## anonymous 5 years ago Hey everyone, need help finding the arc length of the curve y^2=x^3 from (0,0) to (1/4,1/8).

1. anonymous

I know the formula and I think y'=(3x^2)/(2y)

2. anonymous

let me try this, see if it works. $y=x^\frac{3}{2}$ yes?

3. anonymous

ok right, do we only take the +x^3/2?

4. anonymous

$y'=\frac{3}{2}x^\frac{1}{2}$

5. anonymous

y = x^(3/2) y' = 3/2*x^1/2 (y')^2 = 9/4*x integrate sqrt(1+9/4*x) from 1/8 to 1/4 use u-sub if you have to

6. anonymous

Rim would it not be from 0 to 1/4?

7. anonymous

oh sorry yes 0 to 1/4

8. anonymous

we are going from (0,0) to (1/8,1/4) so it is ok yes? maybe i am wrong, but i will graph it. in any case we have to integrate $\int_0^{\frac{1}{8}} \sqrt{1-\frac{9}{4}x} dx$

9. anonymous

oops you are right it is 1/4

10. anonymous

anti derivative is $-\frac{(4-9x)^\frac{3}{2}}{27}$

11. anonymous

and integral is $\frac{8}{27}-\frac{7\sqrt{7}}{216}$

12. anonymous

did you guys get .2824?

13. anonymous

let me get a calculator i will check

14. anonymous

no in fact i got .21055...

15. anonymous

ok, but isnt it also sqrt(1+(9/4)x)?

16. anonymous

17. anonymous

yes it is .2824

18. anonymous

well i guess i messed up hold on

19. anonymous

you are right and i am wrong (0ften the case)

20. anonymous

take advantage of Wolfram alpha it is a very useful tool

21. anonymous

i put - where i should have had +. sorry

22. anonymous

no problem thanks for the help guys

23. anonymous

no problem

24. anonymous

Just why do we pick positive sqrt all the time?