Prep AB Calculus C 2006-07

Calvin Says

2007-05-08T21:14:00.000-07:00

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

2007-03-22T10:26:00.000-07:00

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

2007-03-15T14:37:00.001-07:00

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…

Section 9.2: Direction Fields and Euler's Method

2007-03-12T20:36:00.001-07:00

Hey all...so today's lesson deals with sketching and approximation to figure out the shape of the curve of a function without any real in-depth calculations. According to the book, the definition for a Direction (or Slope) Field is: "If we draw short line segments with the slope F(x,y) at several points (x,y), the result is called a direction field (or slope field)." In other words, you are given an equation set equal to a derivative. For example, you could be given y'=x+xy. Then, you make a table of values divided into three columns: x, y, and y'. Pick a random pair of points for your (x,y) coordinate, plug this pair into your equation, and figure out the slope of the function at that particular point. Draw a short line at the designated point that has approximately the same slope as the slope that you just found. Here is an example:

Sketch the direction field for the differential equation y'=(x+xy)-y.

First, set up a table of values.

x y y'
-1 -1 -1
0 0 0
1 1 1
etc.

Now, draw in the slopes of the function at the given points. The end result should look like:

Of course, this is a very inexact method. Euler's method seeks to make the process of drawing a direction field a little more accurate. "For the general first-order initial-value problem y'=F(x,y), y(x0)=y0, our aim is to find approximate values for the solution at equally spaced numbers x0, x1=x0+h, x2=x1+h, ..., where h is the step size. The differential equation tells us that the slope at (x0, y0) is y'=F(x0, y0. The general equation to express Euler's rule is:

Now, let's apply this concept to a real problem.

Use Euler's method with step size .1 to construct a table of approximate values for the solution of the initial-value problem. y'=x+y, y(0)=1

y1=1+.1(0+1)=1.1
y2=1.1+.1(.1+1.1)=1.22
etc.

So as you can see, Euler's method allows us to draw a much more exact direction field by giving us more exact values for slopes. Here's a fun link for your personal enjoyment: http://tutorial.math.lamar.edu/AllBrowsers/3401/DirectionFields.asp

BRIAN YOU ARE NEXT HAVE FUN!

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."

9.1 Modeling with Differential Equations

2007-02-27T23:23:00.000-08:00

Alright, so in class we learned about the two different models of population growth. The first one is stated as the Rate of Growth for a Population-unbounded on our concept sheets. It is:

This equation shows that population is always growing, and as it gets bigger its growth rate gets bigger as well. As Mr. French said, as the population gets bigger, there's more people to make the population bigger.

The second formula we were taught is dubbed the Rate of Growth for a Population-bounded on our concept sheets. It is:

This equation is also the derivative of a logistic function. It expresses the fact that when we have a small population, the population will grow quickly. But as the population approaches the carrying capacity, it won't be able to grow so quickly and will level off. If P/K were to ever become greater then 1, then the population would start to decline.

The next thing we learned about was the IVP, or the Initial Value Problem. The two steps to these situations are:

1. Solve the differential equation for a general solution.
2. Use the general solution and data point to solve for a specific solution.
For example:

Given and the point y(0)=5, find the IVP.

http://www.math.ohiou.edu/~ashish/h13.pdf That's a nice quick summary straight from our book that Ohio University made.

http://www.mathsci.appstate.edu/~hph/3310/diffeqn/ This site has examples of different kinds of differential situations.

Alex, you get the blog next =D

My favorite quote for the past month has been "You smell funny" from POTC:DMC.

And now for some lame math jokes...

Q: What is the first derivative of a cow?

A: Prime Rib!

Q: What caused the big bang?

A: God divided by zero. Oops!

"A mathematician is a blind man in a dark room looking for a black cat which isn't there."- Charles Darwin

and for my last few words:

A guy gets on a bus and starts threatening everybody: "I'll integrate you! I'll differentiate you!" So everybody gets scared and runs away. Only one person stays. The guy comes up to him and says: "Aren't you scared, I'll integrate you, I'll differentiate you!" And the other guy says: "No, I am not scared, I am e^x."

Thursday's Quiz Topics

2007-02-27T18:04:00.001-08:00

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:

8.1 Arc Length

2007-02-26T19:41:00.001-08:00

Okay, so it's finally my turn again~! I'm here to explain Chapter 8, Lesson 1, which is about finding arc lengths.

When a curve is a polygon, finding the length is easy because all you have to do is add up the sides but when you get a continuous curve, it gets tricky! Remember that a curve is defined by the equation y= f(x) where f is continuous on a (equal to or less than) x (equal to or less than) b. When we estimate the value of a curve, we are taking approximations as if the sides of a polygon were present in it. (page 547 in the textbook)

The length is described using the distance formula (as the limit approaches infinity). The distance formula is not practical to use with a smooth function, so we can derive an integral formula for L where the function has a continuous derivative, because there is only a very small change in f'(x). (Also think of the approximation as taking the Pythagorean theorem to find the hypotenuse with infinitely many tiny triangles, as described in class)

The arc length forumula for the curve y=f(x) where is a is (equal to or less than) x (equal to or less than) b

(if f' is continuous on [a,b])

it can also be notated as:

In the other case, if a curve has the equation x=g(y) with c (less than or equal to) y (less than or equal to) d, and g'(y) is continuous, we get this formula for arc length:

Keeping these two formulas in mind, let us try a problem!

If y=e^4x and 0 (less than or equal to) x (less than or equal to 1), find the arc length.

We begin by writing the integral, which will be from 0 to 1, and plug in the derivative. Remember to use the chain rule in this case, because the derivative of e^4x is 4(e^4x). Don't forget details such as chain or product rule when taking derivatives! Also remember to even TAKE the derivative, and to SQUARE it. But I digress. After setting everything up, it should look like this:

Afterwards, you can either figure out the antiderivative or if you are short on time or just wish for simplicity (in this case), plug it in to your calculator and do fnInt! Oh yeah, another minor detail. Don't forget to write "dx" when doing your problems, and don't get your dx's and dy's mixed. I think it's just me, but it's still a possible mistake.

...aaaannndddd here are a few links:

http://en.wikipedia.org/wiki/Arc_length (history teachers can scorn Wikipedia but this site gives a good explanation of this math concept, provided that no users/pranksters go edit it, but it always gets edited back anyhow)
http://tutorial.math.lamar.edu/AllBrowsers/2414/ArcLength.asp

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

and here's a funny little video for you all. just remember when you decide to bring your kitten in for show-and-tell, don't let it near Mr. French's laptop. (Has anyone posted it on the blog already? Remind me if anyone has)

carpe diem, everyone.

until next time~!

-Sonia

Oh yea, Crystal. You're up next xD

Section 7.7 Approximate Integration

2007-02-21T22:36:00.000-08:00

Hello everyone.
This section is called Approximate Integration. It covers three methods of approximating integrals, but only two of them are covered on the AP Exam.

The first method is the Midpoint Rule. We have covered this before, but just in case you cannot remember it, here it is:

where (for all of the equations).

This calculates the area under a curve using rectangle approximations, using the midpoints as the heights. A more accurate method is trapezoidal approximation. Instead of drawing rectangles under the curve, you can draw trapezoids. Here is the formula for using trapezoidal approximation.

Finally, there is a third method that is not on the AP Exam, and it is known as the Simpson’s Rule. Here it is:

For this to work, n must be even.

Here is an example:

For use the Trapezoidal Rule, the Midpoint Rule, and Simpson’s Rule to approximate the integral if n = 4.

For the Trapezoidal Rule, plug in the values for the equation. First calculate .

Now:

This method gives us an approximation of 44.

Using the midpoint rule, will be the same. Next find the midpoints.

Now plug into the equation:

This gives us an approximation of 21.Finally, for Simpson’s Rule. remains the same.

Plug into the equation:

This gives us an approximation of .

And that's the lesson.

Here is a link to a helpful resource: This website shows the Trapezoidal Approximation and Simpson’s Rule and how to use them: http://archives.math.utk.edu/visual.calculus/4/approx.1/index.html

Sonia you are next.

On a lighter note, here are some comics:

Friday's Quiz Topics

2007-02-21T22:30:00.000-08:00

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.

7.1 Integration of Parts

2007-02-20T18:55:00.000-08:00

You should remember that the product rule is :

By rearranging this equation you see that:

By using the substitution rule and replacing u for f(x), v for g(x), du for f'(x)dx, and dv for g'(x)dx, you will see that the formula for integreation of parts becomes:

Let's look at a couple of examples!

1)
Choose your simplest component for u. In this case that would be x. Make a table of your u, du, v, and dv.

Now plug it into the equation
You get

Don't forget the C!

2)
You can use the tabular method here.

Now try a problem!

Here, you must use integration by parts again

Now plug this equation back into the end of the first integration of parts:

Here is a good site for Integration of Parts:
http://www.sosmath.com/calculus/integration/byparts/byparts.html

Saladang Song: My obsession with food has led me to this great Thai restaurant. I highly recommend it (especially the Yellow Curry) and it's very well priced.
363 S Fair Oaks Ave (Cross Street: W Bellevue Drive)Pasadena, CA 91105

Kyle, you're up next!

Thursday's Test Topics

2007-02-13T20:30:00.000-08:00

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: Average Value of a Function

2007-02-13T18:45:00.000-08:00

Hey guys! Hey Mr. French! I hope you are all having a great Tuesday, I know I am. Anyways, let's get into the lesson. This lesson deals with the definition of an Integral: An approximation of the area bewteen the given curve and the x-axis between two given points. Now let's look at an example. We want to find the area under this cubic function, F(X).

Graph

Ok, now we want to find all the area bewteen the curve and the x-axis between the points a and b. Let's look at a visual depiction of that:

Now we want to find this area using one large rectangle. Now of course a rectange is stright on all sides, so this area is going to be an approximation. We want to draw the rectangle so the area that it extends past the graph seems to be close to the area that is fails to cover in the graph. This will make more sense when looked at graphically. Let's look at that right now!

So, as we can see the rectangle extends above the graph but fails to fill in certain areas under the graph. These areas cancel each other out. We are left with a close approximation of the area under the graph. Remember that any part of the graph below the x-axis represents negative area.

Now that we have looked at this visually, we can look at it in terms of an equation. We have just visually depicted the mean value theorem. The mean value theorem can be expressed mathematically as:

a and b are our boundaries and c is a point between a and b. f(c) is the avaerage value of the integral.

Now let's solve the equation for f(c):

Ok, now let's look at an example of this numerically....

We have the equation y=x+x2 [-1,2]

We don't actually need to graph it. We can use our equation for the mean value theorem.

a is -1, b is 2, and f(x) is x+x^2, so let's put that into our equation:

Now, let's simplify it further....

Next we use the rules of integration to take the antiderivative of the equation...

And lastly, we plug our a and b values (-1 and 2 into the antiderivative equation)...

1.5

1.5 is our "c" value, which is a point between a and b that lies on the graph. Now to find f(c), the average value, we solve our F(x) equation for x, setting it equal to our c value.

x+x^2=1.5.

When we solve this with the quadratic formula, we find that x= .823 and -1.823. Only .823 is within the boundaries, so that is our actual c-value.

Here is a good site about about the average value of a function: http://www.mathwords.com/a/average_value_function.htm. It's short, but i think it shows the concept clearly.

Also, before I go, I wanted to share with you an inspiration quote

"If A is sucess in life, then A equals X+Y+Z. Work is X, Y is play, and Z is keeping your motuh shut"- Albert Einstein

Here's a nice picture for all of you:

Since the order is now random, I have no one to remind. I'm out.

Monday's Quiz Topics and New Posting Order

2007-02-08T22:15:00.001-08:00

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...

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 randomly assigned posting order:

Ryan
Ami
Kyle
Sonia
Crystal
Alex
Brian
Magnus
Jessica
Ismael
Joey P.
Kane
Lauren
Joseph Y.
Jeff
Claire
Isaac

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

2007-02-08T20:36:00.000-08:00

In 6.2 which Claire so wonderfully covered, we learned how to find volumes using cross-sections of solids perpendicular to the axis of rotation. We did this using the disk and washer methods:

Disk: Washer:

However, the volumes of certain solids cannot easily be found using these two methods. Sometimes it is necessary to use another method: the method of cylindrical shells. Such can be used when the the perpendicular R and r of a washer depend on the same curve. Instead of using cross-sections perpendicular to the axis of rotation, we can use cross-sections parallel to the axis of rotation. Rotating the parallel cross sections about an axis create shells....

If this curve is rotated about the y-axis, one shell would look like this:

In this next picture of a shell, you can see that the thickness of the shell is the difference between radii.

So, the formula for the volume of a cylindrical shell is:

, where r is the average of the radii and is the difference of the radii.
In words, Volume = (circumference)(height)(thickness). This is easier to understand if you imagine the shell cut and rolled out to form a rectangular solid with length , height h, and width .

Based of this formula for the volume of one shell, we can come up with a formula for the volume of an entire solid made up of many shells. If f is a continuous function on the closed interval [a, b], then the volume of the solid obtained by rotating the graph of f from x = a to x = b about the y-axis is
(where ).

Again, is the circumference, f(x) is the height, and dx is the thickness of the shells.

Now for a sample problem:

Find the volume of the solid obtained by rotating the region bounded by and y= 0

about the line x = 3.

Solution:

First, visualize or sketch the graph. Either with your mind or with your calculator, you can see that the region bounded by the restrictions goes from x=0 to x=1 and lies between a smooth curve and the x-axis (similar to the example on page 458).

Then, draw or picture a vertical cross-section of the region. Rotated around x= 3, the shell has a radius of 3-x, a circumference of , and a height of y or .

So, the volume would be:

Plug in 1 and we get .

Some websites:

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

http://pear.math.pitt.edu/Calculus2/week6/6_2li1.html

And another awesome photooo and some awesome pollution:

It looks like RYAN is up next.

6.2 Volumes

2007-02-07T10:07:00.000-08:00

In 6.1, we learned about the area between two curves, but now, Mr. French and the James Stewart have decided to make it more interesting by introducing a third dimension. We are now figuring out the volume between two curves, or the volume caused by the rotation of the area between two curves.

The general formula for finding the area between two curves is:

So let's say that we want to find the volume of a circle, using squares. The squares would come out the circle, with the base edge bounded by the circle. It would look something like this:

So, to calculate the volume we would use an infinite amount of infinitely thin boxes to calculate it, like a Riemann sum. Because it's a square we know that:And the formula for a circle is: where R is the radius. So the set up would look like this:

The green side, S, is double what the y value is for that x value, which means thatUsing this information, we can now define A in terms of x, like in the general equation for the volume, and find the volume in terms of R, which is an undefined constant.

Using the equation for the circle, we get that:And according to the general equation for volume:
So,Since the circle is also symmetrical about the y-axis, taking the integral from -R to R is the same thing as doubling the integral of 0 to R.Now, we just do what we know how to do - evaluate the integral, keeping in mind that R is a constant.And we're done!

