Question

If r(t)=cos(−3t)i+sin(−3t)j+7tk, compute the tangential and normal components of the acceleration vector.

Answer 1

is r(t) the position vector?

Answer 2

r'(t) = velocity = <3sin(t),-3cos(t),7>

r''(t) = <3cos(t),3sin(t),0>

maybe

Answer 3

im sure that simplictic and i forgot a few things

Answer 4

Yeah, r(t) is the position vector. But is the tangential just the first derivative and the normal is just the second derivative?

Answer 5

well, i am sure the first is the tangent vector; im shaky as to the 2nd tho

Answer 6

i dont see why not, but still its been too long to remember it

Answer 7

I'll try and let you know. Thanks!

Answer 8

It says it needs a formula that returns a number. But what we gave it returns a list of numbers.

Answer 9

let me read up on it and see what i can recall

Answer 10

you have any idea in what direction we can go with it?

Answer 11

maybe like finding curvature and whatnot?

Answer 12

I mean, I don't feel like it would be curvature, but maybe it is. I'm thinking maybe do a magnitude of the vectors you just gave me?

Answer 13

this is what im thinking of... as from http://omega.albany.edu:8008/calc3/curves3-dir/lecture.html

v = dR/dt = dR/ds * ds/dt = T * ds/dt

a = dv/dt = dT/dt * ds/dt + T * =
= dT/dt * (ds/dt)^2 + T * dT/ds
= K*N, where K is the radius of curvature.

a = K(ds/dt)^2 * N + * T
a = aN * N + aT * T

aN = K*(ds/dt)^2 = K |v|^2 = (|v|^2)/rho, where rho is the radius.

aT = = d|v|/dt

aN from above is the centripetal acceleration.

Answer 14

its prolly more readable from this tho:

http://www.ltcconline.net/greenl/courses/202/vectorFunctions/tannorm.htm

Answer 15

checking it out now. thanks

Answer 16

\begin{array}l
r'(t) = velocity = <3sin(t),-3cos(t),7>\\\\
|r'(t)|=\sqrt{9sin^2(t)+9cos^2(t)+49}\\\\
|r'(t)|=\sqrt{9+49}\\\\
|r'(t)|=\sqrt{58}\\\\
\\\\
r''(t) = <3cos(t),3sin(t),0>\\\\
|r''(t)| = \sqrt{9cos^2(t)+9sin^2(t)}\\\\
|r''(t)| = \sqrt{9}=3\\\\
\end{array}

Answer 17

Neither of those just worked :(. I just tried this for the first part: (cos(-3t)+(sin(-3t))+(7t))/(sqrt(((3sin(t))+(-3cos(t))+(7))))

Answer 18

http://tutorial.math.lamar.edu/Classes/CalcIII/Velocity_Acceleration.aspx

here is another site that helps me out

Answer 19

\[a=a_tT+a_nN\]

where T and N are the unit vectors

Answer 20

\[a_t = \frac{r'*r''}{|r'|}\]
\[a_t = \frac{<3sin(t),-3cos(t),7>*<3cos(t),3sin(t),0>}{|\sqrt{58}|}\]
\[a_t = \frac{9sin(t)cos(t)-9sin(t)cos(t)+0}{\sqrt{58}}\]
\[a_t = \frac{0}{\sqrt{58}}=0\]
but I gotta wonder

Answer 21

you were right!!

Answer 22

That's the first one. Awesome, thank you!

Answer 23

\[a_n=\frac{|r'x\ r''|}{|r'|}\]
\[a_n=\frac{|<3sin(t),-3cos(t),7>x<3cos(t),3sin(t),0>|}{|\sqrt{58}|}\]
\[a_n=\frac{|-21sin(t)+21cos(t)+9|}{\sqrt{58}}\]
\[a_n=\frac{|21[cos(t)-sin(t)]+9|}{\sqrt{58}}\]
if i did the cross right, this is what i get for the normal component

Answer 24

might have to put some ijks in it tho

Answer 25

\[a_n=\frac{|-21sin(t)\ I+21cos(t)J+9K|}{\sqrt{58}}\]

Answer 26

ugh, i think the top there is a cross of magnitudes

Answer 27

yeah, I'm trying different things and what you put up before but it hasn't worked so far.

Answer 28

\[a_n=\frac{\sqrt{(-21sin(t))^2\ +(21cos(t))^2+9^2}}{\sqrt{58}}\]
simplified

Answer 29

sin^2+cos^2 = 1 so we are left up top with:

\[\sqrt{21^2+9^2}\]

Answer 30

forgot yo add 81 up top

Answer 31

\[a_n=\frac{\sqrt{522}}{\sqrt{58}}\]

Answer 32

which is 3 :)

Answer 33

:(...none of those last 5 answers have worked so far.

Answer 34

bummer, cause unless i read it wrong, thats the best I got for it

Answer 35

you've done plenty! I'll keep going at it alone. Thanks for all your help!

Answer 36

good luck with it ;)