tag:blogger.com,1999:blog-337674282014-12-25T07:02:49.384-08:00Prep AB Calculus C 2006-07An interactive log for students and parents in AB Calculus Section C. This site is only as rich as YOU make it.Math Maverickhttp://www.blogger.com/profile/10631468736601964037noreply@blogger.comBlogger91125tag:blogger.com,1999:blog-33767428.post-11441457455479889492007-05-08T21:14:00.000-07:002007-05-08T21:27:12.236-07:00Calvin Says<p align="center"><span style="font-size:180%;color:#006600;"><strong><em>Calvin Says:</em></strong></span></p><p align="center"><a href="http://bp1.blogger.com/_sNG9dzZilv4/RkFNVygHTEI/AAAAAAAAAHI/YA2smZxaueA/s1600-h/cal.gif"><img id="BLOGGER_PHOTO_ID_5062412492870208578" style="CURSOR: hand" alt="" src="http://bp1.blogger.com/_sNG9dzZilv4/RkFNVygHTEI/AAAAAAAAAHI/YA2smZxaueA/s400/cal.gif" border="0" /></a></p><br /><br /><p><strong><em><span style="color:#000099;">Don’t Forget “+C”<br />Check Your Endpoints!<br />Remember Initial Conditions!<br />Remember the Chain Rule! (Especially with implicit differentiation!)<br />Remember the Product Rule! (Especially with implicit differentiation!)<br />The integral of a rate of change is a NET CHANGE!<br />Critical Points are candidates for Maximums and Minimums<br />Critical Points occur where the first derivative equals zero OR IS UNDEFINED!<br />Speed is the ABSOLUTE VALUE of velocity<br />“Speeding Up” means the velocity and the acceleration have the SAME SIGN!<br />Derivative = Instantaneous Rate of Change = Slope of the Tangent Line<br />An Antiderivative is the area between a curve and the x-axis<br /></span></em></strong></p><br /><br /><p><strong><span style="color:#000099;">You have all done a great job this year! I know you're going to do great tomorrow! Thanks for a really fun year...</span></strong></p><br /><br /><p><em><span style="color:#009900;">You can teach a student a lesson for a day; but if you can teach him to learn by creating curiosity, he will continue the learning process as long as he lives."<br />- Clay P. Bedford</span><br /></em></p><br /><br /><p>I hope I made you curious...</p>Math Maverickhttp://www.blogger.com/profile/10631468736601964037noreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-4344101974976765822007-03-22T10:26:00.000-07:002007-03-22T16:42:59.777-07:00Friday's Test TopicsHere’s a list of topics that will be covered on this Friday’s Chapter 7-9 Test.<br /><br /><strong><em><u><span style="font-size:130%;color:#ff0000;">Chapter 7-9 Test Topics</span></u></em></strong><br />Integration by parts (Sec. 7.1, #3,7,21)<br />Arc length (Sec. 8.1, #1,3,5,9,11)<br />Approximate Integration – Midpoint/Trapezoidal Rule (Sec. 7.7, #1,3,7,29)<br />Slope Fields/Differential Equations –Solutions (Sec. 9.2, #11,13)<br />Exponential Growth/Decay – Newton’s Law of Cooling (You knew it was coming!) (Sec. 9.4, 13,15)<br /><br />As always, your homework is a good place to start reviewing, and the book has several other problems to give you more practice!<br /><br />That’s it! I’ll be around after school on Thursday until 3:00 and back after 4:15 (faculty meeting) and in early on Friday. Donut holes and OJ!<br /><br /><em><span style="color:#000099;">I don’t know whether my life has been a success or a failure. But not having any anxiety about becoming one instead of the other, and just taking things as they came a long, I’ve had a lot of extra time to enjoy life.<br />—COMEDIAN HARPO MARX<br /><a href="http://bp0.blogger.com/_sNG9dzZilv4/RgK8he9rBCI/AAAAAAAAAF8/DoDGrEW7lmE/s1600-h/dilbertmath3.gif"><img id="BLOGGER_PHOTO_ID_5044801816041817122" style="CURSOR: hand" alt="" src="http://bp0.blogger.com/_sNG9dzZilv4/RgK8he9rBCI/AAAAAAAAAF8/DoDGrEW7lmE/s400/dilbertmath3.gif" border="0" /></a></span></em>Math Maverickhttp://www.blogger.com/profile/10631468736601964037noreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-46499545939003788882007-03-15T14:37:00.001-07:002007-03-15T14:37:45.564-07:00Friday's Quiz Topics<div>Here’s a list of topics that will be covered on this Friday’s Quiz.<br /><br /><strong><em><u><span style="font-size:130%;">Quiz – Sections 9.2-4<br /></span></u></em></strong>Solve a differential equation (Sec. 9.3, #1,5)<br />Solve a differential equation (IVP) (Sec. 9.3, #11,15)<br />Exponential Growth/Decay – Formulas, Rates, Values, Times and Graphs (Sec. 9.4, #1,3,9)<br />Slope/Direction Fields (Sec. 9.2, #11,13)<br /><br />That’s it for now! I’ll be around after school on Thursday, online Thursday evening/night and in early on Friday – OJ and donut holes!<br /><br /><em><span style="color:#000099;">I like nonsense, it wakes up the brain cells. Fantasy is a necessary ingredient in living, It's a way of looking at life through the wrong end of a telescope. Which is what I do, And that enables you to laugh at life's realities.<br />- Dr. Seuss<br /></span></em><br />And for those of you that didn’t see it, here’s a cute set of <a href="http://www.animalliberationfront.com/News/AnimalPhotos/Animals_91-100/BabyDogHugs.htm">instructions for properly hugging a baby</a>…<br /><a href="http://bp3.blogger.com/_sNG9dzZilv4/Rfm8c6-dlCI/AAAAAAAAAFs/-hv_mB4UiBU/s1600-h/chickweedexam1.jpg"><img id="BLOGGER_PHOTO_ID_5042268462871581730" style="CURSOR: hand" alt="" src="http://bp3.blogger.com/_sNG9dzZilv4/Rfm8c6-dlCI/AAAAAAAAAFs/-hv_mB4UiBU/s400/chickweedexam1.jpg" border="0" /></a></div>Math Maverickhttp://www.blogger.com/profile/10631468736601964037noreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-89946571916253316462007-03-12T20:36:00.001-07:002007-03-12T21:20:27.565-07:00Section 9.2: Direction Fields and Euler's MethodHey all...so today's lesson deals with sketching and approximation to figure out the shape of the curve of a function without any real in-depth calculations. According to the book, the definition for a Direction (or Slope) Field is: "If we draw short line segments with the slope F(x,y) at several points (x,y), the result is called a direction field (or slope field)." In other words, you are given an equation set equal to a derivative. For example, you could be given y'=x+xy. Then, you make a table of values divided into three columns: x, y, and y'. Pick a random pair of points for your (x,y) coordinate, plug this pair into your equation, and figure out the slope of the function at that particular point. Draw a short line at the designated point that has approximately the same slope as the slope that you just found. Here is an example:<br /><br />Sketch the direction field for the differential equation y'=(x+xy)-y.<br /><br />First, set up a table of values.<br /><br />x y y'<br />-1 -1 -1<br />0 0 0<br />1 1 1<br />etc.<br /><br />Now, draw in the slopes of the function at the given points. The end result should look like:<br /><br /><a href="http://photobucket.com" target="_blank"><img src="http://i146.photobucket.com/albums/r266/aconforti7/directionfield1.jpg" border="0" alt="Photo Sharing and Video Hosting at Photobucket"></a><br /><br />Of course, this is a very inexact method. Euler's method seeks to make the process of drawing a direction field a little more accurate. "For the general first-order initial-value problem y'=F(x,y), y(x0)=y0, our aim is to find approximate values for the solution at equally spaced numbers x0, x1=x0+h, x2=x1+h, ..., where h is the step size. The differential equation tells us that the slope at (x0, y0) is y'=F(x0, y0. The general equation to express Euler's rule is: <a href="http://photobucket.com" target="_blank"><img src="http://i146.photobucket.com/albums/r266/aconforti7/yn.jpg" border="0" alt="Photo Sharing and Video Hosting at Photobucket"></a><br /><br />Now, let's apply this concept to a real problem. <br /><br /><em>Use Euler's method with step size .1 to construct a table of approximate values for the solution of the initial-value problem. y'=x+y, y(0)=1</em><br /><br />y1=1+.1(0+1)=1.1<br />y2=1.1+.1(.1+1.1)=1.22<br />etc.<br /><br />So as you can see, Euler's method allows us to draw a much more exact direction field by giving us more exact values for slopes. Here's a fun link for your personal enjoyment: http://tutorial.math.lamar.edu/AllBrowsers/3401/DirectionFields.asp<br /><br />BRIAN YOU ARE NEXT HAVE FUN!<br /><br />Two mathematicians were having dinner in a restaurant, arguing about the average mathematical knowledge of the American public. One mathematician claimed that this average was woefully inadequate, the other maintained that it was surprisingly high. <br /> "I'll tell you what," said the cynic. "Ask that waitress a simple math question. If she gets it right, I'll pick up dinner. If not, you do."<br /> He then excused himself to visit the men's room, and the other called the waitress over. <br /> "When my friend returns," he told her, "I'm going to ask you a question, and I want you to respond 'one third x cubed.' There's twenty bucks in it for you." She agreed.<br /> The cynic returned from the bathroom and called the waitress over. "The food was wonderful, thank you," the mathematician started. "Incidentally, do you know what the integral of x squared is?"<br /> The waitress looked pensive, almost pained. She looked around the room, at her feet, made gurgling noises, and finally said, "Um, one third x cubed?"<br /> So the cynic paid the check. The waitress wheeled around, walked a few paces away, looked back at the two men, and muttered under her breath, "...plus a constant."alex chttp://www.blogger.com/profile/12710827233923109665noreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-25962063264337336252007-02-27T23:23:00.000-08:002007-02-28T09:03:51.210-08:009.1 Modeling with Differential Equations<div align="center"><span style="font-family:verdana;color:#000000;">Alright, so in class we learned about the two different models of population growth. The first one is stated as the Rate of Growth for a Population-unbounded on our concept sheets. It is:</span><br /><br /></div><p align="center"><img id="BLOGGER_PHOTO_ID_5036494014652404050" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_cH8QfroYd9o/ReU4nzC80VI/AAAAAAAAAAk/a4XOM5RD4XI/s200/dpdt.bmp" border="0" /></p><div align="center"><span style="font-family:Verdana;">This equation shows that population is always growing, and as it gets bigger its growth rate gets bigger as well. As Mr. French said, as the population gets bigger, there's more people to make the population bigger. </span></div><div align="center"><br /><span style="font-family:Verdana;">The second formula we were taught is dubbed the Rate of Growth for a Population-bounded on our concept sheets. It is:</span><br /><a href="http://bp2.blogger.com/_cH8QfroYd9o/ReU3QzC80UI/AAAAAAAAAAc/W1TGTzXb4Sc/s1600-h/dp.bmp"><img id="BLOGGER_PHOTO_ID_5036492520003785026" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_cH8QfroYd9o/ReU3QzC80UI/AAAAAAAAAAc/W1TGTzXb4Sc/s200/dp.bmp" border="0" /></a><br /><span style="font-family:Verdana;">This equation is also the derivative of a logistic function. It expresses the fact that when we have a small population, the population will grow quickly. But as the population approaches the carrying capacity, it won't be able to grow so quickly and will level off. If P/K were to ever become greater then 1, then the population would start to decline.</span><br /><br /><span style="font-family:verdana;">The next thing we learned about was the IVP, or the Initial Value Problem. The two steps to these situations are:</span></div><p align="center"><span style="font-family:verdana;">1. Solve the differential equation for a general solution.</span><br /><span style="font-family:verdana;">2. Use the general solution and data point to solve for a specific solution.<br /></span><span style="font-family:verdana;">For example:</span> </p><p align="center"><span style="font-family:Verdana;">Given </span><a href="http://bp2.blogger.com/_cH8QfroYd9o/ReU9TzC80WI/AAAAAAAAAAs/GXC_Onbax1Y/s1600-h/y2.bmp"><img id="BLOGGER_PHOTO_ID_5036499168613159266" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_cH8QfroYd9o/ReU9TzC80WI/AAAAAAAAAAs/GXC_Onbax1Y/s200/y2.bmp" border="0" /></a> and the point y(0)=5, find the IVP.</p><br /><br /><br /><p align="center"><a href="http://bp3.blogger.com/_cH8QfroYd9o/ReVFtDC80XI/AAAAAAAAAA0/3F4ok3bRaEU/s1600-h/dy.bmp"><img id="BLOGGER_PHOTO_ID_5036508398497878386" style="WIDTH: 156px; CURSOR: hand; HEIGHT: 77px" height="56" alt="" src="http://bp3.blogger.com/_cH8QfroYd9o/ReVFtDC80XI/AAAAAAAAAA0/3F4ok3bRaEU/s200/dy.bmp" width="118" border="0" /></a></p><br /><br /><br /><p align="center"><img id="BLOGGER_PHOTO_ID_5036508999793299842" style="WIDTH: 167px; CURSOR: hand; HEIGHT: 83px" height="51" alt="" src="http://bp3.blogger.com/_cH8QfroYd9o/ReVGQDC80YI/AAAAAAAAAA8/MkoltNPBmfw/s200/int.bmp" width="131" border="0" /></p><p align="center"><a href="http://bp0.blogger.com/_cH8QfroYd9o/ReVGsTC80ZI/AAAAAAAAABE/ZoHerzbYH_Q/s1600-h/v.bmp"><img id="BLOGGER_PHOTO_ID_5036509485124604306" style="WIDTH: 137px; CURSOR: hand; HEIGHT: 165px" height="161" alt="" src="http://bp0.blogger.com/_cH8QfroYd9o/ReVGsTC80ZI/AAAAAAAAABE/ZoHerzbYH_Q/s200/v.bmp" width="153" border="0" /></a></p><p align="center"><a href="http://www.math.ohiou.edu/~ashish/h13.pdf">http://www.math.ohiou.edu/~ashish/h13.pdf</a> That's a nice quick summary straight from our book that Ohio University made.</p><p align="center"><a href="http://www.mathsci.appstate.edu/~hph/3310/diffeqn/">http://www.mathsci.appstate.edu/~hph/3310/diffeqn/</a> This site has examples of different kinds of differential situations.<br /><br /><span style="font-family:verdana;font-size:180%;color:#6600cc;"><strong>Alex, you get the blog next =D</strong></span><br /></p><br /><br /><br /><p align="center"><span style="font-family:verdana;">My favorite quote for the past month has been "You smell funny" from POTC:DMC.</span></p><div align="center"><br /><br /></div><p align="center"><span style="font-family:verdana;">And now for some lame math jokes...</span></p><div align="center"><br /><br /></div><p align="center"><br /><span style="font-family:verdana;">Q: What is the first derivative of a cow?</span></p><p align="center"><span style="font-family:verdana;">A: Prime Rib!</span></p><div align="center"><br /></div><p align="center"><span style="font-family:verdana;">Q: What caused the big bang?