Question

Find the arc length of: y=(x^3/3)+(1/4x) 1 13 years ago

Answer 1

arc length has to do with ds right?

Answer 2

\[ds = \sqrt{(x')^2+(y')^2}\]

Answer 3

Arc length \[\int\limits_{a}^{b} \sqrt{1+[f \prime(x)]^2}\]

Answer 4

since x' with regard to x = 1

\[ds = \sqrt{1+(y')^2}\]

Answer 5

y= (x^3/3) + (1/4x) ; is the 4x all a denom?

Answer 6

Yes

Answer 7

x^3/3 derived is: 3/3 x^2 = x^2

1/4 x^(-1) derived is: -1/4 x^(-2)

Answer 8

\[\left(x^2-\frac{1}{4}x^{-2}\right)^2=x^4-\frac{1}{2}+\frac{1}{16}x^{-4}\]

Answer 9

add 1 to that and we have:

\[\sqrt{x^4+\frac{1}{16x^4}+\frac{1}{2}}\]

that doesnt look very nice at all does it

Answer 10

Then complete the square?

Answer 11

i was thinking more like opening up a half gallon of happy tracks icecream, balling up in the corner, and crying :)

Answer 12

There has to be some way to break this down.

Answer 13

maybe swithing it to polars ?

Answer 14

I think that's too much work.

Answer 15

Wouldn't be be easier if we left it as 1+(x^2+(1/16x^2)

Answer 16

dunno, but i dount it

what if we add them first?

Answer 17

what do you mean by that?

Answer 18

add the fractions to get a quotient

\[\frac{x^3}{3}+\frac{1}{4x}=\frac{4x^4+3}{12x}\]

Answer 19

Yeah, I forgot.

Answer 20

\[y'=\frac{12x(16x^3)-12(4x^4+3)}{12*12x^2}\]
\[y'=\frac{12x^4+3}{12x^2}\]
\[y'=\frac{4x^4+1}{4x^2}\]

Answer 21

-1, not +1 forgot to carry the negative all the way thru

Answer 22

that does not look like it's easier

Answer 23

square that for\[(y')^2=\frac{16x^8-8x^4+1}{(4x^2)^2}\]

\[1+(y')^2=\frac{16x^8-8x^4+1+16x^4}{(4x^2)^2}\]
\[1+(y')^2=\frac{16x^8+8x^4+1}{(4x^2)^2}\]
\[\sqrt{1+(y')^2}=\frac{\sqrt{16x^8+8x^4+1}}{4x^2}\]

the top is a quadrquadratic in disguise

Answer 24

Oh, I see that's a lot easier

Answer 25

say u = x^2

\[\frac{\sqrt{16(u^4+\frac{1}{2}u^2)+1}}{4u}\]
\[\frac{\sqrt{16(u^4+\frac{1}{2}u^2+\frac{1}{16})}}{4u}\]
\[\frac{\sqrt{16(u^2+\frac14)^2}}{4u}\]
\[\frac{u^2+\frac14}{u}\]

Answer 26

\[\int_{a}^{b}u+\frac{1}{4u}du\]

Answer 27

since x = 1 to 2

and u = x^2

then u = 1^2 to 2^2

a = 1, b=4

Answer 28

i think i missed a math someplace along the way :/

Answer 29

I'll check

Answer 30

u = x^2

du = 2x dx

dx = du/2sqrt(u)

so maybe:\[\int_{1}^{4}\frac{u}{2\sqrt{u}}+\frac{1}{8u\sqrt{u}}du\]

Answer 31

forgot to change my dx to a proper du part :)

Answer 32

Can't the first one be sqrtu/2?

Answer 33

yes, and that ones chks out better :)

i get 2.4583 for each route

Answer 34

yeah, that's right

Answer 35

yay!! :)