The position of the particle moving in space at time t ≥ 0 is r(t) = (2 + 2 cos(t))i − 2 sin(t)j + (3 − t π )k. Find the first time moment t0 such that the velocity vector v(t0) is orthogonal to the vector i − j.

t0 = π/4

Step-by-step explanation:

The velocity v(t) = r'(t) = <-2sint,-2cost>

Since <1,-1>⊥v, <1,-1>•v=0. So,

(1)(-2sint) + (-1)(-2cost) = 0

cost - sint = 0

uhhh idk u yolo?

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.