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Velocity is the derivative of the function. Find the derivative first.

well for part a), I took the integral of v(t) to find position

good, integral and an initial condition...good

b is the values for t from teh quadratic formula

how do you determine position when t is greater than or equal to 0?

the position is determined by the integral of the function; using at t=1; p=0 as the anchor

F(t) = 4t^3-18t^2+15t +C
0 = 4-18 +15 +C; solve for C :)

ok, that makes sense

so C=-1

giving us the answer as p(t) = 4t^3-18t^2+15t-1?

http://www.wolframalpha.com/input/?i=f%28t%29+%3D+4t%5E3-18t%5E2%2B15t+-1

the derivative tells tha the particle is at rest on the hump and in the valley of that graph right?

I don't understand that part

well, i typed it wrong to begin with :)
when velocity = 0 it is at rest correct?

if you take the derivative=0...which is velocity = 0; it is at rest; if only for a brief moment :)

the max min of the position graph just tells you "where" it is.. not how fast it is moving...

can I find when it is at rest by plugging v(t) = 0?

http://www.wolframalpha.com/input/?i=v%28t%29%3D12t%C2%B2-36t%2B15

yes

ok, according to that graph, it would be at t=0.5 and t=2.5.

yes

which corresponds to the hump and valley of the position :)

That makes sense. Now to find the max velocity between 0

yes

http://www.wolframalpha.com/input/?i=12t%5E2-36t%2B15++from+0+to+2

it appears to be at zero..

how would you find this algebraically? like if it is non calculator

the 2nd derivative tells you concavity points...

we know that 1/2 and 5/2 are critical points; so which one is between 0 and 2?

1/2 is in the bounds

12t^2-36t+15
12(1/2)^2-36(1/2)+15
12/4 -36/2 + 15 = ?? for starters :)

okay we get 0 :D

im remembering this stuff as i go lol

check p(t) at 0 and at 2 as well.... :)

haha I know what you mean :) and hmm why does the (-) give you a MAX?

the maximum velocity of the particle is going to be when v(t) is the biggest..

okay. and to answer c) the max velocity would be v(t)=15?

regardless of position; so the 2nd derivative is actually the 1st derivative of v(t) :)

the MAX velocity is 15 yes :)

okay :D and somebody in the questions scroll down is saying they need you for a minute ;)

if you don't mind, I'd appreciate if you also walked me through part d).

.....lol :) can you hold? or you think you got this?

but go ahead and see what they need :D

could you explain how you got that?

I'm confused at how you "see that the length is just s=sqrt(1+x^2).." etc

the length of a curve is the sum of all its straight parts..... do you agree?

ok that makes sense

calculus allows us to look so close to a curve, that it straightens out right?

all these "straight" parts are just a bunch of pythag identites added togehter

what are pythag identities?

x^2 + y^2 = z^2 is the pythagorean theorum right? use any letters you want, its the same thing :)

okay, I'm associating that with a right triangle though.. ><

you should associate it with a right triangle; because the "s" part is the hypotenuse

and the legs are the derivatives of x and y...squared

let me draw a picture :)

awesome

calculus get us so close to this that x become dx and y becoems dy and s becomes ds

wow, so I'm not crazy for imagining that ^^'

f(x) = x; f'(x) = dx
g(y) = y; g'(y) = dy
h(s) = s; h'(s) = ds

ds^2 = dx^2 + dy^2
[h'(s)]^2 = [f'(x)]^2 + [g'(y)]^2

ds = sqrt( [f'(x)]^2 + [g'(y)]^2)

so whats our length from 0 to 2?
it amounts to ln(sec(t)+tan(t)) from 0 to 2

uhh would you plug in [ln(sec(2)+tan(2))]-[ln(sec(0)+tan(0))]?

yes

resubstitute in for sec(u) and tan(u)

p'(t) = tan(u) is what i mean to get across lol....
p'(t) = v(t) right?

ehh you lost me here. ><

so we're working with the hypotenuse of that?

no, that triangle is how we substitute and re integrate our values for p'(t)

F(0) = ln[ sqrt(226) +15 ]
F(2) = ln[ sqrt(82) - 9 ]

length from 0 to 2 = F(2) - F(0)

-2.89 - 3.40 = |-6.29| = 6.29

wow, and that's the final answer?

i believe that is the answer; as long as my fingers dint trip on the calculator ;)

ok, thank you so much! :)

is there a way to see if its right?

when my teacher grades it, I'll find out

ha!! .... tell your teacher to hurry up ;)

lol :)