Arc Length of a Curve

Arc Length of a Curve

with Anna, Chelsea, and Claire

https://docs.google.com/document/d/1cXAeMFuCnEbcgDYZE16NXayhOy5J3gEyBHdcpVDKm3g/edit?usp=sharing

Scribe Post 1/31/13

To Be Continued...

We started off class today with the beautiful image above, because OB was slacking and gave us no direction. The skilled author Francie Merril, came up with the image in her multitudes of extra time as she and the CHRHS Math Team dominated the competition, coming in second place overall. Well done ;)  Clearly, everyone should join the math team for more fun times like this. 

The class started with a festive Valentines Day exploration (two to be exact), that you can find right here: https://dl.dropbox.com/u/3243156/CHRHS/apcalc/U4%20Definite%20Integral%20%26%20Trapezoidal%20Rule.pdf
...and looks like this:

We were given approximately 10 minutes (uninterrupted except by an interjection from Becca at the back of the room who was offended when Sarah chose her "eye candy" over her best friend for the task of tackling the exploration) with a partner to try out the problems, the first side dealing with general definite integrals, and the second with the Trapezoidal Rule, which will be mentioned in detail latter in the post and was alluded to in IW #6.
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Here are a few notes about the Valentines Sheet:

Side #1

First thing's first; know what type of graph your dealing with! In this case it is velocity. Second thing to check is the units and scale, in this case meters/second and seconds. Also, remember D= S * T  , or the area underneath the function. This is how you can solve the first question. 

1) 1,200
2) estimates between 27-28 squares
3) Each square is = 50ft (Check the scale!) 
4) 1,420

During the discussion about this side of the exploration, Alex W. was on target in explaining 

that symbolically the function in definite integral form would look like: 

with the 'S' being the integral sign, the first part of the integral (lower one) on the bottom, in this case 0 and the second part (higher one) on the top, in this case the 20. Those two values are called the limits. The other parts of the definite integral form are detailed in the image above. 

OB posed an interesting question however (couldn't just let a student get the entirely of a question right without trying to confuse him/her with another part to the question), which was "Why do we need to include the (dt)?" Well, the function needs to take into account B * H . With indefinite integrals (regular integrals) this is unnecessary, but since it is here the (dt) has more meaning because the B * H aspect comes form the elongation because it is a summation of rectangles, a Riemman's sum. 

We also looked at the graph towards the end done on the board:

Checking units again, its inches squared times inches. We discussed how a cross sectional cut of a football would look like...a circle? a sliver? and finally arrived at a diamond shape. When calculating cross sectional area of anything, there won't be much to calculate, but Mr. O'Brien pointed out that the more you take of the football, the more the curve goes down and then down some more until the get to the middle, where the cross sections then get smaller, since the football is symmetrical. 

If you think about it, its like geometry! Think about finding the volume of a cylinder, you would use the formula . In that case of this one, the base is the main variable, and when you sum it all up, the volume is the same as the definite integral.

The last note on the first side was that you should ALWAYS TRY TO GET: graph --->  analytical function ---> calculator. 

Side #2

OB didn't post answers to these explorations (  :(  ), and this one wasn't gone over in as much detail, so solutions can be found on the internet is needed. 

We started a discussion about why you would ever use trapezoids instead of rectangles. Being the lazy senioritis infected seniors most of us are (proudly), we gave the simplest possible answer first that seemed like what OB might want to hear: the trapezoids must be easier to work with. **Wth?** We didn't think that one through....the next attempt however was much better, as we decided that the method must be overall more accurate, which it is. 


***Trapezoids underestimate for functions that are concave down and overestimate for concave up, however still, more accurate most times than rectangles can be!***

Here is where we started playing with calculating trapezoidal sums on GeoGebra, which will be detailed more later in this scribe post. It was important to note however that the heights always stay the same, the base varies. When you sum them all up, you get your answer. 

How do limits tie into all this? The limit is the actual value, which on GeoGebra you can do by typing in the 'integral' command. You can get that, and the trapezoid answer and note the difference in error. 

A few last things to note on the end of this back side:

• If it asks you for the fastest ____? Look at the maximum. 
• Rate of change =derivative
• This function comes slowly to a stop

 After all this, we began...

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DAILY GOALS:

1) Become more comfortable with Riemann's Sums and Definite Integrals

2) Be able to use resources (calculator and GeoGebra) to solve such problems

3) Get a full grasp on the analytical vs. visual (graphic) interpretations of definite integrals 


Let's recap for a second.....


