1. p '(t) = -2(π/4) sin(π/4t) , which is the velocity
to move to the left, p '(t) < 0 or negative, so
sin (π/4t) must be positive.
the period of sin(π/4t) is 2π/(π/4) = 8
so for the first 4 seconds the curve is above the x-axis or it is positive
so for 0 ≤ t ≤ 6, the particle is moving to the left for the first 4 seconds, then to the right for the next 2 seconds.
ARGGGG , it asked for particle R and I found it for P, Oh well, maybe we can use the above later on
r ' (t) = 3t^2 - 12t + 9
zeros are t = 1 and t = 3
so r'(t) is positive for t < 1 or t > 3 and is negative for 1 < t < 3
make a sketch of both derivatives on the same grid
look at when both curves are on the same side of the x-axis ---> going same direction
when both curves are on opposite sides of the x-axis --> same direction
From 0 to 1, R is moving to the right,
from 1 to 3, R is moving to the left
from 3 to 6 , R is moving to the right again.
So from 0 to 1, they are moving in the opposite directions
From 1 to 3, they are moving in the same direction , (both v's are negative
from 3 to 4 , they are moving in opposites
from 4 to 6 , they are moving in same directions
3. r '' (t) = 6t - 12
r ''(3) = 18-12 which is positive, so r ' (t) or the velocity of R is increasing.
4. when is 3t^2 - 12t + 9 = -2
3t^2 - 12t + 11 = 0
t = (12 ▒ √12)/6 = appr 2.577 or 1.423
both t = 2.577 and t = 1.423 fall in the interval 1 ≤ t ≤ 3
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