Prep AB Calculus B 2006-07

Calvin Says

2007-05-08T21:27:00.001-07:00

Calvin Says:

Don’t Forget “+C”
Check Your Endpoints!
Remember Initial Conditions!
Remember the Chain Rule! (Especially with implicit differentiation!)
Remember the Product Rule! (Especially with implicit differentiation!)
The integral of a rate of change is a NET CHANGE!
Critical Points are candidates for Maximums and Minimums
Critical Points occur where the first derivative equals zero OR IS UNDEFINED!
Speed is the ABSOLUTE VALUE of velocity
“Speeding Up” means the velocity and the acceleration have the SAME SIGN!
Derivative = Instantaneous Rate of Change = Slope of the Tangent Line
An Antiderivative is the area between a curve and the x-axis

You have all done a great job this year! I know you're going to do great tomorrow! Thanks for a really fun year...

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."
- Clay P. Bedford

I hope I made you curious...

Friday's Test Topics

2007-03-22T10:28:00.001-07:00

Here’s a list of topics that will be covered on this Friday’s Chapter 7-9 Test.

Chapter 7-9 Test Topics
Integration by parts (Sec. 7.1, #3,7,21)
Arc length (Sec. 8.1, #1,3,5,9,11)
Approximate Integration – Midpoint/Trapezoidal Rule (Sec. 7.7, #1,3,7,29)
Slope Fields/Differential Equations –Solutions (Sec. 9.2, #11,13)
Exponential Growth/Decay – Newton’s Law of Cooling (You knew it was coming!) (Sec. 9.4, 13,15)

As always, your homework is a good place to start reviewing, and the book has several other problems to give you more practice!

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!

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.
—COMEDIAN HARPO MARX

Friday's Quiz Topics

2007-03-15T14:34:00.000-07:00

Here’s a list of topics that will be covered on this Friday’s Quiz.

Quiz – Sections 9.2-4
Solve a differential equation (Sec. 9.3, #1,5)
Solve a differential equation (IVP) (Sec. 9.3, #11,15)
Exponential Growth/Decay – Formulas, Rates, Values, Times and Graphs (Sec. 9.4, #1,3,9)
Slope/Direction Fields (Sec. 9.2, #11,13)

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!

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.
- Dr. Seuss

And for those of you that didn’t see it, here’s a cute set of instructions for properly hugging a baby…

9.4 Exponential Growth and Decay

2007-03-15T10:34:00.001-07:00

Relative Rate of Growth: Things are growing based on what you already have.

Constant change based on what you start with

1. Separate the variables

2. Take the integral of both sides

3. Evaluate the integral (take the antiderivative).
Don't forget to add +C!

4. Come up with an equation for P (Isolate P).

This is a constant and the initial value:

P = population
P sub o = initial
K = constant
t= time

To find value of K:

Half-Life: Rate of decline or breakdown.

The (.5) is the rate at which the substance declines or breaks down.

I am going to use the carbon-14 (C-14) example even though we did it in class because it makes sense to most people after taking Bio and Chem.

Example Problem: EXPONENTIAL GROWTH

Newton's Law of Cooling:
Difference between an object and its surroundings.
Ok now that I totally get this (and so does Genny!) here goes:

Data:
mac and cheese (that is what I am eating right now) at 110 F.
Room at 68 F.
30 minutes later, the mac and cheese is at 100 F.
Find the temperature after 1 hour.
1. Set up specific equation

The ratio inside of ln is (later data point)/(original data point).
The units for time do not matter as long as you are consistent.

2. Solve
Plug in 60 for t.

3. Add T sub o to the answer in part 2. (This is the typical mistake that people make, at least Genny and I did)
Add 68 to the answer in part 2.

Sorry I could not put this up sooner. I realize you may have needed it last night, but I did get it done and I really was not sure if I could. Thank goodness blogger was nice to me today.

Here is a good website. It is the one that Dartmouth uses for its books I think:
www.math.dartmouth.edu/~klbooksite/3.02/302.html

AND MR FRENCH NO ONE IS NEXT...CAN YOU BELIEVE IT?

I have been a die-hard OC fan since the beginning, even when it went kinda crazy. It is strange that today is a Thursday and there is no OC two weeks in a row. In memory of what was once one of the best shows to ever hit television:

9.3 Separable Equations

2007-03-13T23:16:00.000-07:00

Hello friends. I was thinking that we'd talk about Separable Equations today. Doesn't that sound nice? I think so.

Separable equations are differential equations of the first order (i.e. equations that contain a function and its derivative) that we can solve explicitly by separating (wow) the two variables involved into the two sides of the equation and by using integrals on both sides to get rid of the dx's and dy's. (Brainstorm: what if there were a three-sided equals sign? Wouldn't that be cool? Or would it just be pointless?)

There are certain steps we need to follow in order to get full credit on the AP's Free Response Questions. Don't worry, I'll cover them. How about an example? I think so.

Find the solution of the differential equation that satisfies the given condition.

(dy/dx) = y² + 1, where y(1) = 0.

Step One: Separate the variables using algebra.
(dy/ (y² + 1)) = dx

Steps Two and Three: Take the antiderivative of both sides. (⌠= integral sign)
⌠(dy/ (y² + 1)) = ⌠(1dx)
tan^-1y = x

NOTE - The integral of the left side of the equation is one of the rare occurrences of a derivative of an inverse trigonometric function. Yes, it's annoying. (sad face).

Step Four: Recognize the constant of integration.
tan^-1y = x + C

Step Five: Solve for "C" using the given condition (y(1) = 0).
tan^-1(0) = (1) + C
0 = 1 + C
C = -1
Good job.

Step Six: We end up with tan^-1y = x - 1. Our book seems to like solving for y at this point. I'm not sure if this is really necessary, but being the brilliant maths students we are, why not do it?

y = tan (x - 1). This is the answer you've all been waiting for, or, rather, for which you've all been waiting.

SOPHIE. Wake up and take notes; you're posting next.

And, if you couldn't handle my colloquial math tongue, this website presents separable equations in more textbooky way. It also has some sample problems you can try, if by chance you've done all of the ones in our own textbook.

Here is a video that my brother made for his Chinese class at Vassar College. It's about drugs and their effect on the psyche. Not really, but it is about drugs. And it's in Chinese, so it's funny. And it has a cute old Chinese man at the end with a cool voice. I hope you enjoy it, even if their accents sound like an electronic dictionary (Laurie).

More Drugs, More Chinese Problems

Bye bye.

9.2 Direction/Slope Fields

2007-03-10T15:19:00.000-08:00

Direction/Slope Fields!!!!

Direction fields are a way to approximate the solution of the differential equation graphically. (But it is extremely imprecise.)
(Remember, differential equations are equations with the derivative of the function and the function itself.)

It is easiest to learn through an example so....

Given the differential equation y’ = 2x + y, sketch the graph of the equation going through the origin.

1) It is important to remember that slope fields are an extremely imprecise way to find the solution. The first step of solving a problem is to make a chart.

X Y Y’
0 0 0
0 1 1
1 1 3
.
.
.
and so on and so forth. Etc etc.

At each point, draw a small line with the calculated slope until you have a graph something like this....
2) Since the problem is asking for the graph through the origin, so start at the origin and following the slope lines until a graph forms.

This is an extra link to help you! http://www.sosmath.com/diffeq/slope/slope1.html

http://kme.truman.edu/images/difeq.jpg
:) It’s...cute.

TAYLOR YOU’RE UP NEXT.

9.1 Modeling with Differential Equations

2007-02-28T04:04:00.000-08:00

9.1 - Modeling with Differential Equations

REVIEW
Differential equations discuss rates of change, like a population that changes according to what the limits on the population are or what the population already is.

First order: Here, we have a derivative that is equal to the original function times a constant. Thus, a likely equation for P would be Second order: y’’+y’=x+y
As the number of derivatives increases (and by that I mean first derivative to a second to a third...) , the order does, too. So, because there are both a first derivative and a second derivative in the above equation, it is a second order derivative. To solve it, you must not only find the defintion of the first derivative in terms of x for substituion, but also that of the second derviative (I do not think we will have to be able to do this, though).

Logistic growth:

As you might remember from Precalculus, many populations in the real world begin increasing rapidly, and then gradually level off until it appears not to be increasing at all. That is the curve of a logistic function. One can see that if the population P exceeds the carrying capacity C, then what is in the parentheses will be negative, so dP/dt will be less than zero, and the population will decrease. What happens when we see the graph leveling off is that P is approaching C, so what is inside the parentheses will almost equal zero, thus the dP/dt is almost zero, meaning the population is hardly changing.

EXAMPLES
There are two types of problems…

First, we will be provided with a potential solution to test.

This will be tested by finding y’ and plugging it into the equation to, hopefully, find that there are infinite solutions.
Then, we will have initial value problems (IVP) which is a bit like implicit differentiation in reverse. The problem provides us with an equation for which we find the general solution with a C-value, and another equation so we have the capability for solving for y – solve the differential equation for the general solution, then use an initial value to solve for the specific solution.

Provided: 1. Separate the values, isolating x's and y's on different sides of the equal sign 2. Find the antiderivatives of both sides. (You only have to use one C) And, now you have the general equation.3. Solve for C.4. Write the specific equation.

LINKS
Population dynamics and practice

Paul’s (really detailed) math notes

REMINDER
Laurie, you bum, you are next. Unless sum (hah :( ) elaborate switching process has thrown everything off.

PERSONALIZATION
Blind Date - a math love story

CILLIAN!!!!

Thursday's Quiz Topics

2007-02-27T18:02:00.000-08:00

Here’s a list of topics that will be covered on this Thursday’s Quiz.

Quiz – Sections 8.1, 9.1
Arc length – given curve and interval (8.1, #1,3,5,9,11,29)
Arc length – determine setup (8.1, #1,3,5,9,11,29)
Arc length – determine setup (8.1, #1,3,5,9,11,29)
Arc length – given curve and interval (8.1, #1,3,5,9,11,29)
Arc length – given curve, determine interval (8.1, #1,3,5,9,11,29)
Differential equation – analysis and interpretation (9.1, #11)
Differential equation – verification of solution (9.1, #1,5)

That’s it for now! I’ll be around after school on WEdnesday and online Wednesday evening/night.

In any collection of data, the figure most obviously correct,
beyond all need of checking, is the mistake

Corollaries:
(1) Nobody whom you ask for help will see it.
(2) The first person who stops by, whose advice you really
don't want to hear, will see it immediately.

And on another note, look for the simple solution:

Problem....

2007-02-27T01:22:00.000-08:00

Mr. French, for the past 45 minutes, I have been desperately trying to make my blog, but photobucket is "down for maintenance" and I can't post all of my pictures!!! I also tried directly posting them from paint, but the image appeared blank. When I tried to save it as a web page and then convert them to images it said "error: equation.html must support jpeg or gif." I feel like I've done everything in my power to try and fix this, Mr. French!

-James

Ok, this is an update of 10 minutes later. I finally figured out how to upload them as regular images, Mr. French. But in the denominator, the squares aren't showing up because internet explorer doesn't support them or something. So I don't want to post it yet, because it's incorrect information that will confuse people!

-James
Ok, Mr. French it's now 2:37 AM, and I figured out how to get everything to post correctly...except for the graphs. I could do it with photobucket, but that's still down. And I don't want to post it yet, because I don't want to without it being absolutely complete! (I even used winplot like you suggested!)

-James

Ok, Mr. French. It's 3:50 am, and I finally figured out how to get the graphs to come out ok. The numbers are kind of small though. I'm not sure if I should post it. Maybe I should wait for photobucket to come back? Well, I'm sleeping. Night!

-James

8.1 Arc Length

2007-02-26T21:21:00.000-08:00

HIIIIIIII EVERBODY!!!!!!!!!!!!!!
WUZZZZZUPPPPPP?????

Ok, well, today we're going to figure out how to find the the length of an arc. It really is a beautiful thing. Now, remember that we always like to do things in terms of x and y. So, in order to express a constantly curving line, we will simply use an infinite number of hypotenuses (or is it hypoteni?) of right triangles. In order to express the hypotenuse in terms of y's and x's , we use pythagorean theorem:

*IMPORTANT REMINDER*: The legs must always be in terms of dy and dx, or the change in y and the change in x, for the given interval of the hypotenuse. (So you would start at one end of the hypotenuse and go to the other end, measuring how much the y changed and how much the x changed.)

But of course, we need the sum of all these hypotenuses in a given domain. So we use the integral sign and a dx. But....so that we don't change the nature of the function, we also DIVIDE by the dx (dx/dx =1, and thus the value remains unchanged.)

Note the dividing by dx on the bottom.

So now, we want to try and simplify that devilish little formula. So....we put the bottom dx inside the square root. And to do that, we need to square it, so that we don't change the value of the function. (The square root of dx squared is of course dx, what we had in the first place)

Alright. Excellent. Now that we've done that, we can simplify what's inside those parentheses, since dxsquared over dxsquared is 1:

Unfortunately, Gianna, there is no +C when we have it definite like this. But that's ok. You still get it, right?
If you notice, in parentheses, we have (dy/dx), WHICH IS THE SAME THING AS THE DERIVATIVE OF THE FUNCTION!!! ISN'T THAT WILD???? So it looks like this:
*IMPORTANT REMINDER*: f(x) MUST be continuous on [a,b] and x must fall between or equal to a and b.

It's very true that we could have skipped all that and cut to the chase by going right to this formula. But I think that Mr. French and I would both agree that we all should value process over product. PLUS IT'S JUST PLAIN COOL TO SEE HOW WE DERIVED THAT FORMULA!!!

Ok. So let's test this theory.

Let's use this math problem:

Find the length of the following curve.

To sort of illustrate, this is what the graph looks like (Fortunately, unlike vons, my images actually came out pretty decently the first time round):

And this is the specific portion that we have to find the length of.

Let's go ahead and take the derivative. Wait! Just kidding. First we should put the function in more recognizable terms (After all, we don't want it to be so scary-looking that a certain someone would just start frantically trying to change it into slope-intercept form.)

Note that I simply made the coefficeints more recognizable. I changed the divide by 6 and instead multiplied x to the 5th by 1/6 (the same thing). Then I took the xcubed out from the denominator of the second term and multiplied (1/10) by x to the -3 (same thing).

Now take the derivative for real this time:

Let's plug it into our equation:

*IMPORTANT REMINDER*: Dont' forget to add 1 inside the integral!!!! (After all, we know that some people in our class seem to have a bit of a problem with adding.......Go Kristin!)
And plug it into our TI-83's and 84's (Or in John Cynn's case TI-89 Platinum Addition. Man, he thinks he's so cool with it, doesn't he?) with the good ol' fnInt(Y1, X, 1, 2,).

Of course I'm sure you could do the complicated way and do it by hand, taking the antiderivative of that and then plugging in 1 and subtracting it from what you get when you plug in two. But I'm pretty sure the only people arrogant enough to do that would be Mark Chodas. (No offense, Mark. You know we love you.)

This is a great site:
http://www.pinkmonkey.com/studyguides/subjects/calc/chap8/c0808501.asp
There's obviously not much to explain abou this topic. So I found a site that gives a lot of example problems and shows how to solve them!! :)

So as many know, I am a big fan of Desperate Housewives, and I have no shame in telling people that.....

WHO IS YOUR FAVORITE HOUSEWIFE?

a) Gabrielle Soliz

b) Edie Britt

c) Lynette Scavo

d) Susan Meyer

e) Bree VandeKamp

(Feel free to leave a comment about your choice.)

Also.....

CONTEST

Whoever can guess where this is from first will get a prize.....!!!!!! (And it'll be good.)

Genevieve....Teague....you're up!!

Friday's Quiz Topics

2007-02-21T22:38:00.001-08:00

Here’s a list of topics that will be covered on this Friday’s Quiz.

