42SAP Calculus

Friday, April 07, 2006



Woooooooooooo there is my work tell me if it is wrong.

Monday, April 03, 2006

supplementary problems chp.8, #21


alright sorry it's late but it took me a while, should have done it earlier because i have a feeling only the ridiculously hard questions were left, and they were!!! plus, it was all around not a good night for me cause i couldn't figure out how to do it so i did it all on paint, and i don't even think it's the right answer...

Question #23



Excuse my terrible painting job please.

Sunday, April 02, 2006

Missing 4!

Okay where are the other 4? I think we need some peer pressure!
Hope you have a good review class tomorrow. Feel free to post if you have any questions!
See you soon!

Chapter 8 Supplementary Problem Numero 5 (not checked by Ms. German)

Chapter 8 question 9... O_o

Supplementary # 27!! Hope you like my writing =P

Saturday, April 01, 2006

Chapter 8 question #19

Friday, March 31, 2006

Chapter 8 Supplementary Problem #3

Okay, so my computer's being really weird and it won't let me put the pictures in the right order. So the first part is at the bottom. The second part is at the top. The third part is second from the bottom, and the last part is third from the bottom. Sorry about that! If anyone knows how to fix it, please let me know.







For part c), you have to split up the region into two parts as you're revolving it around the y-axis. The reason for this is that from [0,3] on the y-axis, g(x) > f(x), but from [3,4], f(x) > g(x). So you make it into two regions, evaluate each of those and add them up to get the total area.


Tuesday, March 28, 2006

Chapter 8 Question 13 c

c.) 0 = 3t2 – 6t
t = 2,0
test t = 1, v(1) = 3(1)2 – 6(1)
v(1) = -3
test t = 3, v(3) = 3(3)2 – 6(3)
v(3) = 9
therefore, 0 <> 2 is positive
0 - (3t2 – 6t) dt
[-t3 + 3t2]
[(-(2)3 + 3(2)2) – 0]
= 4
3t2 - 6t dt
[t3 – 3t2]
[((4)3 – 3(4)2) – ((2)3 – 3(2)2)]
[16 – (-4)]
= 20
total distance traveled = 20 + 4
= 24 units

Chapter 8 Question 13 a & b

Chapter 8 Question 13

a.) v(t) = 3t2- 6t
x(t) = t3- 3t2 + C
x(2) = 4
4 = (2)3 + 3(2)2 + C
4 = -4 + C
C = 8
x(t) = t3 – 3t2 + 8

b.) M(t) = 1/b-a v(t) dt
M(t) = 1/4-1 3t2 – 6t dt
M(t) = 1/3 [ t3 - 3t2 ]
M(t) = 1/3 [ ((4)3 – 3(4)2) – ((1)3-3(1)2)]
M(t) = 1/3 [64 – 48 - 1 + 3]
M(t) = 1/3 [18]
M(t) = 6
6 = 3t2 - 6t
3(t2 - 2t - 2) = 0
(-b √ (b2-4ac)) / 2a
=(-(-2) √((-2)2 – 4(1)(2))) / 2(1)
=(2 √(4 + 8)) / 2
=(2 2√3) / 2
=1 + √3, 1- √3 , [1,4]
therefore, 1 + √3

Sunday, March 26, 2006

chapter 8 supplementary problem 29

chapter 8 supplementary problem 29

Chapter 8 Supplementary Problem #15