## anonymous 5 years ago 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

1. amistre64

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

2. amistre64

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

3. amistre64

dt was supposed to be dy.....

4. anonymous

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

5. anonymous

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

6. amistre64

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

7. amistre64

dy/dx = the derivative of your function

8. anonymous

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

9. amistre64

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

10. anonymous

Is that right?

11. amistre64

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

12. anonymous

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

13. amistre64

:)

14. anonymous

$y=(x^{4}/4) + 1/(8x^2)$

15. anonymous

A lil neater :D

16. amistre64

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

17. anonymous

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

18. amistre64

4x^3 16x y' = ------ (-) --------- 4 (8x^2)^2 this should be y' = dy/dx

19. amistre64

can reduce as needed :)

20. anonymous

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

21. amistre64

yep, that looks good :)

22. anonymous

haha awesome.

23. anonymous

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

24. amistre64

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

25. anonymous

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

26. amistre64

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

27. anonymous

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

28. anonymous

hold on a sec :)

29. amistre64

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

30. anonymous

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

31. anonymous

nvm, I see what you did haha

32. anonymous

hm, maybe.. let me see real quick

33. amistre64

y' = x^(3) - (1/4x^(3)) get like denominators.... 4x^3 is common

34. anonymous

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

35. amistre64

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

36. anonymous

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

37. amistre64

yep...

38. amistre64

reduce it if you want :)

39. anonymous

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

40. amistre64

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

41. anonymous

42. amistre64

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

43. anonymous

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

44. anonymous

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

45. amistre64

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

46. anonymous

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

47. anonymous

oh you cant do that :( darnnnn

48. anonymous

are you sure?

49. amistre64

watch: sqrt(4*3) = sqrt(4) * sqrt(3) = 2sqrt(3) BUT!! sqrt(4+3) NOT equal sqrt(4) + sqrt(3)

50. anonymous

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

51. anonymous

:(

52. amistre64

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 :)

53. amistre64

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...

54. anonymous

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

55. anonymous

oh yeahhh, trig substitution... ugh

56. amistre64

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

57. anonymous

lol I'll check...

58. anonymous

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

59. amistre64

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

60. amistre64

if y' = tan sqrt(1+tan^2) = sqrt(sec^2) = sec... (S) sec du or something...

61. anonymous

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

62. anonymous

63. amistre64

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

64. anonymous

Thanks again :)