There's another type of volume that Mr. French and James Stewart might ask you to find, and that's a rotation volume. It's when you take an area and rotate it around an axis, creating a solid with circular cross-sections. When the base of the area touches the rotation axis, it creates a disc volume, and when it doesn't, it creates a washer volume.

Take for instance the graph of the square root of x from 2 to 9.

And let's rotate it around the x-axis. And since the base of the area touches the axis of rotation, we have a disc volume, and because the cross sections are circles, the integral for the volume would be:

And the radius of the cross-section circle differs with the function. It is defined by the y-value, so R=Y (at x=9 the radius is 3, the same as the y-value of the function at 9). Since we know that y equals the square root of x, we can substitute the square root of x in for R. This gives us:

And from there, we all know how to solve this problem.

If we had rotated the function around the line x=-3, then we would have had a washer volume, since the base of the function wouldn't have touched the rotation axis. To solve that problem, we would have had to find the volume without the space, and then subtracted the space. There would be two radii, one representing the radius of the empty cylinder in the middle and one representing the radius of the whole volume.

To solve this problem, we would use:

In this case, r would be a constant of 3, while R would be 3+ y-value of the function or

So we can plug all of this into the big equation and get:

And we all know how to solve the problem from there!

If you need any more help on this concept, you should check out these websites:

http://archives.math.utk.edu/visual.calculus/5/volumes.4/index.html

http://archives.math.utk.edu/visual.calculus/5/volumes.1/index.html

http://archives.math.utk.edu/visual.calculus/5/volumes.2/index.html

And finally I just wanted to remind everyone that our favorite bunch of castaways are back on tonight after a long winter hiatus.

Just as a recap, Jack, Kate and Sawyer captured by the Others, and Jack had Benry on the table in the middle of back surgery, when he insisted that Kate and Sawyer be let go or he'd leave Benry on the table to die. So that should make for an interesting episode tonight!

Also, reminder to Lauren for the next post.

Section 6.1: Areas Between Curves

2007-01-10T15:32:00.000-08:00

Yay it's my turn now. As I look at these problems more and more, they seem less scary, so if any of you are worried about this concept, as I was, don't worry because with some practice, I'm sure that you can kick these problems' butts! Before I get into summarizing this lesson, here is a link for any further information: http://tutorial.math.lamar.edu/AllBrowsers/2413/AreaBetweenCurves.asp

Ok so anywho, this lesson takes integrals as we have learned them a step further. Now, instead of finding the area beneath a curve on a given domain, we have to find the area between two different curves on a given domain. It's just a little extra wrinkle that we have to deal with, but no big deal. Let's say that we are given the following graph:

The red curve represents the upper curve, y=f(x), and the blue curve represents the lower curve, y=g(x). a and b represent the given domain, and the shaded yellow area represents what you are trying to find when you solve one of these problems. In order to find this area, you can use the following general formula:

Now we can use that for a fun sample problem!! Brace yourselves...

First, we can draw the graph of this situation. The yellow graph is the area that we are trying to find.

From this picture, we can see that the upper boundary is the blue curve, or y=2x-x(x), and the lower boundary is y=x(x). Therefore, the area is gonna be (upper boundary-lower boundary)dx or (2x-x(x)-x(x)). Also, the region is between x=0 and x=1. Therefore, the total area is

So, your answer is 1/3. YAY JESSICA YOU'RE NEXT HAVE FUN!!

Friday's Test Topics

2007-01-09T21:10:00.000-08:00

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:

Section 5.5

2007-01-09T15:42:00.000-08:00

This section is about the Substitution rule which allows anti-differentiation of complex expressions. The idea is to replace the complex section with a variable, (u), anti-differentiate, and then substitute back in the complex statement: a sort of inverse chain rule.

Substitution Rule for indefinite integrals

If u=g(x) is a differentiable function whose range is an interval I and f is continuous on I then
∫ f(g(x))g'(x) dx = ∫ f(u) du

This rule can also be applied to definite integrals by adjusting the range for u

∫ from a to be of [f(g(x))g'(x) dx] = ∫ from g(a) to g(b) of [f(u) du

Example

∫ (cos√t)/ √t

Given that u = √t then dt = 2du/t^-.5

substituting both expressions in removes the denominator and the roots to give "∫ 2cos(u) du"

antidifferentiate to get 2sin(u) + C and substitute the original expression for x to get 2sin√t + C

This is a helpful site: http://www.sosmath.com/calculus/integration/substitution/substitution.html

Alex you're next.

Happy New Year!

2006-12-31T08:41:00.000-08:00

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

2006-12-19T20:13:00.000-08:00

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

2006-12-18T20:12:00.001-08:00

INDEFINITE INTEGRALS
An indefinite integral is basically the antiderivative of the function. It doesn't have upper and lower bounds because that would make it a definite integral. Indefinite integrals need the +C!

Table of Indefinite Integrals

Sample Problem
Find the general indefinite integral

Using the formula:

THE NET CHANGE THEOREM

The Integral of a rate of change is the net change (displacement for position functions)

Basically this theorem states that the integral of f or F' from a to b is the area between a and b or the difference in area from the postion of F(a) to F(b).

This can be applied to things such as:
volume
concentration
density
population
cost
velocity

So for a velocity function:
To calculate displacement we can use the equation

to calculate total distance traveled we can add the absolute values of the areas of each sector from each x intercept to the next x intercept

Sample Problem
A particle moves along a line so that its velocity at time t is

(m/s)

a) find the displacement from t=[1,4]
b) find the distance traveled during that time period

Finding the displacement:

m.

Finding the total distance traveled during that time period

m.

The total distance traveled and the displacement are the same because the position function does not pass below the x axis therefore there are no negative areas. If there were negative areas the displacement would be a smaller number and the distance would stay the same.

Some LINKS:
http://www.coolschool.ca/lor/CALC12/unit5/U05L04.htm

A Lesser Lesson in Indefinite Integrals but helpful nontheless

Magnus You're Up NeXT!!!

5.3 The Fundamental Theorem of Calculus Part II

2006-12-13T16:06:00.000-08:00

Fundamental Theorem of Calculus Part II

If f is a continuous function on [a,b] then

where F is any antiderivative of f, that is, a function such that F'=f.

Once you find the antiderivative of f(x), you evaluate the end points from a to b and then subtract the antiderivative function.

Lets look at an example.

Once you find the antiderivative, you plug the top number 5 into x and subtract that antiderivative function from an antiderivative function with the number 3 in x.

You can check your answer by using your calculator.
Plug into your calculator.
Go back to the home page. Click Math then 9.
Once fnInt pops up, put fnInt( ,X,3,5). You will get 98/3

Try this problem!

You will see that this antiderivative is:

Thus you plug in 4 into the x and then subtract the antiderivative with 2 in the x value.

This a reminder to Brian to do the next blog!

This website can help out with this concept:
http://math.ucsd.edu/~wgarner/math10b/ftc.htm

Dixie Chicks

Thursday's Quiz Topics

2006-12-13T10:27:00.000-08:00

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.

5.3 The Fundamental Theorem of Calculus

2006-12-12T19:49:00.000-08:00

Hey guys this is Izzy. Today in class we learned about the fundamental theorem of calculus. This theorem establishes a connection between the two branches of calculus: differential calculus and integral calculus. In class we covered how if you multiply the function that you are taking the integral of by a certain factor, that you can factor out that number. Example:

This makes it easier to deal with the original funtion first, without any coefficients to clutter up the process.

In class, we saw how the integral of a function can be a function itself, represented by g(x). Expressing the integral of a function as the function g(x) allows one to actually graph and express the area between the curve of a graph and the x-axis in terms of x:

In this graph,

By using this equation, we can find the area between the graph and the x-axis. Let's say we wanted to find the area between x=0 and x=3. You would multiply the change in x, 3, by f(3), 3, and .5, because it's a triangle. So, g(3)=(3)(3)(.5)=4.5 .

Today we also made the amazing discovery that the antiderivative of a function is the same as the integral of the function. Observe:

As you can see from the graph, the area of the shaded region (teal) is:

We established that

To find g'(x), we have to replace t with x and multiply whatever you get by the derivative of x.

Wait a minute, the derivative of the Area function is also x! Thus A'=g'(x). We just proved that the antiderivative of a function is the same as its integral! i'm so happy.

What we just did relates to the Fundamental theorem of Calculus, Part 1, which states that:
If the function 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)

MOVING ON, we now know how to find the derivative of these types of functions. Let's try this problem: Find

Whenever you see these types of problems, you must first look at the upper bound, or any of the bounds that has a variable in it. In the process of calculating the equation for g'(x), you must replace the variable t with the variable bound, and wherever you see dt, replace it with the derivative of the variable bound. In this case, the upper bound has the variable: x^2.
To solve this problem, we have to use the chain rule (make u=x^2):

There you have it.

This a reminder to Ami to do the next blog!

These are a few websites that help out with this concept:

http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/integration/ftc.html
http://archives.math.utk.edu/visual.calculus/4/ftc.9/

Here are some cool math jokes
Q: What do you get if you divide the cirucmference of a jack-o-lantern by its diameter?
A: Pumpkin Pi!

Q: Why do you rarely find mathematicians spending time at the beach? A: Because they have sine and cosine to get a tan and don't need the sun!

A mathematician is flying non-stop from Edmonton to Frankfurt with AirTransat. The scheduled flying time is nine hours. Some time after taking off, the pilot announces that one engine had to be turned off due to mechanical failure: "Don't worry - we're safe. The only noticeable effect this will have for us is that our total flying time will be ten hours instead of nine." A few hours into the flight, the pilot informs the passengers that another engine had to be turned off due to mechanical failure: "But don't worry - we're still safe. Only our flying time will go up to twelve hours." Some time later, a third engine fails and has to be turned off. But the pilot reassures the passengers: "Don't worry - even with one engine, we're still perfectly safe. It just means that it will take sixteen hours total for this plane to arrive in Frankfurt." The mathematician remarks to his fellow passengers: "If the last engine breaks down, too, then we'll be in the air for twenty-four hours altogether!"

A math student is pestered by a classmate who wants to copy his homework assignment. The student hesitates, not only because he thinks it's wrong, but also because he doesn't want to be sanctioned for aiding and abetting. His classmate calms him down: "Nobody will be able to trace my homework to you: I'll be changing the names of all the constants and variables: a to b, x to y, and so on." Not quite convinced, but eager to be left alone, the student hands his completed assignment to the classmate for copying. After the deadline, the student asks: "Did you really change the names of all the variables?" "Sure!" the classmate replies. "When you called a function f, I called it g; when you called a variable x, I renamed it to y; and when you were writing about the log of x+1, I called it the timber of x+1..."

5.2 The Definite Integral

2006-12-11T11:03:00.000-08:00

So there's some notation for the Riemann sum that we did yesterday. If we take, for example, the area under the curve of y=x^2 from x=1 to x=3 like this:

and we want to have 2 rectangles with the sample points being left endpoints, then we would assign some notation. 1 would become x1, and 2 would be x2. To put this in notation we would use a capital sigma to symbolize sum. Underneath the sigma would be the starting point and over the sigma will be the ending point, both using i as the variable. The sum would be the sum of the rectangles, so it would be the height times the width (change in x). Put all this together and the equation would look like this:

To find the answer to this problem, we would plug in 1 for i then two, and add them. So it would be: f(x1)(x2-x1) + f(x2)(x3-x2) = f(1)(2-1) + f(2)(3-2) = (1)(1) + (4)(1) = 5.

We know from 5.1 that as you have more and more rectangles, the estimate of the area becomes more and more accurate. So ideally we would like infinitely many rectangles. We can write this as :

and to simplify this we use the integral symbol, and it would be :

Because as we get more and more boxes and n approaches infinity, the change in x gets closer and closer to zero so eventually the change in x becomes the derivative at x.

Take this graph and equation for example:

Let's find the area underneath the curve, but above the x axis.

Both the equation and the graph suggest that it's a semi-circle with a radius of 5. We could :

But common sense tells us that since we want want the area under the graph of a semi-circle, then we want the area of a semi-circle. So the area would be

This simple way of looking at the problem can be applied in different ways. Let's suppose we wanted to find the area between the curve and the x-axis if f(x) was 2x from x=0 to x=4 and f(x)=8 from x=4 to x=8 and f(x)=3x-16 from 8 to 10, then you could split that graph into two triangles and one rectangle and work out the area from there.

The area would be (1/2)(4)(8) + (6)(8) + (1/2)(6)(2) = 70.

Another topic covered was the Mid-point estimate. Using the mid-point estimate balances out the over and under estimation, bringing us closer to the actual area underneath the graph.

Using the left side makes an overestimation, but using the right side is an underestimation. Using the midpoint, however, balances out the over and underestimation, making it about right.

There are some properties of integrals as well.

2. If f(x)=c=7, and a=3 and b=5, then:

And:

Therefore:

That's it!

Here's a couple of cute calculus pick-up lines:

1. "Hey, if I was sin^2x and you were cos^2x, together we would be one!"

2. "I wish I was your derivative so that I could lie tangent to your curves."

Here's a website about the properties of integrals:

www.analyzemath.com/calculus/Integrals/integral_properties.html.

Ismael, you're up next!

Section 5.1: Areas and Distances

2006-12-10T21:51:00.000-08:00

In this section, we look at the areas under the curves of functions and the distance an object has travelled, given its velocity.

AREA:
Take a look at this curve:

Finding the area under this curve will be easy for us. All we have to do is multiply the base and the height.

Take a look at these three areas:

It should be easy here too.

Now, we have a curve that is complicated:

We can find the area under this curve by separating it into rectangles with equal intervals.

We find the width of the intervals by using this equation:

Then, a question comes: Which heights do we use?

We can use the heights to the left (LRAM: Left Rectangle Approximation Method) and the heights to the right(RRAM: Right Rectangle Approximation Method).

LRAM: Find height of initial x and extend left to interval.
RRAM: Find height of initial x and extend right to interval.

Let's take a look at y=x^2.

LRAM gives us:

RRAM gives us:

Notice that in this case, the RRAM gives us an overestimate (upper estimate). The LRAM gives us a underestimate (lower estimate).

We'll call the LRAM area Ln and the RRAM area Rn.

There is another method: the Midpoint Method (MRAM)
If we take the y=e^-x graph and use the MRAM, we get:

(There is another method: Trapezoid Method. We won't get into that now. However, if you want to, there is a link near the bottom.)

My interpretation is that these methods are ways to cut up a non-calculatable area into smaller calcultable areas.

These methods all give us approximations of the area under the curve. If we want to find a more accurate area, we can make the intervals infinitely small.

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.

