Scribe Post 2/6

    The past couple days have been treacherous in AP Calc as Mr. O’Brien ditched us in our time of need. We were left to fend for ourselves in the land of integrals and average values and when he finally returned he did not waste a second to throw yet another new lesson at us and the news of a quiz on this Friday, February 8th, on IW#1-8 and a test next Thursday, Happy Valentines Day from O’B to all his suffering calc students! YAY!
    As class today started terror struck as O’B discovered he DESPERATELY needs a hair cut *GASP* and we didn’t have a scribe for the day *AH*

grumpy O'B

but, when all was in despair O’B embraced his shaggy dog, yet still product filled hair, Hayley found his look alike, and I volunteered to be scribe. Oh joy, here we go . . .

happy O'B!

O'B's look alike, I'm pretty sure this is Joseph from one of those biblical movies...

Before I begin on what we actually did in class I’m going to post some things that will be helpful for the quiz on Friday if O’B let’s us use the scribe posts for notes. Here are some good rules and equations to know:

RULES FOR DEFINITE INTEGRALS:

1. Order of Integration: 

2. Zero: 

3. Constant multiple:

4. Sum and Difference:

5. Additivity:

6. Max-Min Inequality: if max-f and min-f are the maximum and minimum values of f on [a,b] then: 

7. Domination:

The AVERAGE (MEAN) VALUE:

 If f is integrable on [a,b], it’s average (mean) value on [a,b] is:

Onward to the beginning to class! We started class off with and ended up spending most on questions 1-4 on page 297 of our textbooks. These four problems are also part of IW#8 so if you don’t have the work and/or answers down already I suggest you do that!

 1.

In question one we discussed how the f(x) in an integral is the height of the rectangles made in the Riemann Sum and the dx, or the change in x, is the base. This means that the functions multiplied by the change in x (base X height) equals a+2b. Now if we add 3 to the height, f(x)+3, we’re shifting the whole function up 3 and adding a rectangle to the area underneath the curve. This added rectangle has a height of 3 and a base of the change of x which in this case is b – a. 

graph showing the added rectangle from changing f(x) to f(x)+3

math using Sum and Difference Definite Integral Rule to solve

 FINAL ANSWER: (D) 5b-2a

2. 

 

    In this question we had to take a similar approach as we did in number 1. When looking at the given expression and thinking about the previous worksheets and questions we’ve done with the Riemann Sum we can see that the “1/20” is the base of all the rectangles or the change in x and all the numbers in the parentheses are the changing heights. Identifying the base of the rectangles, or the change in x, should be one of the first things you do for Riemann Sum.

    When thinking of what we just found out in an integral form that means the 1/20 becomes dx and all the different heights become f(x). There’s nothing out in front of the integral, everything in the expression has now been accounted for, so we can cross out all answer except (A) and (B). We know that the integral is going from 0 to 1 because the square rot of 20/20 = 1, so know we just have to decide if f(x) is the equation in (A) or (B). But, if we know the integral is only going from 0 to 1, which are the x-values, and if you plug in from 0 to 1 as x in the equation in (A) you only end up with √(1/20) not √(20/20). With the equation in (B) we can get √(20/20) so we know know that’s our answer because with this equation the change in x is 1/20 not 1.

graph of Riemann Sum of function

 FINAL ANSWER: (B)

3. 

to solve use zero rule (see above)

 FINAL ANSWER: (C) 2

       Going off this equation we looked at the integral:

 

 and O’B asked why this integral is negative. After a short discussion we realized it’s because the integral is going backwards from 2 to -2 so it’s a negative change in x.

 Then we looked at:

and O’B asked why, if the last integral was negative, is this one positive. This is because it’s a negative height, for being below the x-axis, multiplied by a negative base, for going backwards from 2 to -2, which gives you a positive number.  

4. 

a.) 

b.) 

       from the equation given above we know that:

so when using the given interval and the equation we found in part a we get:

then we can use our calculators and plug in fnINT[f,x,a,b] and get an answer of -6. We then must multiply by1/2 so the final answer is -3.

BUT what if we can’t use our calculators O’Brien asks. Well, then we’d use The Fundamental Theorem of Calculus.

Let’s put the FUN in Calculus! The FUNdamental Theorem of Calculus! WOO!

But before we actually get into the two parts of the theorem let’s do a geogebra exercise to give us a visual of the Fundamental Theorem.

