Part One
Based on my CIN 46728, my equation for part 1 reads as follows:
g (x )=
4 x6
evaluate at x=2
7 x+2
( 7 x +2 )
d
d
( 4 x6 )(4 x6) (7 x +2)
dx
dx
2
(7 x +2)
d
d
( 4 x )=4, (6 ) =0
dx
dx
d
d
( 7 x ) =7, ( 2 )=0
dx
dx
( 7 x +2 ) ( 4 )( 4 x6)(7)
( 7 x
Discussion Board Week 4
CIN 46728
Part 1- Find the second derivative and evaluate at x=3:
2 x +6 4
f ( x )= evaluate at x=3
u= 2x+6
n= 4
First I find the first derivative:
2 x+6 3
2 x +6 3 ( 2 )=8
4
Next I find the second derivative, discarding the 8 for
Mod 8, CIN 46728
7
1. Evaluate
x 6 x 2 +2
1
dx
and round your answer to two decimal places.
7
x 6 x 2 +2 dx ,
1
u=6 x 2 +2, du=12 xdx , dx=
1
dx
12 x
6 x 2+2 0.5 +1
1
u
1
1 u0.5 +1
1
=
x u 12 x du= 12 du= 12
u du= 12
0.5+1 12
6 x 2+2 3 /2
2
6 x +2
3
CIN 46728
1. The x and y coordinates are given by two parametric equations, find the magnitude
2
and direction of velocity for x=4 t +t ; y =27 t at t=3
dx 2
dy
First find the derivatives Vx= dt 4 t + t=8 t+1Vy= dt 27 t=7
Now plug in t=3 so
8(3)+1= 25 and
1) Based on my CIN 46728, to find the slope of the line passing through (c, d)
and (1, b), I first substituted my CIN numbers into the points. This gave me
(7, 2) and (1, 6).
To find the slope of this line from my new points I used the formula
y2 y1
x2 x1
1. s=[b]*[d]+64t-16t2 ,
s = (6)(2) + 64t 16t2,
s ' =6432 t , 6432t=0,64=32 t , t =
s = 12+ 64t 16t2
64
,t =2
32
s=12+64 ( 2 )16 ( 2 )2 , s=12+12864, s=76
The highest the rock can go is 76 feet (using feet as no unit of measure was given).
_
2. You are des
Mod 7 DB, CIN # 46728
2
9x
(+2 x6) dx
1. Find the constant integration, C if y =
and the curve passes through the
point (4,8).
( 9 x 2 +2 x 6 ) dx , 9 x 2 dx+ 2 xdx 6 dx
9 x 2 dx=9 x 2 dx=9
2+1
3 x3 + x 26 x +C
4 3+ 4 26 ( 4 ) +C=8, 184+C=8,
3
C=179
1+1