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 speed of the particle at t=3?

Answer 1

`= sqrt(9 cos^2 (3) + 1)`

Explanation:

`vec r(t) = ((x(t)),( y(t))) = ((3sin t - 3),( t-1)) qquad t in [0, 2 pi]`

`vec v = vec r'(t) = ((x'(t)),( y'(t))) = ((3cos t),(1))`

speed, `dot s = sqrt (\vec v * vec v)`

`= sqrt( ((3cos t),(1)) * ((3cos t),(1)))`

`= sqrt(9 cos^2 t + 1)`

`dot s_(t = 3) = sqrt(9 cos^2 (3) + 1)`