Question

Given x(t)=3sin(t) - 3, y(t)=t-1 for 0 is less than or equal to t is less than or equal to 2pi How do you find the velocity of the particle at t=3?

Answer 1

The particle's velocity is `sqrt(9cos^2(3)+1)approx3.1338`

Explanation:

We can find the `x`- and `y`-velocities by taking the derivatives of each position function.

`{(x'(t)=3cos(t)),(y'(t)=1):}`

So at `t=3`, we have

`{(x'(3)=3cos(3)),(y'(3)=1):}`

The magnitude of the velocity is then the hypotenuse of the right triangle whose legs are the `x` and `y` components of the velocity:

`v(3)=sqrt((x'(3))^2+(y'(3))^2)=sqrt(9cos^2(3)+1)approx3.1338`