Using the right endpoints, we get:

We get the same value if we use left endpoints:

Distance:
When an object is moving along a straight line and we are given its velocity function, we can find the distance it has travelled using the concept we just learned.

The area under the curve of a velocity vs. time graph is the distance travelled.

This is because velocity=distance/time. So, distance= velocity*time If we add up all the changes in position, we can find the distance travelled.

Sample:
What is the area under the y=x^-1 curve between 1 to 4? (Use 3 intervals) Show both LRAM and RRAM. Which is lower? Which is upper?

Here we have the graph:

Using the interval equation:
(4-1)/3 = 1
Therefore, each rectangle must have a base width of 1.

LRAM:

We can find that the total area of the 3 rectangles is:
A=1(1+0.5+1/3)=1.8888... units squared

RRAM:

The total area is:
A=1(0.5+1/3+0.25)=1.083333... units squared

With these two answers, we can easily see that LRAM has given us an upper estimate and RRAM has given us a lower estimate.

Links:
Here is a site that may help you understand this concept further:
http://www.sosmath.com/calculus/integ/integ01/integ01.html

Here is a site that shows all 4 (left, right, midpt, trapezoid) methods and has a lot of pictures:
http://homepage.mac.com/nshoffner/nsh/CalcBookAll/Chapter%206/Ch6DefInt.htm

Here is a good video. This even shows some calculator procedures: (Quicktime required)
Quicktime 6: http://www.math.armstrong.edu/faculty/hollis/calcvideos/SV3/21-area.mov
Quicktime 7: http://www.math.armstrong.edu/faculty/hollis/calcvideos/H.264/21-area-H264.mov

Here is a comic. Enjoy!

Reminds me of a time when I did something like that...

Claire, you are next.

Solution to Question #1, Quiz 4.5-7

2006-12-06T20:29:00.000-08:00

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

2006-12-06T07:41:00.000-08:00

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

Section 4.10: Antiderivatives

2006-12-05T11:26:00.000-08:00

Wassup!

Section 4.10 is all about antiderivatives(Precisely how to find antiderivatives).

Before any specifics, the class should know what the antiderivative exactly is.

-Antiderivative(Book's interpretation): F(x) is an antiderivative of f(x) if F'(x)=f(x), on an interval
-Antiderivative(My interpretation): Opposite of the derivative(The class finds the original function, given the derivative of that function)

Given the definition of the antiderivative, a theorem representing the general equation for antiderivatives can be developed: Antiderivative of the given function=F(x)+C, C being the constant

Example:f(x)=3x^4
F(x)=(3x^5)/5+C, which means F(x)=3(x^5)/5+10000,F(x)=3(x^5)/5+573892

Finding the antiderivative(Finding the original function, given the derivative) is just as easy as finding the derivative.
Simply remember the rule: x^n=(x^n+1)/(n+1), when n does not equal -1
REMEMBER that this rule can be applied to any term in a function

Example:f(x)=5x^4+2x^3-3x^2
F(x)=(5x^5)/5+(2x^4)/4-(3x^3)/3=x^5+(2x^4)/4-x^3 + C

Example:f'(x)=5x^4+2x^3-3x^2, f(0)=2(MADE UP, MAY NOT WORK OUT EVENLY)
f'(x)=x^5+(2x^4)/4-x^3+C, C=1+2-1+2=4(WHY? Add the coefficients of the other terms plus what is given[In this case, coefficients are 1,2,-1, and given of 2])

However, the rule: x^n=(x^n+1)/(n+1) has a problem when n=-1, since the antiderivative ends up undefined(Since the denominator=0).

Example: f(x)=x^-1
F(x)=(x^0)/0=Undefined

Luckily, there is another rule that allows us to find the antiderivative of a function when n=-1. The rule: ln x, when n=-1

Example: f(x)=x^-1=1/x
F(x)=ln x + C

Here is a link that can help you understand antiderivatives even more!
http://v5o5jotqkgfu3btr91t7w5fhzedjaoaz8igl.unbsj.ca/~talderso/UWOnotes/UNIT2.pdf

Now, before I leave...
Here are some pictures of some my favorite breakdance crews/breakdancers

BBoy Hong 10

BBoy Wake-Up

Expression Crew

REMEMBER, YOUR NEXT KANE!!!

4.8: Applications to Business and Economics

2006-12-03T12:01:00.000-08:00

Hey guys i hope you are all having a great weekend. I am in laguna right now writing this on the balcony overlooking the ocean, enjoying the warm breeze and sipping on a mango smoothie. Thought i would make you jealous. Anyway, let's get into it. 4.8 is all about applying derivatives to the business world.

Let's start off by defining some things, since a good portion of this section is about understanding business jargon. You will see that these business terms are simply real world applications of functions and derivatives of functions.

1) c(x) = the cost function. It is the cost of producing "x" amount of a certain product. The cost function tells us how much is spent in relation to how much is produced.

2) c'(x)= marginal cost. This is the derivative of the cost function. From what we learned before about expressing the meaning of a derivative verbally (how something is changing with respect to somethign else), we can say the marginal cost is the rate at which the cost is changing with respect to the number of units produced.

= The average cost. This is the cost divided by the total number of units. Remember from 1) that the cost function is the cost of producing "x" amount of a certain product.

..Now because we are in calculus and examine derivatives, we are of course going to look at minimums and maximums, terms we are all familiar with. The term minimum average cost means that the the marginal cost, defined above as C'(x), or the derivative of the cost, = the average cost defined above as the cost function divided by the number of units produced, expressed in terms of "x".

In mathematic terms, minimum average cost=

4) p(x)= the price function. The price function represents the price per unit a company can charge if it sells "x" units.

5) R(x)= p(x) . x= revenue. Revenue is the product of the price, expressed as(p(x)) and the quantity, expressed as "x". Simply put revenue is the amount of money the company collects.

6) R'(x)= the marginal revenue function. Like the marginal cost function, this is simply the derivative of the revenue function. Once again we can use our prior knoweldge of the meaning of a derivative to verbally state that the marginal revenue is the instantaneous rate of change of revenue with respect to the number of units sold.

7) P(x)= R(x)-C(x)= the profit function. The proft function is the difference between the revenue function (see 5) and the cost function (see 1). RYAN'S WARNING: Note that the profit functuon is represented by a capital P, while price is is represented by a lowercase p.

8) P'(x)= R'(x)- C'(x)= the marginal profit function. P' is the derivative of the profit function (see 7), R'(x) is the derivative of the revenue function, or the marginal revenue (see 5 and 6), and C'(x) is the derivative of the cost function, or the marginal cost (see 1 and 2).

...Once again, because we are in calculus and are using derivatives, we are going to of course have to examine maximums and minimums (since these occur when a derivative equation is equivalent to zero) When marginal revenue equals marginal cost, profit is maximized, or when R'(x)= C'(x) (the dertivative of the revenue function equals the derivative of the cost function), profit is maximized.

Now that we have defined all these thick business terms and can see that if they are flowery terms for things that we all understand, like liner functions and derivatives of functions, let's see if we can do a problem..

Ryan Lamont, after mastering Calculus and no longer failing Mr. French's quizes, opens up a Calculus Art exhibit in which he draws the graphs of derivatives. He puts them on display so the the public can enjoy his masterpieces. He sells 600 tickets at $9 and predicts, through an intense study of current market trends, that if he drops the price to $7, he will sell 800 tickets. At what price should he charge for tickets to maximize his revenue?

ok, so first let's make an equation for the price function. The price function represents a linear relationship so we can use point slope form to make our equation...

Now that we have made our equation, let's solve for m. We have two x values (600 and 800) and two p, or y values (9 and 7) so let's subtract the y's and divide the difference by the difference between the x's. Let's take the slope...

7-9/800-600= -2/200= -1/100= -.01. m= -.01.

Now, since we have two x values and two p values, we use one pair and substitute it into the eqaution with our newly found m also. Let's use ($7, 800).

So.. p-7= -.01(x-800).

Now let's distribute and isolate P by adding 7... p-7=-.01x+ 8. WHen we add 7, we get
p= -.01x+15

Now that we have our profit equation, we know that revenue equals the product of the profit equation and the quantity in terms of x (see 5). So, R(x)= p(x) . x. Since p(x)= -.01x+ 15, R(x)= (--.01x+15) (x), which equals -.01x^2+15x.

So, R(x)= -.01x^2+15x

Now that we have our revenue function, we want to take the derivative, which eauals the marginal revenue. So, using basic power rule properties, we come up with the equation:

R'=-.02x+15

Now, we want to set the derivative of the revenue function to zero, as we always do when finding maximums and minimums, and solve for x in the R' equation. So, we subtract 15 and divide by -.02.

-15/-.02= 750 so x=750.

We know that this x value maximizes our revenue. We can use the first derivative test and test values to the right and left of 750 on the number line, or we can use logic and realize that the graph of R is a parabola which opens downward. Now that we have proven this value gives us a maximum, our last step is to plug it back into the orginial price equation, p=-.01x+15.

...-.01(750)+15= 7.5, so our answer is $7.50. Pricing the tickets at $7.50 Ryan will maximize his revenue.

Ok well that is the end of the lesson.

Since the only new thing in this lesson was new terms ( all the actual calculus involved is review), i found a site that explains those business terms: http://ingrimayne.com/econ/MakeProfit/Overview10mi.html

There are also some great problems in here, not all of them are relevant to this section, but the ones that are are excellent: http://64.233.161.104/search?q=cache:0JHVsGpN1nwJ:www.math.mcgill.ca/~laayouni/Teaching/Math122/Lecture17.pdf+derivatives+calculus+maximize+cost&hl=en&gl=us&ct=clnk&cd=6 eaching/Math122/Lecture17.pdf+derivatives+calculus+maximize+cost&hl=en&gl=us&ct=clnk&cd=6

Before i go, i wanted to share with you an alliteration poem i wrote in 5th grade in Mrs. Rogers' class. For those of you who don't know, alliteration means using the same first letter multiple times in a sentence.

Here is my poem:

Royal Radical Ryan Rode Rough Rhinoceroses to Rhode Island While Ravishing Ruby Red Radishes.

I knew you would enjoy that.

This picture speaks for itself:

Oh, and another reminder that Calculus and alcohol don't mix so please do not drink and derive!

Reminder: Joseph you are next, man. Make it Happen. Make us proud. You are the valiant knight fighting the seemingly invincible blog.

Wait! I'm still not done...

Do you see a duck or a rabbit? Haha isn't that cool.

Ok one more, then I'm out.

Young woman or old hag?

Friday's Quiz Topics

2006-11-30T14:34:00.000-08:00

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

2006-11-29T10:09:00.000-08:00

We can use what we've learned about finding extreme values of functions to solve real-world applications that involve optimization--maximizing or minimizing something for the best results. There are eight main steps in solving these types of problems:

1) Read the question carefully and understand what the goal is (what information you are looking for).

2) Draw a diagram and label everything.

3) Determine the maximum/minimum equation.

4) Determine the constraints.

5) Use the constraint relationships to reduce the maximum/minimum equation to one variable.

6) Use the maximum/minimum equation to identify local maximums/minimums by taking the derivative and finding critical points.

7) Test the critical points and endpoints to determine the absolute maximum/minimum.

8) Answer the question asked.

Here is an example:

You construct mini, customized above-ground pools. One customer requests that you line their pool with red tile. You only have a limited amount of red tile and it is more expensive than regular tile. So, you want to minimize the amount of red tile you use. All your pools must be 400 cubic feet in volume. Your pools have square bases. Find the dimensions that minimize surface area.

So, following our 8 steps:

1) Our goal is: find the dimensions (the width and height) of the pool that yield the minimum surface area and a volume of 400 cubic feet.

2) Draw a diagram: the base is a square so its sides are both w. The height is h.

3) The max/min equation is: , where w is the width and h is the height. Remember the base is a square, so the area of the base is . The total area of the four sides then is 4(wh). So you add and 4wh to get the surface area you need to cover with tile (the bottom of the pool and the four inner sides).

4) Now for the constraints. Based on logical dimensions, you can say w is greater than 5 and less than 20 and that h is greater than 0 and less than 10. Also, the volume must equal 400 cubic feet. So, you can come up with a relationship for the volume: .

5) Using the second constraint, you can reduce the minimum/maximum equation to one variable by solving the constraint for one variable. Choosing h would be easiest, so

. Then, plug this into the min/max equation to get

and then .

6) Now you take the derivative of f to find the maximums/minimums. You can also write f as. The derivative is . Set this equal to zero and solve to find the critical points:

w = 9.283

You can use the First Derivative Test to determine that 9.283 is a minimum. You can test using 1 and 10.

When w is less than 9.283, the derivative is negative and when w is more than 9.283 it is positive. S0, there is a minimum at x=9.283 because the derivative changes from negative to positive.

7) But is 9.283 an absolute minimum? When x is 9.283, f(x) is 258.53. Now test the endpoints of the domain of w--5 and 20. When w is 5, f(x) is 345 and when w is 20, f(x) is 480--both are greater than at w=9.283, so (9.238, 258.53) is the absolute minimum.

8) Our goal was to find the dimensions of the pool that yield the minimum surface area and a volume of 400 cubic feet. Luckily, we didn't lose sight of it, as is so easy to do. For the width, we got 9.283 ft. But we need the height too! We found earlier that

, so plug in 9.283 for w and you get 4.642 feet.

So our answer is:

9.283 ft by 9.283 ft by 4.642 ft.

Some cool links:

http://www.qcalculus.com/cal08.htm

http://tutorial.math.lamar.edu/AllBrowsers/2413/Optimization.asp

And here's a nice picture I took:

RYAN, you are up next.

4.6 Graphing with Calculus and Calculators

2006-11-27T20:23:00.000-08:00

The main point of this section is that when plotting a graph using a calculator, the calculator can often be misleading. If the equation of a graph is highly complicated, the calculator can miss many important parts of the graph of the equation, such as extreme points. In order to fix this, you can adjust the window settings on the calculator to give you a better view of the graph. By using the first and second derivative, you can find the extreme points, the intervals of increase or decrease, inflection points, and the concavity of the graph. These values can help you determine a more appropriate viewing window for the graph.

For example:
Produce graphs of the above equation that reveal all the important aspects of the curve.

When you first graph the equation, you will probably get something like this by using ZoomFit on the calculator:

This graph does not reveal very much about the significant aspects of the curve. A more appropriate viewing window is needed. To make this process easier, find the first derivative.

Now find the extreme points. You can use the graph of the derivative to find them:

The extreme points are (-2.255,) and (1.098,)

Based on this, the intervals of increase are:

while the interval of decrease is: (-2.255,1.098)

In order to find the concavity and points of inflection, first find the second derivative.

Now find the points of inflection. It is easy to do this by graphing the second derivative:

There is only one point of inflection, which is at (-1.4128,)

The graph of f(x) is concave up during:

and concave down during:

From this information we can find a more appropriate view of the graph.