</span><br /></p><p align="center"><span style="font-family:verdana;">A: God divided by zero. Oops! </span></p><div align="center"><br /></div><p align="center"><span style="font-family:verdana;"></span></p><div align="center"><br /><span style="font-family:verdana;">"A mathematician is a blind man in a dark room looking for a black cat which isn't there."- Charles Darwin<br /></div></span><a href="http://bp1.blogger.com/_cH8QfroYd9o/ReW0UjC80aI/AAAAAAAAABs/VidYPXLmcZc/s1600-h/calculuscircus.jpg"></a><br /><br /><p align="center"><a href="http://bp2.blogger.com/_cH8QfroYd9o/ReW04zC80bI/AAAAAAAAAB0/XTBnJm22R_4/s1600-h/calculuscircus.jpg"><img id="BLOGGER_PHOTO_ID_5036630646152024498" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_cH8QfroYd9o/ReW04zC80bI/AAAAAAAAAB0/XTBnJm22R_4/s400/calculuscircus.jpg" border="0" /></a></p><p align="center">and for my last few words:</p><p align="center"><br /><span style="font-family:verdana;">A guy gets on a bus and starts threatening everybody: "I'll integrate you! I'll differentiate you!" So everybody gets scared and runs away. Only one person stays. The guy comes up to him and says: "Aren't you scared, I'll integrate you, I'll differentiate you!" And the other guy says: "No, I am not scared, I am e^x." </span></p>crystalhttp://www.blogger.com/profile/04625658975609725767noreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-76373925556877861602007-02-27T18:04:00.001-08:002007-02-27T18:05:00.347-08:00Thursday's Quiz Topics<div>Here’s a list of topics that will be covered on this Thursday’s Quiz.<br /><br /><strong><em><u><span style="font-size:130%;color:#cc0000;">Quiz – Sections 8.1, 9.1<br /></span></u></em></strong>Arc length – given curve and interval (8.1, #1,3,5,9,11,29)<br />Arc length – determine setup (8.1, #1,3,5,9,11,29)<br />Arc length – determine setup (8.1, #1,3,5,9,11,29)<br />Arc length – given curve and interval (8.1, #1,3,5,9,11,29)<br />Arc length – given curve, determine interval (8.1, #1,3,5,9,11,29)<br />Differential equation – analysis and interpretation (9.1, #11)<br />Differential equation – verification of solution (9.1, #1,5)<br /><br />That’s it for now! I’ll be around after school on WEdnesday and online Wednesday evening/night.<br /><br /><em><span style="color:#3333ff;">In any collection of data, the figure most obviously correct,<br />beyond all need of checking, is the mistake<br /><br />Corollaries:<br />(1) Nobody whom you ask for help will see it.<br />(2) The first person who stops by, whose advice you really<br />don't want to hear, will see it immediately.<br /></span></em></div><br /><div>And on another note, look for the simple solution:</div><a href="http://bp3.blogger.com/_sNG9dzZilv4/ReTjAa78maI/AAAAAAAAAEc/MxMDyLmZvj4/s1600-h/dilbertengineering1.gif"><img id="BLOGGER_PHOTO_ID_5036399879677319586" style="CURSOR: hand" alt="" src="http://bp3.blogger.com/_sNG9dzZilv4/ReTjAa78maI/AAAAAAAAAEc/MxMDyLmZvj4/s400/dilbertengineering1.gif" border="0" /></a><br /><div></div>Math Maverickhttp://www.blogger.com/profile/10631468736601964037noreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-81037690194922638662007-02-26T19:41:00.001-08:002007-02-27T19:01:30.402-08:008.1 Arc LengthOkay, so it's finally my turn again~! I'm here to explain Chapter 8, Lesson 1, which is about finding arc lengths.<br /><br />When a curve is a polygon, finding the length is easy because all you have to do is add up the sides but when you get a continuous curve, it gets tricky! Remember that a curve is defined by the equation y= f(x) where f is continuous on a (equal to or less than) x (equal to or less than) b. When we estimate the value of a curve, we are taking approximations as if the sides of a polygon were present in it. (page 547 in the textbook)<br /><br />The length is described using the distance formula (as the limit approaches infinity). The distance formula is not practical to use with a smooth function, so we can derive an integral formula for L where the function has a continuous derivative, because there is only a very small change in f'(x). (Also think of the approximation as taking the Pythagorean theorem to find the hypotenuse with infinitely many tiny triangles, as described in class)<br /><br />The arc length forumula for the curve y=f(x) where is a is (equal to or less than) x (equal to or less than) b<br /><br />(if f' is continuous on [a,b])<br /><a href="http://photobucket.com" target="_blank"><img alt="Photobucket - Video and Image Hosting" src="http://img.photobucket.com/albums/v600/twistedfate320/Random%20pics/integral1dx.jpg" border="0" /></a><br />it can also be notated as:<br /><a href="http://photobucket.com" target="_blank"><img alt="Photobucket - Video and Image Hosting" src="http://img.photobucket.com/albums/v600/twistedfate320/Random%20pics/integral1dx2.jpg" border="0" /></a><br /><br />In the other case, if a curve has the equation x=g(y) with c (less than or equal to) y (less than or equal to) d, and g'(y) is continuous, we get this formula for arc length:<br /><a href="http://photobucket.com" target="_blank"><img alt="Photobucket - Video and Image Hosting" src="http://img.photobucket.com/albums/v600/twistedfate320/Random%20pics/integral2dy.jpg" border="0" /></a><br /><br /><br /><br />Keeping these two formulas in mind, let us try a problem!<br /><br />If y=e^4x and 0 (less than or equal to) x (less than or equal to 1), find the arc length.<br /><br />We begin by writing the integral, which will be from 0 to 1, and plug in the derivative. Remember to use the chain rule in this case, because the derivative of e^4x is 4(e^4x). Don't forget details such as chain or product rule when taking derivatives! Also remember to even TAKE the derivative, and to SQUARE it. But I digress. After setting everything up, it should look like this:<br /><br /><a href="http://photobucket.com" target="_blank"><img alt="Photobucket - Video and Image Hosting" src="http://img.photobucket.com/albums/v600/twistedfate320/Random%20pics/sampleintegral.jpg" border="0" /></a><br /><br />Afterwards, you can either figure out the antiderivative or if you are short on time or just wish for simplicity (in this case), plug it in to your calculator and do fnInt! Oh yeah, another minor detail. Don't forget to write "dx" when doing your problems, and don't get your dx's and dy's mixed. I think it's just me, but it's still a possible mistake.<br /><br />...aaaannndddd here are a few links:<br /><br /><a href="http://en.wikipedia.org/wiki/Arc_length">http://en.wikipedia.org/wiki/Arc_length</a> (history teachers can scorn Wikipedia but this site gives a good explanation of this math concept, provided that no users/pranksters go edit it, but it always gets edited back anyhow)<br /><a href="http://tutorial.math.lamar.edu/AllBrowsers/2414/ArcLength.asp">http://tutorial.math.lamar.edu/AllBrowsers/2414/ArcLength.asp</a><br /><br /><a href="http://www.math.hmc.edu/calculus/tutorials/arc_length/">http://www.math.hmc.edu/calculus/tutorials/arc_length/</a><br /><br /><br />and here's a funny little video for you all. just remember when you decide to bring your kitten in for show-and-tell, don't let it near Mr. French's laptop. (Has anyone posted it on the blog already? Remind me if anyone has)<br /><br /><embed src="http://us.i1.yimg.com/cosmos.bcst.yahoo.com/player/media/swf/FLVVideoSolo.swf" width="425" height="350" type="application/x-shockwave-flash" flashvars="id=1762130&emailUrl=http%3A%2F%2Fvideo.yahoo.com%2Futil%2Fmail%3Fei%3DUTF-8%26vid%3D033c52938829778cceebb58cb4f5fd8f.1762130%26fr%3Dyvmtf&amp;amp;amp;amp;amp;amp;imUrl=http%25253A%25252F%25252Fvideo.yahoo.com%25252Fvideo%25252Fplay%25253F%252526ei%25253DUTF-8%252526vid%25253D033c52938829778cceebb58cb4f5fd8f.1762130&imTitle=DanceCat&searchUrl=http://video.yahoo.com/video/search?p=&profileUrl=http://video.yahoo.com/video/profile?yid=&amp;amp;amp;amp;amp;amp;creatorValue=YnVkZHlfMjI1Nw%3D%3D&vid=033c52938829778cceebb58cb4f5fd8f.1762130"></embed><br /><br />carpe diem, everyone.<br /><br />until next time~!<br /><br />-Sonia<br /><br />Oh yea, Crystal. You're up next xDevilevilevilhttp://www.blogger.com/profile/09667678604279010437noreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-25480505929587022572007-02-21T22:36:00.000-08:002007-02-24T12:49:43.981-08:00Section 7.7 Approximate Integration<div>Hello everyone.<br />This section is called Approximate Integration. It covers three methods of approximating integrals, but only two of them are covered on the AP Exam.<br /><br />The first method is the Midpoint Rule. We have covered this before, but just in case you cannot remember it, here it is:<br /><br /><div><a href="http://bp1.blogger.com/_GZkjHU3k2Ik/Rd08SUv1zWI/AAAAAAAAACU/EFshKo63VQI/s1600-h/MidpointRule.bmp"><img id="BLOGGER_PHOTO_ID_5034246243974761826" style="CURSOR: hand" alt="" src="http://bp1.blogger.com/_GZkjHU3k2Ik/Rd08SUv1zWI/AAAAAAAAACU/EFshKo63VQI/s320/MidpointRule.bmp" border="0" /></a><br />where <a href="http://bp0.blogger.com/_GZkjHU3k2Ik/Rd08IEv1zVI/AAAAAAAAACM/qFQHyoxykK4/s1600-h/deltaxequals.bmp"><img id="BLOGGER_PHOTO_ID_5034246067881102674" style="CURSOR: hand" alt="" src="http://bp0.blogger.com/_GZkjHU3k2Ik/Rd08IEv1zVI/AAAAAAAAACM/qFQHyoxykK4/s320/deltaxequals.bmp" border="0" /></a>(for all of the equations).<br /></div><br /><div>This calculates the area under a curve using rectangle approximations, using the midpoints as the heights. A more accurate method is trapezoidal approximation. Instead of drawing rectangles under the curve, you can draw trapezoids. Here is the formula for using trapezoidal approximation.<br /></div><br /><div><br /><div><a href="http://bp2.blogger.com/_GZkjHU3k2Ik/Rd077kv1zTI/AAAAAAAAAB8/qn4WPT2AvXM/s1600-h/TrapezoidalRule.bmp"><img id="BLOGGER_PHOTO_ID_5034245853132737842" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_GZkjHU3k2Ik/Rd077kv1zTI/AAAAAAAAAB8/qn4WPT2AvXM/s320/TrapezoidalRule.bmp" border="0" /></a> </div><br /><div>Finally, there is a third method that is not on the AP Exam, and it is known as the Simpson’s Rule. Here it is:</div><br /><div><br /><a href="http://bp3.blogger.com/_GZkjHU3k2Ik/Rd08A0v1zUI/AAAAAAAAACE/J7xmu7-Ebxk/s1600-h/SimpsonsRule.bmp"><img id="BLOGGER_PHOTO_ID_5034245943327051074" style="CURSOR: hand" alt="" src="http://bp3.blogger.com/_GZkjHU3k2Ik/Rd08A0v1zUI/AAAAAAAAACE/J7xmu7-Ebxk/s320/SimpsonsRule.bmp" border="0" /></a> </div><br /><div>For this to work, n must be even.<br /><br />Here is an example:</div><br /><div></div><br /><div>For <a href="http://bp3.blogger.com/_GZkjHU3k2Ik/Rd07d0v1zRI/AAAAAAAAABs/nkdxP3xw6dc/s1600-h/ExampleEquation.bmp"><img id="BLOGGER_PHOTO_ID_5034245342031629586" style="CURSOR: hand" alt="" src="http://bp3.blogger.com/_GZkjHU3k2Ik/Rd07d0v1zRI/AAAAAAAAABs/nkdxP3xw6dc/s320/ExampleEquation.bmp" border="0" /></a> use the Trapezoidal Rule, the Midpoint Rule, and Simpson’s Rule to approximate the integral if n = 4. </div><br /><div><br />For the Trapezoidal Rule, plug in the values for the equation. First calculate <a href="http://bp0.blogger.com/_GZkjHU3k2Ik/Rd07YEv1zQI/AAAAAAAAABk/ZMcdRZayqQc/s1600-h/deltax.bmp"><img id="BLOGGER_PHOTO_ID_5034245243247381762" style="CURSOR: hand" alt="" src="http://bp0.blogger.com/_GZkjHU3k2Ik/Rd07YEv1zQI/AAAAAAAAABk/ZMcdRZayqQc/s320/deltax.bmp" border="0" /></a>.</div><br /><div><a href="http://bp2.blogger.com/_GZkjHU3k2Ik/Rd07Tkv1zPI/AAAAAAAAABc/e1IPIdRoylk/s1600-h/deltaxequalswhat.bmp"><img id="BLOGGER_PHOTO_ID_5034245165937970418" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_GZkjHU3k2Ik/Rd07Tkv1zPI/AAAAAAAAABc/e1IPIdRoylk/s320/deltaxequalswhat.bmp" border="0" /></a><br />Now: <a href="http://bp2.blogger.com/_GZkjHU3k2Ik/ReCkw0v1zXI/AAAAAAAAADk/foj_fH9j-xo/s1600-h/FixedMathEquation.bmp"><img id="BLOGGER_PHOTO_ID_5035205542100192626" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_GZkjHU3k2Ik/ReCkw0v1zXI/AAAAAAAAADk/foj_fH9j-xo/s320/FixedMathEquation.bmp" border="0" /></a><br /></div><div></div><br /><div>This method gives us an approximation of 44.</div><br /><div></div><br /><div>Using the midpoint rule, <a href="http://bp0.blogger.com/_GZkjHU3k2Ik/Rd07KEv1zNI/AAAAAAAAABM/iH4tyIA9eDs/s1600-h/deltax.bmp"><img id="BLOGGER_PHOTO_ID_5034245002729213138" style="CURSOR: hand" alt="" src="http://bp0.blogger.com/_GZkjHU3k2Ik/Rd07KEv1zNI/AAAAAAAAABM/iH4tyIA9eDs/s320/deltax.bmp" border="0" /></a>will be the same. Next find the midpoints.</div><br /><div><a href="http://bp3.blogger.com/_GZkjHU3k2Ik/Rd07F0v1zMI/AAAAAAAAABE/sFkE53gey2I/s1600-h/ExampleMid.bmp"><img id="BLOGGER_PHOTO_ID_5034244929714769090" style="CURSOR: hand" alt="" src="http://bp3.blogger.com/_GZkjHU3k2Ik/Rd07F0v1zMI/AAAAAAAAABE/sFkE53gey2I/s320/ExampleMid.bmp" border="0" /></a><br />Now plug into the equation: <a href="http://bp0.blogger.com/_GZkjHU3k2Ik/Rd062Ev1zKI/AAAAAAAAAA0/FMYPxKvMh_o/s1600-h/ExamplePart2.bmp"><img id="BLOGGER_PHOTO_ID_5034244659131829410" style="CURSOR: hand" alt="" src="http://bp0.blogger.com/_GZkjHU3k2Ik/Rd062Ev1zKI/AAAAAAAAAA0/FMYPxKvMh_o/s320/ExamplePart2.bmp" border="0" /></a> </div><br /><div></div><br /><div>This gives us an approximation of 21.Finally, for Simpson’s Rule. <a href="http://bp2.blogger.com/_GZkjHU3k2Ik/Rd06wkv1zJI/AAAAAAAAAAs/T9RbqqivaDA/s1600-h/deltax.bmp"><img id="BLOGGER_PHOTO_ID_5034244564642548882" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_GZkjHU3k2Ik/Rd06wkv1zJI/AAAAAAAAAAs/T9RbqqivaDA/s320/deltax.bmp" border="0" /></a> remains the same.</div><br /><div>Plug into the equation: <a href="http://bp2.blogger.com/_GZkjHU3k2Ik/Rd069kv1zLI/AAAAAAAAAA8/o3Va5u5bzpY/s1600-h/ExamplePart3.bmp"><img id="BLOGGER_PHOTO_ID_5034244787980848306" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_GZkjHU3k2Ik/Rd069kv1zLI/AAAAAAAAAA8/o3Va5u5bzpY/s320/ExamplePart3.bmp" border="0" /></a><a href="http://bp3.blogger.com/_GZkjHU3k2Ik/Rd06j0v1zII/AAAAAAAAAAk/nOaJWm_3Vjs/s1600-h/ExamplePart3.bmp"></a><br /><br />This gives us an approximation of <a href="http://bp3.blogger.com/_GZkjHU3k2Ik/Rd06d0v1zHI/AAAAAAAAAAc/irI10lhSHtY/s1600-h/Answer3.bmp"><img id="BLOGGER_PHOTO_ID_5034244242520001650" style="CURSOR: hand" alt="" src="http://bp3.blogger.com/_GZkjHU3k2Ik/Rd06d0v1zHI/AAAAAAAAAAc/irI10lhSHtY/s320/Answer3.bmp" border="0" /></a>.</div><br /><div>And that's the lesson.</div><br /><div></div><br /><div>Here is a link to a helpful resource: This website shows the Trapezoidal Approximation and Simpson’s Rule and how to use them: <a href="http://archives.math.utk.edu/visual.calculus/4/approx.1/index.html">http://archives.math.utk.edu/visual.calculus/4/approx.1/index.html</a><br /><br />Sonia you are next.