***START REVIEW***

Def-i-nite In-te-gral 
(Noun)
***An integral expressed as the difference between the values of the integral at specific upper and lower limits of the independent variable.***

Although that definition is the dictionary's interpretation, it nicely and simply sums up the book's definition and the equation below:

Remember, the function f(x) must be continuous and have an interval divided into n sections (which Mr. O'Brien says don't actually have to all be the same width because either way the values will be approaching 0 as they get smaller and smaller, and as any one of the rectangles approaches the limit the values will converge). Also, the  is simply the chosen point. 

• To solve definite integrals on your calculator, refer back to Anna's scribe post. 

• To practice some definite integrals and do some Chocolate-Studded Dream Cookie baking, here is a (yummy) activity! (someone should do this and bring these in...or make OB do it...HINT HINT)

***END REVIEW***


Next, Mr. O'Brien asked everyone to get out their IW #6 and fire up GeoGebra. The topic of discussion, with five main subtypes  was: 

Rectangle Approximation Methods (RAM)

1.  MRAM (Midpoint Sum):  The midpoint method takes the midpoint of a single, normal Riemman's bar and uses it to approximate the height of that bar a MRAM analysis. 
2.  LRAM (Left Hand Sum): Using a left hand sum, the point farthest to the left on each normal Riemman's bar is the determiner of height for the bars. 
3.  RRAM (Right Hand Sum): Using a right hand sum, the point farthest to the right on each normal Riemman's bar is the determiner of height for the bars.
4.  Upper RAM: uses highest possible rectangle values.  These points may be sometimes more left, right, or in the middle of the normal 'Integral' function bars GeoGebra can also generate, depending on where that upper point is.
5.  Lower RAM: uses lowest possible rectangle values.  These points may be sometimes more left, right, or in the middle of the normal 'Integral' function bars GeoGebra can also generate, depending on where that low point is.

There is also the promised trapezoidal sum, which is obviously separate from the above rectangular methods. On GeoGebra, trapezoidal sums can be taken unlike on your calculator, and are found by inputing the following:

TrapezoidalSum[ <Function>, <Number a>, <Number b>, <Number n> ]

We practiced finding the trapezoidal sum graphically on GeoGebra with one of the equations from the exploration. Here is what mine looked like: 

The slider function for n allowed us to come to the conclusion that as n gets larger, the integral gets smaller, until finally if you set your slider to a Max of 100 and approach it, the area under the function appears to be almost completely shaded in by the very close bars.  
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 So taking a step back, what is the difference between rectangle and trapezoidal methods?  

Well, the first three sums in value are picked to make rectangles, whereas in trapezoids you divide into the number and take a point at beginning and end.
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 For midpoints you take the point in the middle of each interval, and thats the value you use to find the height of the rectangle. (see image below)

 The advantage of the midpoint method is  that if the function is increasing or decreasing you get some of the area that is missed and added to balance themselves out, lowering overall error. 

The sum from the left or right, take points to the left or right of interval as evident in the names. Using left it is important to note that you are overestimating when the function is decreasing or underestimating when it is increasing. Consequently, using the right sum (in right it is the opposite: overestimating when the function is increasing and underestimating when the function is decreasing.  Below are, in order from left to right, the images of these sums draw on top of the 'definite integral' function on GeoGebra. 

After this, using the methods for inputting (below) into GeoGebra listed in the same order as above, we took time to play around and try to use all of these functions. To see the first three rectangular approximation methods in an interactive way, click here.

1. RectangleSum[ <Function>, <Start x-value>, <End x-value>, <Number of rectangles>, <Position>]
2. LeftSum[ <Function>, <Start x-value>, <End x-value>, <Number of rectangles>]
3. UpperSum[ <Function>, <Number a>, <Number b>, <Number n> ]

      
4. LowerSum[ <Function>, <Number a>, <Number b>, <Number n> ]

While the interactive applet above this shows the first three, Cal skillfully got all five rectangular approximations not only correctly inputed with a slider, but also interactive..take a look!

Murmurs of #PrettyMath went around. 

Mr. O'Brien added a last comment that rectangles of error never change in height, only in base (EX: entire 0-10 interval divided by number of rectangles).  

Now that we'd visually played with Riemman's Sums, definite integrals, and found the prettier side of math, we moved on to practice problems.

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EX #1: The interval   is partitioned into n subintervals of length . Letdennote the midpoint of the subinterval. (note the correction from the wording from the picture with actual equations below)

EVALUATE: 

The first problem is designed to scare you, Mr. O'Brien explained. “Horribly scary, scarily, scary, AHHHH” says OB to be exact. This is asking for an evaluation of the limit of a sum.... of craziness. 

