anonymous
  • anonymous
P(t) = 1000+360sin((π/6)t)+180sin((π/3)t) Find the derivative and then solve for when P(t)=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
its alot of sums; so deriving is easy
amistre64
  • amistre64
0 +360pi/6 cos(t.pi/6) +180pi/3cos(t.pi/3)
anonymous
  • anonymous
ok i got that part right. now the solving for p(t)=0 part is where im stuck

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amistre64
  • amistre64
looks tricky ;)
amistre64
  • amistre64
360/6 = 180/3 so that all the same
amistre64
  • amistre64
cos(t.pi/6) + cos(t.pi/3) = 0 when.....
amistre64
  • amistre64
you agree with this so far?
anonymous
  • anonymous
yep
amistre64
  • amistre64
we gotta get them at 180 from each other than right?
amistre64
  • amistre64
t.60 = t.30 + 180; or, t.30 = t.60 + 180
amistre64
  • amistre64
t(60-30) = 180 t(30) = 180 t = 60 maybe?
amistre64
  • amistre64
err...6 lol
amistre64
  • amistre64
6.60 = 360 6.30 = 180; thats a good one right?
amistre64
  • amistre64
a(1) + a(-1) = 0 right
amistre64
  • amistre64
so every multiple of 6 should be good for that one...
amistre64
  • amistre64
t.30 = t.60 + 180 t(30-60) = 180 t = 180/-30 t = -6; same thing
amistre64
  • amistre64
so whenever t = a multiple of 6, we should be good
anonymous
  • 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
  • amistre64
given that t not equal zero perhaps; or some such restriction
amistre64
  • amistre64
there are bound to be multples of 6 that make t.60 = t.30; in that case its of no use
amistre64
  • amistre64
ok; we need angles that are 180 from each other in order for there cosines to be opposites of each other right?
anonymous
  • anonymous
uh huh
amistre64
  • 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
  • amistre64
t.60 - t.30 = 180 t(60 - 30) = 180 t = 180/30 t = 6
anonymous
  • anonymous
ooh ok
amistre64
  • 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
  • amistre64
360/30 = 12; so every multiple of 12 wont work.... right?
anonymous
  • anonymous
uh huh
amistre64
  • 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
  • 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
  • anonymous
lol i think im good. thank you.
amistre64
  • amistre64
:) ok
anonymous
  • anonymous
another question if you dont mind? lol
amistre64
  • amistre64
i dunno, im pretty much good for one question and then go stupid lol
amistre64
  • amistre64
whats the q
anonymous
  • anonymous
LOL, it's a position, velocity, acceleration question.
anonymous
  • 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
  • amistre64
is it posted already or you wanna hit me with it
anonymous
  • anonymous
Find the initial position and the time t > 0 at which the particle passes again through the initial position.
anonymous
  • anonymous
i know i should do anti derivative for the first part in order to find the velocity equation.
amistre64
  • amistre64
acceleration is the derivative of velocity is the derivative of position
anonymous
  • anonymous
yep
amistre64
  • amistre64
with a constant acceleration of 2; that means we got linear velocity of 2x+c right?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
would i plug in velocity = 0 since it vanishes or no?
amistre64
  • amistre64
and a position of x^2 +bx +cthen
amistre64
  • amistre64
the only way for a linear function to vanish at some point it to make a hole iin the graph right?
anonymous
  • anonymous
yeah
amistre64
  • amistre64
which means making the velocity function an extention of a rational function
amistre64
  • amistre64
2x +c ------ at best d-2
amistre64
  • amistre64
but with a d-2 up top
anonymous
  • anonymous
what's d-2?
amistre64
  • amistre64
something to make a whole at 2 to make the graph vanish
anonymous
  • 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
  • amistre64
i could be over thinking it tho :)
anonymous
  • anonymous
lmao would it be wrong to set the velocity - 0?
anonymous
  • anonymous
velocity = 0*
amistre64
  • amistre64
cant have a constant acceleration whout velcoty can we?
amistre64
  • amistre64
my keyboard hates me lol
anonymous
  • anonymous
haha its okay
anonymous
  • anonymous
there was a similar question to this, and my sis helped me with that.
anonymous
  • anonymous
and in that other problem, my sis taught me to set velocity = 0
anonymous
  • anonymous
then solve for the constant
amistre64
  • amistre64
cant say that i ever came across the problem before, so it might be plausible :)
anonymous
  • anonymous
and then from that, do another anti derivative to find the position fuction
anonymous
  • anonymous
lol alrighty, thanks anyways! enjoy the rest of your day :]
amistre64
  • amistre64
:) yw
toxicsugar22
  • toxicsugar22
amristre can u help me

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