<br /><br />On a lighter note, here are some comics:<br /><a href="http://bp2.blogger.com/_GZkjHU3k2Ik/Rd06Ykv1zGI/AAAAAAAAAAU/JkWi4r5egjI/s1600-h/Comic1.jpg"><img id="BLOGGER_PHOTO_ID_5034244152325688418" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_GZkjHU3k2Ik/Rd06Ykv1zGI/AAAAAAAAAAU/JkWi4r5egjI/s320/Comic1.jpg" border="0" /></a><br /><a href="http://bp0.blogger.com/_GZkjHU3k2Ik/Rd06VEv1zFI/AAAAAAAAAAM/nCFFTA3kKdk/s1600-h/Comic2.gif"><img id="BLOGGER_PHOTO_ID_5034244092196146258" style="CURSOR: hand" alt="" src="http://bp0.blogger.com/_GZkjHU3k2Ik/Rd06VEv1zFI/AAAAAAAAAAM/nCFFTA3kKdk/s320/Comic2.gif" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /></div><br /><div></div></div></div>Kylehttp://www.blogger.com/profile/04027474156848702171noreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-51204986875447138572007-02-21T22:30:00.000-08:002007-02-21T22:37:59.272-08:00Friday's Quiz Topics<div>Here’s a list of topics that will be covered on this Friday’s Quiz.<br /><br /><span style="font-size:130%;color:#ff0000;"><strong><em><u>Quiz – Sections 7.1,7</u></em></strong></span><br />Integration by Parts – basic (7.1,#3,7)<br />Integration by Parts – definite integral<br />Integration by Parts – f and g won’t go away (7.1,#15 - not assigned, but good practice!)<br />Integration by Parts – tabular method (7.1, #61)<br />Trapezoidal Rule (7.7, #1,3,7,29)<br />Midpoint Rule (7.7, #1,3,7,29)<br /><br />That’s it for now! I’ll be around after school on Thursday (after 3:30) and in early on Friday. Donut holes and OJ...<br /><br /><em><span style="color:#3333ff;">At New York's Kennedy Airport today, an individual, later discovered to be a public school teacher, was arrested trying to board a flight while in possession of a ruler, a protractor, a set square, and a calculator. Attorney General John Ashcroft believes the man is a member of the notorious Al-Gebra movement. He is being charged with carrying weapons of math instruction.<br />Al-Gebra is a very fearsome cult, indeed.They desire average solutions by means and extremes, and sometimes go off on a tangent in a search of absolute value. They consist of quite shadowy figures, with names like "x" and "y", and, although they are frequently referred to as "unknowns", we know they really belong to a common denominator and are part of the axis of medieval with coordinates in every country. As the great Greek philanderer Isosceles used to say, there are 3 sides to every angle, and if God had wanted us to have better weapons of math instruction, He would have given us more fingers and toes.<br />Therefore, I'm extremely grateful that our government has given us a sine that it is intent on protracting us from these math-dogs who are so willing to disintegrate us with calculus disregard.<br />These statistic bastards love to inflict plane on every sphere of influence. Under the circumferences, it's time we differentiated their root, made our point, and drew the line. These weapons of math instruction have the potential to decimal everything in their math on a scalar never before seen unless we become exponents of a Higher Power and begin to appreciate the random facts of vertex.<br />As our Great Leader would say, "Read my ellipse". Here is one principle he is uncertainty of---though they continue to multiply, their days are numbered and sooner or later the hypotenuse will tighten around their necks.<br /><a href="http://bp3.blogger.com/_sNG9dzZilv4/Rd06JP_gw0I/AAAAAAAAAD4/I5tgPnU0XXo/s1600-h/dilbert8c.gif"><img id="BLOGGER_PHOTO_ID_5034243889056236354" style="CURSOR: hand" alt="" src="http://bp3.blogger.com/_sNG9dzZilv4/Rd06JP_gw0I/AAAAAAAAAD4/I5tgPnU0XXo/s400/dilbert8c.gif" border="0" /></a></span></em></div>Math Maverickhttp://www.blogger.com/profile/10631468736601964037noreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-78629491633215238492007-02-20T18:55:00.000-08:002007-02-21T23:02:25.812-08:007.1 Integration of PartsYou should remember that the product rule is :<br /><div><a href="http://bp2.blogger.com/_2G75oahVTGQ/Rdu080wSEeI/AAAAAAAAAAM/VHSAE7wE3Bg/s1600-h/2.GIF"><img id="BLOGGER_PHOTO_ID_5033815965562966498" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_2G75oahVTGQ/Rdu080wSEeI/AAAAAAAAAAM/VHSAE7wE3Bg/s400/2.GIF" border="0" /></a></div><div><br /></div><div>By rearranging this equation you see<span style="font-size:0;"></span> that:</div><div><a href="http://bp3.blogger.com/_2G75oahVTGQ/Rdu16EwSEfI/AAAAAAAAAAU/SRXAVi3y9u0/s1600-h/3.GIF"><img id="BLOGGER_PHOTO_ID_5033817017829954034" style="CURSOR: hand" alt="" src="http://bp3.blogger.com/_2G75oahVTGQ/Rdu16EwSEfI/AAAAAAAAAAU/SRXAVi3y9u0/s400/3.GIF" border="0" /></a><br /><a href="http://bp3.blogger.com/_2G75oahVTGQ/Rdu2NEwSEgI/AAAAAAAAAAc/TOttMwdkrGw/s1600-h/4.GIF"></a><a href="http://bp1.blogger.com/_2G75oahVTGQ/Rdu2pkwSEhI/AAAAAAAAAAk/mJs5AepRQAo/s1600-h/3.GIF"></a><a href="http://bp3.blogger.com/_2G75oahVTGQ/Rdu23EwSEiI/AAAAAAAAAAs/ObnIaR7XRdY/s1600-h/3.GIF"><img id="BLOGGER_PHOTO_ID_5033818065801974306" style="CURSOR: hand" alt="" src="http://bp3.blogger.com/_2G75oahVTGQ/Rdu23EwSEiI/AAAAAAAAAAs/ObnIaR7XRdY/s400/3.GIF" border="0" /></a></div><br /><div><br /></div><div>By using the substitution rule and replacing u for f(x), v for g(x), du for f'(x)dx, and dv for g'(x)dx, you will see that the formula for integreation of parts becomes:<br /><span style="font-size:0;"></span><a href="http://bp1.blogger.com/_2G75oahVTGQ/Rdu3zkwSEjI/AAAAAAAAAA0/4J1P0_r225I/s1600-h/4.GIF"><img id="BLOGGER_PHOTO_ID_5033819105184059954" style="CURSOR: hand" alt="" src="http://bp1.blogger.com/_2G75oahVTGQ/Rdu3zkwSEjI/AAAAAAAAAA0/4J1P0_r225I/s400/4.GIF" border="0" /></a> </div><div><br /><br /></div><div>Let's look at a couple of examples! </div><div>1) <img id="BLOGGER_PHOTO_ID_5033821244077773394" style="CURSOR: hand" alt="" src="http://bp3.blogger.com/_2G75oahVTGQ/Rdu5wEwSElI/AAAAAAAAABE/j7r-5fRUoSw/s400/6.GIF" border="0" /><br />Choose your simplest component for u. In this case that would be x. Make a table of your u, du, v, and dv. </div><div></div><div></div><div><a href="http://bp3.blogger.com/_2G75oahVTGQ/Rd0-IkwSE6I/AAAAAAAAAGA/uLeCuWOyefg/s1600-h/15.GIF"><img id="BLOGGER_PHOTO_ID_5034248275496145826" style="CURSOR: hand" alt="" src="http://bp3.blogger.com/_2G75oahVTGQ/Rd0-IkwSE6I/AAAAAAAAAGA/uLeCuWOyefg/s400/15.GIF" border="0" /></a><br />Now plug it into the equation <img id="BLOGGER_PHOTO_ID_5033821596265091682" style="CURSOR: hand" alt="" src="http://bp1.blogger.com/_2G75oahVTGQ/Rdu6EkwSEmI/AAAAAAAAABM/c5ySHrntlys/s400/4.GIF" border="0" /><br />You get<br /><a href="http://bp1.blogger.com/_2G75oahVTGQ/Rd0-kEwSE7I/AAAAAAAAAGI/kTYDvmCojB8/s1600-h/16.GIF"><img id="BLOGGER_PHOTO_ID_5034248747942548402" style="CURSOR: hand" alt="" src="http://bp1.blogger.com/_2G75oahVTGQ/Rd0-kEwSE7I/AAAAAAAAAGI/kTYDvmCojB8/s400/16.GIF" border="0" /></a><br /><a href="http://bp2.blogger.com/_2G75oahVTGQ/Rd0-0UwSE8I/AAAAAAAAAGQ/ZzrVByw_EyM/s1600-h/17.GIF"><img id="BLOGGER_PHOTO_ID_5034249027115422658" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_2G75oahVTGQ/Rd0-0UwSE8I/AAAAAAAAAGQ/ZzrVByw_EyM/s400/17.GIF" border="0" /></a><br />Don't forget the C!</div><div><br /></div><div>2) <a href="http://bp2.blogger.com/_2G75oahVTGQ/Rdv1V0wSE3I/AAAAAAAAAFY/s2th1VZen-0/s1600-h/21.GIF"><img id="BLOGGER_PHOTO_ID_5033886763803874162" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_2G75oahVTGQ/Rdv1V0wSE3I/AAAAAAAAAFY/s2th1VZen-0/s400/21.GIF" border="0" /></a><br />You can use the tabular method here. </div><div><a href="http://bp2.blogger.com/_2G75oahVTGQ/Rdv1g0wSE4I/AAAAAAAAAFg/i6_wSq_bmEU/s1600-h/20.GIF"><img id="BLOGGER_PHOTO_ID_5033886952782435202" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_2G75oahVTGQ/Rdv1g0wSE4I/AAAAAAAAAFg/i6_wSq_bmEU/s400/20.GIF" border="0" /></a></div><div><a href="http://bp1.blogger.com/_2G75oahVTGQ/Rdv2HkwSE5I/AAAAAAAAAFo/WOM_9bYDxBk/s1600-h/22.GIF"><img id="BLOGGER_PHOTO_ID_5033887618502366098" style="CURSOR: hand" alt="" src="http://bp1.blogger.com/_2G75oahVTGQ/Rdv2HkwSE5I/AAAAAAAAAFo/WOM_9bYDxBk/s400/22.GIF" border="0" /></a><br /><br /><br /></div><div>Now try a problem!</div><div><a href="http://bp2.blogger.com/_2G75oahVTGQ/Rd0_fUwSE9I/AAAAAAAAAGY/Ywd8N9VkJuc/s1600-h/18.GIF"><img id="BLOGGER_PHOTO_ID_5034249765849797586" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_2G75oahVTGQ/Rd0_fUwSE9I/AAAAAAAAAGY/Ywd8N9VkJuc/s400/18.GIF" border="0" /></a><br /></div><div><a href="http://bp1.blogger.com/_2G75oahVTGQ/RdvgzkwSEqI/AAAAAAAAACY/uLtN53pybfY/s1600-h/10.GIF"><img id="BLOGGER_PHOTO_ID_5033864185160798882" style="CURSOR: hand" alt="" src="http://bp1.blogger.com/_2G75oahVTGQ/RdvgzkwSEqI/AAAAAAAAACY/uLtN53pybfY/s400/10.GIF" border="0" /></a><br /><a href="http://bp1.blogger.com/_2G75oahVTGQ/RdvhBkwSErI/AAAAAAAAACg/OLq4eYdqpl8/s1600-h/11.GIF"><img id="BLOGGER_PHOTO_ID_5033864425678967474" style="CURSOR: hand" alt="" src="http://bp1.blogger.com/_2G75oahVTGQ/RdvhBkwSErI/AAAAAAAAACg/OLq4eYdqpl8/s400/11.GIF" border="0" /></a></div><div><a href="http://bp2.blogger.com/_2G75oahVTGQ/RdvkV0wSEvI/AAAAAAAAADQ/CNBI23uoKhY/s1600-h/12.GIF"><img id="BLOGGER_PHOTO_ID_5033868072106201842" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_2G75oahVTGQ/RdvkV0wSEvI/AAAAAAAAADQ/CNBI23uoKhY/s400/12.GIF" border="0" /></a><br /></div><div>Here, you must use integration by parts again<br /><a href="http://bp0.blogger.com/_2G75oahVTGQ/RdvzSUwSE2I/AAAAAAAAAFQ/8JqpRhOjXc4/s1600-h/13.GIF"><img id="BLOGGER_PHOTO_ID_5033884504651076450" style="CURSOR: hand" alt="" src="http://bp0.blogger.com/_2G75oahVTGQ/RdvzSUwSE2I/AAAAAAAAAFQ/8JqpRhOjXc4/s400/13.GIF" border="0" /></a><br /><a href="http://bp0.blogger.com/_2G75oahVTGQ/RdvmOUwSEwI/AAAAAAAAADY/qHbkmEtqHM0/s1600-h/15.GIF"><img id="BLOGGER_PHOTO_ID_5033870142280438530" style="CURSOR: hand" alt="" src="http://bp0.blogger.com/_2G75oahVTGQ/RdvmOUwSEwI/AAAAAAAAADY/qHbkmEtqHM0/s400/15.GIF" border="0" /></a><a href="http://bp0.blogger.com/_2G75oahVTGQ/RdvjMUwSEuI/AAAAAAAAADI/StqWcC2F3xM/s1600-h/14.GIF"><img id="BLOGGER_PHOTO_ID_5033866809385816802" style="CURSOR: hand" alt="" src="http://bp0.blogger.com/_2G75oahVTGQ/RdvjMUwSEuI/AAAAAAAAADI/StqWcC2F3xM/s400/14.GIF" border="0" /></a><br /><a href="http://bp2.blogger.com/_2G75oahVTGQ/Rdvmu0wSExI/AAAAAAAAADg/hnEfOnT6djM/s1600-h/16.GIF"><img id="BLOGGER_PHOTO_ID_5033870700626187026" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_2G75oahVTGQ/Rdvmu0wSExI/AAAAAAAAADg/hnEfOnT6djM/s400/16.GIF" border="0" /></a><br /><a href="http://bp0.blogger.com/_2G75oahVTGQ/RdvnJUwSEyI/AAAAAAAAADo/jepofTtxylI/s1600-h/16.GIF"><img id="BLOGGER_PHOTO_ID_5033871155892720418" style="CURSOR: hand" alt="" src="http://bp0.blogger.com/_2G75oahVTGQ/RdvnJUwSEyI/AAAAAAAAADo/jepofTtxylI/s400/16.GIF" border="0" /></a></div><div><br /></div><div>Now plug this equation back into the end of the first integration of parts:</div><div><a href="http://bp3.blogger.com/_2G75oahVTGQ/RdvoKEwSEzI/AAAAAAAAADw/_walqNMbQ3Q/s1600-h/17.GIF"><img id="BLOGGER_PHOTO_ID_5033872268289250098" style="CURSOR: hand" alt="" src="http://bp3.blogger.com/_2G75oahVTGQ/RdvoKEwSEzI/AAAAAAAAADw/_walqNMbQ3Q/s400/17.GIF" border="0" /></a><br /></div><div><a href="http://bp2.blogger.com/_2G75oahVTGQ/Rdvoz0wSE0I/AAAAAAAAAD4/POwCBf5kDWI/s1600-h/18.GIF"><img id="BLOGGER_PHOTO_ID_5033872985548788546" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_2G75oahVTGQ/Rdvoz0wSE0I/AAAAAAAAAD4/POwCBf5kDWI/s400/18.GIF" border="0" /></a></div><div><br /><br /><br /></div><div>Here is a good site for Integration of Parts:<br /><a href="http://www.sosmath.com/calculus/integration/byparts/byparts.html">http://www.sosmath.com/calculus/integration/byparts/byparts.html</a></div><div><br /><br /></div><div>Saladang Song: My obsession with food has led me to this great Thai restaurant. I highly recommend it (especially the Yellow Curry) and it's very well priced.<br />363 S Fair Oaks Ave (Cross Street: W Bellevue Drive)Pasadena, CA 91105<br /></div><div><a href="http://bp2.blogger.com/_2G75oahVTGQ/Rdvqk0wSE1I/AAAAAAAAAEA/AgXCh52Cl50/s1600-h/saladang.bmp"><img id="BLOGGER_PHOTO_ID_5033874926874006354" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_2G75oahVTGQ/Rdvqk0wSE1I/AAAAAAAAAEA/AgXCh52Cl50/s400/saladang.bmp" border="0" /></a><br /></div><div>Kyle, you're up next!<br /></div><div><br /></div>amihttp://www.blogger.com/profile/02731237501764087842noreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-76665517646340405512007-02-13T20:30:00.000-08:002007-02-14T09:11:26.538-08:00Thursday's Test Topics<div>Here’s a list of topics that will be covered on this Thursday’s Chapter 6 Test.<br /><br /><strong><em><u><span style="font-size:130%;color:#ff0000;">Chapter 6 Test Topics<br /></span></u></em></strong>You will be given a set of functions determining a region. You will need to determine the area of the region, the volume of a solid created by revolving the region around the x-axis, the y-axis, around a line parallel to the x-axis, and around a line parallel to the y-axis. You can use disks, washers or shells to get your answers. (Sections 6.1-3 - all)<br />Determine the volume of solids generated by building shapes with a known cross-section off a given base. (6.2, #55)<br />Determine the average value of a function for a given interval. (6.5, #1,5,7,13)<br />Determine any value(s) c that generate the average value of a function on a given interval. (6.5, #9)<br /><br />For additional practice problems, look at the chapter review (pp. 431-433)<br /><br />That’s it! I’ll be around after school on Wednesday until 3:15 and in early on Thursday. Donut holes and OJ!<br /><br /><em><span style="color:#000099;">It’s kind of fun to do the impossible. – Walt Disney<br /></div></span></em><em><span style="color:#000099;"><p align="center"><a href="http://bp1.blogger.com/_sNG9dzZilv4/RdKQEf_gwyI/AAAAAAAAADg/OdcZE3GLwf8/s1600-h/PythagMickey.jpg"><img id="BLOGGER_PHOTO_ID_5031242140708029218" style="CURSOR: hand" alt="" src="http://bp1.blogger.com/_sNG9dzZilv4/RdKQEf_gwyI/AAAAAAAAADg/OdcZE3GLwf8/s400/PythagMickey.jpg" border="0" /></a></span></em></p>Math Maverickhttp://www.blogger.com/profile/10631468736601964037noreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-78727642464424556632007-02-13T18:45:00.000-08:002007-02-14T07:12:27.147-08:006.5: Average Value of a Function<div><br /><br /><div>Hey guys! Hey Mr. French! I hope you are all having a great Tuesday, I know I am. Anyways, let's get into the lesson. This lesson deals with the definition of an Integral: An approximation of the area bewteen the given curve and the x-axis between two given points. Now let's look at an example. We want to find the area under this cubic function, F(X).