Quiz – Sections 7.1,7
Integration by Parts – basic (7.1,#3,7)
Integration by Parts – definite integral
Integration by Parts – f and g won’t go away (7.1,#15 - not assigned, but good practice!)
Integration by Parts – tabular method (7.1, #61)
Trapezoidal Rule (7.7, #1,3,7,29)
Midpoint Rule (7.7, #1,3,7,29)

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...

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.
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.
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.
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.
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.

7.7 Approximate Integration

2007-02-21T17:23:00.000-08:00

Hey everyone, ready for some fun math? K, well remember Reimann sums? it's backkkkk. but this time it's even more exciting because we use TRAPEZOIDS instead of RECTANGLES!

Let's take a look at a graph:

As you can see, the blue shaded regions are trapezoids, NOT rectangles. Therefore when we find the area, we will need to use the area of the trapezoids. In this example, there are five trapezoids. Their heights are B1, B2, B3, B4, B5 and B6. So now that we know the height, the width and the number of trapzeoids, we can find the total area.

Just in case we forgot geometry, the area of a trapzoid is:

* in this case, our "a" and "b" are B1, B2, B3...and so forth*

Finally we get the general equation which is called the trapezoid rule!

The trapezoid rule is just another way to approximate the area under a curve, but it is more accurate than using midpoints or Reimann sums.

But wait...don't you hate having to add numbers and then mulptify them? Don't you wish there was a simpler way? Well Simpson has the answer! Then using the same graph as above, the Simpson rule states:

*** Don't forget that the Simpson's Rule ONLY applies to even number of intervals.***

Example numero uno:

Use the trapezoid rule and the Simpson's rule to approximate: and also, n=4.

When n=4, Δx = π/4. Therefe when we use the trapezoid rule:

But wait! n=4, which is an even number! that means we can use the Simpson rule!

yay! now we are all gonna ace the quiz on friday right? of courseeee. but if you need more help, i found some really great sites:

http://archives.math.utk.edu/visual.calculus/4/trapezoid.2/index.html Unfortunately we couldn't see the visuals that Mr. French was going to show us, but this site has animations on creating trapezoids to approximate the area undert he curve.

http://tutorial.math.lamar.edu/AllBrowsers/2414/ApproximatingDefIntegrals.asp I don't know who the guy who wrote these Calc notes are, but he is AMAZING! This site offers detailed directions and explanations on how to use the two equations!

JAMES YOU'RE UP NEXT!

TOGA DANCE FRIDAY!!!!!!!!

7.1 - Integration by Parts

2007-02-20T19:25:00.000-08:00

Alright, folks, the Lord Protector of Math is back for some more Calculus action! Huzzah!

This lesson will be about integration by parts. This is somewhat like the substitution rule, but it deals with the Product Rule instead of the Chain Rule.

The basics of integration by parts are pretty easy. Let's start with the basic formula.

If we were to use u and v in place of the functions, the formula would simplify to this.

This is taken from the product rule for derivatives, and it allows us to integrate when two functions are multiplied.

Let's get an example. Evaluate the integral.

First, we go about setting values. There will be u, v, du and dv. It's generally a good idea to set the simplest function as the u value and the more complicated function as the dv value.
So, let's set x as u, and sin2x as dv.
This gives us:Since dx is the derivative of x and -(1/2)cos2x is the antiderivative of sin2xdx.
Are you with me? We're almost done.

Now we plug these equations into the equation.

See? We just plugged in the equations for u, du, v, and dv.
Now we evaluate the indefinite integral.

Don't forget your + C, or I will laugh at you pretentiously.
Anyway, see where we're going with this? Let's finish up.

Done! All evaluated.

There are some other things you need to remember:

1.You may have to repeat the process more than once to get rid of all of the integrals.
2. If the functions keep repeating, continue until you have reached an equation similar to the starting problem. Then, add/subtract it to one side, and solve.

Finally, the ultimate time-saver!

THE TABULAR METHOD

This is pretty straightforward and easy. If you have an equation with a relatively simple u and dv, it might be better to use this method. Let's try it with this function.

Alright, let me explain this tricked-out diagram. Basically, one column shows u and its derivatives and the other shows dv and its anti-derivatives. Drawing those diagonal lines helps you keep track of which u multiplies with the dv, and putting the alternating signs above the lines tells you whether those will be positive or negative. So, x^2 will go with e^x and be positive, 2x will go with e^x and be negative, etc.

So, the final answer turns out like this.

Don't forget the + C.

If you want more practice, this cool little web page lets you solve a bunch of problems step by step in Flash.

http://archives.math.utk.edu/visual.calculus/4/int_by_parts.3/

REMINDER: Lisa, you're up next.

And now I finish my post with this Photoshop edit:

Hope this was helpful, folks. Until next time.

P.s.s

2007-02-13T22:09:00.000-08:00

PS!!!!

EVAN, you're NEXT!!

Thursday's Test Topics

2007-02-13T20:22:00.000-08:00

Here’s a list of topics that will be covered on this Thursday’s Chapter 6 Test.

Chapter 6 Test Topics
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)
Determine the volume of solids generated by building shapes with a known cross-section off a given base. (6.2, #55)
Determine the average value of a function for a given interval. (6.5, #1,5,7,13)
Determine any value(s) c that generate the average value of a function on a given interval. (6.5, #9)

For additional practice problems, look at the chapter review (pp. 431-433)

That’s it! I’ll be around after school on Wednesday until 3:15 and in early on Thursday. Donut holes and OJ!

It’s kind of fun to do the impossible. – Walt Disney

6.5: The Average Value of a Function

2007-02-13T17:30:00.000-08:00

In 6.5, we learn to take our abundant knowledge of the Mean Value Theorem and adjust it in order to apply it to our new best friend: INTEGRALS!

So, for starters, let's just try to state the mean value theorem in its simplest, most "kindergarten-esque," terms:

Assuming that the graph of 'f' is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), there is a number 'c' in (a, b) such that
Now, let's be really intelligent and try to apply it, at the most basic level, to integrals and their importance as the area under a curve. Let's take a look at this graph:

As you can see, I've taken the area under the curve and made an approximating rectangle in order to find it. Now, if we are told that the area under the curve is 18, we could find out what X is...6! That X value, 6, is called the average value of the function (i.e. f(c)).

Now that we know that, I can introduce the mean value theorem for integrals.

if f(x) is continuous on [a, b] then there exists a 'c' value such that
so, now we know how to find that average value of the function numerically, using the mean value of integrals:
Ok, so now let's try a practice problem to see if we really understand the concept

on the interval, [-2, 5]

Step 1: find the area under the curve by evaluating the integral of the function
Step 2: plug that integral value that we just found back into the equation for finding the average value of the equationso f(c)=4.333 or (13/3)

THIS IS THE AVERAGE VALUE OF THE EQUATION

Step 3: plug the f(c) back into the equation for y in order to find the 'c' value that gives the f(c)QUADRATIC FORMULA

YAY!!! We have finally found all the information that we need for 6.5. Let's give ourselves a pat on the back. Now everyone, let's take a second to relax our minds and read a few calculus jokes:

Q: How does a mathematician induce good behavior in her children?
A: "I've told you n times, I've told you n+1 times..."

Q: What is the first derivative of a cow?
A: Prime Rib!


dx    d CABIN
---  =  log x     so,      ------------ = log CABIN
x     CABIN


/
|     d CABIN
|     ------------  = log CABIN + C     (which is a house boat)
|       CABIN
    /

HAHAHA! LAUGH IT UP!!
Peace, Love and Calculus Soul Everyone!!

p.s. here's a fun link

Monday's Quiz Topics and New Posting Schedule

2007-02-08T22:13:00.000-08:00

Here’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!)

Quiz – Sections 6.1-3
Area between 2 curves
Volume of revolution – disk
Volume of revolution – washer
Volume of revolution – cylindrical shell
Volume of a solid with a known cross-section off a specified base

The area question(s) can be in terms of x- or y- relationships.
The volumes of revolution can be revolved around the x- or y-axis, or some other defined line.

Note: I'm not going to post relevant homework problems, because in this case they're all relevant...

That’s it for now! 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.

As promised, here's the new posting order:

Kate
Lauren
Gianna
Evan
Lisa
James
Genevieve
Laurie
Taylor
Sophie
Isabella
Kristin
Mark
Jean
John
Zak
Alison
Tessa (sorry, Tessa!)

In order to attain the impossible, one must attempt the absurd.
- Miguel de Cervantes
(of course, the absurd seems to happen every day in our class…)

Poor study strategies:

6.3 Volumes by Cylindrical Shells

2007-02-08T15:49:00.000-08:00

Ok, so in the last lesson, which Kate has failed to post (jk kate... its funny bc now I have back to back posts! lucky me.... well let's just say I'm done for the year!) we learned how to find volumes by using DISKS and WASHERS. WEll, Luckily enough for us, we now get to learn how to find Volumes by using CYLINDRICAL SHELLS!!!! I know I just made all of your lives complete by informing you of this good news.

The reason for using cylindrical shells is if we are faced with a difficult problem like finding the volume of this shape around the y-axis.

If we had to find π ∫ R^2 - r^2 dy, (using the dashed line)

It would br rather difficult to solve for x in terms of y.

So, instead we can use the bolded line, and instead find the volume of the cylindrical shell made by rotating around the y-axis.

Volume of a cylinder = 2πrh(dx)

Where......

r = x and h = top - bottom = y-values = 0 - ((x^3)-x)

Another example:

Using cylindical shells (sections parallel to the axis of rotation), find the volume of the area enclosed by y = x, y = x^2, about the y-axis

V = 2π ∫ r h (dx) on the interval a to b.

Then, r = x
h = top - bottom = x- x^2

V = 2π ∫ x(x - x^2) dx on the interval 0 to 1.

That's it. Wahooo for Calc blog.

Here is a link......

http://www.geocities.com/pkving4math2tor7/7_app_of_the_intgrl/7_03_02_finding_vol_by_using_cylind_shells.htm

I don't know WHO the Lucky person is that's Up Next....

But, It's GREY's NIGHT!!!!!!

EVERYONE, Turn ON ABC At 9 for the TIME OF YOUR LIFE. TRUST ME ON THIS ONE.

-Lauren

Lesson 6.2: Volumes

2007-02-06T21:14:00.000-08:00

I would like to apologize in advance for my computer being a pain in the neck. I spent like an hour typing up all these amazing equations in Word and the darn thing won't let me upload. Grr. I'm sorry if it gets confusing without graphs and stuff...I really am...

Ok, so here goes. So far with integrals, we have learned how to find the area under a curve (between the curve and the x-axis) and the area between curves. And now, for something completely different...we are going to fine the volume of solids using integrals.

Don't panic! It's not so bad...

Ok so firstly, let's think of the awesome $500 shape thingies that Mr. French showed us. Remember the one that had a circular base and had sides consisting of squares? Visualize it...because I don't have one for you...

Let's think way back to geometry for a moment. Remember how we learned how to find the volume of stuff like cylinders and prisms? When the volumes had fixed side lengths, it was simple. However, since this is calculus, it needs to be a little bit more complex.

With geometry, we could find the area of, say, a cylinder by multiplying the area of the circular base by the height of the cylinder. We can apply this same principle to calculus using integrals.

If our base is a circle, we can use the formula for area (A = pi x radius^2) and multiply it by the height...but what about when our height isn't a constant? This is where integrals come into play. For simple area of a region, the base was a constant, but the height was constantly changing, so we used the small rectangles to compute the area. With solids, it's similar, but we're using prisms.
The book definition for the volume of a solid (S) that lies on the xy - graph between x = a and x = b is as follows:

V = the limit as n approaches infinity is the sum of n where i =1 of A(x sub i) times delta x = integral from a to b of A(x)dx

if the cross-sectional area of S that lies in the plane Px, through x and perpendicular to the x-axis is A(x), where A is a continuous function.

Whew! That's a lot. But what does it mean? Well think back to the solid that was a circle with square cross-sections. The area of the base never changed, but the height was not a constant. The height was a function of the length of the cross section, which was a function of x. So we can find the area of the solid by finding the integral of the height and multiplying it by the base. Then we integrate to find the volume.
V = (integral from a to b) A(x)dx

So let's think back to the shapey thingy. The area of the base is [x^2 + y^2 = r^2] and the area of a square is s^2. s = 2y where y = plus-or-minus (root of r^2 - x^2). So s = 2 times (root of r^2 - x^2), and s^2 = 4(r^2 - x^2)
We then integrate:
integral from -r to r of s^2 dx
integral from -r to r of 4(r^2 - x^2)dx
2 [integral from 0 to r of 4 r^2 - 4 x^2 dx]
2(4 r^2 x - 4/3 x^3) from 0 to r
8 r^3 - 8/3 r^3

And voila!

We also learned about solids of revolution. Let's start with a graph of y = x^2
[sorry about not having the actual graph!]
And let's take this graph from zero to one.
We also need to set a y-boundary, so let's set it at y = 1

To create the solid of rotation, we set an axis of rotation (such as x = 0) and rotate the area of the graph around it. We will get a funky shape that resembles a ring. But it's not exactly cylindrical, so how can we calculate the volume? Same way as we did with the other shapes...we can break it into infinitely small prisms.

So with solids of rotation, we first need to ask one question: does the original shape share a side with the axis of rotation? Usually if the axis of rotation also acts as a boundary of the given area, the answer is yes, then we are dealing in terms of disks.
The general formula for solids of rotations with disks is:
integral from a to b of pi r^2 dx

It's the same principle as that behind the previous solid, just with a different basic formula (pi - r - squared times heights instead of length-times-width-times-height). But what if the graph doesn't use the axis of rotation as a boundary? Like...with the graph of x^3, when it is bound by x = 1 and y = 1

As you can imagine, when you rotate the area it creates a big empty space in the middle, and if you divided it into cross-sections you would end up like something resembling a very well-sliced bagel. That's ok, though, because we can still find the volumes because of the differences in radii length. So, when we're using washers, it looks like this.
integral from a to b of pi R^2 - r^2 dx
when R is the outer radius and r is the inner radius.

See, when finding the volume of the little cylinders, we can subtract the volume of the inner cylinder (aka the hole) from that of the big cylinder. ***Big important note: Subtracting the volumes is NOT AT ALL the same as using the difference between radii to find a volume. You WILL get the wrong answer if you do this*** (This is just in your best interest, because I know I've tried it...)

The last thing I think I need to remind you of is the use of axes and dx vs. dy. You can do this in terms of either, since your formulas are going to be set equal to each other (y = x^3, x^2 + y^2 = 4), but your "slices" (which are expressed by the derivative) should always run perpendicular to the axis of rotation. That will help you determine which one to use.

In closing, since I cannot upload a comic like I usually do, I leave you with this:

http://www.math.hmc.edu/calculus/tutorials/volume/

This is probably a better tutorial than I can accomplish in my current state, so please use this well. It also includes shell stuff...

Oh, and some of you may have overheard me whining about my existential crisis brought on by physics, so as if I can't get any more confused about what else I thought were constants in life, check this out:

http://www.math.hmc.edu/funfacts/ffiles/30001.1-3-8.shtml

Crazy stuff...Lauren, you're up next...oh wait, you already know, never mind...

This is priceless...I couldn't resist...this one goes out to Gianna especially...

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.
"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." He then excused himself to visit the men's room, and the other called the waitress over. "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.
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?"
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?"
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."

6.1 Areas between curves

2007-01-10T19:19:00.000-08:00

OK WELL LET ME START OFF BY SAYING..... BEWARE.... I, LAUREN, AM ON A PC. ENOUGH SAID.

6.1
In this section we will be finding the area between curves that are not necessarily bound by just the x and y axis

Example 1:

Find the area of the bound by f (x) and g (x) on the interval [a,b]

To find the area, all you would need to do is subtract the integral of f (x) on the interval [a,b] from the integral of g (x) on the same interval.

