Question

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

Answer 1

its alot of sums; so deriving is easy

Answer 2

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

Answer 3

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

Answer 4

looks tricky ;)

Answer 5

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

Answer 6

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

Answer 7

you agree with this so far?

Answer 8

yep

Answer 9

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

Answer 10

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

Answer 11

t(60-30) = 180

t(30) = 180

t = 60 maybe?

Answer 12

err...6 lol

Answer 13

6.60 = 360

6.30 = 180; thats a good one right?

Answer 14

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

Answer 15

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

Answer 16

t.30 = t.60 + 180

t(30-60) = 180

t = 180/-30

t = -6; same thing

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

uh huh

Answer 23

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?

Answer 24

t.60 - t.30 = 180

t(60 - 30) = 180

t = 180/30

t = 6

Answer 25

ooh ok

Answer 26

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?

Answer 27

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

Answer 28

uh huh

Answer 29

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

Answer 30

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

Answer 31

lol i think im good. thank you.

Answer 32

:) ok

Answer 33

another question if you dont mind? lol

Answer 34

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

Answer 35

whats the q

Answer 36

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

Answer 37

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.

Answer 38

is it posted already or you wanna hit me with it

Answer 39

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

Answer 40

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

Answer 41

acceleration is the derivative of velocity is the derivative of position

Answer 42

yep

Answer 43

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

Answer 44

yeah

Answer 45

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

Answer 46

and a position of x^2 +bx +cthen

Answer 47

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

Answer 48

yeah

Answer 49

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

Answer 50

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

Answer 51

but with a d-2 up top

Answer 52

what's d-2?

Answer 53

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

Answer 54

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

Answer 55

i could be over thinking it tho :)

Answer 56

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

Answer 57

velocity = 0*

Answer 58

cant have a constant acceleration whout velcoty can we?

Answer 59

my keyboard hates me lol

Answer 60

haha its okay

Answer 61

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

Answer 62

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

Answer 63

then solve for the constant

Answer 64

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

Answer 65

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

Answer 66

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

Answer 67

:) yw

Answer 68

amristre can u help me