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

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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

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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

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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?
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
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?
wait, so if you take the max. and min. of the position graph, the derivative (or velocity) will be 0, 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 :)
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....
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
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
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
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
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?
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
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
the total distance traveled is going to be an integral of sqrt(1-x^2) dx ...but that might be the position function...
could you explain how you got that?
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
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)
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.. ><
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)
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
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
let me draw a picture :)
awesome
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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)
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: \[\int\limits_{a}^{b}\sqrt{ [f'(x)]^2 + [g'(y)]^2}ds\]
since we only have a g(y) = p(t) then x=x giveing us D(x) = 1 the length of the curve is then... \[\int\limits_{a}^{b} \sqrt{1+[p'(t)]^2} dt\]
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
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 :)
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. ><
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so we're working with the hypotenuse of that?
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) ] \[\ln[(\sqrt{(12t^2 -36t+15)^2 +1}) + (12t^2 -36t +15)]\]
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 :)
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 =??
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)]
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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