14-2 Arithmetic, Geometric, and other Sequences

14.2 Arithmetic, Geometric, and other Sequences

Sequence - is a function, with a term of pattern (n term - integer) and value (a number)

Recursion formula- specifies tn as a function of the previous term (tn-1)

Explicit formula - specifies tn as a function of n.

Arithmetic Sequence - Add or Subtract the same number to each preceding term
Common Difference - the constant number being added or subtracted in an arithmetic sequence

Arithmetic Linear Function -
Explicit: Tn = To + D(n-1)
Recursive: Tn = Tn-1 + D and To = x
where Tn is a value in the sequence, To is the first value in the sequence, D is the common difference and Tn-1 is the previous term.

Geometric Sequence - Multiply or Divide the same number to each preceding term
Common Ratio - the constant number being multiplied or divided in a geometric sequence

Geometric Exponential function -
Explicit: Tn = To * R(^n-1)
Recursive: Tn = Tn-1*R and To = x
where Tn is a value in the sequence, To is the initial value in the sequence, Tn-1 is the previous term and R is the common ratio.

Sequence Mode on the Calculator -

1. MODE


2. SEQ


3. Y=


4. nMin = (this is the starting term number..normally 1)


5. u (n) = (your equation goes here: u(n-1) + 5 ("u" can be found by 2ND 7 and "n" is the x, T, Theta, n button)


6. u (nMin) = (this is your starting term value)


7. in order to find a certain term in a sequence go to TBLSET (2ND, WINDOW ) and make the TblStart whatever term you are looking for then go to TABLE (2ND GRAPH ) and the value will be in the table under u(n).

Example Problem:

a. Tell Whether the sequence is arithemetic, geometric

b. Write the next 3 terms

c. Find t76

d. Find the term number of the term after the first ellipsis marks.

1. 26, 41.5, 57... ,6071.

2. 843, 140.5, 23.417... ,0.1087.

HOW TO SOLVE:

a. First you must look for a common difference or ratio by subtraction terms into each other like 41.5-26 = 15.5 and 57 - 41.5 = 15.5 so the common difference is add 15.5. If there is no common difference then you must divide terms into each other like 843 / 140.5 = 6 and 140.5 / 23.417 = about 6 because of rounding. So the common ratio is divide by 6. The first sequence is arithmetic because there is a common difference whereas the second sequence is geometric because there is a common ratio.

b. Use the seq mode TABLE with the TblStart = 1 in the TBLSET to see the next 3 terms.

c. Use the seq mode TABLE with the TbleStart = 76 in the TBLSET.

d. You can search for this value in the TABLE but it is much easier to solve for n in the equation for the sequence. So, since the first sequence is arithmetic use Tn = To + D(n-1) to solve for n: 6071 = 26 + 15.5 (n-1). So subtract 26 from both sides and then divide by 15.5 to both sides and 390 = n-1, then add one to both sides and n=391 meaning 6070 is the 391st term in the sequence. For geometric sequences, using the explicit formula to solve for n, you must use log to get the n out of the exponent. So Tn = To * R(^n-1) : 0.1087 = 843 / 6^(n-1). Divide 843 by 0.1087 (=7755.289). Then take the log of both sides which moves n-1 infront of log6. Then divide both sides by log 6. So log 7755.289 / log 6 = 4.999 = 5 = n-1. Then add one to both sides to get n= 6. So 0.1087 is the 6th term.

ANSWERS:

1. a. Arithmetic

b. 72.5, 88, 103.5

c. 1188.5 is the 76th term.

d. 6071 is the 391st term.

2. a. Geometric

b. 3.9028, 0.65046, 0.10841

c. 4 * 10 ^(-56)

d. 0.1087 is the 6th term.

Additional Information can be found here:

http://home.alltel.net/okrebs/page131.html

Quiz yourself at this neat website: http://www.ltcconline.net/greenl/java/IntermedCollegeAlgebra/ArithGeo/ArithGeo.html

Madison you are up next!!!

Trey Kozacik, Xander Berry and I did a science fair project and we won 2nd place in our division on Saturday in the LA County Science fair. This means that we are going to STATE!!! Who would've thought considering we did this project purely for fun and then were convinced by teachers to turn it into a science fair project!! Our project is called Shooting For Distance (I know, lame right..well I wanted something a little fun..I was going for SHOOTING FOR THE STARS....) We built a 10 ft air cannon that shoots weighted water bottles up to about 900 ft.

Thursday's Test Topics

Here’s a list of topics for Thursday’s test:

Precalculus Chapter 13 Test Topics:
Relationships between polar and Cartesian coordinates
Plot polar points.
Convert polar equation to Cartesian
Transform a polar graph/equation
Parametric equations – linear motion (word problem)
Determine a point on a curve given polar data
Describe the formation/creation of a polar curve
Parametric equations – nonlinear motion (word problems)

I’ve posted the solutions to the review handout on the class website .
See you in class! I’ll be around until 3:00 pm on Wednesday afternoon, and in early on Thursday.