Thus this is a more appropriate view of the graph. The y-window is [-500,500]. The x-window is [-5,5]

Here is a link to a website that can help you with this topic: www2.edc.org/cme/showcase/KY/calcliesarticle.pdf

This site applies this topic, but uses sinusoids rather than polynomials. Otherwise it is the same thing as this section.

Isaac, you are next to post, since Luke is no longer here.

Now, to finish things up, here is a comic. Hopefully, the AP exam will not be like this:

4.5 Slant Asymptotes

2006-11-20T16:34:00.000-08:00

I was having trouble editing my other post, so i created a new one for the stuff we learned today dealing with slant asymptotes. If you're like me, you never learned about long division involving a variable. Its very similar to normal long division.
First, I'll tell you about the application of long division. In rational functions, slant asymptotes occur when the degree(highest exponent) of the numerator is one larger than the degree of the denominator. A slant asymptote is a linear function (following the form y =mx + b) that is never reached by the function.

So a function can have no vertical or horizontal asymptotes, but it might have a slant asymptote.
To find the slant asymptote, use long division. For example: ( 2 - 3x^2)/(x-1)

start by setting the equation up as a normal long division problem.
to learn how to do long division go to this very helpful site (they explain it better than i can): http://www.purplemath.com/modules/polydiv2.htm

the result is -3x-3 + (-1/(x-1))
you only use the -3x-3 part of this because the rest is a remainder.
Therefore, the slant asymptote is y = -3x - 3

4.5 Summary of Curve Sketching

2006-11-16T17:35:00.000-08:00

To start things off, I'd like to remind Kyle to do the next blog.

this lesson is (as its title suggests) a summary of what we have learned thus far. I don't know about you, but for almost everything I learn, I make little devices or words to remember them. For this section, the things you need to remember when sketching a graph, given the equation are:
1. remember any Asymptotes
2. find the Intercepts of the graph -for y, plug in 0 for x, vice versa for x-intercepts
3. keep in mind the Domain
4. For time-saving's sake, see if you can apply Symmetry
5. use the derivative to find Intervals of increase/decrease on the graph
6. find Max's andmin's-- using the first derivative test, if f ' changes from positive to negative, it is a local max.
7. f '' =0 is a pt. of Inflection and where this f '' is greater than 0, it is Concave up.
8. Sketch the graph

You can rearrange the bold letters any way that helps you remember, but I personally like to think : I SAID MISC , or IM AIDS SIC
for any other creative ideas to remember these guidelines, comment on this blog....

for the graph of y = (5x^3)/(x+1)
asymptotes: when x is -1, there will be a vertical asymptote
intercepts: when x=0, y= 0.......... when y=0, x also must =0 place a dot at the origin
domain: when x gets increasingly negative, what happens? a negative over a negative equals a postive. so while the graph has an asymptote at x=-1, the graph will resume on the left of the x-axis and get increasingly large.
symmetry: there is no symmetry because the function is neither odd nor even.
Intervals: the der. is (using the quotient rule) [(x+1)(15x^2)-5x^3] / (x+1)^2

so this derivative starts out negative, but gets closer to zero, where it does some funky stuff, then becomes positive when x is positive. using this info, the graph will start out with a negative slope, then because of the asymptote become positive.

local max, min: the der. is 0 at -1.5 and 0 and undefined at -1
concavity: do some "simplifying" to get [(x+1)^2 (30xx+30x) - (10xxx+15xx)(2x+2)]/ (x+1)^4

the graph of f" is negative only between -1 and 0 so f(x) should be concave up most of the time

pts. of inflection: when f" = 0 x= 0

finally attempt to sketch the curve. You can check your graph on your calculator.

links: http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/apps/graphs.html

http://www.mathgraphs.com/mg_clc7e.html
advice for next time: use an easier equation : )

the red sox just paid $51.1 million for the rights (not even the salary) of Daisuke Matsuzaka, a pitcher from Japan.... here he is......

Friday's Quiz Topics

2006-11-16T13:09:00.000-08:00

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

Section 4.4 Indeterminate Forms and L' Hospital's Rule

2006-11-16T06:26:00.000-08:00

Alright so hopefully that comic got you ready for this section. Because it's on how you can get limits for infinities and zeros.

And just a heads up this blog is going to be slightly messy because I don’t have Equation Editor on my computer.

So first off, L’Hospital’s rule says that:

or, verbally, the derivative of a function f at point c is the same as the function itself if the function’s limit is equal to either zero or positive or negative infinity.

An example would be the equation we used in class

To find the limit, you need to plug in infinity for x.

We can see that it equals infinity over infinity, which is a indeterminate form. However using L’ Hospital’s rule, we can take the derivative of the function and start again. Thus, we can take the derivative of the function.

The derivative comes out to be 1over x. Then if you plug in inifinity, it is obvious that it nears 0. Thus the limit is zero.

Another problem is:

It's solution:

And how we can apply L'Hospital's Rule:

(taken from http://archives.math.utk.edu/visual.calculus/3/lhospital.1/index.html )

And now moving on to the other indeterminate forms. There are seven total that you need to know. The first two have already been mentioned: infinity over infinity and zero over zero. Those are called the indeterminate quotients. The indeterminate product is zero times infinity. The indeterminate difference is infinity minus infinity. The three indeterminate powers are zero to zero, infinity to zero, and one to infinity. In these equations infinity means a really really big number, 0 means a really really small number, and one means...one.

Here are some good sites if you are still confused after my amazing blog:
http://www.sosmath.com/calculus/indforms/otherquotient/otherquotient.html
http://archives.math.utk.edu/visual.calculus/3/lhospital.2/ (These are more problems if you want to test your self some more on L'Hospital's Rule)

Reminder to JOEY to post tomorrow’s notes!

4-3 How Derivatives Affect the Shape of a Graph

2006-11-14T18:51:00.000-08:00

In this section, we learned how to use information about to give us information about .

Increasing/Decreasing Test

Since represents the slope of the curve at the point , tells us when the function is increasing or decreasing. The Mean Value Theorem we learned earlier leads to the conclusions:

1. When is positive, is increasing.

2. When is negative, is decreasing.

We can use these two statements to test whether a function is increasing or decreasing based on the first derivative. Hence, the name “Increasing/Decreasing Test”.

The First Derivative Test

We can also use the first derivative to find the local maximum and minimum of the graph of the function.

We first find the critical numbers of the function. Critical numbers are the x-values in the domain of a function when the derivative is zero or undefined. To find the critical numbers, set and solve for x.

Once the critical numbers have been found, determine whether each critical number is a local maximum, a local minimum, or neither. To do so, draw a number line, mark the number line , and mark the critical numbers on the line.

We know that is equal to zero at the critical numbers. Using the number line, we can find the sign changes for each critical number. These sign changes determine the nature of each critical number. This is called the First Derivative Test:

1. If the sign of changes from positive to negative at a critical value, then has a local maximum at that x-value.
2. If the sign of changes from negative to positive at a critical value, then has a local minimum at that x-value.
3. If does not change sign at a critical value, then there is neither a local minimum nor a local maximum at that x-value.

Example

For , (a) Find the critical values & identify each as a local maximum, minimum, or neither. (b) State the intervals where is increasing.

Solution

Here is an example in which we use the Increasing/Decreasing Test and the First Derivative Test.

(a)

Now that we have the critical values, we create a number line and plot those values.

Using the First Derivative Test, we conclude the following:
Since changes from + to – at , is a local maximum
Since changes from – to + at , is a local minimum

(b) Since is positive on the intervals to the left of -2 and to the right of 0, is increasing on the intervals and .

Concavity & The Second Derivative

The first derivative of a function helps determine the local maximum and minimum and also tells where the function is increasing or decreasing. The second derivative tells us about the shape of the curve, its concavity, and also gives us the Concavity Test.

Look at Curve #1 below. In the graph, the slopes of the tangent lines change from negative to zero to positive. Therefore, the slope of the tangent, , is increasing. When a function is increasing, the derivative of , that is , is positive. This type of shape is concave up.

Look at Curve #2 below. In the graph, the slopes of the tangent lines change from positive to zero to negative. Therefore, the slope of the tangent, , is decreasing. When a function is decreasing, the derivative of , that is , is negative. This type of shape is concave down.

In Curve #1, there is a local minimum. In Curve #2, there is a local maximum. This gives us the Second Derivative Test:

If c is a critical number of , and
, then there is a local minimum at c.
, then there is a local maximum at c.

The Second Derivative Test is easy to use, but it is also limiting because there is no conclusion when . In that case, use the First Derivative Test, which can be applied in every case.

Points of Inflection

Points of inflection are points where the concavity of a function changes. Points of inflection are at values in the domain of a function where changes sign from positive to negative or from negative to positive.

We find points of inflection in a way similar to finding critical numbers. First, find of a function and solve for the x-values when is zero or undefined. These x-values are possible points of inflection. Next, draw a number line, mark the number line , and mark the x-values on the number line. We know that at these possible points of inflection is zero or undefined. Using the number line, we can find the sign changes for each of these points. If changes sign around the x-values on the number line, then that x-value is a point of inflection.

Example: Let us find the points of inflection of .

Solution:
We find the first derivative, .
Then we find the second derivative, .

Since is never undefined for this function, find the points of inflection by solving:

changes from negative to positive around -1. Therefore, the point of inflection for is (-1, 6).

Two links you can check out for further information:

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

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

Reminder: Crystal, you are up to post next.

Here are some pictures I took at the Huntington Gardens:

Test 3 #17

2006-11-14T09:55:00.000-08:00

We know the velocity equation(the derivative of the position equation) =

Thus the zeros are 0 and 3.

The acceleration equation(the derivative of the velocity equation) =

Thus the zeros are 0 and 2.

Plug in numbers less than 0, between 0 and 3, and greater than 3 into the velocity equation for t

Plug in numbers less than 0, between 0 and 2, and greater than 2 into the acceleration equation for t:

The particle is slowing down when the signs are opposite:
and

Test 3 Question #4

2006-11-11T22:56:00.000-08:00

Differentiate the function:

In order to differentiate this type of function use the equation:

Then solve:

Test Chapter 3, #3

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

#3: Differentiate the function :

We know (from our concept sheet on logarithmic functions) that :

because the numerator is g'(x), or the derivative of (found with the chain rule), and the denominator is g(x) or . Then,

And finally, using trig identities (sinx/cosx = tanx)

This is choice a.

Chapter 3 Test Question 6

2006-11-09T22:14:00.000-08:00

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

In order to solve this, you need to use the chain rule. The chain rule states that for function h(x)=f(g(x), h'(x)=f'(g(x))(g'(x)). So if we plug in the values given into this equation, we get:

h'(1)=f'(g(1))(g'(1))

h'(1)=f'(3)(-3)

h'(1)=(9)(-3)

h'(1)=-27

Ch. 3 Test Question #16

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

A particle moves along the x-axis, its position at time t given by x(t) = -(t^4) + 4(t^3) - 2, where t is measured in seconds and x in meters.

#16 When is the particle speeding up?

First, find the velocity by taking the derivative of the position function. You get:
v(t) = -4(t^3) + 12(t^2)

Then, find the acceleration by taking the derivative of the velocity [this is the second derivative of the position function]
a(t) = -12(t^2) + 24t

graph the velocity and acceleration functions using a reasonable window. (I apologize for not being able to show you a picture, but my internet is unusually slow as of now)

the particle is speeding up when the values of the velocity and acceleration functions HAVE THE SAME SIGN. they have the same sign when t is greater than 0 or less than 2 [both positive] and when t is greater than 3 [both negative]. (for some odd reason, because of html tags, blogger will not let me actually put in the greater than and less than signs.)

an alternative method (not mentioned in the book, I believe but correct me if I'm wrong) would be to sketch the graph, and take the absolute values of the graph of the velocity funcition (flip the negative parts by reflecting it over the x-axis). then if the graph is going downwards, it is slowing down. if it is going upwards, it is speeding up.

Chapter 3 test question #9

2006-11-09T21:22:00.000-08:00

Hey guys this is Izzy and I am going to explain how to do problem 9 from the test.

The Question asks:

Differentiate the function:

Ok, first you think: properties of logarithmic functions. Three of these properties are:

So, you apply that to this function and get:

then you differentiate, to make the whole process easier. Because if we had used the product and quotient rules, man, that would have been messy.
So, to differentiate, we remember that

We must remember to take the derivative if whatever u is even after placing a 1 over the u. So, using this property, we get:

or

Chapter 4.2

2006-11-09T21:06:00.000-08:00

Today we discussed Rolle's Theorem and the Mean Value Theorem.

Let's start with Rolle's Theorem!

Rolle's Theorem is applicable if the following three conditions are true:

1.) f(x) is continuous on a closed interval [a,b]
2.) f(x) is differentiable on an open interval (a,b)
3.) f(a)=f(b)

(Note: In these conditions, f(x) is the function and a and b are points on the function)

Basically, Rolle's Theorem states that somewhere between points a and b on the graph of a function f, there exists a third point c that will have a slope of zero. This means that somewhere in between points a and b there is at least one maximum or minimum in the graph. A third way of saying this is that somewhere in between points a and b, there is at least one point that has a horizontal tangent.

Here is a basic graph that demonstrates the main idea of Rolle's Theorem:

As you can see in this graph, all three of the conditions of Rolle's Theorem exist, and, as the theorem states, there is at least point(in this case there are two) in between points a and b that has a slope of zero and whose tangent is horizontal.

Here are some examples of cases in which the conditions of Rolle's Theorem break down:

Condition that breaks down: f(a)=f(b)

As you can see, there is no value c in between a and b whose slope is zero.

Condition that breaks down: f(x) is differentiable on (a,b)

As you can see, at point c, the graph of f(x) is not differentiable, and thus point c does not have a slope of zero

Condition that breaks down: f(x) is continuous on closed interval [a,b]

As you can see, since the graph of f is discontinuous, there is no point c in between points a and b with a horizontal tangent line.

Now Let's Review the Mean Value Theorem!:

In the Mean Value Theorem, the first two conditions of Rolle's Theorem hold true, but not the third, which states that f(a)=f(b). So, in order for the Mean Value Theorem to be applicable, f(x) must be continuous on the closed interval of [a,b] and f(x) must be differentiable on (a,b), but f(a) need not equal f(b).

Basically the Mean Value Theorem states that somewhere between points a and b on the graph of function f, there exists a point c whose tangent line has the same slope of the secant line between the end points.

Here is a geometric proof of the Median Value Theorem:

As you can see, the tangent line at point c, the red line, is parallel to the pink secant line between points a and b in the first graph, and that the tangent lines at point c1 and point c2 are parallel to the blue secant line from point a to point b in the second graph.(The first graph is accurate, while I had trouble making the lines in the second graph parallel, but you get the picture.) Basically, I was trying to portray that the Median Value Theorem is applicable in both scenarios.

