, the solution is simple since we already know how to find the area under the graph via the integral. Be careful, as the integral = SIGNED area, meaning if you take the integral from the left most root to the right most root you'd get the wrong answer because part of that integral is negative and thus that integral would be smaller than the total area that we are looking for. In this case, because the integral from 1  to 2 is negative, when we add them together we add the negative integral instead to get the total area.

That basically sums up today's (or rather, yesterdays...) class, I'll try to expand this post to further explain the ideas we explored (yesterday) today, but as of the moment I am way too tired.

And finally because Im tired and want to go to bed, the next scribe is J-Cap (J+ME).

Good Night, and or good morning.

The Layers of Onions

Why Onions? Well as we work through our project and complete a part of our it, it would mean that we had completed one chapter, or layer if you like, of the Onion dungeon. Here is the outlook through each chapter and sub chapters of the dungeon.

Man, I should really lay-off of those war tactics games D:

Chapter 1- IX (9): The Selection
Decide on the 6 sample questions from the textbook (questions or theme may change)

Chapter 1- XVIV (19): The Contrivance
Brainstorm and discuss about the kinds of questions should we make, using reference of the six sample question and try to make a few of them.






Chapter 2- X (10): Genesis
Create the six questions that will be used for the DEV

Chapter 2- XX (20): Decryption
Solve the 6 created question and edit the ways to solve them, making it easier to understand and cleaner to look at






Chapter 3- V (5): Initiation
We begin the making of the DEV

Chapter 3- XX (20): Sound Chronicle
We begin voice over of the DEV

Chapter 3- XXX (30): Fine Tuning
Edit what we have of the DEV






Chapter 4- VI (6): Scrutiny
Look over the DEV and discuss if it needs more work or not






Chapter 5- III (3): The Final Battle
Complete and submit the final copy of the DEV project

Chapter 5- IV (4): Epilogue
Its over!! However we have to study for the upcoming exam D:

And that is the end of our outlook of the Onion Dungeon

The Second Derivative Test

Hey guys, this is 'Not Paul' scribing for Dec 2nd 2008. Today (or rather, yesterday as of the time this is being written), we continued talking about the Second Derivative Test and how we can use it. Oh and by the way the next scribe is Francis.

Okay, first, what is the second derivative test?

The second derivative test is basically a little trick that lets you know the concavity of your function.

If the value of the second derivative at a critical point (of the original function) is positive, then the original function is concave up at that point and is a local maximum. If it is negative, it is a local minimum.

To help me explain why this is, I have drawn a little squiggly:

Okay, so on this squiggly, we have atleast 4 critical points. For now, we'll deal only with one of the critical points to demonstrate the second derivative test. The easiest point to use is the local maximum at approxmiately (-1,1). I have labelled it C here.

Well, how is this related to the second derivative test? I'll tell, if you'd stop being so impatient.

The second derivative is the derivative of the first derivative right? And what is the first derivative, but the slope of the tangent line at a point in an interval? So our first function here, its on the interval [-3,3]. And the first derivative of our point c (c') is the slope of the tangent line at c.

As you can see, at our critical point c, c' is a horizontal line, which means the slope of c' (and the derivative of c) is 0. This is what the graph of our first derivative would look like (roughly):

Note the roots of the first derivative function at point c.

On a side note, as Benchmen pointed out we can find inflection points by looking for local maximum and minimums in the first derivative. When the first derivative function has a maxmimum or mimimum, that is where the slope of the original function is largest or smallest locally. That is basically the definition of an inflection point.

We can see that in the graph as the local maximums and minimums of the first derivative graph occur at inflection points on the original graphs. 

So anyway, our second derivative is just a derivative of our first derivative. Easy.

Yes, messy, I know, but I promise, this has a point.

See, if we take a look at my rough graph of the second derivative, we can tell three things:

a) At c, c" is negative

b) At c, c' is zero

c) The roots of the second derivative function are where we have inflection points

I'll make another point, g (for grue), and label it on the graph.