<br /><br />Graph</div><br /><br /><p><img id="BLOGGER_PHOTO_ID_5031227911862130754" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://bp3.blogger.com/_js492I1Dy6w/RdKDIRaSmEI/AAAAAAAAAAc/y7hxpJHAHrY/s320/jamia.GIF" border="0" />Ok, now we want to find all the area bewteen the curve and the x-axis between the points a and b. Let's look at a visual depiction of that:<br /><br /><br /><br /><br /><a href="http://bp3.blogger.com/_js492I1Dy6w/RdKFeRaSmHI/AAAAAAAAABE/y0i-FcpUKJk/s1600-h/untitled.GIF"><img id="BLOGGER_PHOTO_ID_5031230488842508402" style="FLOAT: left; MARGIN: 0px 10px 10px 0px; WIDTH: 237px; CURSOR: hand; HEIGHT: 258px" height="248" alt="" src="http://bp3.blogger.com/_js492I1Dy6w/RdKFeRaSmHI/AAAAAAAAABE/y0i-FcpUKJk/s320/untitled.GIF" width="198" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />Now we want to find this area using one large rectangle. Now of course a rectange is stright on all sides, so this area is going to be an approximation. We want to draw the rectangle so the area that it extends past the graph seems to be close to the area that is fails to cover in the graph. This will make more sense when looked at graphically. Let's look at that right now!<br /><br /><a href="http://bp0.blogger.com/_js492I1Dy6w/RdKE6haSmGI/AAAAAAAAAA8/S1yMynTUAgA/s1600-h/chya.GIF"><img id="BLOGGER_PHOTO_ID_5031229874662185058" style="FLOAT: left; MARGIN: 0px 10px 10px 0px; WIDTH: 267px; CURSOR: hand; HEIGHT: 285px" height="263" alt="" src="http://bp0.blogger.com/_js492I1Dy6w/RdKE6haSmGI/AAAAAAAAAA8/S1yMynTUAgA/s320/chya.GIF" width="217" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />So, as we can see the rectangle extends above the graph but fails to fill in certain areas under the graph. These areas cancel each other out. We are left with a close approximation of the area under the graph. Remember that any part of the graph below the x-axis represents negative area.<br /><br /><br />Now that we have looked at this visually, we can look at it in terms of an equation. We have just visually depicted the mean value theorem. The mean value theorem can be expressed mathematically as:<br /><br /><br /><a href="http://bp2.blogger.com/_js492I1Dy6w/RdKJSBaSmII/AAAAAAAAABM/vkW7R4Nqa7g/s1600-h/untitled.bmp"><img id="BLOGGER_PHOTO_ID_5031234676435622018" style="WIDTH: 215px; CURSOR: hand; HEIGHT: 82px" height="76" alt="" src="http://bp2.blogger.com/_js492I1Dy6w/RdKJSBaSmII/AAAAAAAAABM/vkW7R4Nqa7g/s320/untitled.bmp" width="181" border="0" /></a><br /><br /><br />a and b are our boundaries and c is a point between a and b. f(c) is the avaerage value of the integral.<br /><br />Now let's solve the equation for f(c):<br /><br /><br /><a href="http://bp2.blogger.com/_js492I1Dy6w/RdKLIBaSmJI/AAAAAAAAABU/lGaJh1x0wzM/s1600-h/jamal.bmp"><img id="BLOGGER_PHOTO_ID_5031236703660185746" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_js492I1Dy6w/RdKLIBaSmJI/AAAAAAAAABU/lGaJh1x0wzM/s320/jamal.bmp" border="0" /></a><br /><br /><a href="http://bp2.blogger.com/_js492I1Dy6w/RdKLIBaSmJI/AAAAAAAAABU/lGaJh1x0wzM/s1600-h/jamal.bmp"></a><br /><br />Ok, now let's look at an example of this numerically....<br /><br /><br /><br /><br />We have the equation y=x+x2 [-1,2]<br /><br /><br />We don't actually need to graph it. We can use our equation for the mean value theorem.<br /><br /><br /><br />a is -1, b is 2, and f(x) is x+x^2, so let's put that into our equation:<br /><br /><br /><br /><a href="http://bp3.blogger.com/_js492I1Dy6w/RdKT1RaSmKI/AAAAAAAAABs/1DBYE5d26OU/s1600-h/mya.bmp"><img id="BLOGGER_PHOTO_ID_5031246277142288546" style="CURSOR: hand" alt="" src="http://bp3.blogger.com/_js492I1Dy6w/RdKT1RaSmKI/AAAAAAAAABs/1DBYE5d26OU/s320/mya.bmp" border="0" /></a><br /><br /><br /><br />Now, let's simplify it further....<br /><br /><br /><br /><a href="http://bp0.blogger.com/_js492I1Dy6w/RdKUdhaSmLI/AAAAAAAAAB0/8Yb1JOtTQN4/s1600-h/i.bmp"><img id="BLOGGER_PHOTO_ID_5031246968632023218" style="CURSOR: hand" alt="" src="http://bp0.blogger.com/_js492I1Dy6w/RdKUdhaSmLI/AAAAAAAAAB0/8Yb1JOtTQN4/s320/i.bmp" border="0" /></a><br /><br /><br /><br /><br />Next we use the rules of integration to take the antiderivative of the equation...<br /><br /><br /><br /><a href="http://bp3.blogger.com/_js492I1Dy6w/RdKU9RaSmMI/AAAAAAAAAB8/fw6FkMFuYpE/s1600-h/yu.bmp"><img id="BLOGGER_PHOTO_ID_5031247514092869826" style="WIDTH: 190px; CURSOR: hand; HEIGHT: 72px" height="61" alt="" src="http://bp3.blogger.com/_js492I1Dy6w/RdKU9RaSmMI/AAAAAAAAAB8/fw6FkMFuYpE/s320/yu.bmp" width="184" border="0" /></a><br /><br /><br /><br />And lastly, we plug our a and b values (-1 and 2 into the antiderivative equation)...<br /><br /><br /><br /><a href="http://bp0.blogger.com/_js492I1Dy6w/RdKV8haSmNI/AAAAAAAAACE/J_mi8O0Yzss/s1600-h/fjdh.bmp"><img id="BLOGGER_PHOTO_ID_5031248600719595730" style="CURSOR: hand" alt="" src="http://bp0.blogger.com/_js492I1Dy6w/RdKV8haSmNI/AAAAAAAAACE/J_mi8O0Yzss/s320/fjdh.bmp" border="0" /></a>1.5<br /><br /><br />1.5 is our "c" value, which is a point between a and b that lies on the graph. Now to find f(c), the average value, we solve our F(x) equation for x, setting it equal to our c value.<br /><br />x+x^2=1.5.<br /><br />When we solve this with the quadratic formula, we find that x= .823 and -1.823. Only .823 is within the boundaries, so that is our actual c-value.<br /><br /><br />Here is a good site about about the average value of a function: <a href="http://www.mathwords.com/a/average_value_function.htm">http://www.mathwords.com/a/average_value_function.htm</a>. It's short, but i think it shows the concept clearly.<br /><br /><br /><br /><br />Also, before I go, I wanted to share with you an inspiration quote<br /><br /><br /><br /><br /><br />"If A is sucess in life, then A equals X+Y+Z. Work is X, Y is play, and Z is keeping your motuh shut"- Albert Einstein </p><br /><br /><p></p><br /><br /><p>Here's a nice picture for all of you:<br /><a href="http://bp0.blogger.com/_js492I1Dy6w/RdMl0RaSmOI/AAAAAAAAACc/cYd9749PfOQ/s1600-h/yo.jpg"></a><a href="http://bp1.blogger.com/_js492I1Dy6w/RdMmshaSmPI/AAAAAAAAACk/m5GPF2hoMqg/s1600-h/n1041870144_30015657_2787.jpg"><img id="BLOGGER_PHOTO_ID_5031407755027716338" style="CURSOR: hand" alt="" src="http://bp1.blogger.com/_js492I1Dy6w/RdMmshaSmPI/AAAAAAAAACk/m5GPF2hoMqg/s320/n1041870144_30015657_2787.jpg" border="0" /></a><br /><br />Since the order is now random, I have no one to remind. I'm out.</p></div>Ryanhttp://www.blogger.com/profile/01472829230582291277noreply@blogger.com1tag:blogger.com,1999:blog-33767428.post-44325043522228639962007-02-08T22:15:00.001-08:002007-02-09T01:51:47.947-08:00Monday's Quiz Topics and New Posting OrderHere’s a list of topics that will be covered on this Monday’s Quiz. Note that I’m including section 6.1 (area between two curves) since you haven’t been formally tested on the information (except for the midterm!)<br /><br /><strong><em><u><span style="font-size:130%;color:#ff0000;">Quiz – Sections 6.1-3<br /></span></u></em></strong>Area between 2 curves<br />Volume of revolution – disk<br />Volume of revolution – washer<br />Volume of revolution – cylindrical shell<br />Volume of a solid with a known cross-section off a specified base<br /><br /><em>The area question(s) can be in terms of x- or y- relationships.<br />The volumes of revolution can be revolved around the x- or y-axis, or some other defined line. </em><br /><em><strong><u><span style="font-size:130%;color:#cc0000;">Note: I'm not going to post relevant homework problems, because in this case they're all relevant...<br /></span></u></strong></em><br />I’ll be around after school on Friday and in early on Monday. If you have questions over the weekend, send me an email and I’ll respond Sunday evening.<br /><br /><br />As promised, here's the new randomly assigned posting order:<br /><br /><br /><p><br /><span style="font-size:180%;color:#006600;"><em>Ryan<br />Ami<br />Kyle<br />Sonia<br />Crystal<br />Alex<br />Brian<br />Magnus<br />Jessica<br />Ismael<br />Joey P.<br />Kane<br />Lauren<br />Joseph Y.<br />Jeff<br />Claire<br />Isaac</em></span></p><br /><em><span style="color:#000099;">In order to attain the impossible, one must attempt the absurd.<br />- Miguel de Cervantes<br /></span>(of course, the absurd seems to happen every day in our class…)<br /></em><br /><br /><strong>Poor study strategies:</strong><br /><br /><br /><p align="center"><em><a href="http://bp3.blogger.com/_sNG9dzZilv4/RcwX7f_gwrI/AAAAAAAAACU/avfkbSkGxlI/s1600-h/FoxtrotTest0001.gif"><img id="BLOGGER_PHOTO_ID_5029421194833609394" style="WIDTH: 498px; CURSOR: hand; HEIGHT: 150px" height="137" alt="" src="http://bp3.blogger.com/_sNG9dzZilv4/RcwX7f_gwrI/AAAAAAAAACU/avfkbSkGxlI/s400/FoxtrotTest0001.gif" width="454" border="0" /></a></p></em>Math Maverickhttp://www.blogger.com/profile/10631468736601964037noreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-12571697727048381602007-02-08T20:36:00.000-08:002007-02-09T01:51:37.186-08:006.3 Volumes by Cylindrical ShellsIn 6.2 which Claire so wonderfully covered, we learned how to find volumes using cross-sections of solids perpendicular to the axis of rotation. We did this using the disk and washer methods:<br /><br /><br /><br /><div><div><div><div><div><div><div><div><div><div></div><div>Disk: <a href="http://bp3.blogger.com/_bvfDj-E3s7Y/RcwGdxJybpI/AAAAAAAAAA0/WqulypCTk7A/s1600-h/image007.gif"><img id="BLOGGER_PHOTO_ID_5029401992346365586" style="CURSOR: hand" alt="" src="http://bp3.blogger.com/_bvfDj-E3s7Y/RcwGdxJybpI/AAAAAAAAAA0/WqulypCTk7A/s320/image007.gif" border="0" /></a> Washer:<a href="http://bp2.blogger.com/_bvfDj-E3s7Y/RcwHkhJybqI/AAAAAAAAAA8/fPXte8FirP4/s1600-h/image009.gif"><img id="BLOGGER_PHOTO_ID_5029403207822110370" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_bvfDj-E3s7Y/RcwHkhJybqI/AAAAAAAAAA8/fPXte8FirP4/s320/image009.gif" border="0" /></a><br /></div><div></div><div></div><div><br />However, the volumes of certain solids cannot easily be found using these two methods. Sometimes it is necessary to use another method: the method of cylindrical shells. Such can be used when the the perpendicular R and r of a washer depend on the same curve. Instead of using cross-sections perpendicular to the axis of rotation, we can use cross-sections parallel to the axis of rotation. Rotating the parallel cross sections about an axis create shells....</div></div><div></div><div><div></div><div></div><div></div><div>If this curve is rotated about the y-axis, one shell would look like this:<br /><a href="http://oregonstate.edu/instruction/mth252/cq2/Lesson15/ex3.gif"><img style="WIDTH: 320px; CURSOR: hand" alt="" src="http://oregonstate.edu/instruction/mth252/cq2/Lesson15/ex3.gif" border="0" /></a><br /></div><div></div><div></div><div><br /></div><div>In this next picture of a shell, you can see that the thickness of the shell is the difference between radii. </div><div><br /></div><div><a href="http://bp3.blogger.com/_bvfDj-E3s7Y/RcwM4xJybrI/AAAAAAAAABU/RD6OMYD4kJ8/s1600-h/cylShell.eps"></a><a href="http://bp3.blogger.com/_bvfDj-E3s7Y/RcwM4xJybrI/AAAAAAAAABU/RD6OMYD4kJ8/s1600-h/cylShell.eps"></a></div><div><a href="http://bp0.blogger.com/_bvfDj-E3s7Y/RcxBrBJyb7I/AAAAAAAAADs/zTObktdbvBM/s1600-h/cylShell.eps"><img id="BLOGGER_PHOTO_ID_5029467091165671346" style="WIDTH: 118px; CURSOR: hand; HEIGHT: 119px" height="163" alt="" src="http://bp0.blogger.com/_bvfDj-E3s7Y/RcxBrBJyb7I/AAAAAAAAADs/zTObktdbvBM/s320/cylShell.eps" width="171" border="0" /></a> </div><div></div><div></div><div></div><div>So, the formula for the volume of a cylindrical shell is:</div><div><a href="http://bp3.blogger.com/_bvfDj-E3s7Y/Rcv9rBJyblI/AAAAAAAAAAM/otns4poLhmw/s1600-h/image002.gif"><img id="BLOGGER_PHOTO_ID_5029392324374982226" style="CURSOR: hand" alt="" src="http://bp3.blogger.com/_bvfDj-E3s7Y/Rcv9rBJyblI/AAAAAAAAAAM/otns4poLhmw/s320/image002.gif" border="0" /></a> , where r is the average of the radii and <a href="http://bp1.blogger.com/_bvfDj-E3s7Y/RcwPqRJybtI/AAAAAAAAABk/mzpvxnBK_2E/s1600-h/image013.gif"><img id="BLOGGER_PHOTO_ID_5029412102699380434" style="CURSOR: hand" alt="" src="http://bp1.blogger.com/_bvfDj-E3s7Y/RcwPqRJybtI/AAAAAAAAABk/mzpvxnBK_2E/s320/image013.gif" border="0" /></a> is the difference of the radii.<br />In words, Volume = (circumference)(height)(thickness). This is easier to understand if you imagine the shell cut and rolled out to form a rectangular solid with length <a href="http://bp2.blogger.com/_bvfDj-E3s7Y/RcwRihJybvI/AAAAAAAAACE/Cg4IAM9xTRA/s1600-h/image010.gif"><img id="BLOGGER_PHOTO_ID_5029414168578649842" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_bvfDj-E3s7Y/RcwRihJybvI/AAAAAAAAACE/Cg4IAM9xTRA/s320/image010.gif" border="0" /></a>, height h, and width <a href="http://bp2.blogger.com/_bvfDj-E3s7Y/RcwRshJybwI/AAAAAAAAACM/0KXohQogl-c/s1600-h/image013.gif"><img id="BLOGGER_PHOTO_ID_5029414340377341698" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_bvfDj-E3s7Y/RcwRshJybwI/AAAAAAAAACM/0KXohQogl-c/s320/image013.gif" border="0" /></a>.</div><div></div><div></div><div></div><div></div><div></div><div>Based of this formula for the volume of one shell, we can come up with a formula for the volume of an entire solid made up of many shells. If f is a continuous function on the closed interval [a, b], then the volume of the solid obtained by rotating the graph of f from x = a to x = b about the y-axis is<br /><a href="http://bp0.blogger.com/_bvfDj-E3s7Y/Rcv9rRJybmI/AAAAAAAAAAU/jYzHQhyooU8/s1600-h/image004.gif"><img id="BLOGGER_PHOTO_ID_5029392328669949538" style="CURSOR: hand" alt="" src="http://bp0.blogger.com/_bvfDj-E3s7Y/Rcv9rRJybmI/AAAAAAAAAAU/jYzHQhyooU8/s320/image004.gif" border="0" /></a> (where <a href="http://bp1.blogger.com/_bvfDj-E3s7Y/RcwQWRJybuI/AAAAAAAAABs/vsTHrousD6M/s1600-h/image015.gif"><img id="BLOGGER_PHOTO_ID_5029412858613624546" style="CURSOR: hand" alt="" src="http://bp1.blogger.com/_bvfDj-E3s7Y/RcwQWRJybuI/AAAAAAAAABs/vsTHrousD6M/s320/image015.gif" border="0" /></a>). </div><div></div><div></div><div></div><div></div><div>Again, <a href="http://bp2.blogger.com/_bvfDj-E3s7Y/RcwNfhJybsI/AAAAAAAAABc/by9sGQXl-bc/s1600-h/image010.gif"><img id="BLOGGER_PHOTO_ID_5029409718992531138" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_bvfDj-E3s7Y/RcwNfhJybsI/AAAAAAAAABc/by9sGQXl-bc/s320/image010.gif" border="0" /></a> is the circumference, f(x) is the height, and dx is the thickness of the shells.