Here are some helpful formulas for the Mean Value Theorem:

or

Here are some sample problems to help you understand these concepts better:

1.) Find the number, using the Mean Value Theorem, that satisfies this equation within the given domain:

Domain:

Solution: We will find this answer algebraically through a formula
First you must identify four things:
a=-1
b=3
f(a)=13
f(b)= 1

Then, you must plug these four pieces of information into one of the equations for the Mean Value Theorem:
We will use the equation:

When we plug in the values for f(b), f(a), b, and a into this equation, we get:

The next step is to find the derivative of f(x): which is 2x-5.
Now, we must set the derivative of f(x) equal to the derivative of f(c):
2x-5=-3
Once we solve for x, we have our answer!
2x=2
x=1
Don't forget to check to see if your answer lies in the given domain!
Since -1<1<3, our answer is correct.

2.) Suppose that f(0)=-7 and f'(x) <8 and >-3. How large and how small can f(3) possibly be?

Solution: We will also find this answer algebraically using a formula
First identify a,b,f(a), and f(b)
a=0
b=3
f(a)=-7
f(b)=?

To find the largest f(3) can possibly be, set the equation for the Mean Value Theorem equal to 8, which will be our f'(c), because we want to use the steepest slope possible to get the largest value for f(3).
For this problem, we will use the equation:

So, after plugging in
the appropriate numbers into the equation, we get:

So, the largest f(3) can be is 17.

To find the smallest f(3) can be, set the equation for the Mean Value Theorem equal to -3. In other words, f'(c) will be -3.

Solve this like the solution above, except substitute -3 for 8 for f'(c).

Our equation is:

So, the smallest f(3) can possibly be is -16.

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

http://www.maths.abdn.ac.uk/~igc/tch/ma2001/notes/node42.html

This is an official reminder that JESSICA is posting next!

Here are a couple of calculus songs to get you into the upcoming holiday spirit and one song that I thought was funny!

" OH CALCULUS, OH CALCULUS!"
By Denis Gannon (1940-1991) may be sung to "Oh, Christmas Tree"

Oh, Calculus; Oh, Calculus,
How tough are your two branches.
Oh, Calculus; Oh, Calculus,
To pass, what are my chances?
Derivatives, I cannot take,
At integrals my fingers shake.
Oh, Calculus; Oh, Calculus,
How tough are your two branches.
Oh, Calculus; Oh, Calculus,
Your theorems I can't master.
Oh, Calculus; Oh, Calculus,
My proofs are a disaster.
You pull a trick out of the air,
Or find a reason, God knows where.
Oh, Calculus; Oh, Calculus,
Your theorems I can't master.
Oh, Calculus; Oh, Calculus,
Your problems do distress me.
Oh, Calculus; Oh, Calculus,
Related rates depress me.
I cut out boxes in my sleep,
And max and min do make me weep.
Oh, Calculus; Oh, Calculus,
Your problems do distress me.
Oh, Calculus; Oh, Calculus,
My limit I am reaching.
Oh, Calculus; Oh, Calculus,
For mercy, I'm beseeching.
My grades do not approach a B,
They're just an epsilon from D.
Oh, Calculus; Oh, Calculus,
My limit I am reaching.

Calculus
A Calculus Christmas Carol
written by Deborah Alterman, Martin Mohlenkamp, and Gareth Roberts (Sung to the tune of Jingle Bells)

Looking for the tangent
It's really m we seek
With epsilon and delta
Mathematics looks like Greek
Trying to find a limit
Everything gets small
If you can't determine it
You land in L'Hopital
Calculus, Calculus
Let us celebrate
Riemann Sums, so much fun
We can integrate
Calculus, Calculus
Let us celebrate
dx, dy, dz, dt
We love related rates
Calc 2 is coming next
The really fun stuff starts
Integral x sin(x)
Only works in parts
Summing the series harmonic
The terms keep getting small
But isn't it ironic?
It won't converge at all
Calculus, Calculus
Put your mind at rest
Most divergent series fail
The Root or Ratio Test
Calculus, Calculus
While we sing this song
We can sum (1/2)n
It won't take all day long
Now on to Calc 3
Learning spheres and cones
Maximizing functions
With seventeen unknowns
Fancy vector fields
Finding flux of curl
Using Mr. Stokes
Over a circle
Calculus, Calculus
Ain't it really cool
Gauss and Green are really keen
Don't take them for fools
Calculus, Calculus
Ain't it really cool
f and g, composing thee
Time for the chain rule

Chapter 3 Test Question 12

2006-11-09T21:01:00.000-08:00

a) Use linear approximation techniques to estimate the fourth root of 256.8. Leave your answer as a number plus or minus a fraction. Show your work.
b) Determine the calculator-generated value of the fourth root of 256.8
c) To how many decimal places is your estimate similar to the calculator-generated value?

Ok. I'm gonna concentrate on how to do part A, just because part B is putting something into your calculator and part C is subracting your answer from that number.

The trick to the problem is visualization. There's an equation for linear approximation, but it just makes a pretty simple concept really complicated.
NOTE: THIS IS NOT A DIAGRAM OF THE 4TH ROOT OF X. IT IS AN EXAMPLE (because the graph of y=the 4th root of x looks really straight)
This is like the diagram that explains linear approximation in the book. It makes very little sense, so I'll try to explain what's going on. We want to find f(256.8). You may have noticed that f(256) is equal to 4. We can imagine that f(256.8) would be really close to 4, but it should be a tiny bit larger. To find that tiny bit, we're going to get the slope (derivative) at 256 and pretend that it hardly changes between 256 and 256.8. If we imagine a tangent line with that slope that only touches at 256, if we go to 256.8 on that line, the Y value there will be about the same as f(256.8). If the black line in the diagram is f(x) and we set A equal to 256, we can calculate the tangent line (the purple one) at that point. Now, to approximate f(256.8), we're just going to trace that tangent line out .8 units farther and we'll add that change to the original f(256). In the graph above, .8 is represented by the variable X.

To put it mathematically...
Basic Equation
Y=X^.25 (same as fourth root)

f(a)
a=256
Y(256) =4

Derivative Equation
Y'=.25X^-.75 (by the power rule)
Y'(256) = .25 (256^-.75)
Simplifies to...
Y'(256) = 1/256

Distance down the Tangent Line
Since we know f(256) and we want f(256.8), the distance we need to go down the tangent line is .8. Since we know the slope at 256, the change in Y will be the slope times .8, or:
Change in Y = (.8)(.25 (256^-.75))

The Final Answer
Now we just take our original number (f(256)=4) and add the change in Y when we move .8 units to the right (Change in Y = (.8)(1/256)), and we get the answer

Answer
256.8^.25 = (approximately) 4 + (1/320)

Alternate method
An (arguably) easier way to do this question would be with the equation:
f(A+X) = f(A) + X*f'(A)
Where A and X are the same as described above.

Chapter 3 Test Question 2

2006-11-09T20:28:00.000-08:00

Differentiate:

a.

b.

c.

d.

Solution: For this problem, both the chain rule and the derivative of an exponential must be applied.

The answer is b.

Chapter 3 Test: Question #1

2006-11-09T19:27:00.000-08:00

Problem #1: Differentiate y=8^x(sin(x)-cos(x))

Solution:
1) Before we find the derivative of the function, understand the rules that must be applied to find the derivative of this function: Product Rule

2) First, find the two functions that are being multiplied: 8 ^x and (sin(x)-cos(x))

3) Next, find the derivative of both the functions:
d/dx 8^x= d/dx e^((ln 8)x)= e^((ln 8)x) x d/dx(ln a)x= e^((ln 8)x)) x ln 8= 8^x ln 8
AND
(sin(x)-cos(x))=(cos(x)-(-sin x))=(cos(x)+sin(x))= ((sin(x)+cos(x))

4)Simply plug in the numbers that are required for the product rule:
8^x ln 8(sin(x)-cos(x))+8^x(sin(x)+cos(x)) = 8^x(sin(x)+cos(x)) + 8^x(sin(x)-cos(x))ln 8= Choice C

Chapter 3 Test Number 5

2006-11-09T19:08:00.000-08:00

find the derivative of the following function: y = 9^(7x)

recognize that the chain rule must be used

remember the formula: (a^(x))' = (a^(x)) (ln a)

a=9
applying the chain rule to the exponent '7x',
so.... 9^(7x) (7x)'

the der. of 7x is 7

(9^7x)(ln9)(7) = y'(x)

this answer matches the answer choice of 'b'

Chapter 3 Test: Number 8

2006-11-09T18:23:00.000-08:00

#8.)

In order to find the derivative of this function, the quotient rule must be used. The quotient rule is:

Another way to remember the quotient rule is by the following phrase: Low d high minus high d low all over the square of what’s below.

Thus, we will begin solving the problem by first identifying what the derivatives of the top(high) and the derivative of the bottom(low) are.

To find the derivative of the top: The derivative of the top is 1 because u can also be written as:

, and according to the power rule the derivative of x, or in this case, u, raised to
a number is equal to the number, in this case 1, multiplied by x, in this case u, raised to a number that is one less than the original exponent, in this case the new exponent will be 0 because 1-1=0. Thus,

So the derivative of the top is 1.

To find the derivative of the bottom: The derivative of the bottom can be found using both the product rule and the difference rule. We use the product rule, expressed above, to find the derivative of

.
So we have

,

or 2u, as the derivative of the first term of the denominator. The second term in the denominator is 1. According to the derivative of a constant function, the derivative of a constant is always 0. Since the number one is a constant, its derivative is 0. Now, we can use the difference rule to find the overall derivative of the bottom of the function. The difference rule basically states that you can find the derivative of the difference of two numbers by finding the derivatives of each individual number and then subtracting the derivative of the second number from the derivative of the first number. In this case, we take the derivative of the second number, 0, and subtract it from the derivative of the first number, 2u, and we get 2u-0, or 2u as the overall derivative of the denominator of the function.

To find the overall derivative of f(u):
Now, we can use the quotient rule to find the overall derivative of the function.
1.) Take the bottom and multiply is by the derivative of the top:

2.) Now take the top and multiply it by the derivative of the bottom: u(2u)
3.) Next, subtract step 2 from step 1:

4.) Square the bottom:

5.) Divide the numerator(low d high minus high d low), or step 3, by the square of the bottom, step 4:

Thus, as our answer, we have

Chapter 3 Test Question 11

2006-11-09T17:36:00.000-08:00

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

Solution:

Chapter 3 Test: Number 10

2006-11-09T17:24:00.000-08:00

Chapter 3 Test Question 7

2006-11-09T16:44:00.000-08:00

In this question, we first use the Product Rule before using the Chain Rule.

Chapter 3 #14

2006-11-09T16:38:00.000-08:00

Given x(t) = -t^4 + 4t^3 - 2 where x is in meters and t in seconds

Find the acceleration at time t

Solution: Find the second derivitive

Using the power rule you get:

f'(x) = -4t^3 + 12t^2
f''(x) = -12t^2 + 24t

Chapter 3 Test, #13

2006-11-09T15:38:00.000-08:00

1. First, I started by setting up a diagram of this situation:

2. I then saw that from this diagram, I could create a triangle that looked like the following, with the hypotenuse c being the distance between the two ships:

3. I then started to assign numerical values to each of the sides. We are given no value for c, so I left it as c. For b, we have the value of 200 km, which is the horizontal distance between the two starting points of the ships. To find a, I calculated how far each ship travelled by multiplying each of the ships' speeds by 8, because both ships were sailing for 8 hours.

Ship A=(34 km/h) * (8 h)=272 km
Ship B=(21 km/h) * (8 h)=168 km

Then, I added the two distances of Ships A and B, which equals 440 km, and I assigned that value to side a.

4. To find c, we can use the Pythagorean theorem:

We plug in the numbers that we assigned to a and b, and get the following equation to solve for c:

We will also use the Pythagorean theorem for our static equation.

5. Now we have to change the static equation into a dynamic equation by differentiating both sides of the equation. If we differentiate the Pythagorean theorem, we get the following dynamic equation:

6. Now, we plug in the values that we have, solving for dc/dt

a=440
b=200
c=(233,600)^(1/2)
da/dt=rate of ship A+rate of ship B=55
db/dt=0 (b is our constant)
dc/dt=unknown

The answer to the problem, rounding to the nearest thousandth, is 50.070 km/h

Chapter 3 Test Topics - Updated!

2006-11-06T22:20:00.000-08:00

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

2006-11-06T19:26:00.000-08:00

To manually find the maximum and minimum values of a function (without a calculator!), we have to use a combination of basic algebra and derivatives.

There are two types of maximums and minimums: Local and Absolute. Local maximums and minimums are extreme points on a graph that are only maximums in their own area or limited domain. Absolute maximums and minimums, on the other hand, are the very highest and lowest points on a graph (there may be more than one of each, but they would have to share the same Y value). A key indicator of an extreme point is a "kink" in the graph or a point at which the slope is 0.

To find a maximum, you must look at the critical points of a function and at the edges of the domain. For instance, on the graph of y=x^2 on the interval [-1,5], the minimum value would be (0,0) a critical point, while the maximum value would be (5,25), a point on the edge of the domain.

For example, if we were using the equation y= 3x^2 - 2x + 5 on [-2,4] and we wanted to find the extreme points in each direction, we would calculate the y values where the derivative is 0 and at the edges of the restricted domain. The edges of the domain are easy. We just plug -2 and 4 into the equation to get (-2,13) and (4,45). We now need to use the derivative equation to find the critical point(s).

The differentiated equation is y' = 6x-2. This means that at x = 1/3, the slope is horizontal. We now plug 1/3 back into the original equation to find the final critical point, (1/3, 14/3). Using this, we can now say that the maximum in the domain is (4,45) and the minimum is (1/3, 14/3).

Many times, you will also be required to find the local maximums and minimums. They are also critical points and "kinks", but they are not always the absolute minimums or maximums. However, any critical point or "kink" at which the sign (positive or negative) of the slope is different on both sides is a local minimum or maximum.

Here's an interesting website that explains maximums and minimums in more depth.

Lauren's next....But she doesn't have to post until thursday...

