a particle moves along a line so that, at time t, its position is s(t)=8sin2t.
for what values of t does the particle change direction?

Question

a particle moves along a line so that, at time t, its position is s(t)=8sin2t. 
for what values of t does the particle change direction?

anonymous · Answer

at the points of inflection?

anonymous · Answer

no wait, it's at the relative max and minimums right?

anonymous · Answer

yes

agreene · Answer

lol I was going to answer as if this was a displacement vector >.>

anonymous · Answer

its setting the first derivative = 0, and solving for t

anonymous · Answer

s'(t) = 8(sin2t)(cos2t)(2) = 16(sin2t)(cos2t)

anonymous · Answer

sin2t = 0 when t = pi/2
cos2t = 0 when t = pi

anonymous · Answer

i did a similar problem, i forgot roflmao

anonymous · Answer

how would i solve for t?

anonymous · Answer

sin is zero at pi + npi where n is an integer

anonymous · Answer

the answer is tan = pi/4

anonymous · Answer

is there a restriction for t?

anonymous · Answer

um the answer states:
t=pi/4 + pik,3pi/4 for positive integers k

anonymous · Answer

-4cos2t+sin2t=0
-4(1-2sin^2t)(2sintcost)=0
-4(2sintcost-4sin^3tcost)=0
-8sintcost+16sin^3tcost=0
2sin^3tcost=sintcost
sin^2t=1/2
sint=sqrt2/2
t=45,135

anonymous · Answer

does that help?

anonymous · Answer

nope hahah

anonymous · Answer

that was a question u answered, 13 days ago.

anonymous · Answer

http://openstudy.com/study#/updates/4f7b8478e4b0587805a106d1

anonymous · Answer

yeah i'm trying to find that hw problem, but that guy didn't help

anonymous · Answer

he was working in degrees or something

anonymous · Answer

and using identities i think

anonymous · Answer

degrees u can easily convert to radians.

anonymous · Answer

Shouldn't it be 2t=pi/2

anonymous · Answer

yea, but the answer is in tan, and the answer he gave me is both in cos and sin?

anonymous · Answer

yeah, my answer is incomplete

anonymous · Answer

first can u explain how u came up with the derivative.

anonymous · Answer

b/c the derivative i got was s'(t)=16sin2t

anonymous · Answer

16cos2t*

anonymous · Answer

that's right yeah

anonymous · Answer

oops!

anonymous · Answer

i got 16cos2t, not 16sin2tcos2t

anonymous · Answer

waitwait

anonymous · Answer

i almost have it, hang on

anonymous · Answer

ok

anonymous · Answer

are you sure it changes direction at the relative max/min? I wanna say it changes direction at points of inflection..

anonymous · Answer

nvm, I thought I had it, gotta brush up on my trig

anonymous · Answer

um ok