Question

If r(t)=cos(7t)i+sin(7t)j+2tk, compute the tangential and normal components of the acceleration vector.

Answer 1

i know a paul h dunn ...

Answer 2

just did one of these, or tried to :)

Answer 3

\[a_t=\frac{r'*r''}{|r'|}\]
\[a_n=\frac{|r'x\ r''|}{|r'|}\]

Answer 4

I have the 1st part done, I know its 0
I cant figure out the second

Answer 5

http://tutorial.math.lamar.edu/Classes/CalcIII/Velocity_Acceleration.aspx
is what I got to help refresh me memory

Answer 6

http://www.youtube.com/watch?v=EqHNCF8k1Os&feature=related

Answer 7

r(t)=<cos(7t),sin(7t),2t>

r'(t)=<-7sin(7t),7cos(7t),2>

r''(t)=<-49cos(7t),-49sin(7t),0>

Answer 8

< -7sin(7t), 7cos(7t),2>
<-49cos(7t),-49sin(7t),0>
-------------------------
i = 98sin(7t)
j = -98cos(7t)
k = 7^3 sin^2(t) + 7^3 cos^2(t) = 7^3

r'xr'' = <98sin(7t),-98cos(7t), 7^3>

|r'xr''| = sqrt( 98^2sin^2(7t)+98^2cos^2(7t)+7^6)
= sqrt( 98^2+7^6)

Answer 9

Thats correct, I have gotten the same thing so far

Answer 10

I jut have to put it all together haha

Answer 11

as far as i can tell; you divide that by the magnitude of r'

Answer 12

r' = <-7sin(7t),7cos(7t),2>

|r'| = sqrt(49sin^2(7t)+49cos^2(7t)+4)
= sqrt(49+4)
= sqrt(53)

Answer 13

i get 49 ...

Answer 14

that what I am thinking, is the mag of r prime not ((7^2)+(7^2++(2^2))

Answer 15

Wow im dumb haha

Answer 16

So you in college haha?

Answer 17

not that I can algebra out:

\[\sqrt{49sin^2(7t)+49cos^2(7t)+4}\]
\[\sqrt{49(sin^2(7t)+cos^2(7t))+4}\]
\[\sqrt{49(1)+4}\]
\[\sqrt{49+4}\]

Answer 18

yep, in college

Answer 19

thats correct!
I typed in the the computer homework set

Answer 20

I really apprectiate your help

Answer 21

youre welcome, and good luck :)

Answer 22

someplace in florida; home of the lions or something like that