So, as we can see, where g" is positive, g' is zero and g is concave up.

So by this, we know a few things. First, we know that you can easily find the concavity of the function two ways:

By inputting the critical value (value being the x coordinate) into our second derivative function. If the output of the second derivative function is positive at that critical value, we know the original function is concave up. If it's negative, we know it's concave down.

Or by looking at the next value of the first derivative. Since we have value c which is a critical value, and we know c' is zero at c, if the value of the derivative function is negative at c + 1 or c + h where h is any small value (maybe its epsilon!) then the function is concave down (because the slope is becoming negative therefore the original function is decreasing). Conversely, if it's positive then it must be concave up.

Second, we know where the inflections points are (usually, or where they SHOULD be):

They are where you have roots on the second derivative, or where the second derivative changes from negative to positive

Orrr, where the first derivative has a local max/min (makes sense, as we know the derivative has a root at a critical point in the original function and second derivative is the derivative of the first derivative!)

So there you have it, thats what the second derivative test is about. Explained in a really long roundabout way.

Simple right?

So then, this stuff we did today should be a breeze:

We have a function f(x) = 4e^(-x^2).

We find the derivative of it with the chain rule (yes the chain rule!). Think of -x^2 as a function, so we have f(g(x)) = f(x) where g(x) = -x^2. So f(g(x)) = 4e^(g(x)). Simple! The derivative of 4e^(g(x)) is, well, 4e^(g(x)), and the derivative of g(x) is -2x. Multiply! Volia, -8xe^(-x^2) is our first derivative.

Since the only number than will make f'(x) = 0 is x = 0, then at x = 0 we have our critical value.

Then we derive again to get the second derivative.  Input the critical value into the second derivative equation, and f''(0) < x =" 0">

We apply the same process to the second questions. Remember! The derivative of ln(x) is 1/x.

Because of the +1 in the numerator, our function will never be undefined. So we only have to worry about the numerator. At x = 0, the first derivative is 0 so at x = o is where our critical value is.

To find the second derivative, we have to use the quotient rule. LO-DEE HIGH MINUS HI-DEE LO ALL OVER LO LO [In other words, the derivative of the denominator function times the numerator function minus the derivative of the numerator function times the denominator, all divided by the denominator squared].

Input x = 0 into the second derivative equation. It is greater than zero, so it must be a local minimum and the graph is concave up.

We also did some stuff that involved Grade 11 stuff, namely a function that looks like this:

[(x-a)(x-b)]/[(x-c)(x-d)]

Which serves as a lead in to our next topic of INFINITE LIMITS! I'd go into detail but Mr.K didnt post the slides for this, and it's 6:55am. 

Dont forget, homework was exercise 5.2 all odd numbered questions!

And lastly, the next scribe is not Joseph because he'll be away! Nah, the next scribe will not be Lawrence either, because that would be cruel.

Oh, didn't I say who the next scribe is already? I swear I wrote it at the beginning of the scribe post...

Here's something I whipped up as well for the whole scribe post badge thing.

Bobbing about, erm, something

I totally forgot what unit we're on.

So yeah, about that whole math thing.

I get most of the recent stuff we're talking about, but the shaky stuff is mostly is that finding the rate/derivative stuff. I'm still hung up on those, like where you have to find dx/dy given DQ/A&W (silly example that we're given seemingly random other rates like du/dn that somehow relate to the rate/value we're looking for). In fact, the whole derivative/integral/differentiable thing is kind of lost on me still. The last few classes however have been a little clearer and I feel I have a good grasp of the stuff we did today like using the derivative function to try and find critical numbers.

I hope that I'll do well on this test but expect that at best I'll squeak by 50%. More than likely I'll fail it though (I hope not :\).

I hope that with Mr.K back we can clear up whatever people don't understand (and I hope do this before we move on so that we don't encounter problems in the future) because his explainations are much clearer than Dr. Eviatar's.

Good luck tomorrow with the test everyone and get a good nights sleep!

Implicit Differentiables

Okay, late as usual, it is I Not Paul, posting his not scribe post.

