Question

Find the length of the curve

y= (x^5)/6 + 1/(10x^3)

Answer 1

isnt it the integration of sqrt(1+y'^2) ?

Answer 2

Yes it is

So for the derivative I got

\[\frac{ dy }{ dx }=\frac{ 5x^4 }{ 6 } + \frac{ -3x^{-4} }{ 10}\]

Answer 3

i wonder if doing a polar would be simple or not

Answer 4

So, the integral would be

I forgot the limits are 1 to 2

\[\int\limits_{1}^{2} \sqrt{1+(\frac{ 5x^4 }{ 6} + \frac{ -3 }{ 10x^4 })^2}dx \]
?

Answer 5

so far that is to format

can you simplify it any?

Answer 6

Yeah, that's what I was trying to do.

If I did it right I got the derivative to simplify to

\[\frac{ 25x^8-9 }{ 30x^4 }\]

Answer 7

+9 but yes, and that is for x>0

Answer 8

I believe it's -9 because the 3 is negative in the derivative.

Answer 9

d/dx: http://www.wolframalpha.com/input/?i=d%2Fdx++%28x%5E5%29%2F6+%2B+1%2F%2810x%5E3%29

simplified, notice the if x is positive part: http://www.wolframalpha.com/input/?i=sqrt%281%2B%28%2825x%5E8-9%29%2F%2830x%5E4%29%29%5E2%29

Answer 10

So am I supposed to integrate this now

\[\int\limits_{0}^{1}(1+\frac{ (25x^8-9)^2 }{ 900x^8 })^{1/2}dx\]

Answer 11

never mind I get the simplification

Answer 12

then yes, all thats left is to:\[\int\frac{25x^8}{30x^4}+\frac{9}{30x^4}\]
then yes, all thats left is to:\[\int\frac{5}{6}x^4+\frac{3}{10}x^{-4}\]