And so I leave you with yet another beloved Grandpa Simpson quote...
"My Homer is not a communist. He may be a liar, a pig, an idiot, a communist, but he is not a porn star. "

3.11 Linear Approximations and Differentials

2006-11-04T22:12:00.000-08:00

It is possible to calculate the value of f(a) of a function where the graph of f(x) is tangent at the point a, f(a) but calculating for any values that are extremely near a is difficult.

the equation of the linear approximation/tangent line approximation of f at a is
L(x) = f(a) + f '(a)(x - a)

How to get that equation:
Remember the formula f '(a) = (f(x) - f(a))/(x - a) ?
solve for f(x). . .and voila!

you get f(x) = f(a) + f '(a)(x - a). ^_^

For example, suppose the murderer from Ismael's project group's movie wants to bake his victim for dinner (okok, we don't know exactly if he really is a cannibal, but let's just assume that he is one). The temperature of the body is 36 degrees C. Suppose it is baked at 187 degrees C. After an hour, the body is 52 degrees C and after 2 hours it is 66 degrees C. (not 666, haha)
Predict the temperature of the body after four hours. (If you get this right, you won't be the next victim - just kidding)

The equation could be set up with f (x) as the temperature of the body after x hours. Therefore,
f(0) = 36
f(1) = 52
f(3) = 66
we need to find f '(2) in order to make a linear approximation.

f '(2) = lim x->2 (f(x) - f(2))/(x-2)

we can use x = 1 here. (at one hour) so,

f '(2) = (f(1) - f(2))/(1-2) = (52-66)/(-1) = 14 degrees C/hour

this answer is the instantaneous rate of temprature change by the average rate of change between 1 and 2 hours. therefore, the linear approximation is f(4) = f(2) + f'(2)(4-2) = 52+14(2) = 80.

another (more accurate) method is to plot the givein data and estimate the slope of the tangent line (p. 263 of your textbook)

*Tip: when a question asks you to find the linearization of a function which has radicals in it, always change the radical into an exponent (p. 263)

Differentials: If y = f(x) and f(x) is differentiable, then the differential dx is an independent variable. (dx can be any real number) the differential dx is defined by:

dy = f '(x) dx

Refer to the illustration on p. 265.
dx is the change in x from the point at which the tangent line is tangent to another point. dx is the change x.
dy, however is NOT just the change in y.
suppose you are given the points (x1, y1) and (x2, y2). on the original graph. the graph is tangent at (x1, y1) but at x2 the value of the tangent line is b. dx is the change in x from x1 to x2. BUT! change in y is the change from y1 to y2. dy is the change from y1 to b.

sites:
http://archives.math.utk.edu/visual.calculus/2/linear_app.6/index.html
http://www.math.dartmouth.edu/~klbooksite/2.14/214.html
http://tutorial.math.lamar.edu/AllBrowsers/2413/Differentials.asp

Well, hope that helped...and here ends the rant of the evil math student ~ OY!!! JEFF YOU'RE UP NEXT! *poke, poke* =)

What I think about while I play online games (studying for calculus :D)

3.10 Related Rates

2006-11-01T19:51:00.000-08:00

This section has one of the most interesting and most involved of the topics we have covered so far; therefore, a systematic method is suggested for solving related rates problems:

1) Read the problem and take note of units and given values.
2) Draw a picture.
3) Develop a static equation relating all variables.
4) Implicitly differentiate the static equation, usually with respect to time.
5) Substitute the given and/or implied values stated in the problem and solve the equation for the requested unknown.

Sample problem

A television camera is positioned 4000 ft from the base of a rocket launching pad. The mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Assuming that the rocket rises vertically, and that its speed is 600 ft/s when it has risen 3000 ft, how fast is the distance from the television camera to the rocket changing at that moment?

Step 1) All units are in feet and seconds, no conversion necessary. Horizontal distance to rocket is 4000 ft. The height of the rocket = 600T. The time= 5 sec, the height = 3000 ft.

Step 2)

Step 3) (360000T)(T) + (16000000) = N*N

Step 4) 2(600T)*600 = 2N*(dN/dT)

Step 5) (720000T)/2N = dN/dT

3600000/2N = dN/dT

N*N= (3000*3000) + (4000*4000)
N*N = 25000000
N = 5000

36000000/10000 = dN/dT

dN/dT = 360 ft/s

These are two sites which have explanations and examples for related rate problems.

http://www.mathematicshelpcentral.com/lecture_notes/calculus_1_folder/related_rates.htm

http://people.hofstra.edu/Faculty/Stefan_Waner/RealWorld/tutorials/frames4_4.html

If Noah's flood was caused by rain over 40 days and nights, what was the rate at which the rain fell?
SAe = 5.096*10^8 sq km
SA(land) = 1.48*10^8 sq km
SA(water) = 3.616*10^8 sq km
Elevation of Mt. Everest = 8.85 km
Average land elevation = 0.5 km
Volume of water = (3.616*10^8)8.85 + 0.5*(1.48*10^8)
Volume of water = 3.27*10^9 cubic km
average depth rained = V/SA
average depth rained = (3.27*10^9)/(5.096*10^8)
average depth rained = 6.42 km
6.13 km/960 hrs = 6.69 m/hr or .186 cm/s

3.7 Higher Derivatives

2006-10-30T12:19:00.000-08:00

In today’s lesson, we learned that not only can you take a derivative of a function, you can take the derivative of another derivative as well!!! Let’s take a look:

(As usual, we are finding the derivative of the first equation)
(However, we can also find the derivative of the first derivative which is called double primes)
(We can even find the derivative again of double prime and make it triple prime)
(And lastly you can find the fourth prime which in this case is the final derivative possible since you have reached 0!)

Let’s look at some new notation we learned today that might be on the quiz/test/AAAPPP!!!!!

(Now this derivative is finding the slope of the tangent line/ instantaneous rate of change)
(This shows how the derivative is changing)
(This shows how the previous derivative is changing as well)

You can also use the concept of Higher Derivatives on the Trigonometric Functions

As you can see, the 4th prime equation is back to (sinx). This means that every fourth derivative in this situation repeats. Therefore could you find ?

If you divide , you will get 30 remainder 2. Thus you look at your double prime which is the second equation and you see that

Now let’s use this concept on an exponential function:

How about ?

Finally why don’t you try a velocity/ acceleration problem which incorporates the higher derivative concept. If the function is given for the position of a particle, find:
1) Acceleration at time t. What is the acceleration after 4 sec?
2) When is the particle speeding up and slowing down? Show this through a graph for
Velocity function(derivative of the position function):
Acceleration function (derivative of the velocity function) :
For the acceleration after 4 s, just plug it into the previous equation.

To find when the particle is speeding up and down, you must first equal the velocity function to 0.

So at t=1 and 3, the particle is at rest.
Plug numbers in for t (in the equation ) and you will see that if the particle was moving before t=1, the particle has positive velocity, if the particle was moving between t=1 and 3, the particle has negative velocity, and if the particle was moving after t=3, the particle would have positive velocity.

How about acceleration? Make the acceleration function=0:
a=6t-12
0=6t-12
t=2; meaning at 2 seconds, the particle will have 0 acceleration.
Before t=2, the acceleration will be negative and after t=2, the acceleration will be positive.

Therefore, when the signs of velocity and acceleration are the same, the particle is speeding up. If the signs are opposite then it means that the particle is slowing down.

Speeding up:
Slowing down:

For more perspective:
http://tutorial.math.lamar.edu/AllBrowsers/2413/HigherOrderDerivatives.asp

Brian - you're next!

3.8 Derivatives of Logarithmic Functions

2006-10-29T15:48:00.000-08:00

Hey class. Hope all of you enjoyed homecoming. Sorry that the game wasn't much of a show. Poly next week though so keep up the spirits. Well its time for math again, and this concept can be pretty difficult.

There are two basic equations in this section that the others are derived from:
For the first equation it is 1 over both x and ln of the base a

and the second equation is simply one over x when taking the derivative of the natural log of a number

From the second equation, if the the "x" is a polynomial or complex equation that we'll designate as "u" then it is 1 over the equation "u" multiplied by the derivative of that equation:

Logarithmic Differention is basically just applying logarithmic rules to simplify an equation so that its derivative can be taken easier.

Here is a simple example from the book of logarithmic differentiation:

Differentiate:

In the first step we took the natural log "ln" of both sides, and therefore the exponent came down.
In the next step using the product rule we took the derivative of that term. Specifically we took the first term times the derivative of lnx which is 1/x plus lnx times one over twice the square root of x.
In the third step we multiplied y to both sides in order to isolate y' on one side.

To express e as a limit:
As x approaches zero

As n approaches infinity

Additional Links
Another Review
A GREAT place to practice derivatives of logs with sample problems and solutions

MAGNUS you're up next.

This is Maurice Drew, he came out of UCLA last year so this is his rookie season. He's 5'7" and thats football height so he's probably shorter and 200 pounds. His legs as you can see are ginormous and people often mistake him for having pads on when he does not. This season he has 187 yds. and 3 touchdowns.

This is Mr. C who taught geometry here last year as well as coached the 400 team. He is really fast. He holds the prep league record in the 400 and the 200. Mr. C ran track in college at Amherst. He is currently in med school at USC.

3.6 Implicit Differentiation

2006-10-26T19:31:00.000-07:00

Hey guys, this is Ismael if you hadn’t already guessed. Today we learned about implicit differentiation. Implicit differentiation is different from explicit differentiation in that the explicit functions we have dealt with so far only deal with expressing one variable explicitly in terms of another variable. Implicit functions express the relationship between y and x implicitly. For example:

Explicit

Implicit

The implicit equations can be used to draw the graph of a circle. By solving
for y you get .The function is the upper half of the circle and the function is the lower half of the circle. Graphing this function gives us:

The tangent lines represent the derivatives of both functions when x=5.

You can differentiate implicit functions like these by this method:

Use the Chain Rule:
This is the same thing as saying because y=f(x) and . Therefore:

Here is an example of this method of using implicit differentiation to find y' (the slope):

Note: because we are not dealing with x in this case; we are dealing with y, which represents a function. Therefore, it is necessary to take the extra step of taking the derivative of the inside function, which is y. As a result .

Let's continue:

Now move over the and the 6xy' terms to the opposite sides of the equation so that they switch places.

This allows us to factor out y prime

Now we know the slope of the function. Let’s say we had to find the slope of the implicit function at the point (3, 3). We would simply plug in x=3 and y=3 into the equation we derived for y'.

Use the point slope formula:

Let’s say we were to find the points on the graph at which the slope of the tangent line was 0. First, you would set the equation for y' equal to 0.

Now we plug what we got for y into the original equation.

Bring to the other side of the equation:

Now we think: for what values of x would the left side of the equation equal zero? One value, obviously, it would be x=0. To find the second one, we first divide both sides by , giving us:

We then multiply both sides of the equation by eight and get:

Therefore, the values at which the derivative of the graph is zero are and x=0. To find the y values, simply plug in the x values into the equation .

Therefore, the points on the graph at which the tangent line is horizontal to the curve are: (0,0) and ,.

Now, we move on to inverse trig functions!
Let’s say we have
That equation is the equivalent of
By implicit differentiation:

By using our Pythagorean rule for the relationship between sine and cosine
( ) we can come up with:

(Note: We can replace with because of our original equation: )

Therefore,

You can look at page 233 in our text book for the table of all of the derivatives of the inverse trigonometric functions.

By the way, Ami, you are next for the blog posting.

Here are a couple of links supporting the Topic of the Day:
http://archives.math.utk.edu/visual.calculus/3/implicit.7/index.html

http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/implicitdiffdirectory/ImplicitDiff.html

And to end this blog with a bang, here are a few Ben Franklin quotes:

"Anger is never without reason, but seldom a good one"

"Do not fear mistakes. You will know failure. Continue to reach out."

"Either write something worth reading or do something worth writing."

"Employ thy time well, if thou meanest to gain leisure."

"Experience is a dear teacher, but fools will learn at no other."

And some silly quotes:

"Sure there have been injuries and deaths in boxing - but none of them serious." - Alan Minter, Boxer

"I think that the film Clueless was very deep. I think it was deep in the way that it was very light. I think lightness has to come from a very deep place if it's true lightness." - Alicia Silverstone, Actress

"How to store your baby walker: First, remove baby." - Anonymous Manufacturer

"You guys line up alphabetically by height." - Bill Peterson, Florida State football coach

"Men, I want you just thinking of one word all season. One word and one word only: Super Bowl." - Bill Peterson, football coach

"The internet is a great way to get on the net." - Bob Dole, Republican presidential candidate

Friday's Quiz Topics

2006-10-25T20:37:00.000-07:00

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:

3.5 The Chain Rule

2006-10-23T22:09:00.000-07:00

Hey everyone, I know that this may say that Ryan is posting the blog but it is in fact I, Alex, who am posting the blog under Ryan's account. Unfortunately I was rather inept when it came to making my blog account, so here I am. Before I get into the notes taken from today's class, I would like to take this opportunity to remind Izzy that his day as blog master is tomorrow. Anywho, onto the lesson!

Today we learned the chain rule, which, to put it simply, is taking all of the differentiability rules that we have learned so far and applying them to composite functions and power functions. There are a few really important formulas to remember here, one of which is the basic formula for the chain rule. Given that the function F(x)=f(g(x)), then,

In other words, the derivative of the composite function f(g(x)) equals the derivative of f(x) times the function g(x) times the derivative of g(x)

Example problem:

Differentiate the following:

We can take this concept of the chain rule a step further and apply it to the power rule. The power rule's general equation is as follows if n is any real number and u=g(x) is differentiable:

Another way to state this equation is:

Example Problem:

Differentiate the following:

And now we can apply all of this knowledge to find the Derivative of the Exponential!! This may seem somewhat daunting because Oh No! It uses a natural logarithm! (well that's at least how I think-you may think differently), but it is substantially less confusing than it may seem. Here is the general equation for the Derivative of the exponential:

In other words, to find the derivative of a given number "a" raised to any exponent "x", you multiply that number raised to the exponent by the natural log of the number

Example Problem:

Differentiate the following:

Well, that's pretty much it for all the notes and whatnot. Now enjoy my little artistic creation that I have here. I can't help enjoying it, because Grey's Anatomy is the best show on television.Bye everyone!! IZZY YOU GO TOMORROW!! Thought that I should just remind you again :-)

Section 3.4: Derivatives of Trigonometric Functions

2006-10-22T09:58:00.000-07:00

So we've been taking the derivatives of functions for a while now, and I know that a question that's been burning in everyone's mind is if we can take the derivatives of trigonometric functions. Well, the answer is finally here, and of course we can. Just take a look at this graph of sin x.

If you take the derivative of this graph, it comes out looking something like this:

Holy Cow! It kinda looks like the graph of cos x! Imagine that. I'm in shock.

Wait a second, what happens if we draw the derivative of cos x?

Looks sorta like the sin x graph only flipped. Maybe it's the -sin x graph. No way!

So:

and

But there's more! By using a similar way of finding the derivative, you can find that:

and So let's do a problem!

If you want to see the algebraic proofs for this lesson, or if you just need some help understanding the concept, you can check out this website.

http://archives.math.utk.edu/visual.calculus/2/trig.1/index.html

Oh, and I found someone that actually solved the sudoku that was on our test the other day...

P.S. -Reminder to Ismael

Section 3.3: Rates of Change in the Natural and Social Sciences

2006-10-19T06:02:00.000-07:00

Section 3.3 explains to us how derivatives can connect with science and economics. We see in this section how useful it is to find the rate of change of functions in the natural and social sciences.

