A particle moves on the x-axis so that its velocity at any time t is greater than or equal to zero is given by v(t)=12t²-36t+15. At t=1, the particle is at the origin.
a) Fine the position x(t) of the particle at any time t is greater than or equal to zero.
b) Find all values of t for which the particle is at rest.
c) Find the maximum velocity of the particle for 0

Mathematics
- anonymous

A particle moves on the x-axis so that its velocity at any time t is greater than or equal to zero is given by v(t)=12t²-36t+15. At t=1, the particle is at the origin.
a) Fine the position x(t) of the particle at any time t is greater than or equal to zero.
b) Find all values of t for which the particle is at rest.
c) Find the maximum velocity of the particle for 0

Mathematics
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- anonymous

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

- anonymous

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

- amistre64

good, integral and an initial condition...good

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## More answers

- amistre64

b is the values for t from teh quadratic formula

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

ok, that makes sense

- anonymous

so C=-1

- anonymous

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

- amistre64

p(t) = 4t^3-18t^2+15t -1 ; yes unless there is some other interpretation for at t=1 the particle is at the origin. To me that means the particle is at (t,0) wehn t=1

- amistre64

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

- amistre64

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

- anonymous

I don't understand that part

- amistre64

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

- anonymous

wait, so if you take the max. and min. of the position graph, the derivative (or velocity) will be 0, correct?

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

yes

- anonymous

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

- amistre64

yes

- amistre64

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

- amistre64

since the derivative of the position tells the slope at any given point; then yes, the hump and valley of the position are the zeros of the derivative....

- anonymous

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

- amistre64

yes

- amistre64

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

- amistre64

it appears to be at zero..

- anonymous

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

- amistre64

determine the local min/max of the velocity in the interval by the v(t); and use the 2nd derivative to tell you if its a (-)MAX or a (+)MIN

- amistre64

the 2nd derivative tells you concavity points...

- amistre64

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

- anonymous

1/2 is in the bounds

- amistre64

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

- anonymous

okay we get 0 :D

- amistre64

the 2nd derivative is: 24t -36 right?;
so test for "signs" in this one..
24(1/2) - 36
12 - 36 = (-).... this appears to be a MAX
when in doubt just derive x^2 twice since its bowl shaped; you get a (+) for a MIN

- amistre64

im remembering this stuff as i go lol

- amistre64

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

- anonymous

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

- amistre64

its just the nature of the game there....
recall the shape of the x^2 graph? looks like a big "U" whichi means the MIN point is at the bottom of it....
D(x^2) = 2x
D(2x) = 2...which is (+); therefore, the 2nd derivative is (+) we get a MIN

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

the MAX velocity is 15 yes :)

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

but go ahead and see what they need :D

- amistre64

the total distance traveled is going to be an integral of sqrt(1-x^2) dx ...but that might be the position function...

- anonymous

could you explain how you got that?

- amistre64

how i got that? hmmm.... back on my head :)
the way you find the length of a curve is by integrating the position really really close like:
when you get real close; the curve becomes straight and you see that the length is just s = sqrt(1+x^2) which is the pythag thrm; s^2 = x^2+y^2

- amistre64

this pythag is primed for either parameteric equations:
sqrt(f'(x)^2 + g'(x)^2); or regular curves as well, as in:
sqrt(1+[f'(x)]^2)

- anonymous

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

- amistre64

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

- anonymous

ok that makes sense

- amistre64

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

- amistre64

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

- anonymous

what are pythag identities?

- amistre64

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

- anonymous

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

- amistre64

the length of a curve; lets call it "s"
is up close; s^2 = x^2 + y^2
to get up real close, we take a derivative...
ds^2 = dx^2 + dy^2
ds = sqrt(dx^2 + dy^2)

- amistre64

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

- amistre64

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

- anonymous

do you mean that each of the little "straight" segments in a graph are like the hypotenuses of right triangles?? haha I'm a bit confused at this part

- amistre64

let me draw a picture :)

- anonymous

awesome

- amistre64

##### 1 Attachment

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

if the length of one piece is "sqrt( [f'(x)]^2 + [g'(y)]^2)"
then the total length is the sum of all the little pieces of ds.... the integral of:
\[\int\limits_{a}^{b}\sqrt{ [f'(x)]^2 + [g'(y)]^2}ds\]

- amistre64

since we only have a g(y) = p(t)
then x=x giveing us D(x) = 1
the length of the curve is then...
\[\int\limits_{a}^{b} \sqrt{1+[p'(t)]^2} dt\]

- amistre64

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

- anonymous

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

- amistre64

yes

- amistre64

we can say u = v(t) = tan^2
sqrt(1+tan^2) = sec
[S] sec(u) = ln(sec(u)+tan(u)) but we would have to modify the interval if we leave it like this :)

- amistre64

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

- amistre64

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

- anonymous

ehh you lost me here. ><

- amistre64

##### 1 Attachment

- anonymous

so we're working with the hypotenuse of that?

- amistre64

ln[sec(u)+tan(u)] has to be turned back into the original values for p'(t) :)
sec(u) = sqrt[(12t^2 -36t +15)^2 +1]
and
tan(u) = (12t^2 -36t +15)
to amount to.....
ln [ sqrt((12t^2 -36t +15)^2 +1) + (12t^2 -36t +15) ]
\[\ln[(\sqrt{(12t^2 -36t+15)^2 +1}) + (12t^2 -36t +15)]\]

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

-2.89 - 3.40 = |-6.29| = 6.29

- anonymous

wow, and that's the final answer?

- amistre64

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

- anonymous

ok, thank you so much! :)

- amistre64

is there a way to see if its right?

- anonymous

when my teacher grades it, I'll find out

- amistre64

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

- anonymous

lol :)

- amistre64

the other method is to re evaluate the interval to match the substitution...
tan(u) = 12t^2 -36t +15
u = tan^-1(12t^2 -36t +15)
when t = 0 what does u =?
when t = 2 what does u =??

- amistre64

t = 0; u = 86.1859
t = 2; u = -83.6598
soooo...... lets see if thats good ;)
ln[sec(-83.6598) + tan(-83.6598)] - ln[sec(86.1859)+tan(86.1859)]

- amistre64

2.079 - 2.77 ...... i never get that right lol; which is why i always revert it back to the original :)

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