Example #2

Now, find the area bound by the two graphs given below.

Graph it on the calculator to see it

Luckily, we already know that one of the intersection points comes at (0,0). We can use our graphing calculators to find the x-value for the other point. You can go to CALC then to INTERSECT, then magically we find out that the intersection is at the point 1.1807757. Immediately after you find this information out, store the value as the "a" value. Then, with more help form the calculator, you can calculate the area by calculating the integral.

It should look like this on your calculator:

fnInt (Y1 - Y2 , x , 0 , A) = .7853885505

This is because you are caluculating the integral of the difference between the two functions Y1 and Y2, in terms of x, on the interval 0 to A. (You already calculated and stored the value.)

When you have a problem that looks similar to a sinusoid, you have two options:
1. you can add the integrals comprised of one function subtracted from the other on the intervals of the intersection points. So, adding up the segments.

2. you can find the absolute value of the integral of the difference of two funtions, completely disregarding the segments. (Shown below)

EXAMPLE 3

Sometimes, merely subtracting one integral from the other is not the easiest way of doing things. When the rectangles that make the area and integral are horizontal and not vertical, it is easier to put things in terms of y instead of x.

For example, the equations y = x-1 and y^2 = 2x + 6

So, to work horizontally, rework the y equations to x.

x = y + 1

x = .5 y^2 - 3

Now we are left with the integral

Now you basically know how to find an area between curves. I bet you feel a lot smarter now that you know this. I know I do.

Quick link to check out: http://archives.math.utk.edu/visual.calculus/5/area2curves.3/

Mr. French, here's one for you: http://www.teacherschoice.com.au/Maths_Library/Calculus/area_between_two_curves.htm

A friendly reminder that Grey's Anatomy is back on tomorrow, so you better all watch it!

JAMESYWAMSEY YOU'RE UP NEXT......

man i'm excited already!!!!!!

-Lauren

Friday's Test Topics

2007-01-09T21:04:00.000-08:00

Here’s a list of topics that will be covered on this Friday’s Chapter 5 Test.

Chapter 5 Test Topics
Fundamental Theorem of Calculus Part I (Sec. 5.3, #9,13,17)
Fundamental Theorem of Calculus Part II (Sec. 5.3, #13,41)
Substitution (Sec. 5.5, #13,23,278,31,37,41,57)
Evaluate an integral in terms of area (Sec. 5.2, #37)
Riemann sum: sketch, evaluate, explain and interpret (Sec. 5.1, #3,1,13,15)
Definite Integrals (Sec. 5.2, #33)
Net Change Theorem (Sec. 5.4, #47,48)

For additional practice problems, look at the chapter review (pp. 431-433)

That’s it! I’ll be around after school on Thursday and in early on Friday. Donut holes and OJ!

"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."
Ben Stein

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...
http://media.grc.com/mp3/VerizonCantCount.mp3

And on a lighter note:

5.5 The Substitution Rule

2007-01-06T16:31:00.000-08:00

Hello "Wild Calc Class!!!" I hope you have all enjoyed your break. I had an amazing time visiting my camp friends. Happy New Years (a little late, but late's better than never)!!! But to take you all off cloud 9, it really is time to go back to school. We currently have about 40 hours until class resumes, so relax while you can and do all that work you have been saving up, except for calculus because...it is not due until after we get back. That is right, no calc homework over break...woo hooo!!!!!

But if you were planning on getting ahead, here are the notes from 5.5 just in case you need some help:

Substitution Rule = Reverse Chain Rule.

Yup instead of taking the derivative using the chain rule, we will be taking the antiderivative using the Substitution Rule. We will be working backwards. With many things they say working backwards is easier, but unfortunately, with this you are out of luck. The Substitution Rule is tedious and difficult, but with the brilliant minds we all have, completely do-able!

Hey look...the colors are like a watermelon (and that's how I think on vacation)

Let's dive in!

I am going to start with a problem and explain step by step because I think that's the easiest way to grasp the concept. With the reverse chain rule, you need to see it in action to master it.

Step 1:
Write out the integral

Step 2:
Identify u. Always choose u by figuring out which function is the nested function. Properly choosing u will make the whole problem much simpler

Step 3:
Identify du (don't forget to put dx along with the derivative because it is the derivative of u with respect to x)

Step 4:
Insert u and du into the original integral

Step 5:
Find the antiderivative using u

Step 6:
Insert Step 2 into Step 5

Here is a harder problem:
Step 1:

Step 2:

Step 3:
You will need to manipulate the equation of du so that it matches the initial integral. This makes finding du a little tricky sometimes. In the initial equation, there is xdx, but not 2xdx, so the right side of the equation must be what is in the initial integral equation.

Step 4:
When inserting du into the equation, make sure you also insert the coefficient of du, which you found in Step 3. The coefficient of du can be put on the outside of the integral sign.

Step 5:

Step 6:

That wasn't so bad now was it?! The key to those is making sure you take the antiderivative of the trig functions, not the derivative of them...I see that as being the likeliest place for me to mess up.

Speaking of trig functions, another necessary thing to be able to understand is how to use the substitution rule to find the integral of functions using tangent. This does not follow the 6 steps I broke the rest of the integral problems into. It follows its own pattern that will further our understanding of how to take integrals and also help teach us a new way to think about calculus. This is not impossible, it is just new and different and requires practice thinking in a new way.

Write out the integral and break it into usable pieces such as sine and cosine. Change the form to fit trig functions that are easier to work with.

Identify u and du.

Insert u and du into the integral

Find the antiderivative using u.

Insert u.

Remember:

And "reduce" (well this is not so much reducing as putting in the simplest form, which is the form they are asking for.) You need to train your brain to think to finish this last step.

Did you know: a sneeze travels out of a person's mouth at over 100 mph!!! I got a bad cold Friday night after my soccer game and thought I would share that fun little snapple cap fact.

Did you know (#2): Americans eat 18 acres of pizza a day on average!!! (Snapple cap fact again)

Here's an awesome website you can use for some more help on the SUBSTITUTION RULE:
http://tutorial.math.lamar.edu/AllBrowsers/2413/SubstitutionRuleIndefinite.asp

Substitution Rule for Definite Integrals: (I am so sorry I am using the problem you showed us in class, but I had trouble on this part of the homework and I don't want to show a wrong example, so once I understand it, I will change it if you would like...)
Step 1:
Write out the integral.

Step 2:
Find u and du

Step 3:
Insert u and du into integral

Step 4:
Adjust Boundaries to u

u = 2x + 1
u = 2(4) + 1 = 9
u = 2(0) + 1= 1

Step 5:
Rewrite with the boundaries of u

Step 6:
Take antiderivative and solve.

OR

Step 7:
Put x into antiderivative with initial boundaries and solve.

Even Functions: f(x) = f(-x)

Odd Functions: -f(x) = f(-x)

Lauren You Are Next!!!

Happy New Year!

2006-12-31T08:41:00.001-08:00

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…

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!

See you all on the 8th!

Wednesday's Quiz Topics

2006-12-19T20:15:00.000-08:00

Here’s a list of topics that will be covered on this Wednesday’s Quiz.

Quiz – Sections 5.3-4
Fundamental Theorem of Calculus, Part I (Sec. 5.3, #9,13,17)
Fundamental Theorem of Calculus, Part II (Sec. 5.3, #31,41)
Determine general indefinite integrals (+C!) (Sec. 5.4, #9)
Evaluate definite integrals (Sec. 5.4, #19,25,29)
Explain the meaning of a definite integral expression. (Sec. 5.4, #47,48)
Displacement vs. Total Distance Traveled (Sec. 5.4, #55)

Oh, by the way, no calculators on this one...

I’ll be in early Wednesday and I’ll check in tonight online. See you in class!

"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."
-John Wooden

5.4 Indefinite Integrals and The Net Change Theorem

2006-12-18T21:16:00.001-08:00

Hey, it's Zak. I'm filling in for Sophie because she's all bogged down with other stuff. So she'll do 5.5 over break.

Anyway here's 5.4:

An Indefinate Integral is basically just like a definite integral except for the fact that it has no boundaries. A definite integral is limited by the values a and b whereas an indefinite integral is infinite, meaning the boundaries of a and b do not exist. Therefore, the notation for writing an indefinite integral is as follows:

where

Notice how the notation does not have any values for a or b inputed above or below the integral sign. That's how an indefinate integral should be written.

Important: There is a big distinction between definite and indefinite integrals. A definite integral is a number whereas an indefinite integral is a function or family of functions. Also, since indefinite integrals are functions, the "+ C" cannot be disregarded when writing out the function. For example:

When finding the antiderivative of the function:

We get:

Whereas with boundaries, we'd get an exact number and not incorporate the "+ C"

Net Change Theorem: The integral of a rate of change is the net change. In other words, the value of the integral is an object's displacement from the starting point, not the total distance traveled. For example, if I walk 10 feet and come back 9 feet, my displacement is 1 foot but my total distance is 19 feet.

Sample Problem:
Given the equationfind the displacement and total distance an object traveled in the interval

For this we can use our calculators:

Step 1: Enter the equation into Y1
Step 2: Go to the home screen and press (math) then (9) to get "fnInt" on the screen
Step 3: Enter after "fnInt"
Step 4: Press (Enter) and you should get an answer around 1.1715. This is your displacement.

To find the total distance:

Step 5: Go back to the equations screen and insert "abs" in front of the equation. This will prevent the graph from ever dipping below the x-axis, thus resulting in no negative distance and therefore representing the total distance and not the displacement.

Step 6: Follow steps 2, 3, and 4 again but this time you should get an answer around 14.828. This is the total distance traveled.

That's about it, hope that helped.

In the Humorous Spirit of the Holidays:
Mr. Church and Virginia may very well believe in Santa Claus, but some scientists have done some mathematical studies into the physics behind Santa Claus in order to prove that Santa Claus cannot physically exist. Needless to say, their results are very convincing (and humorous). So here they are:

A Scientific Inquiry into Santa Claus:

1) First of all, no known species of reindeer can fly. BUT there are 300,000 species of living organisms yet to be classified, and while most of these are insects and germs, this does not COMPLETELY rule out flying reindeer which only Santa has ever seen.

2) Secondly, there are 2 billion children (persons under 18) in the world. BUT since Santa doesn't appear to handle the Muslim, Hindu, Jewish, & Buddhist children, that reduces the workload to 15% of the total - which is 378 million according to Population Reference Bureau. So at an average (census) rate of 3.5 children per household, that's 91.8 million homes. And of course, one must presume that there's at least one good child in each.

3) Additionally, Santa has 31 hours of Christmas to work with due to the different time zones and the rotation of the earth, assuming he travels east to west (which seems logical). This works out to 822.6 visits/second which means that for each Christian household with good children, Santa has .001 second to park, hop out of the sleigh, jump down the chimney, deliver the presents, eat his milk and cookies, go back up the chimney, get back into the sleigh and move on to the next house. Assuming that each of these 91.8 million stops are evenly distributed around the earth (which, of course, we know to be false but for the purposes of our calculations we will accept), we are now talking about .78 miles/household, a total trip of 75.5 million miles. So Santa's sleigh must be moving at 650 miles/second, 3,000 times the speed of sound. For purposes of comparison, the fastest man-made vehicle on earth, the Ulysses space probe, moves at a poky 27.4 miles/second. A conventional reindeer can run, tops, 15 miles/hour.

4) Moreover, the payload on the sleigh adds another interesting element. Assuming that each child gets nothing more than a medium-sized lego set (2 lb.), the sleigh is carrying 321,300 tons, not counting Santa, who is invariably described as overweight. On land, conventional reindeer can pull no more than 300 lb. Even granting that "flying reindeer" (see #1) could pull 10 TIMES the normal amount, we cannot do the job with 8, or even 9 reindeer. Instead, we need 214,200. This therefore increases the payload - not counting the weight of the sleigh - to 353,430 tons which is four times the weight of the ocean-liner Queen Elizabeth.

5) 353,000 tons travelling at 650 miles/second creates enormous air resistance which will heat up the reindeer up in the same fashion as a spacecraft reentering the earth's atmosphere. Therefore the lead pair of reindeer will absorb 14.3 QUINTILLION joules of energy. Per second. Each. In short, they will burst into flame almost instantaneously, exposing the reindeer behind them thus vaporizing the entire reindeer team within .00426 of a second and all the while creating deafening sonic booms in their wake. Meanwhile, Santa will be subjected to centrifugal forces 17,500.06 times greater than gravity. Therefore, a slim 250-lb Santa would be pinned to the back of his sleigh by 4,315,015 lb. of force.

So if Santa ever DID deliver presents on Christmas Eve, there's no way he could've survived.

Sorry Virginia, sometimes facts are facts, but hopefully you never find out.

Sophie, you're up next, and you have all break to do it...

Mr. Jones is not Mr. French

2006-12-18T18:37:00.000-08:00

I have proven that Mr. Jones does indeed reside in Scotland.

His website is http://www.jonesieboy.co.uk, correct? Some of you have insinuated that Mr. French registered a UK hostname just to fool us. So, I decided to find the IP address of this website. The IP address is 80.68.95.86. This particular IP is assigned to the UK, and presumably Mr. Jones is from Scotland. Just to make even surer, I used the IP to find the host for his website.

As you can see, this host is located in the UK. If you want further proof, I can show you how I did this in class tomorrow. If you want even further proof, I will ping the host myself. There is no way that Mr. French could have done this himself. It defies the laws of computer science. I think it's safe to say that this issue is now resolved. See you all tomorrow!

Evan

MR JONES = MR FRENCH

2006-12-17T20:28:00.000-08:00

Because I have way too much time on my hands, I was checking to see if the YouTube was working, and stumbled across Mr. French's comment on Mr. Jones blog. Thought I'd share with you all!! Hahaha I love our class.

-Lauren

(Mr. French writing on Mr. Jones' blog)

Mr. French
Dec 18th, 2006 at 3:27 am

Hello Mr. Jones! It’s Mr. French from across the sea with the crazy calculus class. I just thought you’d like to know you’ve become something of a cult figure with my students. In fact, you made it into some of their calculus/math holiday songs this year! Here’s a link to one of their videos. The girl on the right is Sophie…
http://www.youtube.com/watch?v=zI2HPAfH5no
Thank you for your comments!
Mr. French

http://www.jonesieboy.co.uk/blog/2006/12/14/blogging-for-beginners-posting-and-commenting-tips/#comment-355

Our Calculus Song on YouTube!

2006-12-17T18:00:00.000-08:00

Hey everyone!!! I think that I have just successfully uploaded our Caluculus Christmas song to YouTube. If you are wondering, it IS Kristin Heintz laughing in the backround because she was too "Embarassed" to be in the video. Oh well... It's pretty much the tightest song ever! Enjoy! (Here is the link below, but if you click on the title, it will take you there too!)

http://www.youtube.com/watch?v=zI2HPAfH5no

-Lauren

5.3: The Fundamental Theorem of Calculus

2006-12-14T21:38:00.000-08:00

Hey guys, this is a brief update on the Fundamental Theorem of Calculus. We learned the first part of it in the previous section, which is:

If f is continuous on [a,b], then the function g defined by

is continuous on [a,b] and differentiable on (a,b) and g'(x) = f(x).

The Fundamental Theorem of Calculus Part II is as follows:

where F is the antiderivative of f so that F' = f.

Let's try a problem now, shall we?

So substitute 0 for a and (pi/2) in for b.

so the area of the graph is pi^2 +5.

sophie, i believe you are next up.

Thursday's Quiz Topics

2006-12-13T10:30:00.000-08:00

Here’s a list of topics that will be covered on this Thursday’s Quiz.