If you have built castles in the air, that is where they should be; now put foundations under them.
- Henry David Thoreau

13.5 - Parametric Equations for Moving Objects

The main idea of this section is to separate what's going on with the x-axis and what is going on with the y-axis

The best way of describing this is using an example. Let's pretend that a firetruck is able to navigate around at 40 mph eastward and 60 mph northward to reach the next city. The firetruck is currently at the point (50, 20)-- 50 miles east and 20 miles north of the origin (we'll say the town center).

First, we'll find a parametric equation for the path of the firetruck, using t for the hours.

To do this, we simply take our coordinates and our velocities to get:

x = 50 + 40t
y = 60 + 20t

When we graph this, we get a line that can help us answer the following:

Predict the time that the firetruck will be 80 miles north of its current location.

To do this, let y = 80:
80 = 60 + 20t

It will take it about an hour to get 80 miles north of its current location.

Here's an image to help visualize this:




The next idea that we learned is: Parametric Equations of a Cycloid

The reason that we learn this is to calculate where a point is on a rolling wheel along the x-axis at any point in time.

In class, we learned how to derive all of this, but here are the equations:

x = a(t - sin t )
y = a(1 - cos t )
where 't' is the number of radians the wheel has rolled since the point was at the origin.

Here's a really really good explanation on how to derive this, if you weren't in class today:

Another Example

Let's take a look at another example of parametrics in action:

Rocky the flying squirrel:

is flying at 1,000 m/hr west and 800 m/hr vertically. At t=0, he is at point (100, 50).

a) Write parametric equations for his flight, using t hours as the parameter.
Solution:
x = 100 - 1000t
y = 50 + 800t

b) When will Rocky reach the top of the eiffel tower, which is 270 m higher than his current location?
270 = 50 + 800t
220 = 800t
t = .275 hours -- which is 16.5 minutes.

Here's two helpful websites:

http://www.libraryofmath.com/parametric-equations.html
is a summary which has this entire explanation reworded on a single page.

and

http://archives.math.utk.edu/visual.calculus/0/parametric.6/

which has the flash animation I posted above as well as a couple others. And something called LiveMath which I don't have, but you may.

In celebration of the recent talent show that we had, here is a clip from it:

http://www.youtube.com/watch?v=4G5XvqmRfSA

Section 13.2: polar equations of conics and other curves

POLAR EQUATIONS OF CONICS AND OTHER CURVES

This section gives you the basic information about polar equations and how to relate polar coordinates (r,θ) with cartesian coordinates (x,y).

Polar coordinates are a different way of identifying a point in space. With polar coordinates we use a radius (r) and an angle (θ) to specify a point. The coordinate plane looks like this:

The pole is the origin. The polar axis is the same as the positive x-axis on a cartesian graph.

To relate r, θ, x, and y, we can use the following trigonometric equations.

x^2 + y^2 = r^2
sin θ = y/r

From these equations we can derive values for x,y, and θ.

y= r*sin θ

x=r*cos θ

θ=tan-1(y/x)

Now, we need to know how to plot polar graphs on our calculators. First, we need to change the mode to polar. When you go to the y= menu it will now say r=. The standard equation for a polar graph is r=a+bsinθ or r=a+bcosθ. These equations will result in a limaςon or cardioid.

Limaςon

Cardioid

The last bit of information in this section is the general polar equations for conic sections.

These equations are r= k / (a+bsinθ) or r = k / (a+bcosθ). With these equations you can make a parabola, ellipse, or hyperbola.

Depending on the relationship of a and b you will get a different conic shape.

abs(a) = abs(b) -> parabola

abs(a) is greather than abs(b) -> ellipse

abs(a) is less than abs(b) -> hyperbola

Note: the focii of a parabola will be at the origin as well as one of the focii of a ellipse or hyperbola.

Thats it for now!

for more info check out http://en.wikipedia.org/wiki/Polar_coordinate_system

Next up is EDWARD

IF YOU DONT KNOW POLAR EQUATIONS THIS WILL HAPPEN TO YOU!

Thursday's Quiz

Here’s a list of topics for Thursday’s quiz:

Precalculus Quiz 13.1-3 Topics
Plotting polar coordinates
Multiple representations of polar coordinates
Plotting a curve given a table of polar coordinates
Interpreting/Reading a polar graph
Determining the equation of a polar graph
True/False intersections of polar graphs
Converting a polar equation to Cartesian

That’s it! The format’s the same as always – ½ Non-calculator, ½ Calculator. There are 7 questions on the Non-calculator portion and 9 questions on the Calculator section. I’ll be around after school on Wednesday, and on campus by 8:00 AM on Thursday.