Today we welcomed back Mr. K again for our official first full day of class that isnt interrupted by assemblies. First thing Mr. K did was show us what he had posted last night on the blog, which was the Wordle. The Wordle was an image that graphically depicted what most of our BOBs had in common and visually highlighted commonly used words such as "understanding" and "Mr.K" by making them large compared the less frequently used words which were small. In this way we all got to see what everyone on the class had in common in regards to the class.

After the wordle we discussed some of the things we thought we needed to know, and Mr. K stressed the importance of understanding both Integrals and Differentiables. He proceeded to prove the Chain Rule ([f o g](x)) = f'(g(x)) * g'(x)) with a circular function. His proof went something along these lines:

We have a equation: 1 = x^2 + y^2

Now while not explicit in this equation, we actually have two other implicit functions in it. We can extract the top half of the circle: y = (4-x)^(1/2) and the bottom half: -y = (4+x)^(1/2)

Now, we'll find the derivative of a circle, or the tangent of any point on a circle:

a = x^2 + y^2 = 1

a' = 2x + 2y * y' = 0

2x = -2yy'

2x/-2y = y'

y' = -x/y

This is true for any function we imply from the original equation.

We then went onto a question about derivatives. The questions is as follows:

Gas is pumped into a spherical balloon at 5 cubic ft/min. Find the rate the radius is changing if the diameter is 18.

So we volume of the sphere, as given by the equation [4(pi)r^3]/3

The radius is given as 9 inches or 3/4 feet.

Since we know dv/dt = 5, and we are looking for dr/dt, we need to find dr/dv because dv/dt * dr/dv = dr/dt.

So we find the derivative of [4(pi)r^3]/3 because that is dr/dv.

= [12(pi)r^2]

So our solution is 12(pi)r^2 * 5 which is 60(pi)r^2.

Sorry if I missed anything, and also sorry this is so late. Also, I'm iffy on whether or not I am correct with the solution to the problem here because we didnt actually finish this question in class today, and I didnt take notes for the entire class like I should have.

Good night/morning, the next scribe is Francis.

I am here

Here I am.

Paul I am.

I like green eggs and ham.

And laptops. I like those too.

Anyway. For the unit so far, I feel I am very unprepared. I understand most of the material in concept but am unclear on most of the application portion of the subject. I also think I'm missing on some key topics like integrals and differentiation (erm, what is that?). I get derivatives mostly, and understand the rules introduced in Chapter 4 the most. I am mostly lost with the stuff in Chapters 2 and 3 as they were primarily graphically "explained" to us, which I had trouble understanding. Part of the problem may have been the reluctance to use the calculator when the book leaned heavily on the calculator.

So yeah, I believe some of the topics us young Jabbamathees don't quite grasp yet could do with some reintroduction and review to ensure we understand the material before we move on.

Welcome back Mr. K. 

October 23rd: Return of the Definite Integral (sort of)

Starring: The Riemann sum

King Kong as Mysterious outline

and Parabolic Building as Opening Picture.

Sure to be a hit this fall, don't miss it!

Anyway back to serious business.

Today (or rather, yesterday if you want to be technical about it), we continued on yesterday's class about the definite integral. This time, instead of getting data from a table to find the mean of the upper and lower limits, we were provided with a function f(x) = x^2 and used that instead.

We'll start on slide 3:

Here we sketched the graph and tried to find the mean of the upper and lower limits.

We created two intervals: [1, 3/2] and [3/2, 2]. We tried to find the upper and lower limits, but give up.

On slide 4, we change the equation to f(x) = x +1 to make things easier.

So we find the lower limit using the Riemann sum (described in detail on the next slide) and by using some basic algebra.  The answer is the same because the function is linear, but obviously the whole solving by subtracting rectangles wont work so well with a wavy and unpredictable function.

I'm going to try to explain the algebraic thing, but it'll likely be unclear and using general terms.