The end result should look like this:

 The Fundamental Theorem of Calculus, Part 1:

If:

        

          then:

Basically, the derivative of an integral is the function (f(x)). This is because taking the derivative of a function and taking the integral of a function are like opposites, they are inverses so they undo each other. This process is better explained my PatrickJMT in this video.

The Fundamental Theorem of Calculus, Part 2:

EXAMPLE:

Find the antiderivative of sinx: -cosx; and find the value for the antiderivative at the given interval:

Don't forget! Quiz next class on IW#1-8, IW#4-8 are also due on Friday! We will be having a text on Thursday (Valentines Day) so study up!

IW#9 is on:

Pg. 297/ Q. 1-4 (answers above)

Pg. 295/ Q. 19, 21, 25, 27, 33, 51

Pg. 206/ Q. 5, 9, 13, 23, 29, 45, 57, 61, 67

Good luck everyone!

P.S. I apologize for any spelling or grammatical errors, or just errors in general

Scribe Post 12/5

The big thing in this scribe post is RELATED RATES, if that’s what you’re looking for, here it is!   

    We started off class with the question, “The side of a square is increasing at 2 cm/sec. How fast is the area increasing when the side is 4 cm long?” This brilliant question by O’Brien given with a wondrous animation on GeoGebra stumped our class. We kept asking questions like “at what time does the area start increasing? 0 seconds?” and “can’t you just tell us how to do it?” After a while of working on it our class decided on our answer: 8 cm^2/sec, which was wrong.
    When O’Brien started actually helping us with this confusingly simple problem he threw our question right back at us: “Does it really matter at what number of seconds you start? 0 seconds? 1 seconds? 16 seconds?” He used a table (shown below) to prove that no, it doesn’t matter, no matter time you start at the volume is increasing at a steady rate.

We further proved this by looking at the rate of change at a moment in time instead of over 2 seconds. We did this by again, using a table (shown below) and looking at the times 1 second and 1.0001 seconds.

We noticed that for  the change between the two times and between the two areas were both reeally close to zero, so we thought zero/zero = zero right? Or undefined? Then we thought, but wait, these two numbers aren’t equal to zero, they’re both just really close which means we’ll actually get a value here:

Once we did this we realized we just found the derivative by using the limit in indeterminate form. The derivative for this particular problem at this moment in time is about 16.0004.
    After all this work O’Brien got his evil smile on to let us know there’s actually a much easier way to do all this. Thanks for letting us know before hand O’Brien. To use this easier way, we must first find the equation to this problem. By looking at our tables and the question asked in this problem it’s easy to see that our equation must be:

To solve for the rate of the area we must differentiate this equation with respect to t (time) (basically implicit differentiation):

Once again we proved that the rate or derivative at a moment in time is 16 cm^2/sec (above when we found the rate at a moment in time is was about 16). Important things to notice in the math above is that once we found the equation and differentiated it we plugged in the values we knew for s and ds/dt which were given in the original problem. It said “the side of a square is increasing at 2 cm/sec” which is ds/dt and that we are to solve for “when the side is 4 cm long” which is s.

Simple enough right?

The basic idea behind related rates that I’ve found O’Brien’s teachings and most of the sites and videos I’ve looked at to agree with is that each variable is now a function of time, or t, just like we did above. That means when we take the derivatives, we won’t be taking them in respect to each other like we used to, but instead with respect to t. This is important to know because some of the variables will not be changing with respect to t. These variables are like constants, and the derivative of them with respect to t is 0. When I say the derivative “with respect to t” I just mean that when you take the derivative of an equation you must multiple the variables in both sides by d/dt, for an example you can see we did this in the problem above.

And seeing as the name of the game is “related rates” it would be important to know if you didn’t already that something per something else, for instance cm/sec, would be a rate. So in an equation, whenever you see anything like that, it’s a rate and most likely important.



On the back of the exploration the Steps for every related rates problem: were given and I’ve found that almost every site or video on related rates uses basically the same steps so take note! They are important and helpful!
1. Label given information (Sketch where appropriate)
2. Note which quantities are variable, and which are constant
3. Identify rates d( )/dt which are given, and which are needed
4. Differentiate
5. Plug in date - ALWAYS AFTER YOU DIFFERENTIATE

The “beautiful man” (as Scotty would put it) Patrick JMT put it another way which I found helpful, he used an acronym: DREDS
Diagram
Rates
Equation
Derivative
Substitute specific info

He did that in this video, the rest was also helpful

Also for future reference Patrick JMT actually has a website and he has videos on there for every math concept imaginable so if you’re ever stuck look here! and *GASP* you can actually kind of see what he looks like!

