P(t) = 1000+360sin((π/6)t)+180sin((π/3)t)
Find the derivative and then solve for when P(t)=0

- anonymous

P(t) = 1000+360sin((π/6)t)+180sin((π/3)t)
Find the derivative and then solve for when P(t)=0

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- amistre64

its alot of sums; so deriving is easy

- amistre64

0 +360pi/6 cos(t.pi/6) +180pi/3cos(t.pi/3)

- anonymous

ok i got that part right. now the solving for p(t)=0 part is where im stuck

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## More answers

- amistre64

looks tricky ;)

- amistre64

360/6 = 180/3 so that all the same

- amistre64

cos(t.pi/6) + cos(t.pi/3) = 0 when.....

- amistre64

you agree with this so far?

- anonymous

yep

- amistre64

we gotta get them at 180 from each other than right?

- amistre64

t.60 = t.30 + 180; or, t.30 = t.60 + 180

- amistre64

t(60-30) = 180
t(30) = 180
t = 60 maybe?

- amistre64

err...6 lol

- amistre64

6.60 = 360
6.30 = 180; thats a good one right?

- amistre64

a(1) + a(-1) = 0 right

- amistre64

so every multiple of 6 should be good for that one...

- amistre64

t.30 = t.60 + 180
t(30-60) = 180
t = 180/-30
t = -6; same thing

- amistre64

so whenever t = a multiple of 6, we should be good

- anonymous

i kinda lost you at t.60 = t.30 + 180; or, t.30 = t.60 + 180.
from cos(t.pi/6) + cos(t.pi/3) = 0
in fact, for that^ i simplified and got 60cos((t x pi/6)+ (t x 2pi/6) = 0

- amistre64

given that t not equal zero perhaps; or some such restriction

- amistre64

there are bound to be multples of 6 that make t.60 = t.30; in that case its of no use

- amistre64

ok; we need angles that are 180 from each other in order for there cosines to be opposites of each other right?

- anonymous

uh huh

- amistre64

so we start at 60 and 30 and need to at least find the angles that are 180 from each other....
t.60 = t.30 + 180 means that for every multiple of t, we move by 60 and 30 till we find angles that are 180 from each other right?

- amistre64

t.60 - t.30 = 180
t(60 - 30) = 180
t = 180/30
t = 6

- anonymous

ooh ok

- amistre64

now when the angels are 360 from eah other they are the same...so we exclude doing the same thing but with +360 instead of +180 right?

- amistre64

360/30 = 12; so every multiple of 12 wont work.... right?

- anonymous

uh huh

- amistre64

so odd multiples of 6 work; but even multiples fail; so we restricct it to when t is an odd multiple of 6

- amistre64

i can also think of one more condition, but havent found a way to describe it yet......... it when they angle to the right matches the angle on the left; like a 60 and a 120 would have opposite cosines

- anonymous

lol i think im good. thank you.

- amistre64

:) ok

- anonymous

another question if you dont mind? lol

- amistre64

i dunno, im pretty much good for one question and then go stupid lol

- amistre64

whats the q

- anonymous

LOL, it's a position, velocity, acceleration question.

- anonymous

Find the position function X(t), of a particle moving along a straight line, with
constant acceleration equal to 2, knowing that the position and velocity vanish at t = 2.

- amistre64

is it posted already or you wanna hit me with it

- anonymous

Find the initial position and the time t > 0 at which the particle passes again through the initial position.

- anonymous

i know i should do anti derivative for the first part in order to find the velocity equation.

- amistre64

acceleration is the derivative of velocity is the derivative of position

- anonymous

yep

- amistre64

with a constant acceleration of 2; that means we got linear velocity of 2x+c right?

- anonymous

yeah

- anonymous

would i plug in velocity = 0 since it vanishes or no?

- amistre64

and a position of x^2 +bx +cthen

- amistre64

the only way for a linear function to vanish at some point it to make a hole iin the graph right?

- anonymous

yeah

- amistre64

which means making the velocity function an extention of a rational function

- amistre64

2x +c
------ at best
d-2

- amistre64

but with a d-2 up top

- anonymous

what's d-2?

- amistre64

something to make a whole at 2 to make the graph vanish

- anonymous

lol i think you might be over thinking the problem. this was suppose to be an easy math problem my prof gave on the test. haha

- amistre64

i could be over thinking it tho :)

- anonymous

lmao would it be wrong to set the velocity - 0?

- anonymous

velocity = 0*

- amistre64

cant have a constant acceleration whout velcoty can we?

- amistre64

my keyboard hates me lol

- anonymous

haha its okay

- anonymous

there was a similar question to this, and my sis helped me with that.

- anonymous

and in that other problem, my sis taught me to set velocity = 0

- anonymous

then solve for the constant

- amistre64

cant say that i ever came across the problem before, so it might be plausible :)

- anonymous

and then from that, do another anti derivative to find the position fuction

- anonymous

lol alrighty, thanks anyways! enjoy the rest of your day :]

- amistre64

:) yw

- toxicsugar22

amristre can u help me

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