This is the slope of the secant line from x2 to x1. As x2 approaches x1 the secant line becomes a tangent line, giving us the instantaneous rate of change at that point. From this, the derivative can be defined as is below.

Average rate of change of y with respect to x:

The book explains the following areas of science that we can apply the derivative to.
Physics:
In physics, the derivative of s = f(t), where s is the position of an object moving in a straight line and t is the time, is the instantaneous velocity.

This is a schematic sketch of the motion of a particle moving back and forth along a line.

http://www.physics.upenn.edu/courses/gladney/mathphys/subsubsection2_1_1_2.html#SECTION02_1_1_2

Chemistry:
In chemistry, the derivative can be used to find the concentration of reactants during a chemical reaction. However, the derivative is often used to find the instantaneous rate of reaction. Compressibility is another area where we can apply the derivative.
http://www.physics.upenn.edu/courses/gladney/mathphys/subsection2_1_4.html
http://www.physics.upenn.edu/courses/gladney/mathphys/subsubsection2_1_1_4.html#SECTION02_1_1_4

Biology:
In biology, we can find the instantaneous rate of growth and velocity gradient. Also, the flow of blood in a blood vessel can be analyzed using the derivative.
http://www.physics.upenn.edu/courses/gladney/mathphys/subsection2_1_2.html#SECTION02_1_2

Economics:
In economics, we can find the instantaneous rate of change of cost, or the marginal cost.

When C is the cost, the marginal cost can be found using the derivative.

Of course, the derivative can be used in many more sciences other than these ones.

Sample:
A particle moves according to a law of motion s = f(t), t > 0, where t is measured in seconds and s in feet.

a.) Find the velocity at time t
b.) What is the velocity after 3 s?
c.) When is the particle at rest?
d.) When is the particle moving in the positive direction?
e.) Find the total distance traveled during the first 8 s.

Solution:
a.) First, we find the derivative of the function.

b.) Then plug in 3 into the derivative function to get:
-9 ft/s

c.) We find the zeroes of the derivative function by either graphing or using the quadratic equation and find that:

Zeroes: x=2, x=6

d.) We find the intervals where the derivative function is positive and we get
0<t<2,>6

e.) This one is a little harder. We have to take in account of when the object is moving backwards. When the derivative function is negative, the object is doing so which is between 2 seconds and 6 seconds.

Then, we plug in the times in the original function.

i.) During the first 2 seconds the object travels 128 ft.
ii.) Then it goes backward between seconds 2 and 6, moving -736 ft.
iii.) Finally it goes forward again after 6 seconds. Between seconds 6 and 8, the object travels 704 ft.

Adding all of these together, we figure out the object’s displacement is 96 ft.

“Respect yourself and others will respect you.”
- Confucius

Reminder to Claire for next blog!

Friday's Quiz

2006-10-19T00:28:00.000-07:00

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

Quiz – Sections 3.1-3
Apply derivative rules to combinations of functions given values of those functions and their derivative at specified values of “x”. (Note – be familiar with the various notations for the derivative and how they are applied.) (Sec 3.2, #31)
Differentiate functions using derivative rules. (Sec. 3.1, #13-31 odd, Sec. 3.2, 3,5,13,15,17)
Find the equation of a tangent line given a curve and a point. (Sec. 3.1, #39, Sec. 3.2, #23)
Find points on a curve where the tangent line is horizontal. (Sec. 3.1, #45,)
Interpret an application of the derivative. (Sec. 3.3, #15,29)
Apply derivative rules to combinations of functions given graphs of those functions. (Sec. 3.2, #35)
Position/Velocity Problem (Sec. 3.3, #3)

I’ll be in early on Friday, and available after school on Thursday until 3:00, when I have to attend a faculty meeting. See you in class!

"There is nothing noble in being superior to someone else. The true nobility is in being superior to your previous self."
-Hindu Proverb

Section 3.2: The Product and Quotient Rules

2006-10-17T19:46:00.000-07:00

Wassup!!!

Section 3.2 is all about using 2 new rules to find derivatives more quickly. These 2 rules are known as the: 1)Product Rule 2)Quotient Rule.

1. The Product Rule: [h'(x) = [f(x) x g(x)]] = [[f(x) x g'(x)] + [g(x) x f'(x)]]

The product rule is quite simple. If one is trying to find the derivative of the product of 2 different functions, finding the product of the 1st function[f(x)] and the derivative of the 2nd function[g'(x)] plus the product of the 2nd function[g(x)] and the derivative of the 1st function[f'(x)] is the same thing.

Answers can be left un-simplified ^_^
THANK THE LORD!!!

Example: f(x)=(x^7+3x^3+4x^2+7)(x^6+5x^4-3x)
f'(x)=(x^7+3x^3+4x^2+7)(6x^5+20x^3+3) + (x^6+5x^4-3x)(7x^6+9x^2+8x)

2. The Quotient Rule: [h'(x) = f(x) / g(x)] = [[g(x) x f’(x) – f(x) x g’(x)] / [[g(x)]^2]
If I can get equation editor to work, I shall fix the equation above to make it look more realistic

The quotient rule is almost as simple as the product rule. If one is trying to find the derivative of the quotient of 2 different functions, finding the product of the denominator function[g(x)] and the derivative of the numerator function[f'(x)] minus the product of the numerator function[f(x)] and the derivative of the denominator function[g'(x)] all over the square of the denominator function[[g(x)]^2] is the same thing.

A good way to remember the quotient rule is the song: "Lo di-Hi minus Hi di-Lo over the square of what's down below"

Answers can be left un-simplified like the product rule ^_^
THANK THE LORD AGAIN!!!

Example: h(x)=(3x^2+e^x) / (4x^3-12x)
h'(x)=[(4x^3-12x)(6x+e^x)-(3x^2+e^x)(12x^2-12)] / [(4x^3-12x)^2]

Here is a good site that can help you ^_^
http://tutorial.math.lamar.edu/AllBrowsers/2413/ProductQuotientRule.asp

Before I...PEACE OUT!!!(That phrase is so unlike me -_-)
Here's a picture of myself doing a chair freeze ^_^

Here's a picture of a very pretty Korean pop-star! BoA ^_^

One joke to top everything off ^_^
Q: What do you get if you add two apples and three apples?
A: A high school math problem!

And last, but not least...
KANE!!! YOUR NEXT ^_^

Chapter 2 Test Topics

2006-10-11T00:23:00.000-07:00

Here’s a list of topics that will be covered on this Thursday’s Chapter 2 Test. I've included relevant homework problems to review:

Chapter 2 Test Topics:
Evaluating limits from a graph. (Sec. 2.2, #5,7; Sec. 2.6, #3)
Evaluating limits at infinity (the Great Battle!) (Sec. 2.6, #13-33 odd,51)
Find horizontal and vertical asymptotes of a rational function. (Sec. 2.6,#37,39,41)
Sketch a discontinuous function and define/explain the discontinuities. (Sec. 2.5,29,39,43)
Find an equation of a tangent line to a curve using a particular method (either x approaches a or h approaches zero). (Sec. 2.7, #5)
Find an equation of a tangent line – any method. (Sec. 2.7, #7,13,17)
Determine “a” to ensure a limit exists for a rational function. (Sec. 2.3, #59)
Determine differentiability of a function at a point. Justify your answer. (Sec. 2.9, #43)
Interpret the meaning of a derivative. (Sec.2.8, #29,31)
Sketch a graph given limit and derivative conditions. (Sec. 2.2, #13; Sec. 2.6, #5,7; Sec. 2.8, #5)
Sketch the derivative of a function given a graph of the original function. (Sec. 2.9, #4,7,9,11)

The last two sketches will be worth a total of 90 points all by themselves, and the test is somewhat long (20 questions). Many of the questions can be answered without a lot of calculations. Be sure to allocate your time wisely!

I’ll be in early on Thursday, and available online after Back to School Night on Wednesday. See you in class!

Here's a site to practice sketching derivatives from a graph.

2.9 The Derivative as a Function

2006-10-09T10:42:00.000-07:00

We have a few definitions for derivative. One is:

f ' (a) = limh→0 ((f(a + h) – f(a))/ h).

This gives the derivative for a specific value of a. However, we can generalize this definition to include any value of a. We use the variable x in place of a:

f ' (x) = limh→0 ((f(x + h) – f(x)) / h).

Here is an example:

f(x) = x^3 - x. Find the formula for f ' (x).

f ' (x) = limh→0 (((x + h)^3 - (x + h)) - (x^3 - x))/h
= limh→0 ((x^3 + 3x^2h + 3xh^2 + h^3 -x - h) - x^3 + x)/h
= limh→0 (3x^2 + 3xh + h^2 - 1)
= 3x^2 - 1
We can call the answer the general rule for the slope of tangent lines to f(x). Graphing the answer would give you a graph of the values of the slopes of the tangent lines to f(x).

Differentiable vs. Not:

There are many notations for the derivative. They include:

f '(x) = y' = dy/dx = df/dx = (d/dx)f(x) = Df(x) = Dxf(x). The D's and d/dx's indicate the operation of differentiation. A function is differentiable if we can find the derivative at a specific point or over a specific interval. Furthermore, if a function is differentiable on an interval, then it is continuous on that interval.

Here are some examples of functions that are not differentiable at certain points. If a graph has a corner (a kink or cusp), a discontinuity, or a vertical tangent at a, then the function is not differentiable at a.

Not differentiable at x=0
(graph has a discontinuity).

Here is one link that has some good sample problems for f ' (x) problems. It shows you how to do them too: http://archives.math.utk.edu/visual.calculus/2/definition.12/. Below those examples is a different interpretation of differentiation.
Here is a differentiation link: http://www.people.hofstra.edu/faculty/Stefan_Waner/RealWorld/calctopic1/contanddiffb.html

Some good quotes:

The whole problem with the world is that fools and fanatics are always so certain of themselves, but wiser people so full of doubts.
Bertrand Russell

Whenever you find that you are on the side of the majority, it is time to reform.
Mark Twain

Look at all the sentences which seem true and question them.
David Reisman

The ultimate measure of a man is not where he stands in moments of comfort and convenience, but where he stands at times of challenge and controversy.
Martin Luther King, Jr.

RYAN, you are next.

Quiz 2.5-8 Topics

2006-10-04T20:44:00.000-07:00

Here’s a list of topics that will be covered on this Friday’s 2.5-8 Quiz. I’ve tried to indicate where a similar homework problem would be helpful.

Quiz 2.5-8
Find the limit of a function as x approaches infinity (The Great Battle) (Sec. 2.6, #13-33 odd)
Find the equation of a tangent line/slope using a specified definition of the derivative of a function at an indicated point. (Sec. 2.7, #5,7,13)
Find a constant to create a continuous function given a piecewise function. (Sec. 2.5, #41)
Sketch a graph given conditions (continuity). (Sec. 2.5, #5)
Sketch a graph given conditions (limits). (Sec. 2.6, #5,7)
Sketch a graph given conditions (derivatives). (Sec. 2.8, #5)
Explain why a function is discontinuous, using the definition of continuity. (Sec. 2.5, #15,17,19)
Use the Intermediate Value Theorem to show a root exists within an interval. (Sec. 2.5, #47)
Find horizontal and vertical asymptotes of a function. (Sec. 2.6, #37,39,41)
Interpret the meaning of a derivative (word problem!) (Sec.2.8, #29,31,34)

We're almost through with Chapter 2. Hopefully you're all thinking about your Calculus Movie Project!

I have a department meeting on Thursday morning, so I won't be available until class or after school. I'll be in early on Friday for any questions...

Making the simple complicated is commonplace; making the complicated simple, awesomely simple, that's creativity.
--Charles Mingus

Section 2.7 Tangents and Velocities

2006-10-03T21:02:00.000-07:00

This section deals with tangents and velocities, which are rates of change. Tangent lines are lines that touch a point on a curve only once. There are two formulas that can be used to calculate the tangent. If a curve has the equation f(x), in order to find the tangent line to the curve at a point P (a, f(a)), then consider a nearby point Q (x, f(x)), and compute the slope of the secant line PQ(a secant line is a line that passes through both points) by using:
mPQ = limx→a f(x) – f(a) / x – a
or if h = x – a then
m = limh→0 f(a + h) – f(a) / h
This is also known as the derivative. The terms slope of a tangent line, derivative, and instantaneous rate of change can be used interchangeably.

Here is an example of a tangent line:

Find an equation of the tangent line to the parabola y=x2.
a=1 , so
m= limx→1 f(x) – f(1)/ x – 1 = limx→1 x2 – 1/ x – 1
= limx→1 (x + 1) (x – 1) / x – 1
= limx→1 (x + 1) = 1 + 1 = 2

Using the point slope form to find the equation of the line:
y – 1 = 2(x – 1)

The average velocity is displacement over time. f(a + h) – f(a) / h Do not confuse this with the derivative.

Here is an example:
Suppose a car drives 10m in 5 sec and 15m in 10 sec. Find the average velocity between the two points.

Take the displacement and put it over the time: 15-10/(10-5) = 5/5 = 1 m / sec.

Here is a link to a useful site:
http://www.sosmath.com/calculus/diff/der01/der01.html

Reminder to Luke for the next posting.

2.6 Limits at Infinity; Horiz. Asymptotes

2006-10-02T18:54:00.000-07:00

This section deals with infinite limits and horizontal asmyptotes. There are two main ways to find a horizontal asymptote.
The first way, which I prefer, is graphically. Plug in the equation to your grapher and visually see where the asymptote is. If it is not obvious, look at the table of values and find the y value that is not quite reached in the graph as x approaches infinity.
The second way is to algebraically find the asymptote(s). To do this with a rational function, you must divide the top leading term by the bottom leading term to find the asymptote- for example: (x^3 +10)/(x^2 -3) the asymptote would be y=x. --[because x cubed divided by x squared equals x....... in the grand scheme of things, that + 10 and -3 will not make a big difference on a graph dealing with large numbers being cubed and squared, so we ignore those].
If there is a leading coefficient and the same degree on top and bottom, you do the same thing, just with the coefficients ... for example: as x approaches infinity, (4x^2 + 7x + 3)/ (2x^2 - x+4) ....4/2= 2 so the asymptote would be [horizontal] at y=2.
An extremely effective way to know which way the graph is heading [in terms of infinity and negative infinity] is to make a 2x2 chart:[the left term is as x approaches neg. inf., the right= inf]

even += up/up
even -= down/down
odd+= down/up
odd-= up/down

so for a problem Mr. French might throw at you on a quiz/test:
find the horizontal asymptote as x approaches +infinity: (sqrt(2x^2 +5))/(3x-5)

to solve this, you must first realize the +5 and -5 are meaningless so scratch them out.
the sqrt of 2x^2= rad2 x
rad2 x divided by 3x = (rad2) / 3 ... [x's cancel]
so the asymptote occurs at y=(rad2)/ 3 when x approaches positive infinity

link: http://www.purplemath.com/modules/asymtote2.htm

quote: "How much the more thou knowest, and how much the better thou understandest, so much the more severely shalt thou therefore be judged, unless thy life be also more holy."
~ Thomas a Kempis in 'Imitation of Christ'

REMINDER: Kyle, please remember to complete the next blog review!

