## anonymous one year ago Find the arc length of y=x^(2/3) for 0<x<1

1. anonymous

2. amistre64

$\int_{}^{}ds\implies \int_{}^{}\sqrt{(x')^2+(y')^2}~dt$

3. amistre64

of instead of dt, we can do with dx, and let x'=1

4. amistre64

what are you getting for your setup?

5. anonymous

im getting this $\int\limits_{0}^{1} \sqrt{(1)^2 + (x^\frac{ 2 }{ 3 })^2} dx$

6. anonymous

oh wait

7. anonymous

|dw:1442357936251:dw|

8. amistre64

thats better yeah

9. anonymous

and then

10. anonymous

|dw:1442358056890:dw|

11. anonymous

then you find the integral?

12. amistre64

then you find the integral ... trying to see if there is a nice form to it tho

13. amistre64

the wolfs integral suggests its a fairly simple u-sub

14. amistre64

or not ...

15. anonymous

how do you get to that? im trying to integrate

16. amistre64
17. anonymous

|dw:1442358829336:dw|

18. amistre64

i just ask the wolf to integrate it ...

19. anonymous

would that be the integral?

20. anonymous

im sure its that just trying to figure out how to get to that point

21. amistre64

how are you sure that is it? i have to wonder since the wolfs assessment is different

22. amistre64

one thing i am thing of is trying to work out a parametric like x = t^3, y=t^2 this way, t = x^(1/3), letting y = x^(2/3) by eliminating the parameter

23. amistre64

t=0 to 1 satisfies the substitution x' = 3t^2, y' =2t sqrt(9t^4 + 4t^2) t sqrt(9t^2 + 4) dt

24. anonymous

|dw:1442359448247:dw|

25. anonymous

would u sub work here?

26. amistre64

i already tried working it like that ... i was getting nowhere. but i believe if we split it into parameters it works out better for us

27. amistre64

t sqrt(9t^2 + 4) dt let u = 9t^2+4 du = 18t dt t sqrt(u) du/18t u^(1/2) du/18 2/3(18) u^(3/2) 1/3(9) u^(3/2) 1/27 u^(3/2)

28. amistre64

u = 4 to 13 when t=0 to 1

29. amistre64

1/27 ((13)^(3/2)-(4^(3/2)), is looking sweet on my end

30. amistre64

i cant make it work if we keep y as a function of x as its defined. so i wondered if parameters would make life easier, and it does

31. anonymous

|dw:1442360116489:dw|

32. anonymous

is that what you mean?

33. amistre64

$y = x^{2/3}~:~x=(0,1)$ $x^{1/3}=t~:~x=t^3~:~t=(0,1)$ $y=t^2$ $x'=3t^2~:~y'=2t$ $\int_{0}^{1}\sqrt{(x')^2+(y')^2}~dt$

34. amistre64

that simplifies to: $\int_{0}^{1}t\sqrt{9t^2+4}~dt$ yes

35. anonymous

aaah i see what you did there

36. anonymous

|dw:1442360353998:dw|

37. anonymous

|dw:1442360476915:dw|

38. anonymous

and then you insert u? and then plug in the 1? and that gets you the answer

39. amistre64

by substitution u = 9t^2 + 4, as t goes from 0 to 1, u goes from 9(0)+4 to 9(1)+4

40. amistre64

so the limits for du stuff is 4 to 13

41. amistre64

youcould replace u witn 9t^2 + 4 if you want, but then the 13 and 4 get calculated in there inways

42. anonymous

umm im a bit lost but i'll try to see where it is that im confused

43. amistre64

k, and dont worry about being lost, my approach was prolly not 'standard' it is just what worked out in my head.