Find speed of particle

[Tubig]

1. Find the speed (As a function of t) of a particle whose position at time t seconds is c(t)=(sint+t,cost+t)
2. speed = [x'(t)^2 + y'(t)^2]^1/2
3.
x'(t) = cost + 1
y'(t)= -sint + 1

speed = [(cos t + 1)^2 +(-sint + 1)^2]^1/2
=[cos^2 t+ 2cost + 1 +sin^2 t - 2sint + 1]^1/2
=[cos^2 t + sin^2 t + 2 cost - 2sint +2]^1/2
=[1 + 2(cost - sint + 1)]^1/2
-Thanks

 

[Dick]

Or (2*(cos(t)-sin(t))+3)^(1/2). That looks ok to me.

 

[HallsofIvy]

No, Dick. It would be (2cos(t)- 2sin(t))+3)^(1/2). The "3" would not be multiplied by 2.

 

[Dick]

No, Dick. It would be (2cos(t)- 2sin(t))+3)^(1/2). The "3" would not be multiplied by 2.

No, Halls. There's an extra set of parentheses around the cos(t)-sin(t). When you cut and pasted you only deleted one of them.