Quiz – Sections 5.1-2
Estimate distance traveled from a velocity graph. (5.1, #15)
Express a Riemann sum as a definite integral. (5.2, #17,19)
Evaluate an integral in terms of areas – show your work! (5.2 #37)
Sketch a graph and estimate the area under the curve using RRAM, LRAM or MRAM (5.1, #3)
Evaluate definite integrals based on a graph (5.2, #33)

I’ll be in early Thursday, available after school this afternoon until 4, and I’ll check in tonight online. See you in class!

In the spirit of the holidays:
Yes, Virginia, There is a Santa Claus
By Francis P. Church, first published in The New York Sun in 1897. [See The People’s Almanac, pp. 1358–9.]
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:

Dear Editor—
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?
Virginia O’Hanlon

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.

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.

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.

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.

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.

About the Exchange
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.
Thirty-six years after her letter was printed, Virginia O’Hanlon recalled the events that prompted her letter:
“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.
“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.
“ ‘Well, I’m just going to write The Sun and find out the real truth,’ I said to father.
“He said, ‘Go ahead, Virginia. I’m sure The Sun will give you the right answer, as it always does.’ ”
And so Virginia sat down and wrote her parents’ favorite newspaper.
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.
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.
“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.
Church married shortly after the editorial appeared. He died in April, 1906, leaving no children.
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.

5.2 The Definite Integral

2006-12-13T04:55:00.000-08:00

Oh my god. This is a huge disaster. My whole blog is teeny tiny! I think if you click on it, and then press the little magnifying glass with the plus in it, it gets bigger. God I hope so.

IN OTHER NEWS...Here is a cool link to a UT website. You can basically plug in the function, your interval, whether you want it to use LRAM, RRAM, or midpoints, the number of rectangles you want, and it will give you both the area of the rectangles as well as the integral value. I'm a big fan.

I generally hate anything that glorifies seniors as somehow superior to other students. I also love the Onion. So combine those two and you get this.

Alison, you are next!

5.1: Areas and Distances

2006-12-08T15:08:00.000-08:00

Here we go with Section 5.1, also known as finding the area on confusing graphs.

The first thing we talked about in class was how to find areas of different geometrical shapes. Of course, we can easily find the area of a triangle or a circle. But wait...how do we find the area of this graph?

Well let's take a look at a more specific example: estimate the area under the parabola y=x^2 from 2 to 4.

Well, we can't use a big triangle to find the area of this parabola because then the hypotenuse side would be curved. And sry Kate, we can't just not do the problem. But we can use rectangles! So let's divide the graph into four different strips (S1, S2, S3 and S4) by drawing vertical lines at x=2.5, x=3 and x=3.5.

Then if we make each of the strips rectangles, they will each have a base of 0.5. The parabola will look like this:

Now we can use these rectangles to estimate the area of the parabola. We know that the width of each rectangle is 0.5. The height is the y-value at each x coordinate. Therefore the heights are (2.5)^2, (3)^2, (3.5)^2 and (4)^2 respectively.
Let's use L4 to represent the sum of the areas of these rectangles.
L4 = (0.5)(6.25) + (0.5)(9) + (0.5)(12.25) + (0.5)(16)
L4 = 21.75
*Note that we don't include 2 as a height, because we can't find the y-value since it is not on the curve.*

However, part of the rectangles are drawn outside of the parabola. Therefore, the exact area of this parabola must be less than 21.75. If we make the rectangles smaller, we can find what the area has to be greater than.

Here the heights are (2)^2, (2.5)^2, (3)^2 and (3.5)^2 respectively.
Let's use R4 to represent the sum of the area of these rectangles.
R4 = (0.5)(4) + (0.5)(6.25) + (0.5)(9) + (0.5)(12.25)
R4 = 15.75
*Be careful NOT to inculude (4)^2 in the heights because the point (4, 16) is outside the rectangle.*

Therefore the area is greater than 15.75, but less than 21.75. If use the same parabola but generalize the interval, we can show that the sum of the areas of the rectangles approach 1/3.

The graph would be:

Each rectangle has a width of 1/n and the heights are the values of f(x)=x^2 at the points 1/n, 2/n, 3/n... The heights are (1/n)^2, (2/n)^2...

R4 = (1/n)(1/n)^2 + (1/n)(2/n)^2 + ......+ (1/n)(n/n)^2 R4 = (1/n)(1/n^2)(1^2 + 2^2 + 3^2+......+ n^2) R4 = (1/n^3)(1^2 + 2^2 + 3^2+......+ n^2)

We can use the formula for the sum of the squares of the first n positive integers:
1^2 + 2^2 +.....+n^2 = n(n+1)(2n+1)/6

After we substitute this expression for limit of Rn, we get:

If you think really really hard, you'll realize that if we keep on decreasing the width of the rectangles, the heights will eventually match the curve. Take a look:

The width of each rectangle in the interval [a,b] is b-a. The formula is:

Defintion: The area A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles:

Alright. Now that we know the formulas, let's try a problem. Find the area under the graph of f(x) = sin x between x = 0 and x = 1.

1) Find width
2) Estimate the area by taking the sample points to be midpoints & use 4 subintervals.

1) Since a=0 and a=1, the width is:

x1 = 1/n, x2 = 2/n, x3=3/n, xi=1i/n. The sum is:

2) n= 4 because we have 4 approximating rectangles. Therefore the width is:

Subintervals: [0, .25], [.25, .5], [.5, .75], [.75,1]

Midpoints are:

Finally we can calculate the sum:

The distance problem. Remember this equation: distance = velocity x time??? of course

But what if the velocity is hard to find? Well then we can find the area of the function! We simply make a table with our x-value as time and our y-value as velocity. (just like on pg. 376)

Now i bet you guys got lost back at the colorful graphs. Well no worries because this site is awesome. There are like 9 pages of explanations on how to find the area of any curve. http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/integration/area.html GO CHECK IT OUT! This site is good too: http://www.counton.org/alevel/pure/purtutintcur.htm

Kristin H! I believe you are up next!

Now for some Chistmas PUNS!: Which elf sings "Love me tender?" Santa's little Elvis!

Oh and this is just funny:

Solution to Question #1, Quiz 4.5-7

2006-12-06T20:34:00.000-08:00

Greetings all!

I've put together a solution to the first question on Quiz 4.5-7 and posted it to our class website. Before we spend time going over it online or in tutoring, take a look at my solution...

See you in the morning!

And remember, math doesn't have to be boring:

Thursday's Test Topics

2006-12-06T07:39:00.000-08:00

Here’s a list of topics that will be covered on this Thursday’s Chapter 4 Test.

Chapter 4 Test Topics
Identify absolute extrema (4.1, #29,41,49(
Mean Value Theorem (4.2, #9,11, bonus - #34)
Limits – Indeterminate form (4.4. #15,21,27)
Maximize Profit/Revenue/Cost (4.8, #9,11,13)
Antiderivatives (#4.10, #9,17,19)
Marginal Profit/Revenue/Cost (4.8, #9,11,13; 4.10, #35,37)
Interpretation/meaning of derivative concepts (Sections 4.1.,4.3.4.5,4.10)
Optimization (4.7, #7,11,15)
More antiderivatives(4.10, #35,37)
More antiderivatives (position/velocity/acceleration) (4.10, #61)
Interpretation of the graph of f ‘ (4.3, #31)

That’s it! I’ll be in early on Thursday, and available after school on Wednesday. See you in class!

"I shall be telling this with a sigh
Somewhere ages and ages hence:
Two roads diverged in a wood, and I --
I took the one less traveled by,
And that has made all the difference."

- Robert Frost
The Road Not Taken

4.10: Antiderivatives

2006-12-05T16:18:00.000-08:00

Hey guys!! So, I'm back to talk about the last section of Chapter 4: ANTIDERIVATIVES!! YAY! ok so let's go...

The book definition of an antiderivative:
A function F is called an antiderivative of f on interval I if

Now, let's take that definition and apply it to this next situation:

1.
if this is my derivative equation, what is the original f(x) equation?
YES!! it would be, its general form,

Now, you might be wanting to ask me, "Gianna, what does the antiderivative have to do with this? And, furthermore, what the heck does 'general form' mean?" To those questions, I would respond by saying that firstly, the antiderivate is the original f(x) equation so that answers your first question. As for the second one, an antiderivative's general form is, simply put, the generic one that contains the + C, which needs to become your new best friend by the way. Furthermore, the general form( general antiderivative) can be found easily. LOOK:

Now that we have this extra tool to help us, let's do some examples.

Ok, FANTASMIC!! Now that we are very comfortable with, or should be at least, the concept of the antiderivative, let's apply it to the real world questions about acceleration, velocity and position equations.

Ok so now that we have a full knowledge of the antiderivative, we can call ourselves COMPLETE PEOPLE!! YAY!! So, Kristin, good luck and here's a comic to help you get going:

4.8: Applications to Business and Economics

2006-12-04T11:19:00.000-08:00

If any of you used to think that calculus does not have to do with the real world you were mistaken!!! This lesson talks about using calculus concepts and applying them to economics, business, and money- things that can relate to our every day lives!

There are the different functions that were represented in this lesson- cost, marginal cost, average cost, demand (or price), revenue, marginal revenue, profit, and marginal profit. I know all of these sound really similar, like, what's the difference between revenue and marginal revenue? Well, the word marginal should make you think of a derivative. Because the revenue function is represented as R, the marginal revenue function is R'(x) (or the derivative of R). The same thing applies to the rest of the equations as well- ie. if profit is P marginal profit is P'(x) (or the derivative of P).

So here are the actual equations for the money functions:

Cost: C(x)
Marginal cost: C'(x)
Average cost: C(x)/ (x)
Demand: p(x) --- note that this p is lower case!
Revenue: R= xp(x)
Marginal revenue: R'(x)
Profit: P(x) = R(x) - C(x)
Marginal profit: P'(x)

Now that we know all the equations we can start to understand them!

Important things to remember:

-If the average cost is a minimum (if you want to minimize the cost of an item), the marginal cost must equal the average cost, so C'(x) = C(x) / x

-When you're talking about the marginal cost, you're talking about how much $ you are making by selling or producing items.

-The average cost is the total cost in terms of units.

-If the profit is a maximum (you want to maximize the profit), the marginal cost must equal the marginal revenue, so R'=C' and R'-C'=0.

A problem that might be asked:

The cost of producing x amount of ribbon is C(x)= 2000 + 10x + .005x^2.

A. Find the cost (C(x)) the average cost, (c(x)) and the marginal cost (C'(x))of producing 50, 500, and 5000 ribbons?

B. When will the average cost of the ribbons be the lowest?

First you have to find the equations for the cost, av. cost, and marginal cost.

You know the equation for the cost is C(x)=2000 + 10x + .02x^2.
The equation for the average cost is C(x) / x, so when you divide the cost equation by x you get: 2000/x + 10 + .005x. So c(x)= 2000/x + 10 + .005x.
The marginal cost is just the derivative, so C'(x)= 10 + .01x.

Now, because there are so many equations and you have to find different values, plug in 50, 500, and 5000 into each equation. You can even make table to make it easier.

So now we have part A answered! (The answers would be the numbers in the table)

On to part B.

The average cost of the ribbons will be the lowest when the marginal cost= average cost! So this means the equation we found for c(x) = C'(x).

2000/x + 10 + .005x = 10 + .01x

Subtract 10 from both sides.

2000/x + .005x = .01x

Subtract .005x from .01x.

2000/x = .005x

Multiply x to both sides.

.005x^2 = 2000

x= 632.45

Now, plug 632.45 back into the c(x) equation.

2000/ 632.45 + 10 + .005(632.45)

You get 16.32.

So the minimum average cost per ribbon is $16.32.

Now we're done with the question!

NEXT QUESTION:

This is another question you'll be asked- its like a word problem.

There's a Joshua Radin concert coming up next week at the Troubador. If the ticket price is $12, 2,000 tickets will be sold. But if the price is dropped to $10, 5,000 tickets will be sold. How many tickets should be sold to maximize the revenue?

Find the slope of (2000,12) and (5000,10) and put this into slope intercpet form.

(The slope is 2 / -3000 which simplifies to 1 / -1500)

p- p1 = m(x- x1)

p - 12 = - 1 / 1500 (x - 2000)

Distribute.

p-12= -1/1500x + 4 / 3

Add 12 to both sides.

p = -1/1500x + 40/3

Now that you have your demand function you have to find the revenue function.

The revenue function (R) = xp(x). So multiply the equation we just found by x. You get:

R= -1/1500x^2 + 40/3x

Find the marginal revenue function (take the derivative)

R' = - 1/750 x + 40/3

The maximum revenue is when the marginal revenue is equal to zero.

So, 0= R'= -1/750x + 40/3

-40/3= -1/750x

x= 10000

Now plug 1000 back into the equation that we have for p.

p = -1/1500(10000) + 40/3

p= 20

So to maximize ticket sales the price of the tickets must be $20!

(I can't believe that with a problem I made up the numbers for I got whole numbers for my answer! Yay!)

Here's an amazing link that explains marginal cost. It even talks to you and it really interactive, theres information on the site on fixed and variable costs & I don't think we're learning that, so just skip down to the bottom to the part on marginal cost. http://hspm.sph.sc.edu/COURSES/ECON/Cost/Cost.html

This site is good too: http://ingrimayne.com/econ/elasticity/RevEtDemand.html

I know it's a little hard to read so i'll just tell you what it says:
"What matters most in your pricing decision, average or marginal cost?"
And the guy in the car is saying: "Officer, my trip to Chicago only took 2 and 1/2 hours which equates to 60 miles per hour!" But the officer is saying he went 90 miles per hour. You can see this page to see how the speed the car is going relates to the marginal cost/revenue etc.
http://www.farin.com/media/postcards/TT_MustangSpeed.pdf

Gianna you're next!!!

Friday's Quiz Topics

2006-11-30T14:31:00.000-08:00

Here’s a list of topics that will be covered on this Friday’s Quiz. I’ll include the relevant homework problems as soon as I get a chance…

Quiz – Sections 4.5-7
Optimization (4.7, #7,11,28)
Calculus and Calculators (4.6, #37)
Analyze a graph based on equation (sketch, etc.) (4.5, #45,59)

I’ll be in early Friday, available after school this afternoon and I’ll check in tonight online. See you in class!

There's no point in being grown up if you can't be childish sometimes.
-Dr. Who

4.7: Optimization Problems!

2006-11-28T22:02:00.000-08:00

Hi, everybody. I hope the second day back from cozy break hasn’t been too jarring. Now, it’s that happy time again…time for me to do my blog, and this time it’s on Optimization Problems (which are quite similar to Related Rates, my previous blog).

Optimization are all about maximizing and minimizing something. To untangle all the information given to us in those tedious word problems, we can use a series of steps (Related Rates, anyone?):

1. Read the question carefully. Understand the information you are given and, most importantly, understand what is being asked.

2. Draw a diagram. Label everything. This is not absolutely necessary, but it’s usually helpful.

3. Set up a maximum or minimum equation. This step is kind of like identifying the static equation in related rates. This is the equation that you will modify (Step 5) and differentiate (Step 6).

4. Identify the constraints presented in the problem.

5. Use the constraints to rewrite the equation in terms of a single variable.

6. Find the derivative of the equation. Use this to identify critical points. Remember, critical points are wherever the derivative is equal to zero OR where the derivative is undefined.

7. Test the critical points you found, and the endpoints (these will usually have to do with the constraints), to determine which gives the absolute maximum or minimum.

And, of course, don’t forget to answer the question specifically being asked.

All right, since these are application problems, examples are always extremely helpful.

EXAMPLE 1
Find the point on the parabola y² = 2x that is closest to the point (1,4).

We’re looking for a point (x,y). Since we’re looking for closest distance, we can set up a minimum equation using the distance formula:

d = rad[(x -1)² + (y – 4)²]

This is the distance between the given point (1,4) and the unknown point (x,y).