People are like stained-glass windows. They sparkle and shine when the sun is out, but when the darkness sets in, their true beauty is revealed only if there is a light from within.
—PHYSICIAN ELISABETH KÜBLER-ROSS

My job:

Chapter 13-3: Intersections of Polar Curves

Ok, it's a nice and easy 2 page lesson. A break for all you lucky juniors and seniors in time for prom.


This lesson is discusses the basics on finding the intersections points when given two polar curves.
Basically to find the number of intersections, and the point they intersect, we insert the equation into the regular function mode. After that, we just do the regular calc -> intersect stuff...

For those that do not understand what I am saying...
I'll point out the different steps
1) Get out your calculator
2) Press Mode
3) Select Func if it is not already selected. (If you realized that Func stands for function, then you are really smart. =])
4) Press Y=
5) Insert equations!!!
6) Go to Window and set Xmin = 0 and Xmax =360 (this is because the polar coordinate system is 360 degrees.) The Ymin and max vary depending on the graph
7) Now, press graph!
8) OMG, a graph. Now we can not only find the number of times the two graphs intersect each other, but we can also find out WHERE they intersect each other.
9) I'm not going to explain the intersect stuff...we did it a million times already.
Ok, now that we have the intersection point...
The X-Coordinate is the degree, and the y is the r...
so the format will be (r, theta)

Alright...now to put all of this information into use...

We are given 2 equations, and we are required to find the intersection points between the two graphs.

Now, we graph them on our calculator. In polar mode, the two equations should look like this...

Ok...Now that we see the equations in Polar mode, now we switch to function mode. (For those who don't know how to do this...scroll up.)

Alright, now insert the equations the EXACT same way into y=

The resulting graph should look something like this.


Awesome...right? Now we use the Calc-> intersect powers of our calculator and find the intersections between the two graphs.

The resulting answers should be...
(117.531, -.387) (193.264, -1.919) (235.510, -.699) (287.063, 1.880)

Now, we have to change them into polar coordinates. Which is NOT that difficult guys.
All we have to do is swap them around and add a degree symbol to the x-coordinate...
(-.387, 117.531°) (-1.919, 193.264°) (-.699, 235.510°) (1.880, 287.063°)
Hurray! We solved the problem!!!

Since there I can't find a website truly discussing the intersections of Polar Curves...I'll post one up about some of the things we already learned for last lesson. Besides, this lesson was the easiest one we have had in a long time.
http://mathforum.org/dr.math/faq/formulas/faq.polar.html

********** Katie You're Up Next. ***********

Of course everybody knows that the Lakers are IN the playoffs right? Well, I hope you all remember last year, when we were up 3-1 against the Suns and LOST. Tomorrow night, they are up against the Sacramento Kings (remember the rivalry?) and if they beat them...they clinch the seventh seed. Yes, and if they clinch the seventh seed, they are once again against the Suns. Except this time...they'll win. So tomorrow, when they're up against Sacramento, I hope that all of you will be rooting for the Lakers...and Kobe Bryant.

Thursday's Test Topics

Here’s a list of topics for Thursday’s test:

Precalculus Chapter 10 Test Topics:
Unit vectors
Planes – Standard Equation vs. normal vectors and points
Points on (or not on) a plane
Lines in Space – points and unit vectors
Direction Angles
Direction Cosines – properties
Vectors between two points
Angles between vectors
Scalar and vector projections
Cross products

I’ve posted a sample test and review on the class website which I’ll distribute in class tomorrow. Review 10-d will not be as helpful as 10-b, 10-c and 10-e.

See you in class!

Here’s a little something that shows the beauty of mathematics:

Math Art: The Beauty and Symmetry of Mathematics.

1 x 8 + 1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
123456 x 8 + 6 = 987654
1234567 x 8 + 7 = 9876543
12345678 x 8 + 8 = 98765432
123456789 x 8 + 9 = 987654321

1 x 9 + 2 = 11
12 x 9 + 3 = 111
123 x 9 + 4 = 1111
1234 x 9 + 5 = 11111
12345 x 9 + 6 = 111111
123456 x 9 + 7 = 1111111
1234567 x 9 + 8 = 11111111
12345678 x 9 + 9 = 111111111
123456789 x 9 +10= 1111111111

9 x 9 + 7 = 88
98! x 9 + 6 = 888
987 x 9 + 5 = 8888
9876 x 9 + 4 = 88888
98765 x 9 + 3 = 888888
987654 x 9 + 2 = 8888888
9876543 x 9 + 1 = 88888888
98765432 x 9 + 0 = 888888888

Brilliant, isn't it? And finally, take a look at this symmetry:

1 x 1 = 1
11 x 11 = 121
111 x 111 = 12321
1111 x 1111 = 1234321
11111 x 11111 = 123454321
111111 x 111111 = 12345654321
1111111 x 1111111 = 1234567654321
11111111 x 11111111 = 123456787654321
111111111 x 111111111 = 12345678987654321

A unique approach to homework: