Question

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

Answer 1

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

Answer 2

let me try this, see if it works.

\[y=x^\frac{3}{2}\] yes?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

oh sorry yes 0 to 1/4

Answer 8

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
\[\int_0^{\frac{1}{8}} \sqrt{1-\frac{9}{4}x} dx\]

Answer 9

oops you are right it is 1/4

Answer 10

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

Answer 11

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

Answer 12

did you guys get .2824?

Answer 13

let me get a calculator i will check

Answer 14

no in fact i got .21055...

Answer 15

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

Answer 16

yes it is .2824

Answer 17

well i guess i messed up hold on

Answer 18

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

Answer 19

take advantage of Wolfram alpha it is a very useful tool

Answer 20

i put - where i should have had +. sorry

Answer 21

no problem thanks for the help guys

Answer 22

no problem

Answer 23

Just why do we pick positive sqrt all the time?