If I read you question correctly, the position is given by
x = t^3
y = 2sin(t)
so when t = 1, x = 1, y = 2sin1, so it is at (1, 2sin(1))
when t = π , x = π^3 , y = 2sinπ = 0 , so it is at (π^3, 0)
so don't you just want the distance between (1, 2sin(1)) and (π^3 , 0) ?
Use your distance formula
If you want the arc-length, then see the last part of ...
www.wolframalpha.com/input/?i=plot+x+%3D+t%5E3,+y+%3D+2sin(t)+from+1+to+%CF%80
The components of a position vector of a particle moving in the plane are components t^3,2sin(t)). What is the distance traveled by the particle from
t = 1 to t = π?
Type your answer in the space below and give 3 decimal places. If your answer is less than 1, place a leading "0" before the decimal point (ex: 0.482)
Arc Lenght= integral from pi to 1 of √[(t^3)^2+(2sin)^2(t)] dt
I tried with this but I didn't get the answer. please help me
1 answer