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.
To find the velocity vector, we need to differentiate each component of the position vector, rt, with respect to t.
Given that rt = <sin(t) - tcos(t), cos(t) + tsin(t), t^2 * sqrt(3)> / 2
Differentiating each component:
d(rt)/dt = d/dt(<sin(t) - tcos(t), cos(t) + tsin(t), t^2 * sqrt(3)> / 2)
= <d(sin(t) - tcos(t))/dt, d(cos(t) + tsin(t))/dt, d(t^2 * sqrt(3))/dt> / 2
= <cos(t) + tsin(t), -sin(t) - tcos(t), 2t * sqrt(3)> / 2
= <cos(t) + tsin(t) / 2, -sin(t) - tcos(t) / 2, t * sqrt(3)>
So, the velocity vector is V(t) = <cos(t) + tsin(t) / 2, -sin(t) - tcos(t) / 2, t * sqrt(3)>
To find the velocity at t = pi, substitute t = pi into the velocity vector:
V(pi) = <cos(pi) + pi*sin(pi) / 2, -sin(pi) - pi*cos(pi) / 2, pi * sqrt(3)>
= <-1 + 0 / 2, 0 - (-pi) / 2, pi * sqrt(3)>
= <-1, pi/2, pi * sqrt(3)>
So, the velocity at t = pi is V(pi) = <-1, pi/2, pi * sqrt(3)>
For the magnitude (speed) of the velocity vector, we need to calculate the square root of the sum of the squares of each component of the velocity vector:
|V| = sqrt((-1)^2 + (pi/2)^2 + (pi * sqrt(3))^2)
= sqrt(1 + (pi^2)/4 + 3pi^2)
= sqrt((4 + pi^2 + 12pi^2)/4)
= sqrt((pi^2 + 48pi^2)/4)
= sqrt(49pi^2/4)
= |7pi/2|
So, the magnitude (speed) of the velocity vector at t = pi is |V(pi)| = |7pi/2|.
There doesn't seem to be a simplified form for the magnitude of the velocity vector at t = pi.