Question

Hi,
I wanted to know how wanted to know if i was doing this step right for my homework. the question is Given rt=<sin(t)-tcos(t), cos(t)+ tsin(t),t^2 sqrt 3 and the whole thing is divided by two.
The question is find the velocity vector and find V(pi)
for the first one i got cos(t)- 1(cos(t)-tsin(t) so that would give -tsint
the second integral would be sint+1*sin(t)+tcost so here it would be tcost. i don't know what the integral is for t^2sqrt 3/2 i want to say that its 2 t^2.
The second part asks to find the speed which is calculated by taking magntiude of the velocity vector.
which happents to be sqrt of t^2cos^2t+t^2sin^2t+4t^4you get that by using 2t^2.
so that becomes sqrt of t^2+ 4t^4 is there any way to simplify it further.
Thank you


I cannot interpret your question. You write:
<<the question is Given rt=<sin(t)-tcos(t), cos(t)+ tsin(t),t^2 sqrt 3 and the whole thing is divided by two.
The question is find the velocity vector and find V(pi)
>>
Is rt supposed to be r as a function of t?
Why are there three commas on right side of your equation? Do they separate different components?
What is pi?


yes r is the function of r. the 3 commas are for x,y,z. it says to find V(t) and then use it to find V(pi)


Thanks for explaining. I will assume that V(pi) mean the velocity when t = pi (seconds). I or someone else will post again later. I need to take a loser look at the problem.


I don't agree with your calculations.
I get
Vx = +t sin t
Vy = - 2 sin t - cos t
Vz = 2 t sqrt 3
Substitute t = pi for the velocity components when t = pi
Vx = 0
Vy = 1
Vz = 2 pi sqrt 3
For the speed take the sqrt of the sum of the squares.
|V| (at t = pi) = sqrt [1 + 12 pi^2] = 10.93
Check my math; there could be some errors but you need to differentiate the three r components more carefully. There is no integration involved here.