CROSS YOUR FINGERS its a Riemman sum, and you can use the definite integral and your calculator to solve.

But how do you know if its a Riemman sum? If the function, representing the height, and the  

  representing the base are present and  the variables can be substituted out for simpler 'x' s, can you find the interval? In this case, can you find the definite integral on the interval from  of  ? Yes!!

The fastest way to solve this now that its been translated from gross math slang to true english is on the calculator using the wonderful (fnInt)  button. It would look like this typed in: MATH, fnInt (f(x), -2,1), which yields a final answer of  -4.5 , no approximations necessary. All in all, after practicing a few of these and learning to see the real question behind the information given, these types of problems are calculator friendly and should only take a few seconds to complete on the AP exam! As OB says, "DONE". 

If you are unlucky enough to get problems, like the final three examples from class today, that are non-calculator so (fnInt) isn't available...well that sucks...therefore, we will deal with those tomorrow.

HOMEWORK: Super Corrections for Quiz #2 (NB: grading will be different than usual, so don't focus on the original work and rather make sure your corrections are thorough and use multiple resources!)

Also, all IWs collected Friday of next week!

**The links from the last IW are now fixed and easily accesible. Answers to the last IW are also now available here.

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Class 40 minutes:

VIP: Super Corrections due by 12:00am Friday the 1st!! Below are the new rules!

What can you look at graphically with velocity, position, and acceleration function together?

Make a spreadsheet: times vs. velocity vs. position vs. acceleration in a table.

ALL GRAPHS AND TABLES SHOULD BE PRINTED (not optional)

EX: 

Annotate it all together; make sure you don't forget to include the steps to getting the answer if you didn't have technology again. Add a SHORT part about what you did in the first 15 minutes. 

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Class Work

The first thing to ask yourself when dealing with these problems in what do I know? For example, in the first problem you should note that the equation is the equation of a circle in disguise. 

BAM, circle. We also know that the definite integral is the area bound by the x-axis (gotten from rectangular sum). Going off these two concepts, we can set y equal to the height portion of the equation (  ) and solve as follows:

Now it is easier to see the circle connection. Therefore, the definite integral of the first problem is .

For the second problem, you have to solve without technology (yikes). A good guess would be 1.5, since everything from 2-3 cancels out due to the negative heights, which cannot be had, and the fact that the error from above and below cancel each other out, as discussed above when summing all the areas up:

Even though that was a non-technology problem, we used one to look at the canceling that was occurring, so Anna asked if it was possible to do these without one at all. What if the equation isn't a nice geometric one? Mr. O'Brien said you did, and that, again, fnInt is your best friend. However after solving the third question with this method, he admitted that he LIED and that we did not know how quite yet, but that we will know how to think cleverly, look for a possible graph, and determine how to add up all the area to get the definite integral sin technology in a few classes...oh joy!! ( -__- )

Here is the lie in action on camera, and OB showing how quickly and skillfully his fnInt method works for the third problem:

We finished class by looking at the exploration on page 283 and figuring out the answers as a class. Here's what we came up with, and a brief reason why. 

1.  -2
2.  0 (+, - cancel out)
3.  1 (symmetry of sine)
4.  2+2pi
5. 4 (double area)
6. 2 (shift over to 2, but go to 2+pi = no change)
7. 0 (-a to a cancels)
8. 4 (base times two ,length same)
9. 0 (+1, -1 cancel)
10. 0 (odd function, cancels out)

For a good reference for what may be in our future about how to find the definite integral manually, take a look at this extremely easy to understand video!

The next scribe will be Cole Ellison.

Scribe Post 9/12

FYI: Continual changes will be made :)

A brief reminder: Check iCal!! All the IW's thus far (#1-4) are due on friday! If you have any questions Mr.O'Brien is happy to assist.

 We started class off today by finishing going over the problems from IW #4 and correcting any errors or missing concepts on Anna's scribe post for 9/10 based on this work. Mr. O'Brien noted that the problem he did from the homework last period was not actually number 77, but rather the "OB Version" of 77. We learned that this was due to the fact that in the diagram he drew the bottom point of the smaller triangle should be labeled 'Q'.
 We had said that , but with this new piece of information we now know that the 'P' at the beginning should be a 'Q'! This makes the new inequality read: . Note that the book asks for , which we determined would equal .  

