Question

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 15 years ago

Answer 1

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

Answer 2

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

Answer 3

good, integral and an initial condition...good

Answer 4

b is the values for t from teh quadratic formula

Answer 5

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

Answer 6

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

Answer 7

F(t) = 4t^3-18t^2+15t +C

0 = 4-18 +15 +C; solve for C :)

Answer 8

ok, that makes sense

Answer 9

so C=-1

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

I don't understand that part

Answer 15

well, i typed it wrong to begin with :)


when velocity = 0 it is at rest correct?

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

yes

Answer 22

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

Answer 23

yes

Answer 24

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

Answer 25

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....

Answer 26

That makes sense. Now to find the max velocity between 0<t<2 would be where the velocity graph is highest? and the max velocity would be the highest y value?

Answer 27

yes

Answer 28

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

Answer 29

it appears to be at zero..

Answer 30

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

Answer 31

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

Answer 32

the 2nd derivative tells you concavity points...

Answer 33

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

Answer 34

1/2 is in the bounds

Answer 35

12t^2-36t+15

12(1/2)^2-36(1/2)+15

12/4 -36/2 + 15 = ?? for starters :)

Answer 36

okay we get 0 :D

Answer 37

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

Answer 38

im remembering this stuff as i go lol

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

the MAX velocity is 15 yes :)

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

but go ahead and see what they need :D

Answer 50

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

Answer 51

could you explain how you got that?

Answer 52

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

Answer 53

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)

Answer 54

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

Answer 55

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

Answer 56

ok that makes sense

Answer 57

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

Answer 58

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

Answer 59

what are pythag identities?

Answer 60

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

Answer 61

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

Answer 62

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)

Answer 63

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

Answer 64

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

Answer 65

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

Answer 66

let me draw a picture :)

Answer 67

awesome

Answer 68

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

Answer 69

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

Answer 70

f(x) = x; f'(x) = dx

g(y) = y; g'(y) = dy

h(s) = s; h'(s) = ds

Answer 71

ds^2 = dx^2 + dy^2

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

Answer 72

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

Answer 73

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:

\[\int\limits_{a}^{b}\sqrt{ [f'(x)]^2 + [g'(y)]^2}ds\]

Answer 74

since we only have a g(y) = p(t)

then x=x giveing us D(x) = 1

the length of the curve is then...
\[\int\limits_{a}^{b} \sqrt{1+[p'(t)]^2} dt\]

Answer 75

so whats our length from 0 to 2?

it amounts to ln(sec(t)+tan(t)) from 0 to 2

Answer 76

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

Answer 77

yes

Answer 78

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 :)

Answer 79

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

Answer 80

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

p'(t) = v(t) right?

Answer 81

ehh you lost me here. ><

Answer 82

so we're working with the hypotenuse of that?

Answer 83

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) ]

\[\ln[(\sqrt{(12t^2 -36t+15)^2 +1}) + (12t^2 -36t +15)]\]

Answer 84

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

Answer 85

F(0) = ln[ sqrt(226) +15 ]

F(2) = ln[ sqrt(82) - 9 ]

Answer 86

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

Answer 87

-2.89 - 3.40 = |-6.29| = 6.29

Answer 88

wow, and that's the final answer?

Answer 89

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

Answer 90

ok, thank you so much! :)

Answer 91

is there a way to see if its right?

Answer 92

when my teacher grades it, I'll find out

Answer 93

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

Answer 94

lol :)

Answer 95

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 =??