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anonymous
 5 years ago
P(t) = 1000+360sin((π/6)t)+180sin((π/3)t)
Find the derivative and then solve for when P(t)=0
anonymous
 5 years ago
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
 5 years ago
Best ResponseYou've already chosen the best response.0its alot of sums; so deriving is easy

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.00 +360pi/6 cos(t.pi/6) +180pi/3cos(t.pi/3)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok i got that part right. now the solving for p(t)=0 part is where im stuck

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0360/6 = 180/3 so that all the same

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0cos(t.pi/6) + cos(t.pi/3) = 0 when.....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you agree with this so far?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we gotta get them at 180 from each other than right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0t.60 = t.30 + 180; or, t.30 = t.60 + 180

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0t(6030) = 180 t(30) = 180 t = 60 maybe?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.06.60 = 360 6.30 = 180; thats a good one right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0a(1) + a(1) = 0 right

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so every multiple of 6 should be good for that one...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0t.30 = t.60 + 180 t(3060) = 180 t = 180/30 t = 6; same thing

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so whenever t = a multiple of 6, we should be good

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0given that t not equal zero perhaps; or some such restriction

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0there are bound to be multples of 6 that make t.60 = t.30; in that case its of no use

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ok; we need angles that are 180 from each other in order for there cosines to be opposites of each other right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0t.60  t.30 = 180 t(60  30) = 180 t = 180/30 t = 6

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0now 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
 5 years ago
Best ResponseYou've already chosen the best response.0360/30 = 12; so every multiple of 12 wont work.... right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so odd multiples of 6 work; but even multiples fail; so we restricct it to when t is an odd multiple of 6

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0lol i think im good. thank you.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0another question if you dont mind? lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i dunno, im pretty much good for one question and then go stupid lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL, it's a position, velocity, acceleration question.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Find 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
 5 years ago
Best ResponseYou've already chosen the best response.0is it posted already or you wanna hit me with it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Find the initial position and the time t > 0 at which the particle passes again through the initial position.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i know i should do anti derivative for the first part in order to find the velocity equation.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0acceleration is the derivative of velocity is the derivative of position

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0with a constant acceleration of 2; that means we got linear velocity of 2x+c right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would i plug in velocity = 0 since it vanishes or no?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and a position of x^2 +bx +cthen

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the only way for a linear function to vanish at some point it to make a hole iin the graph right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0which means making the velocity function an extention of a rational function

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02x +c  at best d2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0but with a d2 up top

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0something to make a whole at 2 to make the graph vanish

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol 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
 5 years ago
Best ResponseYou've already chosen the best response.0i could be over thinking it tho :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lmao would it be wrong to set the velocity  0?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0cant have a constant acceleration whout velcoty can we?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0my keyboard hates me lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there was a similar question to this, and my sis helped me with that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and in that other problem, my sis taught me to set velocity = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then solve for the constant

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0cant say that i ever came across the problem before, so it might be plausible :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and then from that, do another anti derivative to find the position fuction

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol alrighty, thanks anyways! enjoy the rest of your day :]

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0amristre can u help me
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