Section 2.5 Continuity

2006-10-01T00:41:00.000-07:00

Section 2.5 Continuity

Well the title of this section tells all. As normal math section titles should.

Continuity's definition is lim x--> a f(x)=f(a). Basically, continuity is when you are able to follow a specified section of a graph without having any holes or jumps.

So how do you determine whether or not a graph is continuous?
First off, it has to follow three guidelines:

It has to have an existing limit for any value a.
The function has to have a value when x=a.
The limit and value have to be equal for any value a.

Removable discontinuities are basically those things they taught to us as "holes" in algebra.

Infinite discontinuites basically have no limit for the value a.

Jump discontinuities "jump" from value to value, like step functions.

Certain functions have domains which are all real numbers:

Polynomials
Trigonometric (only sin and cos)
Inverse Trigonometric (only tan-1 and cot-1)
Exponential
Roots (only odd ones)

Other functions have special domains:

Rationals
Trigonometry (only tan)
Inverse Trigonometric (only sin-1, cos-1, sec-1, and cosec-1)
Logarithmic

Intermediate Value Theorem (IVT)

Here's a link to another site with information on Continuity, though it has some extra info that isn't taught in our book:

http://www.sosmath.com/calculus/limcon/limcon05/limcon05.html

And here are some continuity problems:

http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/continuitydirectory/Continuity.html

And now for a photo I took while in New York this summer...

Reminder to JOEY to blog Monday's class notes!!

I'm out.

Quiz 2.1-4 Topics

2006-09-27T19:47:00.000-07:00

Here’s a list of topics that will be covered on this Friday’s 2.1-4 Quiz. I’ve tried to indicate where a similar homework problem would be helpful.

Calculate the slope of a secant line to a given level of accuracy. (2.1, #9)
Find an average rate of change from a data table. (2.1, #1)
Determine limits from a graph (2.2, #5,7)
Determine infinite limits given a function (2.2, #25)
Determine a value “a” to create a limit in a rational function (2.3, #59)
Apply the limit laws to determine a limit. (2.3, #1)
Determine delta given x, f(x), L and a (algebraically) (2.4, #1)
Determine delta given x, f(x), L and a (graphically) (2.4, #5)

That's it! I'll be in early on Friday. Don't forget the makeup test for chapter 1 has to be completed before the weekend.

2.4 The Precise Definition of a Limit

2006-09-27T18:03:00.000-07:00

The intuitive definition of a limit gives us estimates that are as close to the exact answer as we might need. However, a more precise definition of the limit gives us the ability to get good estimates.

The precise definition of a limit says that if f is a function on an open interval that contains the number a, except possibly at a itself, then the limit of f(x) as x approaches a is L. We can write this as

Verbally, we can say that f(x) is close to L when x is close to a because we can make the values of f(x) within any distance from L by taking the values of x within a distance from a, but not equal to a. This is illustrated in the graph directly below.

One sided limits can also be precisely defined.

Infinite limits can also be precisely define by saying that if f is a function is on an open interval that contains the numbers around a, but excluding a, then
* Functions that become largely negative as x gets close to a have this definition: Let f be a function defined on an open interval that contains the number a, except possibly at a itself. Then

*For limits, one-sided limits, and infinite limits, we can use the precise definitions to prove the limits convincingly instead of vaguely as we do with intuitive definitions.

Additional Links:

Oregon State Limits

NCTU Limits

Reminder: Crystal, your turn next.

2006-09-26T14:06:00.000-07:00

Today's lesson focused on "Limit Laws" and their uses:

There are eleven limit laws:

1.) lim x--> a[f(x)+g(x)] = lim x-->af(x) + limx-->ag(x)
*Verbally, this means that the limit of a sum is equal to the sum of the limits

2.) limx-->a[f(x)-g(x)] = limx-->af(x) - limx-->ag(x)
*Verbally, this means that the limit of a difference is equal to the diffenences of the limits

3.) limx-->a[cf(x)] = c limx-->af(x), where c is a constant
*Verbally, this means that the limit of a constant times a function is equal to the constant times the limit of that particular function.

4.) limx-->a[f(x)g(x)] = (limx-->af(x)) (lim x-->ag(x))
*Verbally, this means that the limit of a product is equal to the products of the limits

5.) limx-->af(x)/g(x) = (limx-->af(x))/(limx-->ag(x)) -if limx-->ag(x) is not equal to zero.
*Verbally, this means that limit of a quotient is the quotient of the limits(when and if the limit of the denominator is not zero)

6.) limx-->a[f(x)]^n = [limx-->af(x)]^n where n is a positive integer
*Verbally, this means that the limit of a function raised to the power n is equal to the entire limit raised to the nth degree

7.) limx-->ac = c
*Verbally, this means that the limit of a constant is equal to the constant

8.) limx-->ax = a
*Verbally, this means that the limit of x as x approaches a is equal to a,

9.) limx-->ax^n = a^n
*Verbally, this means that the limit of x to the nth degree as x approaches a is equal to a raised to the nth degree.

10.) This law is difficult to type of the computer, so I am going to verbally explain it. Basically this law states that the limit of the nth root of x as x approaches a is equal to the nth root of a.

11.) I will also explain this law verbally. It states that the nth root of f(x) as x approaches a is equal to the nth root of the entire limit of f(x) as x approaches a where n is a positive integer since you cannot take the root of a negative number.

Example Problems: What is another way to express the following limits using the laws listed above?
a.) limx-->2[f(x) + 5g(x)]
b.) limx-->1[f(x)g(x)]
c.) limx-->75
*Answers and Solutions:
a.) Using law #1 above, we can see that the limit of a sum is equal to the sum of the limits. Therefore, limx-->2[f(x) + 5g(x)] is equal to limx-->2f(x) + limx-->2 [5g(x)]. Then, according to law #3, we can see that that the limx-->2[5g(x)] is equal to 5 limx-->2g(x). Therefore, the answer is limx-->2f(x) + 5 limx-->2g(x).
b.) By using law #4, we can see that limx-->1[f(x)g(x)] is equal to the products of the limits of both f(x) and g(x) as x approaches zero. So, the answer is (limx-->1f(x))(limx-->1g(x)).
c.) According to law #7, the limit of a constant as x approaches a is the constant. Therefore, the solution is the constant 5.

Direct Substitution Property:
limx-->af(x) = f(a)
-functions with this property are continuous at a.

The Squeeze/Sandwich/Pinching Theorem:
-If f(x)<g(x)<h(x) when x is near a (except possibly when x is at a) and the limx-->af(x) =
limx-->ah(x) = L, then limx-->ag(x) = L.

*If there is a function whose limit is difficult to find, and you know of two functions that surround the original function around the limit, then you can find the limits of the two surrounding functions, and if they are equal, you can assume that the limit of the original function is the same as well.

Example Problem:
Find the limit of (x^2) sin(1/x) as x approaches 0.

Solution and Answer:
Since the graph of limx-->0 (x^2)sin(1/x) does not clearly show a limit, it is necessary to find two graphs whose limits would equal that of the original problem. So, we know that -1<sin1/x<1, and by multiplying each side by x^2, we get (-x^)2<x^2sin(1/x)<(x^2). From this we can find the limits of the graphs of (-x^2) and (x^2) as x approaches zero. We find that the limits of these two graphs as x approaches zero is equal to zero, then limx-->0(x^2)sin(1/x) is also equal to zero.

Other Important Theorems:
1.) limx-->af(x) = L if and only if limx-->a-f(x) = L = limx-->a+f(x)
*Verbally, this means that the limit of f(x) as x approaches a is the limit if an only if the limits of f(x) as x approaches a from both the negative and positive sides is equal to the limit.

2.) If f(x)<g(x) when x is near a(except possibly at a) and the limits of f and g both exist as x approaches a, then limx-->af(x)<limx-->ag(x).

Websites to help you understand the concepts of 2.3 more!

http://www.calculus.net/ci2/search/?request=category&code=1243&off=0&tag=9200438920658

http://www.geocities.com/mathdepot/squeeze.htm

Reminder to Jessica to write tomorrow's blog!

A couple jokes to keep you guys on your feet!

Q: What's the integral of (1/cabin)d(cabin)?

A: A natural log cabin!

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..."

Section 2.2

2006-09-25T23:23:00.000-07:00

The Limit of a Function

- limits occur when the x value approaches a certain value, f(x) also approaches a certain value
- the general notation for limits: lim x->a f(x) = L,"the limit of f(x), as x approaches a, equals L"
- f(x) can be arbitrarily close to a but will never equal a
- calculators can often make mistakes! (refer to example 2 on pages 94-95) on a calculator, if t is sufficiently small, you will get the value 0.
- when a graph shoes infintely many values for x that approach 0, like on the graph of lim x->0 sin (pi/x). the limit does not exist.
- when guessing a limit using a calculator, you cannot be 100% sure because if you can persevere using smaller and smaller values of x, results may be different.

One-Sided Limits
- Definition: written as lim x-> a- f(x) = L, the left hand limit of f(x) as x appraoches a [or the limit of f(x) as x approaches a from the left] is equal to L if f(x) is arbitrarily close to L, and x sufficiently close but LESS THAN a.
- "the right hand limit of f(x) as x approaches a is equal to L" is written lim x-> a+ f(x) = L, and in this case we are only considering when x>a. this leads to the following definition:
lim x->a f(x) = L if and only if lim x->a- f(x) = L and lim x-> a+ f(x) = L

Infinite Limits
- occurs when y values become infinitely large when x approaches a.
- do not regard infinity as a number!!!!
- the limit of f(x) as x approaches a is negative infinity when the values of x decrease without bound.
- the line x = a is a vertical asymptote of the curve y= f(x) if at least one of the following is true:
*can't insert infinity symbol, so I will represent using the word*

lim x->a f(x) = infinity
lim x->a- f(x) = infinity
lim x->a+ f(x) = infinity
lim x->a f(x) = negative infinity
lim x->a- f(x) = negative infinity
lim x->a+ f(x) = negative infinity

4 steps in determining the limit of a function:

try plugging in values
simplify algebraically
look at a graph
if all of the above don't work, then the limit D.N.E. (does not exist)

Sample Problem: (a simple, yet important concept)

(my internet lags so i apologize for not being able to post pictures)

what is the limit as x approaches 2 from the left? the right? does limx->2 exist? (refer to graph in link)

http://www.math.uncc.edu/~bjwichno/spring2005_math1241_002/Review_Calc_I/Images/1_over_x_2.gif

answer: no, because as x gets arbitrarily close to 2, the values become infinitely small (as x approaches from left) and infinitely large (as x approaches from right). there is, however, an asymptote at x=2.

links:

your ol' classic wiki favorite:
http://en.wikipedia.org/wiki/Limit_of_a_function

and this one, which most likely sums it up better than i do!http://www.math.hmc.edu/calculus/tutorials/limits/

and THIS ONE, is a cool simulator graph for limits. it's quite addicting... o.O http://www.math.psu.edu/courses/maserick/limit/limit.html

I hope that helped...and I believe it is the Lauren's turn?

Ending with a quote today...but I've always liked this quote from the "Father of Modern Mathematics" (pretty sure we'll deal with his concepts later on)...

"I think, therefore I am..." - Rene Descartes.

well, that's it. jaa ne...mata ne! (cya, and until next time!)

Section 2.1

2006-09-24T10:34:00.000-07:00

There are lots ways to say "derivative" - Instantaneous Slope, Velocity, etc. The first section in chapter two focuses on finding it by means of a tangent line to a curve. As we know, the slope of a line is equal to the "rise" divided by the "run". or the change in Y divided by the change in X. But while this system works for finding the slope of a straight line, we have to use it in a very specific way to find the slope of a given point on a curve. For instance, to find the slope at X=1 on the line Y=X^2...

We must have two points to find the slope of any line. Since we are looking for the slope at X=1, one of our points will be (1,1)
To find the second point, we should try some different points. First, let's use (2,4)
The slope between (1,1) and (2,4) is 3/1. Let's see how it changes as our second point gets closer to (1,1)
If our second point is (1.5, 2.25), the slope is 1.25/.5, or 2.5
If we get ridiculously close to (1,1) then our point could be (1.0001, 1.00020001), in which case our slope would be .00020001/.0001, or almost exactly 2.
As we can see, the closer our second point gets to 1, the closer the calculated slope is to 2
This can be expressed in the equation: Slope f(x) x->a = (f(x)-f(a))/(x-a) , where a is the point at which you want to find the slope, and x is a point that is (ideally) infinitely close to a.

Two interesting websites that I found that illustrate the concept of tangent lines in relation to instantaneous slope (both require the Java plugin).
http://www.ies.co.jp/math/java/calc/doukan/doukan.html
, and a much more technical version,
http://www-math.mit.edu/18.013A/HTML/tools/tools04.html

And now for the first of my series of the greatest and most meaningful quotes of all time...
“Then after World War Two, it got kinda quiet, ‘til Superman challenged FDR to a race around the world! FDR beat him by a furlong, or so the comic books would have you believe. The truth lies somewhere in between…”
—Abe Simpson (Better known as Grandpa Simpson)
And on that note, Lauren has the next post...

Test 1 Topics

2006-09-19T10:23:00.000-07:00

Here’s a list of topics that will be covered on this Thursday’s Chapter 1 Test. I’ve tried to indicate where a similar homework problem would be helpful.

Interpolation – given data or a graph, estimate/determine new points in the relationship/function. (1.1 - #17, 1.2 - #11,19)

Extrapolation – given data or a graph, estimate/determine new points in the relationship/function. (1.1 - #17, 1.2 - #11,19)

How does a graph change when certain values within the function vary? (1.4 - #31,34)

Given f(x) and g(x) (algebraically, numerically or graphically), determine values for f(g(x)) given x. (1.3 - #55)

Determine f(x) given a graph. (1.5 - #17)

Determine the range of a function. (1.1 - #23)

Determine the domain of a function. (1.1 - #23)

Determine an appropriate viewing window for a given function. (1.4 - #7,13)

Determine g(x) as a result of transformations of f(x). (1.3 - #3)

Work with inverse functions. (1.6 - #19)

Sketch a graph illustrating a given functional relationship. (1.1 - #11)

Sketch a graph given a transformed function. (1.3 - #3)

Prep AB Calculus C 2006-07

2006-09-19T07:51:00.000-07:00

no, god is not a mathematician. To be a mathematician requires black and white rationality. To create the world we live in requires quite the opposite. Things don't always make sense in today's world and not everything can be answered with a formula.