<br /></div><div></div><div></div><div></div><div></div><div></div><div></div><div>Now for a sample problem:</div><div></div><div>Find the volume of the solid obtained by rotating the region bounded by <a href="http://bp1.blogger.com/_bvfDj-E3s7Y/Rcw40RJybxI/AAAAAAAAACc/FmBIAx24lAM/s1600-h/image016.gif"><img id="BLOGGER_PHOTO_ID_5029457354474811154" style="CURSOR: hand" alt="" src="http://bp1.blogger.com/_bvfDj-E3s7Y/Rcw40RJybxI/AAAAAAAAACc/FmBIAx24lAM/s320/image016.gif" border="0" /></a> and y= 0</div><div>about the line x = 3. </div><div></div><div></div><div></div><div></div><div></div><div></div><div></div><div>Solution:</div><div></div><div></div><div>First, visualize or sketch the graph. Either with your mind or with your calculator, you can see that the region bounded by the restrictions goes from x=0 to x=1 and lies between a smooth curve and the x-axis (similar to the example on page 458). </div><div></div><div></div><div></div><div></div><div>Then, draw or picture a vertical cross-section of the region. Rotated around x= 3, the shell has a radius of 3-x, a circumference of <a href="http://bp0.blogger.com/_bvfDj-E3s7Y/Rcw8HBJybyI/AAAAAAAAACk/MjFBYbIGScE/s1600-h/image020.gif"><img id="BLOGGER_PHOTO_ID_5029460975132241698" style="CURSOR: hand" alt="" src="http://bp0.blogger.com/_bvfDj-E3s7Y/Rcw8HBJybyI/AAAAAAAAACk/MjFBYbIGScE/s320/image020.gif" border="0" /></a>, and a height of y or <a href="http://bp1.blogger.com/_bvfDj-E3s7Y/Rcw8URJybzI/AAAAAAAAACs/RVAd_jvv2wc/s1600-h/image017.gif"><img id="BLOGGER_PHOTO_ID_5029461202765508402" style="CURSOR: hand" alt="" src="http://bp1.blogger.com/_bvfDj-E3s7Y/Rcw8URJybzI/AAAAAAAAACs/RVAd_jvv2wc/s320/image017.gif" border="0" /></a>.<br /></div><div></div><div></div><div>So, the volume would be:<br /><br /><a href="http://bp1.blogger.com/_bvfDj-E3s7Y/Rcw-1RJyb2I/AAAAAAAAADE/lRwybwewwW0/s1600-h/image004.gif"><img id="BLOGGER_PHOTO_ID_5029463968724447074" style="CURSOR: hand" alt="" src="http://bp1.blogger.com/_bvfDj-E3s7Y/Rcw-1RJyb2I/AAAAAAAAADE/lRwybwewwW0/s320/image004.gif" border="0" /></a><br /><a href="http://bp0.blogger.com/_bvfDj-E3s7Y/Rcw-WBJyb0I/AAAAAAAAAC0/UaQRwHpzy8M/s1600-h/image022.gif"><img id="BLOGGER_PHOTO_ID_5029463431853535042" style="CURSOR: hand" alt="" src="http://bp0.blogger.com/_bvfDj-E3s7Y/Rcw-WBJyb0I/AAAAAAAAAC0/UaQRwHpzy8M/s320/image022.gif" border="0" /></a><br /></div><div></div><div><a href="http://bp0.blogger.com/_bvfDj-E3s7Y/Rcw-WBJyb1I/AAAAAAAAAC8/Daiu-G5_bsE/s1600-h/image024.gif"><img id="BLOGGER_PHOTO_ID_5029463431853535058" style="CURSOR: hand" alt="" src="http://bp0.blogger.com/_bvfDj-E3s7Y/Rcw-WBJyb1I/AAAAAAAAAC8/Daiu-G5_bsE/s320/image024.gif" border="0" /></a><br /></div><div>Plug in 1 and we get <a href="http://bp2.blogger.com/_bvfDj-E3s7Y/Rcw_ohJyb4I/AAAAAAAAADU/n2JEO8SJMBc/s1600-h/image023.gif"><img id="BLOGGER_PHOTO_ID_5029464849192742786" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_bvfDj-E3s7Y/Rcw_ohJyb4I/AAAAAAAAADU/n2JEO8SJMBc/s320/image023.gif" border="0" /></a>.<br /></div><div></div><div></div><div>Some websites:<br /></div><div><a href="http://www.geocities.com/pkving4math2tor7/7_app_of_the_intgrl/7_03_02_finding_vol_by_using_cylind_shells.htm">http://www.geocities.com/pkving4math2tor7/7_app_of_the_intgrl/7_03_02_finding_vol_by_using_cylind_shells.htm</a></div><div><br /><a href="http://pear.math.pitt.edu/Calculus2/week6/6_2li1.html">http://pear.math.pitt.edu/Calculus2/week6/6_2li1.html</a> </div><div></div><div><br /></div><div></div></div><div></div><div>And another awesome photooo and some awesome pollution: <div></div><div><br /><a href="http://bp3.blogger.com/_bvfDj-E3s7Y/RcxAxxJyb6I/AAAAAAAAADk/218CS87JMjs/s1600-h/P1010380.JPG"><img id="BLOGGER_PHOTO_ID_5029466107618160546" style="CURSOR: hand" alt="" src="http://bp3.blogger.com/_bvfDj-E3s7Y/RcxAxxJyb6I/AAAAAAAAADk/218CS87JMjs/s320/P1010380.JPG" border="0" /></a> </div><div></div><div>It looks like <span style="font-size:180%;color:#ff0000;">RYAN</span> is up next. </div><div></div><div></div><div></div></div></div></div></div></div></div></div></div></div>Isaachttp://www.blogger.com/profile/17082667399749679013noreply@blogger.com1tag:blogger.com,1999:blog-33767428.post-28619576844707683332007-02-07T10:07:00.000-08:002007-02-07T12:08:51.632-08:006.2 VolumesIn 6.1, we learned about the area between two curves, but now, Mr. French and the James Stewart have decided to make it more interesting by introducing a third dimension. We are now figuring out the volume between two curves, or the volume caused by the rotation of the area between two curves. <div><div><div><div><div><div><div><div><div><div><div><br /><div>The general formula for finding the area between two curves is:<br /></div><img id="BLOGGER_PHOTO_ID_5028858835931497266" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://bp3.blogger.com/_qe2U5o3AYAY/RcoYd51IAzI/AAAAAAAAAAc/S84T8kiS5J4/s320/2blog+1.bmp" border="0" />So let's say that we want to find the volume of a circle, using squares. The squares would come out the circle, with the base edge bounded by the circle. It would look something like this:</div><br /><div></div><img id="BLOGGER_PHOTO_ID_5028871476020249554" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://bp2.blogger.com/_qe2U5o3AYAY/Rcoj9p1IA9I/AAAAAAAAABs/47gAUsak7Os/s320/2blog+11.bmp" border="0" /> So, to calculate the volume we would use an infinite amount of infinitely thin boxes to calculate it, like a Riemann sum. Because it's a square we know that:<img id="BLOGGER_PHOTO_ID_5028863165258531666" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://bp3.blogger.com/_qe2U5o3AYAY/RcocZ51IA1I/AAAAAAAAAAs/1RI6-3tVqzc/s320/2blog+3.bmp" border="0" />And the formula for a circle is:<img id="BLOGGER_PHOTO_ID_5028863783733822306" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://bp3.blogger.com/_qe2U5o3AYAY/Rcoc951IA2I/AAAAAAAAAA0/xrF8KqBMInU/s320/2blog+4.bmp" border="0" /> where R is the radius. So the set up would look like this:<br /><br /><div><p><img id="BLOGGER_PHOTO_ID_5028862460883895106" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://bp3.blogger.com/_qe2U5o3AYAY/Rcobw51IA0I/AAAAAAAAAAk/yKHsnZ69bIc/s320/2blog+2.bmp" border="0" />The green side, S, is double what the y value is for that x value, which means that<img id="BLOGGER_PHOTO_ID_5028864105856369522" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://bp2.blogger.com/_qe2U5o3AYAY/RcodQp1IA3I/AAAAAAAAAA8/X42YqyMFCRc/s320/2blog+5.bmp" border="0" />Using this information, we can now define A in terms of x, like in the general equation for the volume, and find the volume in terms of R, which is an undefined constant.</p><p>Using the equation for the circle, we get that:<img id="BLOGGER_PHOTO_ID_5028866133080933266" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://bp2.blogger.com/_qe2U5o3AYAY/RcofGp1IA5I/AAAAAAAAABM/INePnMb5j5Y/s320/2blog+7.bmp" border="0" />And according to the general equation for volume:<br /><img id="BLOGGER_PHOTO_ID_5028864990619632514" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://bp0.blogger.com/_qe2U5o3AYAY/RcoeEJ1IA4I/AAAAAAAAABE/Yr_z0cXuRuY/s320/2blog+6.bmp" border="0" />So,<img id="BLOGGER_PHOTO_ID_5028866682836747170" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://bp2.blogger.com/_qe2U5o3AYAY/Rcofmp1IA6I/AAAAAAAAABU/HpGk1ByBORQ/s320/2blog+8.bmp" border="0" />Since the circle is also symmetrical about the y-axis, taking the <span class="blsp-spelling-corrected" id="SPELLING_ERROR_0">integral</span> from -R to R is the same thing as doubling the integral of 0 to R.<img id="BLOGGER_PHOTO_ID_5028867597664781234" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://bp3.blogger.com/_qe2U5o3AYAY/Rcogb51IA7I/AAAAAAAAABc/NUbS6Aurz3k/s320/2blog+9.bmp" border="0" />Now, we just do what we know how to do - evaluate the integral, keeping in mind that R is a constant.<img id="BLOGGER_PHOTO_ID_5028869229752353730" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://bp3.blogger.com/_qe2U5o3AYAY/Rcoh651IA8I/AAAAAAAAABk/aO1XbXQHaz4/s320/2blog+10.bmp" border="0" />And we're done!</p><p>There's another type of volume that Mr. French and James Stewart might ask you to find, and that's a rotation volume. It's when you take an area and rotate it around an axis, creating a solid with circular cross-sections. When the base of the area touches the rotation axis, it creates a disc volume, and when it doesn't, it creates a washer volume.</p><p>Take for instance the graph of the square root of x from 2 to 9.</p><img id="BLOGGER_PHOTO_ID_5028873683633439714" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://bp0.blogger.com/_qe2U5o3AYAY/Rcol-J1IA-I/AAAAAAAAAB0/b0OCeXDyADA/s320/2blog+12.bmp" border="0" />And let's rotate it around the x-axis. And since the base of the area touches the axis of rotation, we have a disc volume, and because the cross sections are circles, the integral for the volume would be:</div><img id="BLOGGER_PHOTO_ID_5028875324310946802" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://bp2.blogger.com/_qe2U5o3AYAY/Rcondp1IA_I/AAAAAAAAAB8/qgZOn1Sl_Vg/s320/2blog+13.bmp" border="0" />And the radius of the cross-section circle differs with the function. It is defined by the y-value, so R=Y (at x=9 the radius is 3, the same as the y-value of the function at 9). Since we know that y equals the square root of x, we can substitute the square root of x in for R. This gives us:</div><br /><div><img id="BLOGGER_PHOTO_ID_5028876883384075266" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://bp1.blogger.com/_qe2U5o3AYAY/Rcoo4Z1IBAI/AAAAAAAAACE/RMSQT71ayIc/s320/2blog+14.bmp" border="0" />And from there, we all know how to solve this problem. </div><br /><div>If we had rotated the function around the line x=-3, then we would have had a washer volume, since the base of the function wouldn't have touched the rotation axis. To solve that problem, we would have had to find the volume without the space, and then subtracted the space. There would be two radii, one representing the radius of the empty <span class="blsp-spelling-corrected" id="SPELLING_ERROR_1">cylinder</span> in the middle and one representing the radius of the whole volume.</div><br /><div><img id="BLOGGER_PHOTO_ID_5028879404529878034" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://bp0.blogger.com/_qe2U5o3AYAY/RcorLJ1IBBI/AAAAAAAAACM/Nx79SVwog6U/s320/2blog+15.bmp" border="0" /></div><div>To solve this problem, we would use:</div><div><img id="BLOGGER_PHOTO_ID_5028880572760982562" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://bp0.blogger.com/_qe2U5o3AYAY/RcosPJ1IBCI/AAAAAAAAACU/ikBjpbm9ySA/s320/2blog+16.bmp" border="0" /></div><div>In this case, r would be a constant of 3, while R would be 3+ y-value of the function or<br /></div><div><img id="BLOGGER_PHOTO_ID_5028881182646338610" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://bp2.blogger.com/_qe2U5o3AYAY/Rcosyp1IBDI/AAAAAAAAACc/tw5eVvKBzgc/s320/2blog+17.bmp" border="0" />So we can plug all of this into the big equation and get:</div><br /><div></div><img id="BLOGGER_PHOTO_ID_5028882574215742546" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://bp2.blogger.com/_qe2U5o3AYAY/RcouDp1IBFI/AAAAAAAAACs/o_wXQDylNNE/s320/2blog+18.bmp" border="0" /><br /><div>And we all know how to solve the problem from there!</div><br /><div>If you need any more help on this concept, you should check out these websites:</div><div></div><div></div><div></div><div><a href="http://archives.math.utk.edu/visual.calculus/5/volumes.4/index.html">http://archives.math.utk.edu/visual.calculus/5/volumes.4/index.html</a><a href="http://archives.math.utk.edu/visual.calculus/5/volumes.4/index.html"></a></div><div><a href="http://archives.math.utk.edu/visual.calculus/5/volumes.1/index.html">http://archives.math.utk.edu/visual.calculus/5/volumes.1/index.html</a></div><div><a href="http://archives.math.utk.edu/visual.calculus/5/volumes.2/index.html">http://archives.math.utk.edu/visual.calculus/5/volumes.2/index.html</a></div><div></div><div>And finally I just wanted to remind everyone that our favorite bunch of castaways are back on tonight after a long winter hiatus.<br /></div><img style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 474px; CURSOR: hand; HEIGHT: 294px; TEXT-ALIGN: center" height="233" alt="" src="http://msnbcmedia4.msn.com/j/msnbc/Components/Photos/070205/070205_TV_Lost_hlarge_3p.hlarge.jpg" border="0" /><br />Just as a recap, Jack, Kate and Sawyer captured by the Others, and Jack had <span class="blsp-spelling-error" id="SPELLING_ERROR_2" onclick="BLOG_clickHandler(this)">Benry</span> on the table in the middle of back surgery, when he insisted that Kate and Sawyer be let go or he'd leave <span class="blsp-spelling-error" id="SPELLING_ERROR_3" onclick="BLOG_clickHandler(this)">Benry</span> on the table to die. So that should make for an interesting episode tonight!</div><div> </div><div>Also, reminder to Lauren for the next post. </div></div></div></div></div></div></div></div></div>Clairehttp://www.blogger.com/profile/02680938518615745163noreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-50602323083427844182007-01-10T15:32:00.000-08:002007-01-10T17:15:43.674-08:00Section 6.1: Areas Between CurvesYay it's my turn now. As I look at these problems more and more, they seem less scary, so if any of you are worried about this concept, as I was, don't worry because with some practice, I'm sure that you can kick these problems' butts! Before I get into summarizing this lesson, here is a link for any further information: <a href="http://tutorial.math.lamar.edu/AllBrowsers/2413/AreaBetweenCurves.asp">http://tutorial.math.lamar.edu/AllBrowsers/2413/AreaBetweenCurves.asp</a><br /><br />Ok so anywho, this lesson takes integrals as we have learned them a step further. Now, instead of finding the area beneath a curve on a given domain, we have to find the area between two different curves on a given domain. It's just a little extra wrinkle that we have to deal with, but no big deal. Let's say that we are given the following graph:<br /><br /><a href="http://photobucket.com/" target="_blank"><img alt="Photobucket - Video and Image Hosting" src="http://i146.photobucket.com/albums/r266/aconforti7/section6.jpg" border="0" /></a><br /><br />The red curve represents the upper curve, y=f(x), and the blue curve represents the lower curve, y=g(x). <em>a</em> and <em>b</em> represent the given domain, and the shaded yellow area represents what you are trying to find when you solve one of these problems. In order to find this area, you can use the following general formula:<br /><br /><a href="http://photobucket.com/" target="_blank"><img src="http://i146.photobucket.com/albums/r266/aconforti7/section6-2.jpg" border="0" alt="Photobucket - Video and Image Hosting"></a><br /><a href="http://photobucket.com/" target="_blank"><img src="http://i146.photobucket.com/albums/r266/aconforti7/section6-1.jpg" border="0" alt="Photobucket - Video and Image Hosting"></a><br /><br />Now we can use that for a fun sample problem!! Brace yourselves...