The point of all of this was to show that the process used to solve would be the same as shown last class for #77, but the variables used in Mr. O'Brien's would differ slightly from those placed in the book. He then reviewed the steps of the proof,assured that we recognized the function as having origin symmetry, and reminded us that a squeeze creates a situation where the limit of ALL of the functions involved is the same. For a review of the original process of the above problem click here! 

The last review item we handled was a warning from Mr. O'Brien about problems such as numbers 33 and 72 listed below:

Q33)

Q72)

Mr. O'Brien commented that the issues students run into on problems like these is simply taking them at face value and thinking they're too easy! 
In this case, both involve the limit of .     Both require breaking apart using the limit rules. 
                                                                                                                                                                

Why is the limit  simply one? The only value for 'X' which will not give one is 0/0, which is irrelevant because with limits, we do not care about the value at 0/0, we simply care about the values on either side a.k.a  - 0.00001 or 0.00001.  

Using this information, we were able to solve the problems accordingly:

A33) -----> .

A72) -----> . 

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*Deep breath* Finished! At this point we moved on to the new work, after a few points to remember:
1) We will be moving onto derivatives next!
2) The homework is on iCal and is 12 problems long. (You can also find it here.)
3) Note that we've been working on page 66 up to this point! This is all very basic, but very important stuff! If you don't fully understand anything we've done so far, ask for help or see any of the following links at the bottom of the 'Questions for the Class' page in the blog, or go to this helpful website for interactive practice problems: http://curvebank.calstatela.edu/limit/limit.htm. 
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We took a quick break to fill in the google doc info sheet for Mr. O'Brien before starting to tackle today's goal.

Objective:
- Infinite limits!! (for an introduction style example, see my favorite online help source Patrick!)

The object of today's warm up was to refrain from using technology and attempt to solve limit problems, with and without infinity, by hand by sketching a graph and using it to answer the conditions below. 

Warm Up:

a) Sketch (showing all relevant features)

b)                                             c) 

d)                                            e) 

After giving us about 10 minutes to attempt the warm up with our partners, Mr. O'Brien graphed the equation on GeoGebra as follows to prove why having the skill set to do these types of limit problems by hand is important:

Upon inspecting the graph versus the function, we noticed that the function should have a gap at the point (-1,2/3), but Mr. O'Brien shows us that one is not visible on the image. Likewise, if you tell GeoGebra to plot the point (-1,2/3), it will without error. (See free object A in blue below)

While you can get it to tell you that there is no value, a.k.a the function is undefined at that point by plugging in f(-1) (See dependent object above) rather than (-1,2/3), if you were not able to tell that there should be a gap, or that you would have to input just the 'X' value into the function to get the program to tell you there is this undefined value, you would be in trouble! This example illustrated a situation in which graphing by hand and solving by using the graph, or by algebra would be more useful. When graphing by hand you ensure the graph give you all the information needed. 

Side Note: Wolfram Alpha will graph and give you the limit, and answer to the limit. Will posed the following question though: Can Wolfram Alpha answer limit problems involving infinity? Answer: YES!  Proof: 

                         

How cool is that! Unfortunately, it was back to the old school method to cover how to solve by hand.

Too make it easier, here is the function in question again: 

We determined that the function from the warm up has a few simple relevant features: 

1)Asymptotes: -2,2

2)Domain: All Real Numbers EXCEPT -2,2,-1 

3)Range: All Real Numbers 

If finding any of the above, or the hole in a function tends to trip you up, review these skills using the following site.

Next, we decided we were ready to solve part (b) of the problem since it appeared to be doable algebraically. We began by factoring both the numerator and denominator of the function, in hopes of giving us a new function in which -1 is defined:

Expansion: 

Simplification:

Cross Canceling: .

If we labeled this new function g(x), we found that evaluating for g(-1) gave us -2/-3=2/3 as the answer to (b)! (Note For Future scribes: parenthesis in the denominator of the fraction should be one set around the whole set of numbers, not multiple sets. See change to "expansion" line for proper formatting.)

For parts (c), (d), and (e) Mr. O'Brien wanted us to utilize the sketches we made, so we went over what should go into making a good one. We determined that for this function, a good sketch would note that the function is undefined or has a gap at (-1,2/3), has two asymptotes, one at 2 and one at -2, and has zeros that can be solved for using simple algebra. (the zero we found to be (1,0). We were reminded that having the y-intercept was also useful, and that to find that is x--> 0 we could always substitute that in for all of the 'X's' to get it. This gave us (0,1/4). From these points you can sketch the center part of the function by connecting the dots(here was mine):

The only thing Mr. O'Brien made sure to add about the asymptotes, or the other two pieces of the function effected by them, is that you must ensure they are not volcanic asymptotes which occur where there is a double zero in the denominator of the function. We determined this was not the case in our function however, since to get a double zero we would have to see the following function instead:

  ----> .

