Question

Anyone know how to find the arc length of the curve: y=(((x^(4))/(4))+(1/(8x^(2))) on the interval 1less than or equal to x less than or equal to 2

Answer 1

ds = sqrt(dx^2 + dt^2) if I recall correctly

Answer 2

since y is a function of "x" dx^2 = 1

Answer 3

dt was supposed to be dy.....

Answer 4

\[L=\int\limits_{a}^{b}\sqrt{1+(dy/dx)^2 dx}\]

Answer 5

a=1
b=2
but what is (dy/dx)?

Answer 6

looks right.... but I think it has a "ds" on the end, not that it matters much :) and make sure your "dx" is out of the radical

Answer 7

dy/dx = the derivative of your function

Answer 8

I have (dy/dx)= x^(3) - (1/(4x^(3)))

Answer 9

the length thing is the square root of the sum off the derivatives of x and y; since y is a function of x; x'=1

Answer 10

Is that right?

Answer 11

id have to try to decipher the original equation you posted and maybe implicit it for y'...

Answer 12

okay hold on let me write it again, sorry for it being sloppy hahah

Answer 13

:)

Answer 14

\[y=(x^{4}/4) + 1/(8x^2)\]

Answer 15

A lil neater :D

Answer 16

neater yes; but make sure that this is the correct equation.... double check it :)

Answer 17

Yes, lol. And I have to find the arc length of the curve from \[1\le x \le 2\]

Answer 18

4x^3 16x
y' = ------ (-) ---------
4 (8x^2)^2

this should be y' = dy/dx

Answer 19

can reduce as needed :)

Answer 20

y' = x^(3) - (1/4x^(3))
reduced?

Answer 21

yep, that looks good :)

Answer 22

haha awesome.

Answer 23

now I have to square that. That is when I got : x^(6) - 1/2 + 1/(16x^(6))

Answer 24

as long as you kept track of all the stuff, ill trust you on that :)

Answer 25

let me double check real quick, I'm not sureee..

Answer 26

can we get that into 1 term? get like denominators be easier?

Answer 27

I think that might make everything a bit more confusing... lol

Answer 28

hold on a sec :)

Answer 29

4x^6 - 1
-------- ??
4x^3

Answer 30

ummmm... I'm lost. Sorry!
What did you do?

Answer 31

nvm, I see what you did haha

Answer 32

hm, maybe.. let me see real quick

Answer 33

y' = x^(3) - (1/4x^(3))

get like denominators.... 4x^3 is common

Answer 34

ahhhh, smart move, that helped. let me make that change real fast

Answer 35

16x^12 -8x^6 +1
------------------ = y'^2
16x^6

Answer 36

Yeah! now I have to put that back into the equation for arc length..

Answer 37

yep...

Answer 38

reduce it if you want :)

Answer 39

\[\int\limits_{1}^{2} x^6 + 1/16x^6 + 1/2 dx\]

Answer 40

(S) sqrt[1 + x^6 + (1/16x^6) - (1/2)] dx

Answer 41

oh thats not too bad

Answer 42

dont forget its the "square root" of 1 + y'^2 ....

Answer 43

ohhhhh, this is where I got caught up...

Answer 44

yeah, thats where I didn't know what to do next. Should I take the square root of each term separately?

Answer 45

cant do that; they aint multiplied together, it just a one lump sum under that radical and you cant break that up....

Answer 46

x^(3) + 1/4x^(3) + 1/sqrt2 ?

Answer 47

oh you cant do that :( darnnnn

Answer 48

are you sure?

Answer 49

watch:

sqrt(4*3) = sqrt(4) * sqrt(3) = 2sqrt(3)

BUT!!

sqrt(4+3) NOT equal sqrt(4) + sqrt(3)

Answer 50

ohhhhh, you're right. soooo... what do I do?

Answer 51

:(

Answer 52

I dont really know..... thats were I usually have to open up the book and re-read what they have in there for that section :)

Answer 53

it looks like you might be able to do some sort of trig substitution, but i just aint confident about my abilities in that yet...

Answer 54

ahhh, is there a quadratic function we can make out of those three terms?

Answer 55

oh yeahhh, trig substitution... ugh

Answer 56

sqrt(1+tan^2) = something or another.... but I always mess it up :)

Answer 57

lol I'll check...

Answer 58

thanks for helping me on such an exhausting problem! :D

Answer 59

lol ..... I hope I learned something :)

Answer 60

if y' = tan

sqrt(1+tan^2) = sqrt(sec^2) = sec...

(S) sec du or something...

Answer 61

hahaha, ummm... I dont know if its a trig substitution. But if you wanted to move to another problem or anything thats fine. I dont want to bother you all night with just this problem

Answer 62

totally your choice :)

Answer 63

well, library is closing soon, which takes my internet with it :) So I will be heading out anywhoos :)

It was fun... Ciao

Answer 64

Thanks again :)