9/28/12 The Overview of Unit 2

We started off class with OB telling us about derivatives. We looked at four different perspectives of the derivative: verbal, numeric, graphic and algebraic.

Verbal:


The derivative is the slope of the tangent line. Use local linearity to find the tangent line. The slope is a rate of change. The derivative is also known as instantaneous rate of change, it is the speed at that instant. Any quantity per quantity is a rate of change: cm^3/min, $ spent/year. Anything that is changing is calculus. Calculus: the ultimate applicable mathematics.

Numeric:


See question 4 from the test. The derivative is the slope of a set of points that are really really close to each other. To do this take a set of data such as:

Now you use

and sub in each value for x-final and x-initial where x-initial is 1 and x-final is 1.0001. This gives



m=6 so the derivative is 6.

This process is called the difference quotient, aka the slope.

The symmetric difference quotient takes just a little bit above and just a little bit below x when you wish to find f'(x) of function f(x). Use

From here use the difference quotient to find f'(x).

Graphing calculators can do this for you in a much simpler way by using nDeriv. It can be found by hitting math --> 8. Make sure you are not using MathPrint otherwise it is confusing. There are 4 things which need to be defined: the function, variable, the value for the variable, and some small number h. If nothing is entered for h then the default for h is 0.001

nDeriv(f(x)* ,x, 2, 0.0001)

*on your calculator use y1 from the YVARS menu.

Graphic:

take the graph of the function f(x) in this case:

Add the point (2, f(2)) which is point A then use the add point tool and make a random point on the graph so that it can be dragged around this is point B. Next put a line between these two points. This line is called a secant line.

Drag B so that it is on top of A which is supposed to give the tangent line at point A. The problem is that this is undefined 0/0, fortunately we have been learning about limits for a month! In this case to find the tangent line of A* we write our own limit:

*remember that A is really (2,4)

This gives us the slope of the tangent line which is the derivative at point A. Now we get to the more interesting part: we realize that there is a derivative for every point of f(x). We need to figure out how to find this function which gives the derivative for every point of f(x). This is incredibly easy on Geogebra, in the bottom text box you simply write f'(x) after you have graphed f(x). Unfortunately, I'm doubtful that we will be allowed to use this on opportunity days, there is an algebraic way. Before we get to that look at the graph of the derivative function (the blue line):

It is possible to have f'(x) traced by a point by entering (x(B), b) into geogebra. Where B is the point which can be dragged and b is the slope of the tangent line to B. b can be found by typing slo and selecting Slope[ <Line> ].


Algebraic:


We've done the hard work while learning about limits, the theoretical underpinning of integrals and derivatives. Find the derivative of

start x at x=c


It is not possible to simply sub in because of the removable discontinuity. Multiply by the conjugate FUFOO of the numerator. Also, this is (above) is the definition of a derivative. PatrickJMT has a helpful video

Which gives:

The (x-c)s cancel

So,

A general thing to know for the function f(x) at x is:

Another way to look at question 16 from the test is

So,

In part of problem 16 we were looking for when x=2. If it is plugged into the derivative the answer is

IW 1


p. 105/7, 9, 13-17, 25, 26, 27, 37

Scribe is Cooper

UPDATE:
There are much easier ways to derive functions. These ways include the power rule, quotient rule, and product rule.

The power rule: 

This is extremely useful when deriving polynomial functions.

The quotient rule:
Supposed there are two functions u and v so that

Once you have put in the correct values simplifying is not even necessary! Although it is recommended that we become used to simplifying this so that we improve our algebra and do better on the AP test.

The product rule:
Similar to the quotient rule because there are two functions joined together. Still using functions u and v.

Again, it is not necessary to simplify beyond this but we should become used to it because the multiple choice part of the AP test will not leave the possible answers as "messy" this, according to OB.

These ways of taking the derivative of the function are much easier than solving a limit to figure out the derivative.

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.