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
To calculate the distance traveled by the particle from t = 1 to t = π, we can use the arc length formula. The arc length is given by the integral of the magnitude of the velocity vector over the interval [1, π].
The position vector of the particle is given as r(t) = (t^3, 2sin(t)), where the components represent the x and y coordinates, respectively.
To find the magnitude of the velocity vector, we need to differentiate the position vector with respect to time (t) and calculate its magnitude.
Differentiating the x and y components of the position vector:
dx/dt = 3t^2
dy/dt = 2cos(t)
The velocity vector is then given by v(t) = (dx/dt, dy/dt) = (3t^2, 2cos(t)).
Now, to find the magnitude of the velocity vector, we use the formula:
|v(t)| = √[(dx/dt)^2 + (dy/dt)^2]
= √[(3t^2)^2 + (2cos(t))^2]
= √[9t^4 + 4cos^2(t)]
To calculate the arc length, you need to integrate the magnitude of the velocity vector from t = 1 to t = π:
s = ∫[1,π] √[9t^4 + 4cos^2(t)] dt.
However, this integral doesn't have a nice closed form solution. We need to approximate it numerically using methods like numerical integration or software like MATLAB or Wolfram Alpha.
Using numerical integration or a mathematical software tool, the distance traveled by the particle from t = 1 to t = π is approximately 10.079 units (rounded to three decimal places).
To find the distance traveled by the particle from t = 1 to t = π, you need to integrate the magnitude of the velocity vector over that interval.
The velocity vector is the derivative of the position vector, so let's find the velocity vector first.
Given that the position vector components are t^3 and 2sin(t), we can differentiate these components with respect to t to get the velocity vector components:
Velocity vector V(t) = (d/dt)(t^3) i + (d/dt)(2sin(t)) j
= 3t^2 i + 2cos(t) j
The magnitude of the velocity vector is given by:
|V(t)| = √[(3t^2)^2 + (2cos(t))^2]
= √[9t^4 + 4cos^2(t)]
Now, to find the distance traveled, we need to integrate this magnitude of the velocity vector from t = 1 to t = π:
Distance traveled = ∫[1 to π] √[9t^4 + 4cos^2(t)] dt
Unfortunately, this integral does not have an elementary antiderivative, so we have to rely on numerical methods to approximate the value.
Using numerical integration, we can calculate the distance traveled as approximately 3.503 units (rounded to 3 decimal places).
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