Due to this, we were able to assume the regularly shaped ones could be drawn and our graph was complete. We then solved parts (c) and (d) without much difficulty.

(c) At    the graph shows no value being approached from either the right or left side of the limit, thus we determined the answer to this part was D.N.E.            

(d) At    the graph shows that tracing form the left side of the function leads off endlessly. We would generally call this another D.N.E, however we learned a new concept that helps us correctly label this answer from Mr. O'Brien:

When a value is leading off infinitely in different directions, like in part (c) when  , we say that the limit does not exist because there will never be a value reached on either side of the limit to record. When you are coming from only one side with the limit however, like in (d)  where you come from the left or if d were  and you were coming from the right, although the limit still will not exist and will be going off into infinity (as proven by the answer to part (c)), we can still say that the limit is approaching either negative of positive infinity. In this case, the limit is infinite and technically doesn't exist, however because we can define it as approaching either negative or positive infinity, we will. 

With this information, we determined the final answer to part (d) to be D.N.E; . 

                                                                     

Part (e) really represented the new concept of the day. The evaluation of the function at , means imagining what the function would look like when you inputed larger and larger values as they approached infinity. We decided that the larger the value of 'X', the closer the function would get to 0. 

*Done!*

An interesting thing to note: either the limit can have the infinity part like this: , or the answer to the limit problem can be .  

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Now that we had had the bulk of the lesson, we tackled a few practice problems:

Ex/ 

Ex/ 

Ex/ 

Ex/ 

Our guesses were all over the place! 

For the first two Mr. O'Brien steered our minds towards a parent function of use, which in this case was f(x)= 1/x. We talked about the fact that the only difference between that and the function in the first two examples was a few shifts. This allowed us to realize, through visualizing the parent graph and shifted graph that the answers did not exist (we still expressed them as -/+ infinity though). Here is what this would have looked like on GeoGebra:

The second two problems posed more of a challenge, and we decided it best to try tacking them numerically. At this point we were introduced to the concept of an end behavior model. Mr. O'Brien explained an end behavior model by using the first equation as an example. He pointed out that as 'X' got infinitlily large, 4x+1 would begin to look a lot like a 4x line graphically, with the same occurance on the bottom with 7x+5 looking more and more like 7x as values increased toward infinity. So, as a whole as 'X' approaches infinity and our theoretical graph gets farther and farther zoomed out, we can say that:

 is moving towards being about equal to  . 

The equation on the left of the equal sign above before simplification would be the end behavior model for this function. This means the horizontal asymptote is also 4/7, which tells us that the answer to the third practice problem is 4/7 as well.

Using the same method of end models, we determined the end model for the last practice problem would look and simplify as follows:

 . 

If we're looking at the limit of X as the values get smaller and smaller(due to the negative), we know that  would be approached. 

Finally, Mr. OB referred us to pg. 71 in our books back to the yellow box with notes on the limit rules, which still apply to infinite limit problems, and to an explanation of why the squeeze theorem still works as well. He also pointed out on pg. 72 an interesting exploration problem in a blue box that reminds us that both the numerator and denominator of the function must have existent limits, or else we cannot apply the limit rules that allow us to break them down as they both move towards infinity because you will get 0/0. The following three problems in the box that we did continued to allude to the main point of the exploration: you must know when the limit exists and, if it doesn't, if you can manipulate the function into something that does exist when the function approaches infinite limits. This means that if you find the numerator and denominator to have existent limits, but an undefined value when you plug things in, you may break apart the function using the multiplication or addition rules and the take the simplified form and solve from there. 

HW: Finish IW #1-4 for friday. Tonight pg. 76 #3,7,15,27,35-38,53,55,59,61.

P.S. The next scribe will be...SARAH!

P.P.S Update Below!!! 

In this lab period we made corrections to this post and then went on the answer homework questions. The few we covered were 36,55,,and 7c. Here is the work for 36&55 and then 7c:

Two important questions stemming from this were: Can you touch the asymptotes in these problems? YES it is a limit for the formation the the graph, not one of your answers on a homework problem for example. and Can you have more than one horizontal asymptote? You can have more than one horizontal asymptote based on the definition in the book: .




Here is a great site to help you practice infinite limits! 

 http://17calculus.com/calc02-infinite-limits.php