# Thread: Length of a curve

1. ## Length of a curve

Find the length of a curve given by:

y= (x^4)/4 + 1/8x^2 for 1<x<3

I understand length of curves, but I am confused how to find the square root of:

sqr(x^6 + 1/2 x^4 + 1/16 x^2 + 1)

help? Am I writing this term wrong in the first place?

thanks!

2. I know how these problems are rigged to work.
They are quite artifical and what happens is the +1 changes a -.5 to a +.5

Hence MY guess is that the second power is -2 not +2

y= (x^4)/4 + 1/8x^{-2} for 1<x<3

I GUESS you meant to put 1/(8x^2), which I know works.
But your math showed that you wanted (1/8)x^2.

3. Originally Posted by klooless
I understand length of curves, but I am confused how to find the square root of:

sqr(x^6 + 1/2 x^4 + 1/16 x^2 + 1)

help? Am I writing this term wrong in the first place?
Hmm... how did you end up with the +1 at the end of the equation, I just get sqrt. (x^6 + 1/2 x^4 + 1/16x^2).
Anyway, it'd equal:
x^3 + sqrt. 1/2 x^2 + 1/4 x
You would then integrate this to get:
x^4 / 4 + x^3 / 3sqrt(1/2) + x^2 / 8
Length between 3 and 1 would equal:
(20.25 + 12.73 + 1.13) - (0.25 + 0.47 + 0.13)
= 33.26

4. Originally Posted by klooless
Find the length of a curve given by:

y= (x^4)/4 + 1/8x^2 for 1<x<3

I understand length of curves, but I am confused how to find the square root of:

sqr(x^6 + 1/2 x^4 + 1/16 x^2 + 1)

help? Am I writing this term wrong in the first place?

thanks!
$\displaystyle y = \frac{{{x^4}}}{4} + \frac{1}{{2{x^2}}},{\text{ }}1 \leqslant x \leqslant 3,{\text{ }}{L_y} = ?$

$\displaystyle {L_y} = \int\limits_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } .$

$\displaystyle \sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} = \sqrt {1 + {{\left( {\frac{d}{{dx}}\left( {\frac{{{x^4}}}{4} + \frac{1}{{8{x^2}}}} \right)} \right)}^2}} = \sqrt {1 + {{\left( {{x^3} - \frac{1}{{4{x^3}}}} \right)}^2}} =$

$\displaystyle = \sqrt {{x^6} + \frac{1}{2} + \frac{1}{{16{x^6}}}} = \sqrt {{{\left( {{x^3} + \frac{1}{{4{x^3}}}} \right)}^2}} = {x^3} + \frac{1} {{4{x^3}}}{\text{ }}\left( {1 \leqslant x \leqslant 3} \right).$

So we have

$\displaystyle {L_y} = \int\limits_1^3 {\left( {{x^3} + \frac{1}{{4{x^3}}}} \right)dx} = \left. {\left( {\frac{{{x^4}}}{4} - \frac{1}{{8{x^2}}}} \right)} \right|_1^3 =$

$\displaystyle = \frac{{81}}{4} - \frac{1}{{72}} - \left( {\frac{1}{4} - \frac{1}{8}} \right) = \frac{{1457}}{{72}} - \frac{1}{8} = \frac{{1448}}{{72}} = \frac{{181}}{9}.$

See picture