We are given the equation for the parabola (this is a constraint). We can use this information to solve for x and get rid of the x variables in the equation so we can have an equation in terms of a single variable. (You could also solve for y and substitute that into the minimum equation). Solving for x:

y² = 2x
x = (y²) / 2

We can substitute this value of x into the minimum equation:

d = rad[((1/2)y² – 1)² + (y – 4)²]

To make this equation a little more manageable, we can square it to get rid of the radical:

d² = [(1/2)y² – 1]² + (y – 4)²= f(y)

Now we can differentiate:

f’(y) = 2[(1/2)y² – 1](y) + 2(y – 4)(1) = y³ – 8

Don’t forget about the Chain Rule!!

Now, we have to find our critical points. y³ – 8 is not undefined anywhere. We can look for where it equals 0, though.

y³ – 8 = 0
y³= 8
y = 2

Now we can use the first derivative test to determine whether y = 2 is an absolute maximum or minimum. You can graph the derivative, or simply think it through. When y is negative, y³ – 8 is less than 2. When y is positive, y³ – 8 is greater than 2. By the first derivative test for absolute extreme values, this means that 2 must be the absolute minimum of f(y).

We have our y value for the point closest to (1,4). To find our x value, we plug in our y value into the equation we solved for x.

x = (2²) / 2 = 2

So (answering the original question): the point on y² = 2x closest to (1,4) is (2,2).

EXAMPLE 2
Find the area of the largest rectangle that can be inscribed in a semicircle of radius r.

We can draw a picture to make the semicircle the upper half of a circle x² + y²= r² with its center being the origin. The rectangle inside would have a width of y on either side and a length of 2x on either side. Thus, the area is:

A = 2xy

Now we want only one variable. We can solve for y using the equation of a circle:
y² = r² – x²y = rad(r² – x²)

Substituting this back into our area equation:

A = 2x[rad(r² – x²)] domain: [0, r]

Now we differentiate the area equation:

A’ = 2 rad(r² – x²) – (2x)(1/2)(r² – x²)^(-1/2)(2x)
A’ = 2 rad(r² – x²) – (2x²)(r² – x²)^(-1/2) = 2 rad(r² – x²) – (2x²) / [rad(r² – x²)]
A’ = [2(r² – x²) – 2x²]/ [rad(r² – x²)]
A’ = [2(r² – 2x²)] / [rad(r² – x²)]

To find the critical points, we look at where A’ equals 0:
2(r² – 2x²) = 0
2r²– 4x² = 0
4x² = 2r²
x = r / (rad2)

Plug this value of x into the original area equation (remembering that we also solved for y earlier, so we have something to plug in for x and y):

A = 2[r / (rad2)][rad(r² – x²)]
A = 2[r / (rad2)][rad(r² – r²/2)]
A = r²

Before we assume this is our maximum value, we observe the behavior of the endpoints:

A(0) = 2(0)[rad(r² – 0²)]
A(0) = 0
And
A(r) = 2(r)[rad(r² – r²)] = 2r(0) = 0

This confirms that x = r / (rad2) gives us a maximum value of A. Now, to answer the question: the area of the largest inscribed rectangle is r².

The examples we did in class show the optimization problem procedures using more numbers (like the fence area one), whereas these show optimization problems in the coordinate plane, in a slightly more abstract way. It's useful to see this side of the optimization problems, but it's also good to keep in mind that optimization problems can be applied to real world situations. Our homework definitely gives us a chance to work with some, doesn't it?

Here's a link to a website with a lot of sample problems: good practice.

So, if you thought we had icky math tests...take a look at this:
VELOCIRAPTORS!!!!
Ah, yes.

Tessa, you're up next.

Ave atque vale, everyone. ~8)

4-6 Graphing with Calculus and Calculators

2006-11-27T18:06:00.000-08:00

Alright, this section is very similar to the previous section in that it deals with the graphs of functions and how calculus allows you to explore those functions. However, this section also talks about the presentation of a graph. Your calculator does not always give you the ideal window, for example. You can use calculus to make sure that you have included all the important parts of your graph in your final window.

Now for a sample problem!
Graph 3x^2-x+2 in a window that includes all important parts of the function.

Our first step is to find out all we can about the function using its first and second derivatives, following the procedure that Jean outlined so neatly in her post.

Increasing on: (0,inf)
Decreasing on: (-inf,0)
Local Maximum: none
Local Minimum: (.167,1.92)
Concave Up: (-inf,inf)
Concave Down: nowhere
Inflection Point(s): (.167,1.92)

You can see that the rather simple graph can be contained in a relatively small window. The ideal window would be something like x=[-2,2] and y=[+0 or -0 or 0,10]. Sorry about that; you got a compliment earlier. Here is the final graph:

Isabella, you are next in line for the joy that is blogging!

For those of you that are wondering, here are the next twenty digits:
pi=...62643383279502884197...

4.5: Summary of Curve Sketching

2006-11-19T14:19:00.000-08:00

'Ello, all! God willing, this will be a fairly short post, because, as the title should hint, this section is little more than a "summary" of the previous sections. While 4.1-4.4 laid down the guidelines to drawing and identifying graphs, 4.5 condenses thes guidelines into a nice, easy checklist. Ready for it? Okay:

When sketching a curve...

Determine the Domain - It's really helpful to start out knowing where x is and is not defined, so do that first--it'll lay the groundwork for everything else.
Determine the x and y intercepts - first, find the y-intercept by plugging in 0 for x. Then, solve for x when f(x) = 0, if the equation isn't too difficult to do so fairly quickly.
Determine the symmetry - There are three types of symmetrical functions: even functions, odd functions, and periodic functions. Even functions occur when f(x) = f(-x) and are symmetrical about the y-axis (like cosine graphs and even power functions). If we have an even function, our work is cut in half. We only have to determine the shape of the curve when x is positive. Then we reflect it across the axis and voila! Odd functions, like sine and odd power graphs, occur when f(x) = -f(x) and are symmetrical about the origin. Thus, when dealing with odd functions, we simply determine the shape of the graph when x is positive, then rotate it 180 degrees about the origin. Periodic functions, occur when f(x + p) = f(x) where p is a positive constant called the period. With this graph, we simply have to know what the graph looks like for one interval of length p. Then we can translate that interval to sketch the entire graph. Both sine and cosine graphs are periodic.
Find the asymptotes - The three types of asymptotes (only two of which we have seen before) are: horizontal asymptotes, vertical asymptotes, and slant asymptotes. A review of asymptotes can be found in 2.6. Find and draw in the asymptotes.
Test for intervals of increase and decrease - To compute intervals of increase and decrese, you need the I/D test, or first derivative test. Find the intervals at which f ' is positive to determine when x is increasing and find the intervals at which f ' is negative to determine when x is decreasing.
Calculate local maximums and minimums - By the same first derivative test, you can find the value for which f ' = 0 and note wheter the value is going from negative to positive (minimum) or positive to negative (maximum) or not changing signs (neither local max nor min).
Find concavity and points of inflection (POI) - Use the second derivative test to dertermine concavity. When f " is greater than 0, the graph is concave upward, and when f " is less than 0, the graph is concave downward. The point at which f " changes from concave up to concave down (or vice versa) is the POI.
SKETCH! - once you've figured out all this information, DRAW away!

Okay, so there's the checklist. Now, let's walk through it with some real numbers. Here's the problem:

Step 1 - Domain/Undefined Values: x is undefined at +/- 1

Step 2 - X/Y Intercepts: the x and y intercepts are both 0

Step 3 - Symmetry: f(-x) = f(x), so the curve is even, which means it is symmetric about the y-axis.

Step 4 - Horizontal and Vertical Asymptotes: this function has both horizontal and vertical asymptotes.

SO, y = 2 is a horizontal asymptote. Now let's look at vertical asymptotes:

This means that x =1 and x = -1 are vertical asymptotes.

Step 5 - Intervals of Increase and Decrease: Using the first derivative test, we find...

Since f '(x) is greater than 0 when x is less than 0 (and not equal to -1!), and f '(x) is less than 0 when x is greater than 0 (and not equal to 1!), it can be determined that f (x) is increasing between negative infinity and -1, and between -1 and 0, and is increasing between 0 and 1, and 1 and infinity.

Step 6 - Local Maximums and Minimums: f ' (x) changes from positive to negative at 0 and only 0, so that is the only critical number. It is determined, by the first derivative test, that it is also the local maximum.

Step 7 - Concavity and Points of Inflection:

Thus, the curve is concave upward between negative infinity and -1 and between 1 and infinity, and concave downward between -1 and 1. Because 1 and -1 are not in f's domain, it has no points of inflection.

Step 8 - SKETCH!: now, take all of the bolded conclusions drawn from the previous 7 steps and sketch a graph that works with those conclusions. Got it? Good!

For more practice or a second opinion, a this is a great site.

Yay for short blogs! Actually, it got rather long, didn't it? Long but all summary, a nice simple section. Have a nice weekend, you guys!

I can't help but wonder... was this what Mr. French was like when he was little?

Your turn, MARK!

Friday's Quiz Topics

2006-11-16T13:08:00.000-08:00

Here’s a list of topics that will be covered on this Friday’s Quiz. I’ll include the relevant homework problems as soon as I get a chance…

Quiz – Sections 4.1-4
Find critical numbers of a function (Sec. 4.1, #41)
Verify a function satisfies conditions of Rolle’s Theorem or the Mean Value Theorem, then solve for “c” (Sec. 4.2, #1,11)
L’Hopital’s Rule – evaluate limits (Sec. 4.4, #15,21,47)
Sketch a graph given continuity and max/min conditions (Sec. 4.1, #7,11)
Analyze a function given an equation: determine increasing/decreasing intervals, max/min values, concavity intervals, points of inflection (Sec. 4.1, #29,49, and any of the questions in Sec. 4.5)
Analyze and draw a graph of a function given the graph of the derivative. (Sec. 4.3, #5,7,31)

I’ll be in early Friday. See you in class!

"Seven days without laughter make one weak."
-Joel Goodman

4.4 Indeterminate Forms and L'Hospital's Rule

2006-11-15T17:52:00.000-08:00

Hey everyone. I guess I’m here to help you all learn about the wonders of section 4.4 – Indeterminate Forms and L’Hospital’s Rule.

The point of this lesson is to find out what’s going on near the ends of the graph at infinity.

Okay, I guess we’ll be doing an example problem to learn this lesson quickly.

Example Problem:

We’re going to solve this using L’Hospital’s Rule:

***L’Hospital’s Rule***

Suppose f and g are differentiable and

g’(x)≠ 0 near a (except possibly at a). Suppose that:

and

or that

and

(In other words, we have an indeterminate form of typeor

If the limit on the right side exists (or is infinite or negative infinite).

Now, you always have to remember that the limit at infinity depends on how fast the top grows compared to how fast the bottom grows. And another thing to keep in mind, L’Hospital’s Rule can only be used when the limits are indeterminate.

There are 7 indeterminate forms of limits that we know of:

Although there are 7 indeterminate forms of limits, it’s usually best to try to change any the last 5 forms to one of the first two forms :

Now…you may be asking, “HOW do we do this?” It’s simple. First of all, we can see that if we had left the equation (ln x) / (x-1) alone and tried to find the limit by simply plugging “infinity” into ‘x’, the limit would become infinitely large. Eventually, “ln x” would become “infinity,” and “x-1” would also become “infinity.” Hopefully you realize that is one of our indeterminate forms. Now we know that we can use L’Hospital’s Rule!

So using L’Hospital’s Rule, we rewrite the derivative of “ln x” as (1/x)(1) and the derivative of “x-1” as 1.

To find the derivative of this whole limit of a function, we don’t use the quotient rule. Instead, we take the derivative of the top (ln x) and divide it by the derivative of the bottom (x-1).

Eventually, we can find that the limit, as x approaches infinity, is zero.

To simplify the definition of L’Hospital’s Rule, here’s an equation to help you figure this thing out:

Well, I hope you understand all of this…but in case you don’t, we’ll try some different problems that use The Rule (sorry, I just don’t want to constantly type out L’Hospital’s Rule…which I just did…again. AHHHHHH!!!).

Example:

Using THE Rule, we find that the derivative of (3x² + 4) is 6x, and the derivative of (2x² + 7) is 4x. So we just put the top derivative over the bottom derivative, like this , cancel out the x’s, do some simplifying…and PRESTO! We find that the limit is: Another problem like this:

We used the Rule to find that the limit, as x approaches infinity, is 4! Huzzah!

Okay, last sample problem…this one’s a little different from the last two we’ve had.

Example:

Without using the Rule, we would find that the limit would be:

Now, don’t make this mistake. Although you might think that zero multiplied with negative infinity would equal zero, it doesn’t. Zero multiplied with negative infinity is actually an indeterminate form of limits. That means that we can use the RULE!

So we will re-do this whole thing using L’Hospital’s Rule:

First, let’s start off by showing that the limit becomes an indeterminate form of limits.

We can rewrite x so that it does not end up equaling zero.

X =

Now that this is in an indeterminate form that we like, we use L’Hospital’s Rule:

Derivative of (ln x) divided by the derivative of (x^-1) So we see that as x approaches 0 from the right side, the limit becomes -(0).

And no, Jean, this does not “become negative zero”…it is simply zero.

Hopefully you all understand this lesson now because I know I DO!

BUT, in case you still need help, you can go to these sites:

Site#1

Site#2

Even though I had such a fun time using Equation Editor 41 times and uploading 35 pictures multiple times, it is definitely time to stop before I get addicted to blogging. So here’s a reminder to Jean that she is up next! Hurrah!

sorry about the blacked out equation...you can just click on it to see the equation...o, and sorry about this blue-underlined sentence...I don't know why it's doing this, but it's really annoying and I want it to stop...but it won't. So....I can't do anything.

4.3: How Derivatives Affect the Shape of a Graph

2006-11-14T15:36:00.000-08:00

The title for Section 4.3 is, in my opinion, quite vague. The concepts, however, are specific, and they clear up a lot of things, such as what derivatives and second derivatives actually mean with respect to a graph's appearance. Let's get started. We'll talk first about, well, first derivatives.

The Increasing/Decreasing Test is another one of those things that's quite simple and easy once you hear it, like the Intermediate Value Theorem. It states that

1) If f ' > 0, then f is increasing, and
2) If f ' <>

This makes perfect sense, because we know that f ' represents the slope of the line tangent to f. So, if the slope of f is positive, the y-values must be increasing as the x values increase, and if the slope is negative, the y-values must be decreasing as the x-values increase.

Now, we have the First Derivative Test, which is pretty self-explanatory. For this test, we must first have c represent a critical point on the function f. Reminder: a critical point is a point on the graph of f where either f '(c) = 0, or f '(c) does not exist.

The First Derivative Test says ...
1) If f ' changes from positive to negative at c, then a local maximum of f is at c.
2) If f ' changes from negative to positive at c, then a local minimum of f is at c.
3) If f ' doesn't change signs at c, then there is no local maximum nor local minimum of f at c.

All right then, let's move on to second derivatives. We know that while the first derivative tells us where a function increases and decreases, the second derivative (the derivative of the first derivative) tells us how the slope of the original function changes.

Before talking about how the second derivative affects a function, we must first define concavity, because, as I’ll explain momentarily, it relates to second derivatives.

On an interval, the graph of f is concave up when its tangents lines lie below the curve,
And the graph of f is concave down when its tangent lines lie above the curve.

That being said, we can use the second derivative to determine where a graph is concave up or concave down. The Concavity Test says …

1) If f ">0, then the graph is concave up.
2) If f "<0,>

As a side note, a point of inflection, a concept we learned in Precalc whose origins were somewhat fuzzy, is a point on a graph f at which the curve changes concavity, or, in other words, where f " changes signs.

The Concavity test leads us to our Second Derivative Test, which states ...