<br /><br /><a href="http://photobucket.com/" target="_blank"><img src="http://i146.photobucket.com/albums/r266/aconforti7/sampleproblem.jpg" border="0" alt="Photobucket - Video and Image Hosting"></a><br /><br />First, we can draw the graph of this situation. The yellow graph is the area that we are trying to find. <br /><br /><a href="http://photobucket.com/" target="_blank"><img src="http://i146.photobucket.com/albums/r266/aconforti7/sampleproblemgraph.jpg" border="0" alt="Photobucket - Video and Image Hosting"></a><br /><br />From this picture, we can see that the upper boundary is the blue curve, or y=2x-x(x), and the lower boundary is y=x(x). Therefore, the area is gonna be (upper boundary-lower boundary)dx or (2x-x(x)-x(x)). Also, the region is between x=0 and x=1. Therefore, the total area is <br /><br /><a href="http://photobucket.com/" target="_blank"><img src="http://i146.photobucket.com/albums/r266/aconforti7/answersampleproblem.jpg" border="0" alt="Photobucket - Video and Image Hosting"></a><br /><br />So, your answer is 1/3. YAY JESSICA YOU'RE NEXT HAVE FUN!!<br /><br /><a href="http://photobucket.com/" target="_blank"><img src="http://i146.photobucket.com/albums/r266/aconforti7/calvinandhobbes.jpg" border="0" alt="Photobucket - Video and Image Hosting"></a>alex chttp://www.blogger.com/profile/12710827233923109665noreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-17396944545732293162007-01-09T21:10:00.000-08:002007-01-09T21:11:31.283-08:00Friday's Test Topics<div>Here’s a list of topics that will be covered on this Friday’s Chapter 5 Test.<br /><br /><span style="font-size:130%;color:#ff0000;"><strong><em><u>Chapter 5 Test Topics<br /></u></em></strong></span>Fundamental Theorem of Calculus Part I (Sec. 5.3, #9,13,17)<br />Fundamental Theorem of Calculus Part II (Sec. 5.3, #13,41)<br />Substitution (Sec. 5.5, #13,23,278,31,37,41,57)<br />Evaluate an integral in terms of area (Sec. 5.2, #37)<br />Riemann sum: sketch, evaluate, explain and interpret (Sec. 5.1, #3,1,13,15)<br />Definite Integrals (Sec. 5.2, #33)<br />Net Change Theorem (Sec. 5.4, #47,48)<br /><br />For additional practice problems, look at the chapter review (pp. 431-433)<br /><br />That’s it! I’ll be around after school on Thursday and in early on Friday. Donut holes and OJ!<br /><br /><em><span style="color:#000099;">"It is inevitable that some defeat will enter even the most victorious life. The human spirit is never finished when it is defeated...it is finished when it surrenders."<br />Ben Stein</span></em></div><br /><div><br />Here’s someone who refused to surrender to incompetence – I admire his patience, but I probably wouldn’t be able to last this long. It’s a fairly long audio clip (about 25 minutes) and I just want to say how thankful I am that MY students understand the importance of units...<br /><a title="http://media.grc.com/mp3/VerizonCantCount.mp3" href="http://media.grc.com/mp3/VerizonCantCount.mp3">http://media.grc.com/mp3/VerizonCantCount.mp3</a><br /><br />And on a lighter note:</div><br /><div><a href="http://bp1.blogger.com/_sNG9dzZilv4/RaR1c-wQ7vI/AAAAAAAAABk/ffF8vY_Szw0/s1600-h/integral5.gif"><img id="BLOGGER_PHOTO_ID_5018265025539796722" style="CURSOR: hand" alt="" src="http://bp1.blogger.com/_sNG9dzZilv4/RaR1c-wQ7vI/AAAAAAAAABk/ffF8vY_Szw0/s320/integral5.gif" border="0" /></a></div>Math Maverickhttp://www.blogger.com/profile/10631468736601964037noreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-16731962267222054812007-01-09T15:42:00.000-08:002007-01-10T11:15:30.248-08:00Section 5.5This section is about the Substitution rule which allows anti-differentiation of complex expressions. The idea is to replace the complex section with a variable, (u), anti-differentiate, and then substitute back in the complex statement: a sort of inverse chain rule.<br /><br /><strong>Substitution Rule for indefinite integrals<br /><br />If u=g(x) is a differentiable function whose range is an interval I and f is continuous on I then<br />∫ f(g(x))g'(x) dx = ∫ f(u) du </strong><br /><br /><br /><p><strong>This rule can also be applied to definite integrals by adjusting the range for u</strong><br /><br /></p><p><strong>∫ from a to be of [f(g(x))g'(x) dx] = ∫ from g(a) to g(b) of [f(u) du</strong></p><p></p><em>Example<br /></em><br />∫ (cos√t)/ √t<br /><br />Given that u = √t then dt = 2du/t^-.5<br /><br />substituting both expressions in removes the denominator and the roots to give "∫ 2cos(u) du"<br /><br />antidifferentiate to get 2sin(u) + C and substitute the original expression for x to get <strong>2sin√t + C</strong><br /><br /><br /><br /><br />This is a helpful site: <a href="http://www.sosmath.com/calculus/integration/substitution/substitution.html">http://www.sosmath.com/calculus/integration/substitution/substitution.html</a><br /><br /><br /><br /><br /><br /><br /><a href="http://bp1.blogger.com/_0MCYVp2NwRs/RaU6IBfa3ZI/AAAAAAAAAAM/DKF1IbvJJ58/s1600-h/162070_paper_crane.jpg"></a><p><a href="http://bp2.blogger.com/_0MCYVp2NwRs/RaU66Rfa3aI/AAAAAAAAAAU/AQ5zmW31gzg/s1600-h/30981image1[1].jpg"><img id="BLOGGER_PHOTO_ID_5018482132576296354" style="CURSOR: hand" alt="" src="http://bp2.blogger.com/_0MCYVp2NwRs/RaU66Rfa3aI/AAAAAAAAAAU/AQ5zmW31gzg/s320/30981image1%5B1%5D.jpg" border="0" /></a> </p><p>Alex you're next.<br /><br /></p>Magnus Hawnoreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-52892556711631172832006-12-31T08:41:00.000-08:002006-12-31T08:45:21.389-08:00Happy New Year!<a href="http://bp3.blogger.com/_sNG9dzZilv4/RZfoWodImLI/AAAAAAAAAAs/nyFVxPFmKog/s1600-h/baby-newyear.jpg"><img id="BLOGGER_PHOTO_ID_5014732185614719154" style="FLOAT: left; MARGIN: 0px 10px 10px 0px; CURSOR: hand" alt="" src="http://bp3.blogger.com/_sNG9dzZilv4/RZfoWodImLI/AAAAAAAAAAs/nyFVxPFmKog/s320/baby-newyear.jpg" border="0" /></a><br /><div>Happy New Year! I hope you all had a great holiday. Seniors, I hope all those essays were finished. Juniors, I hope you all enjoyed not having to write them (this year!). I spent the last week having a great time up in Oregon with my family, and if you thought it was cold here…<br /><br />You will notice when you log on to our blogs to creat a post that we have switched over to the “new Blogger.” It’s supposed to make things easier and run smoother (we’ll see!) but it will require each of you to switch over as well (and create a Google account) before you can edit or create new posts. Nothing serious, but I wanted to give you a heads up before your turn came around and you panicked!<br /><br />See you all on the 8th!</div>Math Maverickhttp://www.blogger.com/profile/10631468736601964037noreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-1166588074141371122006-12-19T20:13:00.000-08:002006-12-19T20:18:40.276-08:00Wednesday's Quiz TopicsHere’s a list of topics that will be covered on this Wednesday’s Quiz.<br /><br /><strong><em><u><span style="font-size:130%;color:#ff0000;">Quiz – Sections 5.3-4<br /></span></u></em></strong>Fundamental Theorem of Calculus, Part I (Sec. 5.3, #9,13,17)<br />Fundamental Theorem of Calculus, Part II (Sec. 5.3, #31,41)<br />Determine general indefinite integrals (+C!) (Sec. 5.4, #9)<br />Evaluate definite integrals (Sec. 5.4, #19,25,29)<br />Explain the meaning of a definite integral expression. (Sec. 5.4, #47,48)<br />Displacement vs. Total Distance Traveled (Sec. 5.4, #55)<br /><br />Oh, by the way, no calculators on this one...<br /><br />I’ll be in early Wednesday and I’ll check in tonight online. See you in class!<br /><br /><em><span style="color:#000099;">"Success is a peace of mind which is a direct result of... knowing that you did your best to become the best you are capable of becoming."<br />-John Wooden<br /><br /></span></em><em><span style="color:#000099;"></span></em>Math Maverickhttp://www.blogger.com/profile/10631468736601964037noreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-1166502459065519172006-12-18T20:12:00.001-08:002006-12-19T06:41:58.423-08:005.4 Indefinite Integrals and the Net Change Theorem<span style="font-weight: bold;font-family:arial;font-size:180%;" >INDEFINITE INTEGRALS</span><span style="font-weight: bold;"><br />An indefinite integral is basically the antiderivati</span><span style="font-weight: bold;">ve of the function. It doesn't have upper and lower bounds because that would make it a de</span><span style="font-weight: bold;">finite integral. Indefinite integrals need the +C!<br /></span><br /><span style="font-weight: bold;">Table of Indefinite Integrals</span><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/820322/10.gif"><img style="cursor: pointer; width: 248px; height: 37px;" src="http://photos1.blogger.com/x/blogger/1934/3814/400/288100/10.gif" alt="" border="0" /></a><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/67711/7.gif"><img style="cursor: pointer; width: 187px; height: 36px;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/444480/7.gif" alt="" border="0" /></a><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/40441/6.gif"><img style="cursor: pointer; width: 186px; height: 37px;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/605830/6.gif" alt="" border="0" /></a><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/904176/8.gif"><img style="cursor: pointer; width: 206px; height: 35px;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/312303/8.gif" alt="" border="0" /></a><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/127692/2.gif"><img style="cursor: pointer; width: 291px; height: 40px;" src="http://photos1.blogger.com/x/blogger/1934/3814/400/41001/2.gif" alt="" border="0" /></a><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/69589/3.gif"><img style="cursor: pointer; width: 155px; height: 39px;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/316675/3.gif" alt="" border="0" /></a><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/867559/4.gif"><img style="cursor: pointer; width: 160px; height: 42px;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/411329/4.gif" alt="" border="0" /></a><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/793359/5.gif"><img style="cursor: pointer; width: 217px; height: 40px;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/313618/5.gif" alt="" border="0" /></a><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/67871/1.gif"><img style="cursor: pointer; width: 165px; height: 33px;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/618052/1.gif" alt="" border="0" /></a><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/880093/12.gif"><img style="cursor: pointer; width: 216px; height: 42px;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/918917/12.gif" alt="" border="0" /></a><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/583699/13.gif"><img style="cursor: pointer; width: 207px; height: 39px;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/551669/13.gif" alt="" border="0" /></a><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/110636/9.gif"><img style="cursor: pointer; width: 249px; height: 40px;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/697136/9.gif" alt="" border="0" /></a><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/862417/11.gif"><img style="cursor: pointer; width: 121px; height: 37px;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/760791/11.gif" alt="" border="0" /></a><br /><br /><span style="font-size:130%;">Sample Problem<br />Find the general indefinite integral</span><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/818696/8.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/797151/8.jpg" alt="" border="0" /></a><br /><br /><br />Using the formula:<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/69589/3.gif"><img style="cursor: pointer; width: 155px; height: 39px;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/316675/3.gif" alt="" border="0" /></a><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/950578/9.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/804243/9.jpg" alt="" border="0" /></a><br /><span style="font-weight: bold;"><span style="font-size:180%;"><br /><br /></span></span><br /><span style="font-weight: bold;"><span style="font-size:180%;">THE NET CHANGE THEOREM<br /><br /><span style="font-size:100%;">The Integral of a rate of change is the </span></span></span><span style="font-weight: bold;font-size:100%;" >net change (displacement for position functions)</span><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/803676/netchangetheorem.gif"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 240px; height: 47px;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/150049/netchangetheorem.png" alt="" border="0" /></a><br /><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;"><br /><br /></span></span></span><br /><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;">Basically this </span></span></span><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;">theorem states that the integ</span></span></span><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;">ral of f or F' from a to b is the area between a and b or the difference in area f</span></span></span><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;">rom the postion of F(a) to F(b).