And then while looking for other sites and videos on Related Rates which I thought would be helpful I’ve found this is actually quite the enjoyable experience because the people who make youtube videos on math problems are all so...interesting especially since most, unlike Patrick JMT who you only shows he’s hand in the videos, actually have their whole body in it. So there’s the expected nerdy, wonderfully awkward people trying to explain a math concept in a youtube video and then there’s this guy, Dr. Bob, who looks like he could be a body builder or a wrestling coach or something and is instead, on youtube being a math nerd, who, and I quote, said that he has a method for related rates problems that will “let us tear them down from start to finish.” Enjoy.

I hope I did a decent job explaining Related Rates and you all enjoyed my scribe post! The next scribe will be Sarah Mayberry, good luck!

9/20 Scribe Post

Okay guys, time to start the painful and slightly awkward scribe post seeing as I won't be in your class when you're criticizing it. Also I apologize in advance for any errors, especially grammatical errors, which I guarantee, there will be some.  

Starting off class today I walked in late with a plate full of pumpkin chocolate chip cookies and suddenly, everyone was my best friend. Even O'B said he'd give me extra credit, and yes, I'm going to hold you to that O'Brien. Then we started our warm-up which can be found here, along with IW#7, #7's solutions, IW#8, and #8's solutions. We worked on the warm-up either alone or with a partner for a while and while we were doing this both Francie and Noah also brought in food. It's easy to say we were a happy class today. Francie brought in quinoa burgers and offered one to everyone, and even asked O'Brien directly if he wanted one but he shut her down. That was kind of harsh of him. And then not too much later Noah walked in with a plate full of peanut butter chocolate chip cookies and offered some to everyone as well. By this time we had already spent about 30 minutes of class working on our warm-up and chowing down on food.

So now onto some real math (instead of food):

Here's a quick summary of everything we did in class today:
1. Warm-up on continuous and discontinuous function
2. went over quiz #3, were reminded we have a test on Monday, and IW#8 was handed out
3. went over the questions in IW#7 we didn't get to on Wednesday
4. took notes and learned new ways to find the derivative of a function

Starting with the warm-up . . .

1. Plot the graph of f for k=1. 

The discontinuity seen at x=2 is a jump discontinuity, or a non-removable discontinuity when k=1 (definitions of both of these can be found in Lexi's scribe post here.) As we learned earlier this year, for a function to be continuous at a point not only must the right-hand limit and left-hand limit be the same to ensure that the limit does, in fact, exist, but the limit and the y-value, in this instance f(2), must also be the same.


Due to this discontinuity we then looked at the left- and right-hand limits as x approaches 2. But when we're looking at when x approaches 2 positively we must look at the limit in terms of k:

 

*NOTE: both of the above equations should have an "f(x)" before the equals sign. My bad*

For us to make f(x) a continuous function we must make the left- and right-hand limits the same. To do so we have to find the value of k and to do that we can set the two limits equal to each other.
 

Now that we know that for this function to be continuous k=1/3 we can sketch the new graph:

with this new graph we can find a single limit as x approaches 2 and by looking at the graph we can see:

We can also see that the new graph has a cusp at x=2. A cusp, for those of you that don't know, comes from the latin cuspis which means "a point or apex," and cusp's mathematical definition is: a point where two branches of a curve meet, end, and are tangent. By looking at the graph at the point (2,3) we can see that cusp is a very fitting word to use in this context. Although this cusp means that this function is still not locally linear at x=2.


Once we finished defining and talking about what a fine word cusp is we learned a couple helpful GeoGebra keyboard shortcuts, such as:
- command - drag to zoom in on a specified area of your graph
and - command "m" to return you to the standard view

We then moved onto talking about and reviewing the quiz. 

The big thing we learned  today while going over the quiz was all about this wonderfully useful limit called the Golden Limit:

and how to use our knowledge of it to solve problems like 2 through 4 in our quiz. 

In question #1 we mainly had to rely on our knowledge of factoring polynomials, if you could do that than you would be able to find a limit of 8.

Questions 2-4 got a bit trickier as we had to use our knowledge of the golden limit. In question #2 we had to find the limit of:

First we tried just plugging 0 in as x in our equation and we ended up with 0/0 which means this is an indeterminate function and we therefore can use L'Hospital's Rule. L'Hospital's Rule is explained here, along with how the golden rule can be proved. Down to it's simplest form L'Hospital's Rule is find the limit of an indeterminate function as the ration of the slopes. By using L'Hospital's Rule and by just looking at the slopes of the numerator and denominator we can find the limit of this function is 3/7. This is shown below:

first we take the original equation:

 

then we let

by letting this be true we can manipulate the original function to try to get it to a form of the golden limit. Before we do that we must first make that applicable to all of the function by multiplying it by a FUFOO: 3/3

by doing this the equation now becomes:

and seeing as the golden rule states that sin(x)/x=1 we have now proved that the limit of this function is 3/7.

