Question

find the arc length of the curve y=x^(2/3) for 1<= x <= 27

Answer 1

i know the formula of the arc length, just couldn't integrate!

Answer 2

what was the formula for arc length?

Answer 3

wait...

Answer 4

\[s_{ab}=\int_{a}^{b} \sqrt{1+[f'(x)]^2} \text{d}x\]

Answer 5

\[L = \int\limits_{b}^{a} (\sqrt{(1+[f'(x)]^{2}}) dx\]

Answer 6

yeah, that.

Answer 7

so, it becomes:

Answer 8

\[\int_{1}^{27} \sqrt{1+\frac{4}{9} x^{-2/3}} \ \text{d}x\]

Answer 9

yeah, that's what i got

Answer 10

i just don't know how to integrate

Answer 11

maybe trig sub or integration by parts...idk
@Mimi_x3

Answer 12

would y^3/2 be easier?

Answer 13

maybe u-sub

Answer 14

y = x^(2/3)

y^(3/2) = x

integrate with repect to y instead if its simpler

Answer 15

oh...

Answer 16

wow...nice idea

Answer 17

\[y^{3/2} = x\]\[\frac32y^{1/2}=x'\]
\[\int_{a}^{b}\sqrt{1+(x')^2}dy\]

Answer 18

1<= x <= 27

1<= y <= 9
a b

are your new a and b i believe

Answer 19

thx

Answer 20

youre welcome; but how to integrate sqrt(1+y^3) ... hmmm

Answer 21

:)

Answer 22

also try to do this in vector form, or as a polar equation ... might make it simpler as well

Answer 23

u substitution?

Answer 24

thats y or y^3

Answer 25

?

Answer 26

lol, i am sooo blind :)

Answer 27

lol...:)

Answer 28

\[\int \sqrt{1+ky}~dy\]
\[\int (1+ky)^{1/2}~dy=\frac{2}{3k}(1+ky)^{3/2}\]

Answer 29

got the answer!
(85 sqrt(85) -13sqrt(13))/27

Answer 30

yay!!!

Answer 31

thx so much

Answer 32

good luck :)