<br /></span></span></span><br /><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;">This can be applied to things such as:<br />volume</span></span></span><br /><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;">concentration</span></span></span><br /><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;">density<br />population<br />cost</span></span></span><br /><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;">velocity<br /></span></span></span><br /><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;">So for a velocity function:</span></span></span><br /><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;">To calculate displacement we can u</span></span></span><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/903757/1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/67126/1.jpg" alt="" border="0" /></a><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;">se the</span></span></span><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;"> equation<br /></span></span></span><br /><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;"><br /></span></span></span><br /><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;">to calculate total distance traveled we can add the absolute values of the areas of each sector from each x int</span></span></span><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/blogger/1934/3814/1600/2.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/954107/2.jpg" alt="" border="0" /></a><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;">ercept to the next x intercept</span></span></span><br /><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;"><br /></span></span></span><br /><br /><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;"><span style="font-size:130%;">Sample Problem<br /></span></span></span></span><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;"><span style="font-size:130%;"><span style="font-size:100%;">A particle moves along a line so that its </span></span></span></span></span><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;"><span style="font-size:130%;"><span style="font-size:100%;">velocity at time t is</span></span></span></span></span><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;"><span style="font-size:130%;"><span style="font-size:100%;"><br /></span></span></span></span></span><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/28530/4.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/304164/4.jpg" alt="" border="0" /></a><br />(m/s)<br /><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;"><span style="font-size:130%;"><span style="font-size:100%;"><br />a) find the displacement from t=[1,4]<br />b) find the distance traveled during that time period<br /></span></span></span></span></span><br /><span style="font-weight: bold;"><span style="font-size:180%;"><span style="font-size:100%;"><span style="font-size:130%;"><span style="font-size:100%;">Finding the displacement:</span></span></span></span></span><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/539431/3.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/147051/3.jpg" alt="" border="0" /></a><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/115450/7.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/110400/7.jpg" alt="" border="0" /></a><br /><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/225747/5.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/33297/5.jpg" alt="" border="0" /></a><br />m.<br /><br /><br /><span style="font-weight: bold;">Finding the total distance traveled during that time period<br /><br /></span><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/660318/6.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/514548/6.jpg" alt="" border="0" /></a><br />m.<br /><br /><br />The total distance traveled and the displacement are the same because the position function does not pass below the x axis therefore there are no negative areas. If there were negative areas the displacement would be a smaller number and the distance would stay the same.<br /><br />Some LINKS:<br />http://www.coolschool.ca/lor/CALC12/unit5/U05L04.htm<br /><br /><a href="http://prepcalcabb0607.blogspot.com/">A Lesser Lesson in Indefinite Integrals but helpful nontheless</a><br /><span style="font-size:130%;"><br /></span><span style="color: rgb(255, 0, 0);"><span style="font-weight: bold;"><span style="font-size:130%;">Magnus You're Up NeXT!!!<br /></span><br /></span></span><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/x/blogger/1934/3814/1600/629510/10.gif"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://photos1.blogger.com/x/blogger/1934/3814/320/802392/10.png" alt="" border="0" /></a>brianhttp://www.blogger.com/profile/11529058779388570553noreply@blogger.com1tag:blogger.com,1999:blog-33767428.post-1166056158052978962006-12-13T16:06:00.000-08:002006-12-15T08:36:39.600-08:005.3 The Fundamental Theorem of Calculus Part II<strong>Fundamental Theorem of Calculus Part II<br /></strong><br />If f is a continuous function on [a,b] then<br /><a href="http://photos1.blogger.com/x/blogger/6629/3811/1600/658619/1.gif"><img style="CURSOR: hand" alt="" src="http://photos1.blogger.com/x/blogger/6629/3811/400/545080/1.gif" border="0" /></a><br />where F is any antiderivative of f, that is, a function such that F'=f.<br /><br />Once you find the antiderivative of f(x), you evaluate the end points from a to b and then subtract the antiderivative function.<br /><br />Lets look at an example.<br /><a href="http://photos1.blogger.com/x/blogger/6629/3811/1600/48644/2.gif"><img style="CURSOR: hand" alt="" src="http://photos1.blogger.com/x/blogger/6629/3811/400/6259/2.gif" border="0" /></a><br />Once you find the antiderivative, you plug the top number 5 into x and subtract that antiderivative function from an antiderivative function with the number 3 in x.<br /><br /><a href="http://photos1.blogger.com/x/blogger/6629/3811/1600/201673/3.gif"><img style="CURSOR: hand" alt="" src="http://photos1.blogger.com/x/blogger/6629/3811/400/864384/3.gif" border="0" /></a><br /><br />You can check your answer by using your calculator.<br />Plug <a href="http://photos1.blogger.com/x/blogger/6629/3811/1600/768453/4.gif"></a><a href="http://photos1.blogger.com/x/blogger/6629/3811/1600/155305/4.gif"><img style="CURSOR: hand" alt="" src="http://photos1.blogger.com/x/blogger/6629/3811/400/652342/4.gif" border="0" /></a>into your calculator.<br />Go back to the home page. Click Math then 9.<br />Once fnInt pops up, put fnInt( <a href="http://photos1.blogger.com/x/blogger/6629/3811/1600/381357/5.gif"><img style="CURSOR: hand" alt="" src="http://photos1.blogger.com/x/blogger/6629/3811/400/111312/5.gif" border="0" /></a>,X,3,5). You will get 98/3<br /><br /><br />Try this problem!<br /><a href="http://photos1.blogger.com/x/blogger/6629/3811/1600/290289/6.gif"><img style="FLOAT: left; MARGIN: 0px 10px 10px 0px; CURSOR: hand" alt="" src="http://photos1.blogger.com/x/blogger/6629/3811/320/172114/6.gif" border="0" /></a><br /><br /><br /><br />You will see that this antiderivative is:<br /><a href="http://photos1.blogger.com/x/blogger/6629/3811/1600/918060/7.gif"><img style="CURSOR: hand" alt="" src="http://photos1.blogger.com/x/blogger/6629/3811/400/159690/7.gif" border="0" /></a><br /><br />Thus you plug in 4 into the x and then subtract the antiderivative with 2 in the x value.<br /><a href="http://photos1.blogger.com/x/blogger/6629/3811/1600/311132/8.gif"><img style="CURSOR: hand" alt="" src="http://photos1.blogger.com/x/blogger/6629/3811/400/385037/8.gif" border="0" /></a><br /><br /><br />This a reminder to Brian to do the next blog!<br /><br />This website can help out with this concept:<br /><a href="http://math.ucsd.edu/~wgarner/math10b/ftc.htm">http://math.ucsd.edu/~wgarner/math10b/ftc.htm</a><br /><br /><a href="http://photos1.blogger.com/x/blogger/6629/3811/1600/893375/Dixie%20Chicks.jpg"><img style="CURSOR: hand" alt="" src="http://photos1.blogger.com/x/blogger/6629/3811/400/627131/Dixie%20Chicks.jpg" border="0" /></a><br /><strong>Dixie Chicks</strong>amihttp://www.blogger.com/profile/02731237501764087842noreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-1166034622040516552006-12-13T10:27:00.000-08:002006-12-13T10:30:22.060-08:00Thursday's Quiz TopicsHere’s a list of topics that will be covered on this Thursday’s Quiz. <br /><br /><strong><em><u><span style="color:#ff0000;">Quiz – Sections 5.1-2<br /></span></u></em></strong>Estimate distance traveled from a velocity graph. (5.1, #15)<br />Express a Riemann sum as a definite integral. (5.2, #17,19)<br />Evaluate an integral in terms of areas – show your work! (5.2 #37)<br />Sketch a graph and estimate the area under the curve using RRAM, LRAM or MRAM (5.1, #3)<br />Evaluate definite integrals based on a graph (5.2, #33)<br /><br />I’ll be in early Thursday, available after school this afternoon until 4, and I’ll check in tonight online. See you in class!<br /><br />In the spirit of the holidays:<br /><br /><span style="font-size:130%;color:#000099;"><strong><em>Yes, Virginia, There is a Santa Claus<br /></em></strong></span><span style="font-size:85%;">By Francis P. Church, first published in The New York Sun in 1897. [See The People’s Almanac, pp. 1358–9.]<br /></span>We take pleasure in answering thus prominently the communication below, expressing at the same time our great gratification that its faithful author is numbered among the friends of The Sun:<br /><br />Dear Editor—<br />I am 8 years old. Some of my little friends say there is no Santa Claus. Papa says, “If you see it in The Sun, it’s so.” Please tell me the truth, is there a Santa Claus?<br />Virginia O’Hanlon<br /><br />Virginia, your little friends are wrong. They have been affected by the skepticism of a skeptical age. They do not believe except they see. They think that nothing can be which is not comprehensible by their little minds. All minds, Virginia, whether they be men’s or children’s, are little. In this great universe of ours, man is a mere insect, an ant, in his intellect as compared with the boundless world about him, as measured by the intelligence capable of grasping the whole of truth and knowledge.<br /><br />Yes, Virginia, there is a Santa Claus. He exists as certainly as love and generosity and devotion exist, and you know that they abound and give to your life its highest beauty and joy. Alas! how dreary would be the world if there were no Santa Claus! It would be as dreary as if there were no Virginias. There would be no childlike faith then, no poetry, no romance to make tolerable this existence. We should have no enjoyment, except in sense and sight. The external light with which childhood fills the world would be extinguished.<br /><br />Not believe in Santa Claus! You might as well not believe in fairies. You might get your papa to hire men to watch in all the chimneys on Christmas eve to catch Santa Claus, but even if you did not see Santa Claus coming down, what would that prove? Nobody sees Santa Claus, but that is no sign that there is no Santa Claus. The most real things in the world are those that neither children nor men can see. Did you ever see fairies dancing on the lawn? Of course not, but that’s no proof that they are not there. Nobody can conceive or imagine all the wonders there are unseen and unseeable in the world.<br /><br />You tear apart the baby’s rattle and see what makes the noise inside, but there is a veil covering the unseen world which not the strongest man, nor even the united strength of all the strongest men that ever lived could tear apart. Only faith, poetry, love, romance, can push aside that curtain and view and picture the supernal beauty and glory beyond. Is it all real? Ah, Virginia, in all this world there is nothing else real and abiding.<br /><br />No Santa Claus! Thank God! he lives and lives forever. A thousand years from now, Virginia, nay 10 times 10,000 years from now, he will continue to make glad the heart of childhood.<br /><br /><em><strong>About the Exchange<br /></strong></em>Francis P. Church’s editorial, “Yes Virginia, There is a Santa Claus” was an immediate sensation, and went on to became one of the most famous editorials ever written. It first appeared in the The New York Sun in 1897, almost a hundred years ago, and was reprinted annually until 1949 when the paper went out of business.<br /><br />Thirty-six years after her letter was printed, Virginia O’Hanlon recalled the events that prompted her letter:<br />“Quite naturally I believed in Santa Claus, for he had never disappointed me. But when less fortunate little boys and girls said there wasn’t any Santa Claus, I was filled with doubts. I asked my father, and he was a little evasive on the subject.<br />“It was a habit in our family that whenever any doubts came up as to how to pronounce a word or some question of historical fact was in doubt, we wrote to the Question and Answer column in The Sun. Father would always say, ‘If you see it in the The Sun, it’s so,’ and that settled the matter.<br />“ ‘Well, I’m just going to write The Sun and find out the real truth,’ I said to father.<br />“He said, ‘Go ahead, Virginia. I’m sure The Sun will give you the right answer, as it always does.’ ”<br />And so Virginia sat down and wrote her parents’ favorite newspaper.<br />Her letter found its way into the hands of a veteran editor, Francis P. Church. Son of a Baptist minister, Church had covered the Civil War for The New York Times and had worked on the The New York Sun for 20 years, more recently as an anonymous editorial writer. Church, a sardonic man, had for his personal motto, “Endeavour to clear your mind of cant.” When controversal subjects had to be tackled on the editorial page, especially those dealing with theology, the assignments were usually given to Church.<br />Now, he had in his hands a little girl’s letter on a most controversial matter, and he was burdened with the responsibility of answering it.<br />“Is there a Santa Claus?” the childish scrawl in the letter asked. At once, Church knew that there was no avoiding the question. He must answer, and he must answer truthfully. And so he turned to his desk, and he began his reply which was to become one of the most memorable editorials in newspaper history.<br />Church married shortly after the editorial appeared. He died in April, 1906, leaving no children.