Since f(x) = x +1, we know its linear, meaning its a straight line. No bumps, curves or waves. Because we want to find the area beneath the function, we can simply get the values by using an interval, in this case [0,4]. So we make a rectangle from x = 0 to x=4 and it goes as high as the function does at this interval (5). So we have a length and width, and thus an area (20). Now, because we want the lower limit, we dont want this extra space thats above the function (because the function is not rectangular, rather it is trapezoidal). So we subtract the extra space by finding the length and width of it (4) and dividing that area by two (because we want to keep half of it). So we get 20 (the whole area) - 16 (the area around the function on the interval [0,4]) / 2 (because we want the half of it that is still under the function) = 12

Which is the same result as we get when we used the Riemann sum.

On slide 6 we tried some more with the f(x) =  x^2 problem. I believe here we compared the accuracy of using smaller intervals vs larger intervals. 

Slide 7 is just talking about intervals in formulas. Here delta t represents the difference between two values on an interval while delta v is the output difference related to delta t.

Slide 8 is an ad.

Well, sorry this post is so late and mostly likely incomprehensible. The next scribe is benofschool.

Orange.

And because we all love this video now, I give you:

Tom Lehrer!

This is a great song to write blog posts to. It's so catchy.

Differentiables

Hey guise, its me Not Paul making my blog post for today because I'm Not* Scribe.
*Truthiness may vary

Anyway, onto actual Mathematics, those darn derivative deriving derivates and what in the heck a differentiable function is. I cannot however tell you with absolute certainty what a differential is, but I have a good idea we're going to find out soon enough.

The basic definition of a differentiable is a function where the derivative can be found for any point of a defined segment of that function.

Which is why f(x) = x^2 is differentiable, because we can find the derivative for every point in the function or any portion of the function.

If you get a function with "kinks" or very pointy points when you graph it, it won't have derivatives at those pointy points and thus is not differentiable for the entire function.

However, if you define a segment of that function that doesn't have pointy points then you can say that the function is differentiable for this segment of the function.

The teacher's definition is as follows:

"A function (fcn) is said to be differentiable at variable a if at variable a f '(a) exists. It is called differentiable on an interval if it is differentiable for every number in the interval."


Here's the example from yesterday that we did today [Ex 1] (which we also did yesterday), and another example we did [Ex 2]. Both of these are differentiable, because the derivative f '(x) exists at x = 0. (in Example 1 replace variable a with x)

Ex 1)


(Click to view larger)

Ex 2)


(Click to view larger)

This is an example where f '(0) doesn't exist, therefore it is not differentiable.

Ex 3)



(Click to view larger)

As shown in the image, the function f(x) = |x| is not differentiable because when x = 0, its derivative does not exist/cannot be determined tangentially. If graphed, y = |x| would show it has a "kink" or pointy point at x=0.

At the end of class we started talking about the different people who invented Calculus, and importantly Joseph Louis Lagrange's contribution of f '(x) = dy/dx, which will come up later in the course I'm sure of it, because of this Wikipedia article.

The next scribe shall be...

Zeff (zeph)

Sorry this post is uncredibly late, you can blame Wired and NeoGAF for their unfairly up-to-date news. I started working on this post at 9pm and finished 4 hours later. Of course atleast one of those hours was spent trying to figure out OpenOffice...

Hey guys, I'm Not Paul, formerly known as Paul, and I'm scribing today's class of September 8th 2008.

So today something amazingly confusing happened that I'll never forget.


I entered the room and was greeted by Not Kuropatwa, formerly known as Mrs. Ingram. Apparently something had come up with Mr. K and Mrs. Ingram had to take over!!11 ZOMG, AMIRITE!?

I know.

Anyway, Mrs. Ingram will be teaching the class instead, and Mr. K will return in November to teach Mrs. Ingram's classes.

After that whole big thing (announcement), we searched for some sort of lesson for the class. Everyone was handed textbooks and we started working on the exercises in the book (Exercise 1 - Odd numbered questions + #4) until the end of class.

And that's what happened today.

And the next scribe is....

Francis