A particle moves in the xy-plane so that its velocity for time is given by the parametric equations and . An expression for the distance traveled by the particle on the interval is
A)
B)
C)

Question

​A particle moves in the xy-plane so that its velocity for time  is given by the parametric equations  and ​  .​ An expression for the distance traveled by the particle on the interval  is ​
A)​ 
B)​ 
C)​

Quizplus · Accepted Answer

To find the distance traveled, we need to integrate the magnitude of the velocity vector over the given interval.

The magnitude of the velocity vector is given by:

sqrt((3t^2)^2 + (2t)^2) = sqrt(9t^4 + 4t^2)

Thus, the distance traveled is:

integral from 0 to 2 of sqrt(9t^4 + 4t^2) dt

This integral can be evaluated using trigonometric substitution (letting t^2 = (2/3)tan^2(u)), which gives:

(3/2) * integral from 0 to arctan(sqrt(3)/2) of sec^3(u) du

This integral can be evaluated using integration by parts, which gives:

(3/4) * (sqrt(3)/2) * (3 + sqrt(3))

Simplifying, we get:

(9/8) * (3 + sqrt(3))

Which is approximately 4.60.

Thus, the correct answer is C.

​A Particle Moves in the Xy-Plane So That Its Velocity

A Particle Moves in the Xy-Plane So That Its Velocity