February 18, 2007

scribe post

hey everyone as you all can see its charlene here and yes i am scribe for today. today we started off with a quiz just like every single class from now on. so the quis consisted of four questions and we got like eight minutes to do it. we did go over some of the questions on the quiz so here they are...

A) given f(3)=8 and f'(3)=4 find f(3.02)

y - y1 = m ( x - x1)
y - 8 = -4 (x - 3)
y = -4x + 20

y(3.02) = -4 ( 3.02) + 20
Y(3.02) = -12.08 + 20
Y(3.02) = 7.92

B) given y=1 and xy2 - 2y 3 4y3 = 6

solve for x
x - 2 + 4 = 6
x = 4

differentiate
y2 + 2xyy' - 2y' + 12y2y' = 0

y' ( 2xy -2 + 12y2 ) + -y2
y' = -y2<.sup> / ( 2xy - 2 + 12y2

y'(1 , 4) = -1 / 18

after reviewing some of the questions on the quiz mr.k gave us a table to fill out that consisted of the trapazoid sum, reiman sum, and the simpson sum

i wasn't allowed to upload the table so im just gonna type it on

N ..Tn ....... error ..... Mn ....... error ..... Sn .....error
4 ...0.2656. 0.0156 ... 0.2422 ..-0.0078 .. 0.25 .. 0
8 .. 0.2539 .0.0039 ....0.2480 ..-0.0200 ...0.25 ...0
16 .0.2510 .0.0010 ....0.2495 ..-0.0005... 0.25 ...0

Tn = (B + B)H / 2

Mn = REIMANN SUM
*use 0.5 as the x-choice

Sn = T + 2M / 3
* WE DO NOT HAVE TO KNOW THE SIMPSON SUM FOR ANY TESTS OR EXAM

ERROR = approximation - actual

unforntunately i do not understand the table as much as i should so i will just wait till monday to explain more clearly what is going on in the table. sry guys

by the way the next scribe is CRYSTAL

christian said...

Hey Charlene!

Haha no one's commented in a while. How's everyone been doin?? Anyhow, about the table.

I think on Friday Mr. K just introduced the "Simpson Sum". We did the table to show that the Riemann sum and Trapezoid sum can be refined. As our calculations have shown, the Simpson sum is more accurate than the two. Now that we know how to do this, our integration would be better! Where before our answers were relatively inaccurate, using the Simpson sum we'd get a better approximation. See ya guys tomorrow!

Manny said...

Just a little quick add on to Christians comment above.

Unlike the Riemann sum which uses many rectangles under the curve to approximate the intergral, the Simpson sum uses many parablas. So on a curvy graph, using parabolas is more accurate than rectangles.

Lani said...

Hi there!!

It's great to see you back in the comments supporting each other's learning again!!!! I've really missed that!!! Do you think it might be a good strategy to help you prepare for "The Test"?

Best,
Lani