<br />Virginia O’Hanlon went on to graduate from Hunter College with a Bachelor of Arts degree at age 21. The following year she received her Master’s from Columbia, and in 1912 she began teaching in the New York City school system, later becoming a principal. After 47 years, she retired as an educator. Throughout her life she received a steady stream of mail about her Santa Claus letter, and to each reply she attached an attractive printed copy of the Church editorial. Virginia O’Hanlon Douglas died on May 13, 1971, at the age of 81, in a nursing home in Valatie, N.Y.Math Maverickhttp://www.blogger.com/profile/10631468736601964037noreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-1165987303229942862006-12-12T19:49:00.000-08:002006-12-17T23:50:06.326-08:005.3 The Fundamental Theorem of CalculusHey guys this is Izzy. Today in class we learned about the fundamental theorem of calculus. This theorem establishes a connection between the two branches of calculus: differential calculus and integral calculus. In class we covered how if you multiply the function that you are taking the integral of by a certain factor, that you can factor out that number. Example:<br /><a href="http://photos1.blogger.com/x/blogger/7311/3807/1600/153765/23.jpg"><img style="CURSOR: hand" alt="" src="http://photos1.blogger.com/x/blogger/7311/3807/320/738401/23.jpg" border="0" /></a><br />This makes it easier to deal with the original funtion first, without any coefficients to clutter up the process.<br /><br />In class, we saw how the integral of a function can be a function itself, represented by g(x). Expressing the integral of a function as the function g(x) allows one to actually <u>graph</u> and express the area between the curve of a graph and the x-axis in terms of x:<br /><a href="http://photos1.blogger.com/x/blogger/7311/3807/1600/151253/ddd.jpg"><img style="CURSOR: hand" alt="" src="http://photos1.blogger.com/x/blogger/7311/3807/320/486087/ddd.jpg" border="0" /></a><br />In this graph,<br /><a href="http://photos1.blogger.com/x/blogger/7311/3807/1600/915895/24.jpg"><img style="CURSOR: hand" alt="" src="http://photos1.blogger.com/x/blogger/7311/3807/320/646984/24.jpg" border="0" /></a><br />By using this equation, we can find the area between the graph and the x-axis. Let's say we wanted to find the area between x=0 and x=3. You would multiply the change in x, 3, by f(3), 3, and .5, because it's a triangle. So, g(3)=(3)(3)(.5)=4.5 .<br /><br /><br /><br />Today we also made the amazing discovery that the antiderivative of a function is the same as the integral of the function. Observe:<br /><a href="http://photos1.blogger.com/x/blogger/7311/3807/1600/435662/reallycool.jpg"></a><a href="http://photos1.blogger.com/x/blogger/7311/3807/1600/685311/reallycool.jpg"><img style="CURSOR: hand" alt="" src="http://photos1.blogger.com/x/blogger/7311/3807/320/387274/reallycool.jpg" border="0" /></a><br /><br />As you can see from the graph, the area of the shaded region (teal) is:<br /><a href="http://photos1.blogger.com/x/blogger/7311/3807/1600/311913/1.jpg"><img style="FLOAT: left; MARGIN: 0px 10px 10px 0px; CURSOR: hand" alt="" src="http://photos1.blogger.com/x/blogger/7311/3807/320/947777/1.jpg" border="0" /></a><br /><br /><br />We established that<br /><a href="http://photos1.blogger.com/x/blogger/7311/3807/1600/547054/24.jpg"><img style="CURSOR: hand" alt="" src="http://photos1.blogger.com/x/blogger/7311/3807/320/239958/24.jpg" border="0" /></a><br />To find g'(x), we have to replace t with x and multiply whatever you get by the derivative of x.<br /><a href="http://photos1.blogger.com/x/blogger/7311/3807/1600/554075/vvv.jpg"><img style="CURSOR: hand" alt="" src="http://photos1.blogger.com/x/blogger/7311/3807/320/514741/vvv.jpg" border="0" /></a><br />Wait a minute, the derivative of the Area function is also x! Thus A'=g'(x). We just proved that the antiderivative of a function is the same as its integral! i'm so happy.<br /><br />What we just did relates to the Fundamental theorem of Calculus, Part 1, which states that:<br />If the function f is continuous on [a,b], then the function g defined by<br /><a href="http://photos1.blogger.com/x/blogger/7311/3807/1600/325193/ddddd.jpg"><img style="CURSOR: hand" alt="" src="http://photos1.blogger.com/x/blogger/7311/3807/320/135241/ddddd.jpg" border="0" /></a> <a href="http://photos1.blogger.com/x/blogger/7311/3807/1600/931981/sdsd.jpg"><img style="CURSOR: hand" alt="" src="http://photos1.blogger.com/x/blogger/7311/3807/320/160664/sdsd.jpg" border="0" /></a><br />is continuous on [a,b] and differentiable on (a,b), and g'(x)=f(x)<br /><br />MOVING ON, we now know how to find the derivative of these types of functions. Let's try this problem: Find <a href="http://photos1.blogger.com/x/blogger/7311/3807/1600/685309/gfdur.jpg"><img style="CURSOR: hand" alt="" src="http://photos1.blogger.com/x/blogger/7311/3807/320/672178/gfdur.jpg" border="0" /></a><br /><br />Whenever you see these types of problems, you must first look at the upper bound, or any of the bounds that has a variable in it. In the process of calculating the equation for g'(x), you must replace the variable t with the variable bound, and wherever you see dt, replace it with the derivative of the variable bound. In this case, the upper bound has the variable: x^2.<br />To solve this problem, we have to use the chain rule (make u=x^2):<br /><br /><a href="http://photos1.blogger.com/x/blogger/1189/2644/1600/457553/solution.jpg"><img style="CURSOR: hand" alt="" src="http://photos1.blogger.com/x/blogger/1189/2644/320/668337/solution.jpg" border="0" /></a><br />There you have it.<br /><br /><br />This a reminder to Ami to do the next blog!<br /><br />These are a few websites that help out with this concept:<br /><br /><a href="http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/integration/ftc.html">http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/integration/ftc.html</a><br /><a href="http://archives.math.utk.edu/visual.calculus/4/ftc.9/">http://archives.math.utk.edu/visual.calculus/4/ftc.9/</a><br /><br />Here are some cool math jokes<br />Q: What do you get if you divide the cirucmference of a jack-o-lantern by its diameter?<br />A: Pumpkin Pi!<br /><br />Q: Why do you rarely find mathematicians spending time at the beach? A: Because they have sine and cosine to get a tan and don't need the sun!<br /><br />A mathematician is flying non-stop from Edmonton to Frankfurt with AirTransat. The scheduled flying time is nine hours. Some time after taking off, the pilot announces that one engine had to be turned off due to mechanical failure: "Don't worry - we're safe. The only noticeable effect this will have for us is that our total flying time will be ten hours instead of nine." A few hours into the flight, the pilot informs the passengers that another engine had to be turned off due to mechanical failure: "But don't worry - we're still safe. Only our flying time will go up to twelve hours." Some time later, a third engine fails and has to be turned off. But the pilot reassures the passengers: "Don't worry - even with one engine, we're still perfectly safe. It just means that it will take sixteen hours total for this plane to arrive in Frankfurt." The mathematician remarks to his fellow passengers: "If the last engine breaks down, too, then we'll be in the air for twenty-four hours altogether!"<br /><br />A math student is pestered by a classmate who wants to copy his homework assignment. The student hesitates, not only because he thinks it's wrong, but also because he doesn't want to be sanctioned for aiding and abetting. His classmate calms him down: "Nobody will be able to trace my homework to you: I'll be changing the names of all the constants and variables: a to b, x to y, and so on." Not quite convinced, but eager to be left alone, the student hands his completed assignment to the classmate for copying. After the deadline, the student asks: "Did you really change the names of all the variables?" "Sure!" the classmate replies. "When you called a function f, I called it g; when you called a variable x, I renamed it to y; and when you were writing about the log of x+1, I called it the timber of x+1..."Ismaelhttp://www.blogger.com/profile/08690241157327789431noreply@blogger.com0tag:blogger.com,1999:blog-33767428.post-1165866088162303752006-12-11T11:03:00.000-08:002006-12-12T13:49:30.990-08:005.2 The Definite IntegralSo there's some notation for the Riemann sum that we did yesterday. If we take, for example, the area under the curve of y=x^2 from x=1 to x=3 like this:<br /><br /><img style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 299px; CURSOR: hand; HEIGHT: 228px; TEXT-ALIGN: center" height="125" alt="" src="http://photos1.blogger.com/x/blogger/2120/3805/320/153154/blog%201.jpg" width="205" border="0" /><br /><p>and we want to have 2 rectangles with the sample points being left endpoints, then we would assign some notation. 1 would become x1, and 2 would be x2. To put this in notation we would use a capital sigma to symbolize sum. Underneath the sigma would be the starting point and over the sigma will be the ending point, both using <em>i</em> as the variable. The sum would be the sum of the rectangles, so it would be the height times the width (change in x). Put all this together and the equation would look like this:<img style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://photos1.blogger.com/x/blogger/2120/3805/320/93857/blog%20e%201.jpg" border="0" /></p><br /><p>To find the answer to this problem, we would plug in 1 for <em>i</em> then two, and add them. So it would be: f(x1)(x2-x1) + f(x2)(x3-x2) = f(1)(2-1) + f(2)(3-2) = (1)(1) + (4)(1) = 5.</p><p>We know from 5.1 that as you have more and more rectangles, the estimate of the area becomes more and more accurate. So ideally we would like infinitely many rectangles. We can write this as :</p><p><img style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://photos1.blogger.com/x/blogger/2120/3805/320/260320/blog%20e%202.jpg" border="0" /></p><br /><p>and to simplify this we use the integral symbol, and it would be :</p><p><img style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://photos1.blogger.com/x/blogger/2120/3805/320/132983/blog%20e%203.jpg" border="0" /></p><br /><p>Because as we get more and more boxes and <em>n</em> approaches infinity, the change in x gets closer and closer to zero so eventually the change in x becomes the derivative at x. </p><p>Take this graph and equation for example:</p><p><img style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://photos1.blogger.com/x/blogger/2120/3805/320/374827/blog%202.jpg" border="0" /></p><br /><p>Let's find the area underneath the curve, but above the x axis. </p><p>Both the equation and the graph suggest that it's a semi-circle with a radius of 5. We could :</p><p><img style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://photos1.blogger.com/x/blogger/2120/3805/320/99268/blog%20e%204.jpg" border="0" /></p><br />But common sense tells us that since we want want the area under the graph of a semi-circle, then we want the area of a semi-circle. So the area would be <img style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://photos1.blogger.com/x/blogger/2120/3805/320/250266/blog%203.jpg" border="0" /><br /><p>This simple way of looking at the problem can be applied in different ways. Let's suppose we wanted to find the area between the curve and the x-axis if f(x) was 2x from x=0 to x=4 and f(x)=8 from x=4 to x=8 and f(x)=3x-16 from 8 to 10, then you could split that graph into two triangles and one rectangle and work out the area from there. </p><p><img style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://photos1.blogger.com/x/blogger/2120/3805/320/666272/blog%204.jpg" border="0" /></p><br /><p>The area would be (1/2)(4)(8) + (6)(8) + (1/2)(6)(2) = 70.</p><p>Another topic covered was the Mid-point estimate. Using the mid-point estimate balances out the over and under estimation, bringing us closer to the actual area underneath the graph.</p><p><a href="http://photos1.blogger.com/x/blogger/2120/3805/1600/883929/blog%207.jpg"><img style="FLOAT: right; MARGIN: 0px 0px 10px 10px; WIDTH: 190px; CURSOR: hand; HEIGHT: 156px" height="209" alt="" src="http://photos1.blogger.com/x/blogger/2120/3805/320/897359/blog%207.jpg" width="253" border="0" /></a><a href="http://photos1.blogger.com/x/blogger/2120/3805/1600/94054/blog%205.jpg"><img style="FLOAT: right; MARGIN: 0px 0px 10px 10px; WIDTH: 211px; CURSOR: hand; HEIGHT: 161px" height="201" alt="" src="http://photos1.blogger.com/x/blogger/2120/3805/320/790353/blog%205.jpg" width="64" border="0" /></a><img style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 171px; CURSOR: hand; HEIGHT: 158px; TEXT-ALIGN: center" height="204" alt="" src="http://photos1.blogger.com/x/blogger/2120/3805/320/631224/blog%206.jpg" width="191" border="0" /></p><br /><p>Using the left side makes an overestimation, but using the right side is an underestimation. Using the midpoint, however, balances out the over and underestimation, making it about right.</p><p>There are some properties of integrals as well.</p><p>1.<img style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://photos1.blogger.com/x/blogger/2120/3805/320/867134/blog%20e%205.jpg" border="0" /></p><br /><p>2. If f(x)=c=7, and a=3 and b=5, then:<img style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://photos1.blogger.com/x/blogger/2120/3805/320/621195/blog%20e%206.jpg" border="0" /></p><p>And:</p><p><img style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://photos1.blogger.com/x/blogger/2120/3805/320/4043/blog%20e%207.jpg" border="0" /></p><br /><p>Therefore:</p><p><img style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://photos1.blogger.com/x/blogger/2120/3805/320/189695/blog%20e%208.jpg" border="0" /></p><br />3.<img style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://photos1.blogger.com/x/blogger/2120/3805/320/211079/blog%20e%209.jpg" border="0" /><br /><p>That's it!</p><p>Here's a couple of cute calculus pick-up lines:</p><p>1. "Hey, if I was sin^2x and you were cos^2x, together we would be one!"</p><p>2. "I wish I was your derivative so that I could lie tangent to your curves."</p><p>Here's a website about the properties of integrals:</p><p><a href="http://www.analyzemath.com/calculus/Integrals/integral_properties.html">www.analyzemath.com/calculus/Integrals/integral_properties.html</a>.</p><p>Ismael, you're up next!</p>Clairehttp://www.blogger.com/profile/02680938518615745163noreply@blogger.com0