Please Help! Finding Arc Length

[jstrobel]

I have been having a hard time solving this problem. Any help would be much appreciated.

Find the length of the indicated curve:
x = y^4/16 + 1/(2y^2) between y = -3 and y = -2

I found:
dx = [4y^3/16 - y^(-3)] dy

I know that i must integrate (dx^2 + dy^2)^(1/2) but have been unable to determain how to integrate this.

Any help anyone could provide would be appreciated.

Thank You! Jenny

 

[tkhunny]

I'd be interested to know what your expression is for dx^2 + dy^2.

It seems 1 + (dx/dy)^2 would be simpler, since dx/dy is pretty much staring at you after "dx =", no?

 

[jstrobel]

the integral

the format I have for what I am trying to integrate is:

integral of [1 + (4y^3/16 - y^(-3))^2]dy
= integral of [ 1 + 16y^6/256 - y^(-6)]dy

Is it possible to easily integrate this by using the chain rule? or is it necessary to know the more advanced formulas for an integral? Thanks

 

[tkhunny]

I see, you didn't really do what you said you were doing. You already did what I suggested. Good call.

As far as evaluating the integral, I think you just gave up too soon. All that mess under the square root is a perfect square. I'm thinking you should be able to figure it out.

Let us know.

 

[jstrobel]

factoring

is there are certain way to go about factoring something like that? I have no idea where to even begin.

do you have to find a common denominator and then factor y? but i don't think y will be in all the terms.

 

[pka]