Franco and Guillen's Scribe Post

Introduction
The purpose of this scribe post is to inform all of our classmates about the precalculus lesson learned on Tuesday, May 08, 2007. Mr. Alcantara taught how to get a large series term.
For example, yesterday we learned how to get S(7) or S(5), today we were learning on how to get higher series terms like S(22), it is a more complex process but its still not that hard. Our class consisted of our day by day POD and two other different example problems. For the POD, we found out that there are different ways to solve the problem. MAC, Susy DC, and Mr. Alcantara all showed their different methods to reach the same answer. Throughout the lesson we used Gauss’s updated formula, which is: Sn=n/2 (T1+Tn)
Gauss’s method works for the 1st n terms of any arithmetic series.

Lesson:
POD 05-08-07:
A 50 foot ramp, with one end resting on the ground, climbs 30 vertical feet. The ramp is supported every 2.5 feet by a vertical post. What is the combined length of the posts?

Instructions: First make a drawing of the ramp, this takes a triangle form that enables us to solve for the other side of the ramp, and later on find the number of posts. (look at Drawing 1 ABOVE)

We know two sides of the triangle, therefore we use the Pathagorean Theorem to find the unknown side
30^2 ft + X^2 = 50^2 ft
900 ft+ X^2= 2500 ft
1600 ft = X^2
Then you take the square root of both sides and we get:
X = 40 ft.

We know that our unknown side is 40ft. and that the distance between each post in the ramp is 2.5 ft. Knowing what we are given, we do the following to find out the number of posts. 40 / 2.5 = 16, there are 16 posts in the ramp. This lets us do the following drawing (Look at the first drawing above, note: it is the one not titled as Drawing 1)


The Distance between each post is 2.5 ft, we know that the ramp increases between posts as well, what we need to find out is how much it increases to make a series. MAC, DC, and Mr. Alcantara had different methods to reach the answer.

Here they are:
DC’s method:
“From the drawing, we know that 50 ft. is also divided by 16 posts, therefore that distance between posts is 3.125 ft. because 50/16 is 3.125, then I can use the smallest triangle made by the posts in the ramp and use the pathagorean theorem to solve for what we are looking for.

2.5^2 + x^2 = 3.125^2
6.25 + x^2 = 9.765625
X^2 = 3.515625
Take Square root of both sides
X = 1.875 – this is the first term of the increasing sequence in the ramp.

MAC’s Method: MAC also used the smallest triangle and the bigger triangle as well, but instead of looking for the sides she looked for the angles using the sides she knew. @ = pheta
Sin@ = 30/50
@ = 36.87 degrees

Tan@= P/2.5
P = 1.875

Mr. Alcantara’s method:
All the triangles created in the figure are similar triangles. Therefore we can use the ratio of each to find the variable. LOOK!
30/40 = X/ 2.5
30/40 reduces to 3/4
Cross multiply
3(2.5) = 4x
X = 1.875

The difference between posts is 1.875 and it is going from 0 to 30, it also contains 16 jumps.

Now we use Gauss’s updated formula to find out the combined length of the posts. We are trying to find the term S(16).
S (16) = 16/2 (1.875 + 30)
S(16) = 255 ft.

Notice that the 16th term is 30, which is the highest point in the ramp, and our first term is the difference between posts while the post increases.

The other two class problems we talked about are the following:

1. Find the sum of the 1st 30 terms of the arithmetic series: -5 + -1 + 3+ 7 + 11

First we need to find the rate and T(30) so we can latter use Gauss’s updated formula
The difference between -5 and -1 is 4, the same applies from -1 to 3, and so on, therefore we know the distance is 4.
Tn= T1 + (n-1)d
T(30) = (-5) + (30-1) 4
T(30) = 111
Sn= n/2 (t1 + Tn)
S(30) = 30/2 (-5 + 111)
S (30) = 1590

2. Given S is arithmetic, T1 = 18 and T(5) = -2 , FIND S (22)
First thing to look for is the total jump
Total jump = (-2) – 18 = - 20 ---- change of first term to 5th term
# of jumps from 1 to 5 = 4
Rate = total jumps / # of jumps
Rate or distance = -20 / 4 = -5
Then we need to find T(22)
T(22) = 18 + 21 (-5)
T(22) = -87
Since we now know all the variables, we can use Gauss’s updated formula
Sn = n/2 (t1 +tn)
S(22) = 22/2 (18 + -87)
S(22) = -759


Our homework so far still is pg. 489 #’s 1-8. However we recommend you to study every night so we can get better test results and go on faster in our pre calculus classes.

Congratulations to Abraham and Mau, we thought the scribe you guys did was well organized, and explained all the information clearly. We tried to do the same thing for our scribe because we thought the scribe you guys posted was very accurate.

WE ARE ONE DAY BEHIND 11B :( lets work harder!

The people who we choose to do the next scribe is Susy DC and Bee.. GOOD LUCK!