At a critical point c, with f '(c) = 0,
If f "> 0, then a local minimum of f is at c, and
If f "<>f is at c.

Mr. French gives us a helpful and, I must admit, quite fantastic way to remember this. If f "> 0, or, in other words, if f" is positive, then it is happy, and makes a Smiley Face. The Mouth of a Smiley Face is concave up, and therefore has a local minimum. Huzzah for pneumonic devices!

EXAMPLE

With the information I just explained, we can easily sketch the graph of a function.
Let's try y = (x^5) - 5(x^4)

First, find the first a second derivatives.

y' = 5(x^4) - 20(x^3)
y" = 20(x^3) - 60(x^2)

Next, factor the first derivative and find the zeros to find the graph's critical points. Don't forget to check for place where f ' does not exist. We know there aren't any critical points like that with this function, because it is a polynomial.

y' = 5(x^4) - 20(x^3) = 5(x^3)(x - 4) ==>The critical points are at x=0,4.

Next use either the first or second derivative test to determine whether they are maximums or minumums. Personally, I advocate the second derivative test, as it is much faster and simpler. Plug in the x values to the equation for the second derivative.

y" = 20(x^3) - 60(x^2) = (20(0)^3) - (60(0)^2) = 0. This is neither positive nor negative, so there is neither a maximum nor a minimum.
y" = 20(x^3) - 60(x^2) = (20(4)^3) - (60(4)^2) = 320. This value is positive, so there is a local minimum.

Now, let's plug 0 and 4 into our original equation to find the coordinates. When we do this, we get the coordinates (0,0) and (4, -256).

Next, factor the equation for the second derivative to find the critical points. In the second derivative, the critical points are places where the original function may change concavity, having a point of inflection.

y" = 20(x^3) - 60(x^2) = 20(x^2)(x - 3) ==> The critical points are at x= 0,3. By plugging in values between and outside of those points, we find that before zero, the second derivative is negative, between zero and three, it is still negative, and beyond three, it is positive. Thus, the graph of y changes from concave down to concave up at x=3. By plugging into the original equation, we find that the specific point of inflection is (3,-162).

Now that we know our points of inflection, the specific local minimum, where the graph is decreasing and increasing, and where it is concave up and concave down, we may sketch the function. Using these new processes, we can break down complicated functions to their essential pieces of behavioral information, and use those essential pieces to sketch.

Here is a website that explains quite well how second derivatives relate to concavity and points of inflection.

John the Prophet, you are now John the Scribe for Lesson 4.4 tomorrow. Don't forget!

Here's some advice that I think we could all do good to hear every once in a while. It comes from the hilarious Douglas Adams.

Chapter 3 Test, #10

2006-11-14T15:28:00.000-08:00

Sorry I'm late, guys.

Anyway, the problem gives us the function

and asks us to find

In order to do this, let's first find the higher order derivatives of the function. Remember the chain rule!

The derivative of cos(2x) is:

Find the next derivative:

And the next:

Finally, it repeats itself, but with a higher coefficient.

So, since it repeats at the fourth derivative, we divide the derivative we are looking for, 1428, by 4. And, conveniently, 4 goes into 1428 evenly. This means that we have to use the 4th derivative equation. If you will notice, with each derivative, the coefficent is two to the the power of the number of the derivative.

So, the equation is

And there you have it!

Formal Complaint

2006-11-13T21:12:00.000-08:00

I'd like to file a Formal Complaint, Mr. French. Why is it that the AB Calculus C Blog gets to have BACKGROUND MUSIC and ours doesn't? WHAT TYRANNICAL INJUSTICE.

Chapter 3 Test, problem 13

2006-11-11T11:17:00.000-08:00

At noon, ship A is 200 km west of ship B. Ship A is sailing south at 34 km/hr and ship B is sailing north at 21 km/hr. How fast is the distance between the ships changing at 8:00 pm? Round the result to the nearest thousandth if necessary.

Step 1: Draw a picture and label all the parts.
The picture will be inserted on Tuesday...oh the mac fun.

Step 2: Find the static equation. Then plug in all constants.
(x^2) + (y^2) = (d^2)
(200^2) +(y^2) = (d^2)

Step 3: Find the information provided.
x = 200 km = constant
t = 8 hrs
y = (21 km/hr)(8 hrs) + (34 km/hr)(8 hrs) = 440 km
(d^2) = (200^2) + (440^2)
d = 483.3218 km
dy/dt = 34 km/hr - (-21 km/hr) = 55 km/hr

Step 4: Find the dynamic equation. Then plug in all known information.
0 + 2y(dy/dt) = 2d(dy/dt)
2(440 km)(55 km/hr) = 2(483.3218 km)(dd/dt)
dd/dt = 50.070 km/hr

Answer = 50.070 km/hr

SEE YOU ALL AT PREPSTOCK I HOPE!!!

Chapter 3 Test # 2

2006-11-10T00:01:00.000-08:00

Differentiate:
y = e^cos(3x)

Solution: Remember to use the CHAIN RULE!
1) Differentiate the outside function.
e^cos3x
2) Multiple solution from #1 by the derivative of the inside function.
(e^cos3x)(-sin3x)
3) We must also use Chain Rule for -sin3x. So....
(e^cos3x)(-sin3x)(3)
4) So...
y' = -3(e^(cos3x)) * sin(3x)

Answer to Chapter 3 Test, #8

2006-11-09T23:26:00.000-08:00

This is a quintessential Quotient Rule problem. We all remember Quotient Rule?

So first let’s find all of the pieces we need:

Numerator (“HI”): u
Denominator (“LO”): u2-1
Derivative of Numerator (“dHI”): 1
Derivative of Denominator (“dLO”): 2u

Now it's just simple plug-and-chug.

Yahoo!

Chapter 3 test, Question # 12

2006-11-09T22:58:00.000-08:00

a) Use linear approximation techniques to estimate 256.8^(1/4). Leave your answer as a number plus or minus a fraction. Show your work.

b) Determine the calculator-generated value of 256.8^(1/4).

c) To how many decimal places is your estimate similar to the calculator-generated value?

Solutions
First make a general equation: f(x) = x^(1/4)
Find the derivative of that function: f’(x) = (1/4)x^(-3/4)
Choose an a close to the original x-value that gives a nice answer: a = 256
Substitute the a-value into f(x): f(a) = 4
Substitute the a-value into f’(x): f’(a) = (1/4)256^(-3/4) = 1/256
Use the equation for linear approximation: L(x) = f(a) + f’(a) (x-a)
L(x) = 4 + (1/256)(.8) = 4 + (1/320)

a) Answer: 4 + (1/320)

Enter in calculator: 256.8^(1/4) = 4.00312…
b) Answer: 4.00312…

Compare the two answers
c) Answer: 5 decimals

Test Question #7

2006-11-09T22:51:00.000-08:00

Test Question #7

U(x) = (8x + 4)^11 (7x2 + 9x – 4)^3

This problem requires both the product rule and the chain rule.

U’(x) = ((8x + 4)^11) (3(7x2 + 9x – 4)^2 (14x + 9) + ((7x2 + 9x – 4)^2) (11(8x + 4)^10) (8)

So, In words…

It is the first half (8x + 4)^11

Times the derivative of the 2nd half (7x2 + 9x – 4)^3 [Remember chain rule!!]

Added to

The 2nd half (7x2 + 9x – 4)^3

Times the derivative of the fist half

ALL SET!!!

O, by the way, all of these equations looked absolutely brillant on my incredible MAC, but seeing as blogger is biased against MACs, and gives undeserved preferential treatment to PCs, my equation now looks like I am a 5th grade computer skills failure. Just thought you might want to know.

-Lauren

Chapter 3 Test, Question # 5

2006-11-09T22:47:00.000-08:00

5) Find the derivative of the following function: y = 9^7x

Ok so first we should realize that there is a variable in the exponent. So, we need to bring the x variable down by using logarithmic properties.

Step #1: Take the ln (natural log) of both sides of the equation. The property states: lnb(x^n) = n lnbx. Then we have the equation: ln y = (7x)(ln9).

Step #2: Take the derivative of new equation. The derivate is: (1/y)(y') = (7)(ln9)

Step #3: Isolate or solve for the y' by multiplying both sides by y. Then we get: y' = 7(ln 9)(y')

Step #4: We're almost there! Now dont forget to replace the y with the original function. The final answer is y' = (7)(9^7x)(ln9). The answer was B!!!

Be careful in this problem! DONT USE THE POWER RULE! Choice number A was the answer if we had taken the power rule...which is wrong wrong wrong!

Hope that helped everyone! Good luck.

Test Question #17

2006-11-09T22:24:00.000-08:00

#16
So, once you have completed #14, you have the two equations v=-4t^3+12t^2 and a= -12^2+24t. You graph those two equations and find all the zeroes because these points are the points at which the particle is changing from speeding up or slowing down to slowing down or speeding up.

When the particle is slowing down, the velocity will be positive and the acceleration will be negative, or the velocity will be negative when the acceleration is positive. For t<0, velocity is positive and acceleration is negative, so this is part of our final domain. At t=2, acceleration becomes negative while velocity is still positive. But at t=3, velocity becomes negative. So, this can be written 2<t<3. The correct notation for the final domain is t<0 U 2<t<3.

Chapter 3 Test, Question #1

2006-11-09T21:59:00.002-08:00

Differentiate:

Here we must first recognize that we are dealing with the product rule, meaning that we must multiply the derivative of the outside function by the inside function and then add the outside function times the derivative of the inside function. So:

Derivative of is by the derivative of exponential rule.

Derivative of is by the definitions of trigonometric derivatives.

Therefore by putting the information we have into the product rule, we get:

which, rearranged, looks just like answer (c) which is:

Chapter 3 Test #11

2006-11-09T21:58:00.000-08:00

11. (25 points) Use implicit differentiation to determine all points on the ellipse where the tangent line is undefined.

x² + xy + y² = 12

First, we have to differentiate the given equation (since what we’re going to observe is the slope of the tangent line, which is the derivative of the equation):

2x + (xy’ + y (1)) + 2yy’ = 0

(Remember to use the PRODUCT RULE for xy)

Get all y’ terms on one side:

xy’ + 2yy’ = -2x – y

Solve for y’:

y’ = -2x – y

x + 2y

We’re looking for when the slope of the tangent line is undefined. That is, when is y’ undefined? Since y’ is equal to a fraction, it would be undefined when the denominator is equal to 0:

x + 2y = 0

Solving for x: x = -2y

Now we can substitute –2y for all x values in the original equation:

(-2y)² + (-2y)(y) + y² = 12

Solving for y:

4y² + (-2y²) + y² = 12

3y² = 12

y² = 4

y = + or – 2 (don’t forget that square roots are going to be + and –)

These are the values of y of the points where the slope of the tangent line is undefined. To find the x values at these points, we can plug each y value into the x = -2y equation.

x = -2(2) = -4 Point #1: (-4, 2)

x = -2(-2) = 4 Point #2: (4, 2)

These are the points where the slope of the tangent line (i.e. the derivative of the original equation) is undefined.

Hope that helps everybody. Tomorrow's Friday, party up the wazoo! ~8)

Chapter 3 Test: #16

2006-11-09T21:52:00.000-08:00

A particle moves along the x-axis, its position at time t given by where t is measured in seconds and x in meters.

To find when the particle is speeding up, you first have to find both the velocity and the accelleration equations.

To find the velocity equation, you take the first derivative of which is because of the power rule.

To find the accelleration equation, you take the second derivative of which is

Then, you have to find the zero's for each equation. What I did was plug it into my Quadratic equation solver in my calculator. So, do that and you'll find that the zeros for are 0 and 3, and the zeros for are 0 and 2.

Now, you have to figure out whether you get positive or negative values between each zero. So, I drew two lines like this:

(Pretend that there is a line connecting 0 and 3.... I dont know what happened to it)

Where did those positive and negative signs come from? Look:

Just focus on the velocity equation, and plug in a number between its zeros (between 0 and 3) and you'll get a POSITIVE number. Now plug in a number that's greater than 3. You'll get a NEGATIVE number.

Look at the accelleration equation now. Plug in a number between its zeros (0 and 2) and you'll see that its POSITIVE. Plug in a number greater than two, and it's NEGATIVE.

The particle is speeding up when BOTH accelleration and velocity is positive, or when both are negative. As you can see, both are positive between 0 and 2. And both are negative when a number is greater than 3. So, your answer is

2 > t > 0 or t > 3

I hope that helped!

PS- I found out a really easy way to upload equations/pictures onto here!!! You just press "add image" and then you can add one directly from your computer! You don't even need photobucket. I'm sure everyone else already knew this but I didn't, so yay it was an exciting realization for me.

-Tessa

Chapter 3 Test: Question #15

2006-11-09T21:37:00.000-08:00

Having gone through Kate’s process of determining the acceleration, which is also the second derivative, we now know that the equation for the acceleration is:

In order to find the acceleration in meters per second at time(t) = 3 seconds, we simply plug in 3 for the t-value:

Then, of course, we simplify:

Eventually, we’ll find that the acceleration at 3 seconds is -36 meters/second.

Chapter 3 Test #6

2006-11-09T20:47:00.000-08:00

A table of values for f, g, f', and g' is given. If h(x)=f(g(x)), find h'(1).
x...f(x)...g(x)...f'(x)..g'(x)
1....-5.....3.......-2......-3
2.....4......2........1.....-10
3....10.....6........9......-9

so, since you are trying to find the derivative of h'(1), the general equation for the h'(x) is:

then, plug in 1 for x to get:

next, look at the chart above to find g(1), g'(1), and f'(whatever g(1) equals). you get:

knowing these values, plug in those values for g(1), g'(1), and f'(whatever g(1) equals) respectively:

thus...h'(1) = -27

and theree you go.

Chapter 3 Test Question 9

2006-11-09T20:16:00.000-08:00

Differentiate the function:

y = ln( ((6x^9 + 6x)^2(5x^3 + 9))/(5x^9 + 5x)^(1/2) )

Sorry about the super-primative notation. Equation editor is suddenly nonexistent on my computer. But okie dokie...

First, use the laws of logarithms to make this messy thing simpler to deal with:

Reminder: When one number (or expression) in the ln is divided by a number, you can subtract the ln of the bottom by the ln of the top. When one number (or expression) in the ln is multiplied by another, you can add the ln of one expression to the ln of the other. Finally, when a number is in the exponent of an ln, you can multiply the entire ln by that number. Thus...

y = ln( (6x^9 + 6x)^2 (5x^3 + 9) ) - ln( (5x^9 + 5x)^(1/2) )
y = ln( (6x^9 + 6x)^2) + ln(5x^3 +9) - ln( (5x^9 + 5x)^(1/2) )
y = 2ln(6x^9 + 6x) + ln(5x^3 +9) - (1/2)ln(5x^9 +5x)

THEN, now that you have a nice, straightforward (albeit rather ugly) equation, you can take the derivative, keeping in mind that when dealing with ln, y' = (g'(x))/(g(x)). So, let's break it up...

g(x) = (6x^9 + 6x).....SO.....g'(x) = (54x^8 + 6)
h(x) = (5x^3 + 9).....SO.....h'(x) = (15x^2)
k(x) = (5x^9 + 5x).....SO.....k'(x) = (45x^8 + 5)

Put that all together into the equation -- g'(x)/g(x) -- then factor in the coefficients using the Chain Rule, and what do you get?

y' = 2( (54x^8 + 6) / (6x^9 + 6x) ) + ( (15x^2) / (5x^3 + 9) ) - (1/2)( (45x^8 + 5) / (5x^9 + 5x) )

Yay for being finished!

Chapter 4, Lesson 2

2006-11-09T20:09:00.000-08:00

I have honestly pondered this for years...since the first time I read it...

Ready for Round Two with Tooch?

So we went over two new theorems: The Mean Value Theorem ("mean" as in average, not "mean" as in not nice), and Rolle's Theorem. Let's review Rolle's theorem first, shall we?

Rolle's Theorem has three conditions:
Firstly, the function f must be continuous on the closed interval [a, b].
Secondly, the function f must be differentiable on the open interval (a,b)-->remember, a differentiable function means that every point is approachable from both sides, so this part must be an open interval
Thirdly and Lastly, f(a) = f (b) (This part is pretty definitive for the theorem)

If all these conditions are found to be true, then there exists at least one value c where f'(c) = 0

So, can we all see how that works? I can't get the picture to upload right now, but I plan on conferring with Mr French to fix that by Tuesday...

But think about it: if the end points of your graph have the same y-value, and the graph is continuous and differentiable, wouldn't there have to be a point where the slope is zero, because it has to return to the origianl y-value at some point.

All together now: Ooooooh, yeah...

So how would this theorem break down?
1.) If it is discontinuous (removable, asymptotic, or jump discontinuity).
2.) If it is differentiable (there is a cusp or kink). On a completely unrelated note, we missed you in class today, James. Hope you're enjoying St. Louis!
3.)

Let's look at this last one a bit. Suppose the y-values for a and b were different. Then what? If we look at the first theory, we can figure out that the line segment connecting a and b is 0, the same as the derivative at point c. So they are parallel, yes? (Do we remember Geometry? We have all three of the Geometry Senior Leaders in our section, so you should!) Anywho, if the endpoints' y-values are different, there's still a line connecting them, yes? Would it not be logical to assume that there is at least one point with a tangent line parallel to this new line?

My friends, welcome to the Mean Value Theorem.

If function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one value c such that

In other words, the slope of f'(c) is equal to the secant line between a and b.

I found this website that re-explains everything and provides sample problems to test your wit. Give it a shot if you're still confused.

Excellent...let's put it to use!

Here is a COMPLETELY ORIGINAL (not to brag or anything) sample problem, complete with solution.

I'm going to go through it in words, and then I'll give you the numbers. How do you like them apples?

We have f(x) = 2x^4 + 3x on the closed interval [0, 2]. How can we find (c, f(c)), the point where the tangent line is parallel to the secant line?

Let's start by plugging in both endpoint values We find that f(0) = 0 and f(2) = 38. From here, we do two things. Firstly, we find the derivative of x, using all of our miraculous rules and such. Then, we find the slope of the tangent line at point c, which we know is equal to that of the secant line through a and b. We then set the two equations equal to each other and solve for x, which will give us the x-value for c. We can then plug c back into our original equation to solve for y. Yay!

And our final answer: (1.259, 8.819) Yeah, I know, it's not very pretty, but deal with it. Did you make up the question?

One last thing before I go. Here's another way to conceptualize it. We have an unknown function, but we know that f(0) = 2, and that f'(x) is greater than or equal to 3 for all values of x. What is the smallest value possible for f(5)?

Here's the guidelines: You have the values for a, b, and f(a). Set f'(c) equal to 5 and solve for f(b).

I challenge you to solve it, and the first person who posts back the correct answer will get a small prize and tons of bragging rights (No, Mr French, you can't win, but I will be checking my answer with you.)

Taylor, you're up next! Have a good weekend, and everybody come to PrepStock!

Chapter 3 Test, Question 14

2006-11-09T18:31:00.000-08:00

A particle moves along the x-axis, its position at time t given by where t is measured in seconds and x in meters.

-->>14. Find the acceleration at time t.

The position equation simply tells us where an object is after t seconds with respect to the origin/starting point. The derivative gives us the rate of change of a function. Therefore, the derivative will give us the rate of change of position, or velocity.

We can take the derivative of the position equation to find the velocity equation.

The rate of change in velocity is acceleration. If we take the second derivative of the position equation (the derivative of the derivative), then we can find the acceleration equation.

Ta-da!

Chapter Three Test, Question Three

2006-11-09T17:11:00.000-08:00

3. Differentiate the function: f (theta) = ln(cos 7(theta))

a) f ' (theta) = -7 tan (7(theta))

b) f ' (theta) = sec (7(theta))

c) f ' (theta) = -7 cot (7(theta))

d) f ' (theta) = 7 sec (7(theta))

So, let's differentiate. Because we have three different functions inside of each other, we must use the Chain Rule twice when taking the derivative. For the sake of brevity, I will represent theta using "t".

Step 1) f (t) = ln(cos 7t)

We know that the derivative of ln(x) is simply (1/x). Also, we know that the derivative of cos(x) is -sin(x). Therefore, by the chain rule, we get ...

Step 2) f ' (t) = (1/cos 7t)(-sin 7t)(7)

Now, simplify. Keep in mind that (sin(x) / cos(x)) = tan(x)

Step 3) f ' (t) = -7((sin(7t)/(cos(7t)))

Step 4) f ' (t) = -7 tan(7t)

Thus, the correct answer is choice "A."

Chapter 3 Test Question #4

2006-11-09T16:39:00.000-08:00

Differentiate the function:

a)

b)

c)

d)

The answer to this question is A. Let me show you how i got that answer:

As we all know from our extensive knowledge of Calculus,

So, let's apply this knowledge to this problem. In this case, our a is 10 and our x is

Now, let's put this into the logarithmic equation we have above.

GOOD!! ok, but wait the tricky part is that this is not the answer. We must now apply the Chain Rule so, by definition of the chain rule we multiply the derivative of the whole thing by the derivative of the inside function.

since the derivative of the inside function(3x²+4x-8) is 6x+4. Now, we just do the multiplication and end up with answer A,

THERE YOU GO!! GOOD LUCK REVIEWING!!
--Gianna ;)

Chapter 3 Test Topics - Updated!

2006-11-06T22:18:00.000-08:00

Here’s a list of topics that will be covered on this Wednesday’s Chapter 3 Test. It’s late, so I’ll try to add which homework problems are especially relevant tomorrow…

Chapter 3 Test Topics:
Chain Rule – Power/Polynomial (Sec. 3.5, #7,43)
Chain Rule – Exponential (e) (Sec. 3.1,3.5, #23)
Quotient Rule – Polynomials (Sec. 3.2, #13,23)
Product Rule – Exponentials/Trig (Sec. 3.4, #7, Sec. 3.5, #23)
Chain Rule – Natural Logs/Trig (Sec. 3.8, #3)
Derivative of a Natural Log (properties of logs, chain rule) (Sec. 3.8, #41)
Chain Rule – Exponential (not e) (Sec. 3.5, Calc Concept 13)
Derivative of a Logarithmic Function (not natural log) (Sec. 3.8, #23)
Higher Order Derivatives – Trig (Sec. 3.7, #39)
Higher Order Derivatives – Postion/Velocity/Acceleration/Speed (Sec. 3.7, #49)
Linear Approximation. Estimate a value and compare to calculator. (Sec. 3.11, #31,35)
Implicit Differentiation (Sec. 3.6, #27,31,35)
Related Rates (Sec. 3.10, #13)
Chain Rule – Table (Sec. 3.5, #53)

I’ll be in early on Wednesday, and available after school on Tuesday. See you in class!

An Old Cherokee describes an experience going on inside himself....
"It is a terrible fight and it is between two wolves.
One is evil - he is anger, envy, sorrow, regret,
greed, arrogance, self-pity, guilt, resentment,
inferiority, lies, false pride, superiority, and ego.

The other is good - he is joy, peace, love, hope,
serenity, humility, kindness, benevolence, empathy,
generosity, truth, compassion, and faith. This same
fight is going on inside you - and inside every
other person, too."

The grandson thought about it for a minute and then asked his grandfather, "Which wolf will win?"

The old Cherokee simply replied, "The one you feed."

4.1: Maximum and Minimum Values

2006-11-06T19:10:00.000-08:00

I, the Lord Protector of Math, have returned. Unfortunately, my beast of a custom-built computer has died and I am reduced to working on this Mac. I don't have as much technical ability with Macs, so bear with me.

Well, thankfully for you guys, this section is pretty straightforward.

Let's start with a definition.

A function f has an absolute maximum (or global maximum) at c if f(c) is greater than or equal to f(x) for all x in the domain of f. The number f(c) is called the maximum value of f on D. Similarly, f has an absolute minimum at c if f(c) is less than or equal to f(x) for all x in D and the number f(c) is called the minimum value of f on D. The maximum and minimum values of f are called the extreme values of f.

Whew, that's quite a mouthful. Let me try to break this down.

An absolute maximum (also called the global maximum) would be the highest point on a graph in its domain. The number f(c) is called the maximum value of f on its domain.

An absolute minimum (also called the global minimum) would be the lowest point on a graph in its domain. This number f(c) is called the minimum value of f on its domain.

The maximum and minimum values are collectively known as the extreme values of f.

Pretty straightforward, wouldn't you say?

Let's get a diagram going.

Hey, look, a diagram!
Here we can see that a and b are local maximum and minimum, respectively, for this function. This diagram also leads me to our next theorem, named:

FERMAT'S THEOREM

This theorem states that if a graph has a local maximum and mininum and it is possible to take a derivative, that derivative equals 0. You can see that illustrated in the diagram, where the tangent lines are drawn at the local max and min.

So, let's define local max and min:

A function f has a local maximum (or relative maximum) at c if f(c) is greater than or equal to f(x) when x is near c. Similarly, f has a local minimum at c if f(c) is less than or equal to f(x) when x is near c.

More technobabble. Let's define this more simply.
If a point is higher or lower than its immediate surroundings, such as in the above diagram, then it can be defined as a local maximum or minimum.

Just a little bit more. Stay with me, guys.

CRITICAL POINTS
I could make a bad pun and say I'm at a critical point with JPD coming and all that. But I'll spare you the agony and give you the mathematical definition.

A critical number of a function f is a number c in the domain of f such that either f'(c) = 0 or f'(c) does not exist.

To reiterate:

A critical point is when

1. f'(c) = 0
OR
2. f'(c) doesn't exist

Pretty simple.

Also,
If f has a local maximum or minimum at c, then c is a critical number of f.

ONE MORE THING!

The Extreme Value Theorem. So extreme, its inner workings can hardly be contained in this post.

If f is continuous on a closed interval [a,b], then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some numbers c and d in [a,b].

So, basically, if the function is:

1. Continuous
2. On a closed interval

Then there absolutely has to be an absolute maximum and minimum within that interval. Logical, right?

Want an example problem? Of course you do! (Unfortunately, due to the aforementioned dead computer, I don't have access to my photo editing programs. My bad.)

Try finding the absolute maximum and minimum, as well as the local maximums and minimums.

Abs max and min: Oops! Trick question! There is no absolute max or min, because the graph has an unlimited range.

Local max: There is a local maximum at (1,2)
Local min: There is a local minimum at (2,1)

Those are local because they are higher and lower, respectively, than the points around them.

Here's a cool link that explains this stuff pretty well.
Clicky clicky!

And remember, KATE, you're up next.

And, now, time for my obligatory picture from my archives of awesomeness.

Penguins are already great. Pirate penguins are just out of this world.

Well, I hope this post was as informative as it was long. I also hope you are not collapsed in front of your computer dying slowly of boredom.

Until next time, my children,

Evan, the Lord Protector of Math

3.11 Linear Approximations and differentials

2006-11-02T16:52:00.000-08:00

Linear Approximations

According to the graph, f' (x) = dy / dx
dy = f' (x) dx, dy = f' (a) dx
y-y₁ = f' (x) (x-x), y =y₁+ f' (x) (x-x),
y = f(a) + f' (a) (x-a),
L(x) = f(a) + f' (a)(x-a) or Δy = f(a + Δx) – f(a)

When used over short periods of time, these approximations can be very accurate.

It was very exciting in class to finnally see why it is dy / dx - the slope of the tangent line.

This tangent line approximation or linear approximation of a at b is also referred to as linearization.

Differentials

Linear approximations are often formulated via differentials.

Let y = f(x) be a differentiable function.
The differential of x, dx, is an independent variable.
The differential of y is defined as

dy = f' (x) dx

The variable dy depends on the values of x and dx

Below is a geometric and visual way of approaching differentials

Δy = f(x + Δx) – f(x)

Tangent line PR is, as we know, the derivative of f' (x). So, dy is the rise, the distance from S to R. We can see dy ≠ Δy. Δy is the amount that the curve rises when the x changes by the amount of dx.

Relative error is found by dividing the error by the total (ΔV / V). When multiplied by 100, we can find the percentage error.

Example

Find the value of y = √(25.001) through approximation.
y = √(x + 25) .001 ≈ 0 so our a-value = 0 (that makes it easy!)
f(x) = √(x + 25)
f' (x) = (1/2) (x + 25)^(-1/2)

f(a) = √(0 + 25) = 5
f' (a) = (1/2) (0 + 25)^(-1/2) = 1/10 (*For these equations go to the f(x) and f' (x) equations found above and plug in x with a and the a-value. Do not find the derivative of f' (a) by differentiating the f(a)-value - that will just be zero.)

L(x) = 5 + (1/10)(.001-0)
L(x) = 5 + .0001 = 5.0001
√(25.001) ≈ 5.0001

Another type of problem you might see will ask for a prediction (probably extrapolating) after giving some data. With the data, we can plot a graph or make a table. When the question asks for the y-value after giving an x-value, the most accurate estimate will be found by finding the derivative of the closest given x-value. Then the y-value of the new x will approximately equal the y-value of our old x plus the derivative of the new x times the difference between the new and the old: f(x) ≈ f(a) + f' (a) (x - a) (which is the same as the definition for L(x))

Additional Internet Sources
Flash
Visual Calculus - Differentials
Visual Calculus - Linear Approximations
Powerpoint
Linear Approximations and Differentials

Reminder: Evan, it is your turn again... (but not until Monday, I believe)

Personalization

Although math puns are the first sine of madness, I do enjoy making them.

I am quoted in this article about Degrassi!

Rocko knows to be hot, naughty, and courteous.

3.10: Related Rates

2006-11-01T16:22:00.000-08:00

Hi everyone. This is Isabella, not Laurie; we traded places a while back, so I am officially our last scribe this first time around. Exciting.

Well, for the last few days, we’ve been battling with…RELATED RATES. These are all word problems, which have a tendency to be daunting. But if we break them down, they aren’t as monumentally frightening.

The key with related rate problems is to establish what information we’ve got and what information we are looking for. We also need to set up a static equation, which we will then differentiate to isolate our unknown.

So, before we look at some sample problems, let’s set up our rules:

Draw a picture. This helps you understand how the different components of the problem affect the overall situation.
Know what information is given and what information is unknown.
Set up a static equation, which presents the information when it is NOT CHANGING. The static equation defines the relationship between your variables.
Differentiate your static equation, which presents the information when it is changing with respect to time.

A very important thing to remember with differentiating is that we must still use our differentiation rules! Never forget about Product Rule, Quotient Rule, and particularly Chain Rule (that’s the one we always seem to mess up a tad). Precise differentiation is what gets us going when doing related rates problems.

All right, to make things clearer, let’s go to our examples.

EXAMPLE 1: The Ladder Problem

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft/s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall?

Drawing a picture, we can see that the ladder, 10 feet long, is the hypotenuse of a triangle. The vertical wall is one leg, y feet long. The distance from the wall to the bottom of the ladder is the second leg, x feet long.

The information we are given is:

1. The length of the ladder (hypotenuse) = 10 ft

2. The distance (x) from the wall to the bottom of the ladder = 6 ft

3. The rate of the ladder sliding away from the wall = 1 ft/s

The word “rate” immediately tells us this will be a derivative. Since it is the rate of change of the horizontal component, x, this derivative is (dx/dt).

The information we are looking for is the rate of the ladder sliding down the wall—the rate of change of the vertical component, y, with respect to time, or (dy/dt).

Looking at the relationship between x and y, we can come up with our static equation:

x² + y² = 10² (A simple Pythagorean relationship)

Next, we differentiate EACH SIDE of the static equation (with respect to t = time).

2x (dx/dt) + 2y (dy/dt) = 0

Notice the use of the Chain Rule to differentiate x² and y².

Since we are looking for (dy/dt), we solve our equation for this unknown:

(dy/dt) = - (x/y)(dx/dt)

Now, we can substitute in our given information to find the unknown. We are given x and (dx/dt). To find y, we can just plug x into our original Pythagorean equation.
6² + y² = 10²
y² = 64
y = 8

We now have all information except the unknown. Plugging everything in:

(dy/dt) = - (6/8)(1 ft/s) = -(3/4) ft/s

So, the top of the ladder is sliding down the wall at a rate of (3/4) ft/s. Remember that the negative merely indicates that the distance from the top of the ladder to the ground is decreasing.

Not too bad, right? These problems can get more complicated, but as long as you follow the same procedure and keep track of your information, you can figure it out.

EXAMPLE 2: The Cone Problem

A water tank has the shape of an inverted circular cone with base radius 2 m and height 4 m. If water is being pumped into the tank at a rate of 2 m³/min, find the rate at which the water level is rising when the water is 3 m deep.

Sketching the cone, we see that the radius and the height are always legs of similar triangles.

Now, let’s identify our given information:

1. Radius (leg of triangle) = 2 m

2. Height of cone (leg of triangle) = 4 m

3. The depth of water (height in the cone) = 3 m

4. The rate of water being pumped into the tank is 2 m³/min.

Again, “rate” indicates a derivative. This time, it’s the derivative of the volume of the cone. Notice how “m³” hints that we’re dealing with a volume (just like “m²” would indicate area).

What’s our unknown? The rate at which the water level is rising. Since this is the rate of the vertical component, we know that is it the derivative of h with respect to time, or (dh/dt).

All right, now that we’ve got all that established, we can set up our equations.

Our static equation relates radius and height—the equation for the volume of a cone.

V = (1/3)(pi)r²h

It’s useful to work with V as a function of one variable, not two. Because we have more given information about h, we should eliminate r. Using the given information:

(r/h) = (2/4) so r = (h/2)

Now we can plug this into our static equation:

V = (1/3)(pi)(h/2)²h = (1/3)(pi)(h²/4)h = (pi/12)h³

Next we differentiate both sides with respect to t = time.

(dV/dt) = 3 (pi/12)h²(dh/dt) (CHAIN RULE)

(dV/dt) = (pi/4)h² (dh/dt)

Solving for (dh/dt), our unknown:

(dh/dt) = (4/(pi)h²) (dV/dt)

Now we can substitute the information we were given:

(dh/dt) = [4/(pi)(3)²](2) = (8/9(pi))

So, the water level in the cone is rising at a rate of (8/9(pi)) m/min.

Let’s not forget that our variables can also involve angles and trig functions!

EXAMPLE 3: The Theta Problem

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a speed of 2 ft/s, how fast is the angle between the top of the ladder and the wall changing when the angle is pi/4 radians?

Drawing the triangle, we see that the vertical wall is one leg of the triangle and that the 10-ft ladder is the hypotenuse of the triangle.

What information are we given?

1. The hypotenuse of the triangle = 10 ft.

2. The rate at which the bottom of the ladder is sliding away from the wall = 2 ft/s. This is the derivative of the horizontal component, x, of our triangle, or (dx/dt).

3. The angle between the wall and the ladder = pi/4 rad.

And our unknown? The rate at which the angle between the ladder and the wall is changing. This is the derivative of the angle, or (d(theta)/dt).

Looking at the triangle, we see that we can set up a relationship between the angle tbeta and one of the legs and the hypotenuse of the triangle. Because we know that x is the leg opposite theta, and we have a value for the hypotenuse, we can use sine to describe the relationship.

sin (theta) = (x/10) (opposite leg over hypotenuse)

This is our static equation. Differentiating, we have:

cos (theta)(d(theta)/dt) = (1/10) (dx/dt)

Remember that differentiating sin(theta) involves the Chain Rule, and that (x/10) is another way of saying (1/10)x—that is, (1/10) is the coefficient of x.

Now that we have our equation, we can plug in our information:

cos (pi/4) (d(theta)/dt) = (1/10)(2)

((rad2)/2) (d(theta)/dt) = (2/10)

We want to isolate (d(theta)/dt):

(d(theta)/dt) = (4/10)/(rad2) = 2 / (5rad2) = (2rad2)/10 = (rad2)/5

So, the rate at which the angle between the wall and the top of the ladder is changing is (rad2)/5 radians/sec.

Voila. Tricky, but not impossible. Related rates may seem scary, but as long as you organize (and don't panic), Related Rates will smile on you and perhaps befriend you. And we all know that everyone needs a friend, even one that tends to be fickle.

Here's a website that provides its own version of a procedure for a related rates problem. It also has two useful problems, including an example of a cone problem:
http://www.mathematicshelpcentral.com/lecture_notes/calculus_1_folder/related_rates.htm

I don't know how many of you know this bit of trivia, but...I am very fond of frogs. I think they're awesome. So, I thought I'd share with you guys a froggy image that is close to my heart.

Just barely hanging on...we've all had those moments.

We're back to the beginning of our scribe list, so Jenny, it's your turn tomorrow. Have a good night, everyone! ~8)

3.8 Derivatives of Logarithmic Functions

2006-10-28T16:24:00.000-07:00

Hi, guys. So right now, we're just finding a bunch of ways in which we can make finding derivatives easier. This section, we find out how to find derivatives of logarithmic functions. The lesson itself is pretty short. If you do the derivative of:

you get:

From this, we can generate the general theorem that:

In verbal terms this means that the derivative of the function ln of x equals one divided by x
Don't forget, however, that the chain rule still applies. So if you have this problem for example:

The answer is:

Notice how we first just differentiated the ln of x squared as a whole. Then we differentiated what was inside the parentheses--the x squared itself.

These are very helpful rules because they help us solve many problems that otherwise we wouldn't know how to solve! Like this one!

Find the deriviative of:

How do we do that? With the natural log derivative theorem of course! There are several steps to applying this rule to a problem. These steps are also a general rule of thumb that you can use with any logarithmic equation.

1) Take the natural log of both sides
So:

All we did was take the natural log of both sides

2) Simplify with the help of logarithm laws.
Don't forget from last year that there are several other rules about derivatives that we can use to help us. First off, remember that:

Oh look! We just happen to have the log of a quotient! Perfect!

So this means that our new result is:

The other logarithm rule we should remember is:

This helps us to since we have a a logarithm of a product. So now we have:

Now don't forget the rule where you can take the exponent of a logarithmic function and put it in front of the logarithm. Like this:

So since each of our three terms have exponents, we can just put those exponents in front of each term like so:

3) Use implicit differentiation for x

We know that:

So now we can apply this to our problem. Remember, don't forget to use the chain rule to those that have more than just x inside the ln ( ).

See where we had to use the chain rule? The 2x is the derivative of the x squared plus one and the 3 is the derivative of 3x + 2.
Notice also that we had to differentiate the y also. We had to use chain rule for that too, since 1 over y is the first term, and then we're still left with the y, which we had to differentiate by itself, giving us y prime.

4) Solve for y prime

Oh, but what does y equal? We know from the original equation. So we can go ahead and substitute that in, giving us our final equation:

Yayyy! We did it!
Here's something cool, guys. It's a very simply step-by-step proof of how they "derived" the formula for finding derivatives of logarithmic functions. Enjoy!
http://ltcconline.net/greenl/courses/116/ExpLog/logDerivative.htm

I thought these comics were funny. Mr. French, I couldn't remember if you had the amazing plusman on your wall or not. So I put it up anyway.

You're next, Laurie. And then the cycle starts all over!

Friday's Quiz Topics

2006-10-25T20:39:00.000-07:00

First, I goofed – the last “5” on the TWS should be problem #59…

Here’s a list of topics that will be covered on this Friday’s Quiz. I’ve included the relevant homework problems…

Quiz – Sections 3.4-7
Derivatives of Trigonometric Functions (Sec. 3.4, # 5,7,29)
Chain Rule (Sec. 3.5, #13,53)
Implicit Differentiation (Sec. 3.6, #7,17)
Higher Order Derivatives (Sec. 3.7, #39)
Position/Velocity/Acceleration/Speed (Sec. 3.7, #49)

Remember that these topics can be mixed and matched, along with applications of the Product and Quotient rules!
I’ll be in early tomorrow and Friday. After school on Thursday, I have to leave at 3:00. See you in class!

"I believe in looking reality straight in the eye and denying it."
- Garrison Keillor (1942-) US author, producer

And remember, speed is relative:

3.7 Higher Differentiation

2006-10-25T20:09:00.000-07:00

There were three main concepts in this section:

There were three main concepts in this section:
1. learning how to take second, third, fourth (etc) derivatives
2. New Notation for Higher Differentiation (dy/dx)
3. Relating the 2nd derivatives to motion and acceleration

Second, Third and Fourth Derivatives

Example:

Higher Derivatives:

Using dy/dx notation
Example:

Another Example:
y = sin (x)
y ′ = cos (x)
y ′′ = -sin (x)
y ′′′ = -cos (x)
y(4) = sin (x)
y(121) = cos (x)

The Second Derivative explains how the slope is changing. It tells us whether the graph is getting steeper or flattening out. In motion, the second derivative is used to measure acceleration.
s′ = v - the derivative of the distance equation is the velocity
v′ = a - the derivative of the velocity equation is the acceleration
s′′ = a - the second derivative of the distance equation is also the acceleration

Motion Example:
Initial Velocity--after one second-- Final Velocity--------- acceleration
10 mph------------------------------------ 20 mph--------------- +10 mph/sec
10 mph------------------------------------ 0 mph----------------- (-10) mph/sec
-10 mph---------------------------------- (-20) mph------------- (-10) mph/sec

Speed vs. Velocity
Speed = the magnitude of the velocity vector. It lacks direction.

When the change in velocity and the acceleration have the same sign (positive or negative), then the speed is increasing

Example:
Find the velocity and acceleration equation:

Find when the particle is speeding up:
-First find when the particle is at rest.
v = 3(t – 1)(t – 3)
-the velocity is equal to zero at 1 second and then at 3 seconds
-Then, test the intervals of velocity
- between 0 and 1: positive
- between 1 and 3: negative
- greater than 3: positive
-Finally, find when the acceleration is going to be 0
- a = 0 at 2 seconds
- Compute when acceleration is going to be positive or negative
- below 2 seconds: negative
- greater than 2 seconds: positive
-When the signs match, the particle is speeding up
So: 1 < t < 2 and t > 3

Here is a link for more information:
http://www.intmath.com/Diff/9_HiDe.php

SO there's the end of the blog for today folks.
-Lauren

Here's something to make you all very happy that it's Thursday:

DOESN'T IT MAKE YOU FEEL ALL WARM AND FUZZY INSIDE??

JAMES, have fun with the next one.

3.5 The Chain Rule

2006-10-25T00:23:00.000-07:00

Sorry Mr. French....it said my html code was wrong so I had to start from scratch. Fingers crossed that this works!

Hello Section B.

Just as I like changing colors on Mr. French's interactive whiteboard, I also like changing blog colors...quite innovative don't you think?

If this goes anything like the chain rule homework, I will spend plenty of time doing unnecessary things only to realize that the solution to the problem is much simpler than I thought, redoing the hour and a half of work (which so far is like 3 and a half for the blog cuz it is having some compatibility issues) in about ten minutes.

ok, let's dive in:

The Chain Rule: Deals with composite functions. A composite function is a function within a function. There are 3 steps to follow when using the chain rule:

Step 1: Take the derivative of the inside function.
Step2: Take the derivative of the outside function by finding the value of the inside function with respect to x (do NOT, I repeat, do NOT take the derivative of the inside function in this step).
Step 3: Multiply Step 1 and Step 2.

Now let's put all that English into Mathematical Notation:

F(g(x))
F'(g(x)) = F'(g(x))g'(x)
I always like to write out these two lines because I find it keeps the problem in order and ensures I will remember the whole chain rule and fully apply it.

Example:
F'(6) = 7
g(3) = 6
g'(3) = 4
Step 1:
g'(3) = 4
Step 2:
g(3) = 6
F'(g(3)) = F'(6) = 7
Step 3:
4x7 = 28

Another way to apply the chain rule:
Fill in u for the inside of the function (this helps so that you don't forget steps because the u's should all be gone by the end of the problem):

Example:
F(x) = (3(x^2)+4x)^2 = (u)^2
F'(x) = (2u)(6x+4)
(The derivative of the outer)x(the inner as is)x(the derivative of the inner)
Plug in the inner function for the value of u:
F'(x) = 2(3(x^2)+4x)(6x+4)
F'(x) = 2(18(x^3)+36(x^2)+16x)
F'(x) = 36(x^3)+72(x^2)+32x

You can also foil the equation if you forget how to apply the chain rule on a test, but when the exponent is larger than w, it is an inefficient use of time.

The Power Rule Combined with the Chain Rule:
d/dx (u^n) = n(u^(n-1))x(u')

Example:
F(x) = (6(x^3))+2(x^2))(^100)
Step 1: Multiply the coefficient by the exponent from F(x):
F'(x) = 100(6(x^3)+2(x^2))
Step 2: Subtract 1 from the exponent of F(x) and insert the new exponent (this uses the power rule):
F'(x) = 100(6(x^3)+2(x^2))(^99)
Step 3: Multiply the whole quantity by the derivative of the inside function:
F'(x) = 100(6(x^3)+2(x^2))(^99)x(18(x^2)+4x)

Derivative of an Exponential Function:
d/dx (a^x) = (a^x)(ln a)
The derivative of an exponential function = (the function)x(the natural log of the base)

Example 1:
F(x) = (2^x)
F'(x) = (2^x)x(ln 2)

Example 2:
F(x) = (e^x)
F'(x) = (e^x)x(ln e) = (e^x)x1 = (e^x)

One more thing that is not new, but is a helpful reminder:
F(F'(x)) = x
(e^ln x) = x

If you need a little extra help, here is a good website to visit. It has really good examples and provides a full explanation of the chain rule. The home of the website has many more math resources, but this is the link to the chain rule page only (the link is not yet active...you will just have to wait until tomorrow morning):
http://www.math.hmc.edu/calculus/tutorials/chainrule/

As much as I adored posting my blog, I guess I will relinquish the duties of 3.6 to you ZAK.

Quote of the day:
"Yesterday is history,
Tomorrow is a mystery,
Today is a gift,
That's why we call it
The Present."

3.6 Implicit Differentiation

2006-10-24T20:08:00.000-07:00

So far this chapter we have dealt with functions that express one variable in terms of the other. Or in other words, fuctions that explicitly defined. For example:

or

But now we must learn to deal with functions that do not express one variable in terms of the other, aka implicit functions. For example the fuctions:

are implicit functions because y is not described in terms of x.

So what does this mean? It means that we have a whole new way of finding the derivative of y without needing to solve for an equation for y in terms of x. Instead we now can use the method of Implicit Differentiation.

The method of implicit differentiation involves differentiating both sides of the equation with respect to x and then solving the equation that's left for y'.

For example let's take the equation:

and find the derivative.

1) First we must find the derivative of y squared

We already know that

So therefore by the Chain Rule

2) Now we must differentiate the entire equation

Here we must use the power rule, the product rule, and the known derivative of y squared to solve

3) Now we must solve for y'

factor out the y'

and that's our answer. Not too bad I'd say. I hope this implicit differentiation sample problem will help you guys in the future. Lauren Vons, you're up next!

The following are just a bunch of funny signs that don't pertain to math at all. They're just stupid so I thought you guys might like 'em.