A similar approach is used to solve question #3.

Question #4, on the other hand, had an interesting and useful way to solve it. The question asked:

               Find the limit:

Since we are looking at the "end behavior" of this function (as x approaches infinity) we only have to look at the highest powers because as we get closer to infinity the limit is determined by the higher powers because they overpower values like 3x and +1. Once we are left with only the highest powers of the numerator and denominator we get a slope of -5/2 and a limit of -5/2.

We quickly went over questions 5 through 8 which both mainly reviewed the rules of limits, continuity, and discontinuity. Then we took a look at the bonus and we learned that whenever you're unsure how to approach a question, use a conjugate, it's the easiest way to get yourself started on a problem.

Once we finished going over the quiz and answering any questions on it we moved onto IW#7 and went over our solutions. O'Brien explained the intermediate value theorem by saying "if you evaluate a little above and a little below the value you're looking for it's nice because you know your answer is in between there somewhere." While reviewing IW#7 we spent most of our time looking at question 4 which asked us to find the limit of a function graphically, numerically, and algebraically.

When we first looked at this question we noticed that both the numerator and the denominator have a slope of zero, when we zoomed in close enough both of the graphs flat lined (as seen in the graph below), meaning we can't use the slopes to find the limit.

Although, when we only expanded the y-axis we began to see a difference between the two graphs, the numerator's and denominator's. By expanding the y-axis we are able to look at the slopes of the slopes.

By looking at the above graph we can see that the blue line, the denominator, approaches zero twice as fast as the red line, the numerator. In other words, we can find the limit by looking at the ratio of the red line's y-values to the blue line's values and we get a ratio of 1/2 for the limit.

We then looked at the same function and tried to solve it algebraically, this solution is shown below. To find the limit of this function algebraically we used our knowledge of conjugates, trig identities, and the golden limit. And, as you can see, we ended up with the same answer as the limit, 1/2.

After we solved this we skipped over 5, 6, and 7 seeing as we went over them last class and went on to question 8 in which we looked at the instantaneous rate of change. This is where the derivative of a function differs from its slope because the derivative is the instantaneous rate of change while the slope is the average rate of change. Question 8 asks to us find the slope between the points x=0 and x=2 for the function g(x) and to find the derivative at both those points. The math for this problem is shown below.

*NOTE: another my bad - the last equation should "= 5" not "= lim5"*

*NOTE: last thing - in all the above equations having to do with when x=0 the limit should be when x is approaching 0, not 2, sorry!*

This method of finding the derivative, by finding the change in y over the change in x can be notationally written as either dy/dx or g'(2) and can be used to find every limit.

After we were taught this was to find a derivative we took a look at all five ways:

How to find a Derivative:
1. function explained above: use the function below to find the limit as x approaches c:

2. local linearity - examples in questions 3 and 5 of IW#7
3. nDeriv on our calculators - go to MATH #8 and you'll get the nDeriv option, click enter and then enter in (function, x, point) to get the derivative - this is better explained in this video along with the option to use the #1's function on our calculators
4. Wolframalpha
5. GeoGebra

Here's an extra site with a bunch of helpful videos explaining L'Hospital's Rule, Indeterminate forms. She rambles a bit but overall she cleared a few confusing parts up for me. Or this video, which also explains L'Hospital's Rule, indeterminate forms, derivative, and limits pretty well. The second link also has more videos to the left explaining and proving some of the things we've gone over already and might be helpful to some in clearing up past confusions!

At the end of class we received IW#8 which is basically just a practice test and then we have A TEST ON MONDAY, so don't forget!

And the next scribe will be....Connor! Have fun Connor :)

***UPDATE***

  Since my scribe post we have learned many new ways to find the derivative of a function, such as the power rule, the product rule and the quotient rule all shown below:

The Power Rule:

The Product Rule:

The Quotient Rule:

Although I know many people are still having problems finding the derivative using the h --> 0 approach so here's another helpful link that shows you a step by step process for varying difficulties of functions.

And another helpful link explaining the derivatives of the trig functions because, I don't know about everyone else, but I was still having some trouble understanding how we got the derivatives.