42SAP Calculus

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


Tuesday, March 21, 2006

Chapter 8 Supplementary Problem #1 c and d

Chapter 8 Supplementary Problem #1 a & b