Stuck on my answer on a length of curves problem?

**Kimberly** · March 29th 2011, 02:17 AM

Calculate the length of the curve:

$y=\frac{x^4}{4} + \frac{1}{8x^2}$ from $1\leq{x}\leq3$

I'll base myself on the formula: $\int_{a}^{b}\sqrt{1-(f'(x))^2}dx$

Step 1: Derive the function.

$y=\frac{x^4}{4} + \frac{1}{8x^2}$

$y' = x^3 - \frac{1}{4x^3}$

Step 2: Square the derivative.

$(x^3 - \frac{1}{4x^3})^2$

$x^6 + \frac{1}{16x^6} - \frac{1}{2}$

Step 3: Subtract the derivative from $1$

$1 - (x^6 + \frac{1}{16x^6} - \frac{1}{2})$

$\frac{3}{2} - \frac{1}{16x^6} - x^6$

This yields a monster of a integral...

$\int_{1}^{3}\sqrt{\frac{3}{2}-\frac{1}{16x^6}-x^6}dx$

And THAT I need help with...

EDIT: Sorry for the double post, Admin... my internet acted up. Please delete the other post. Thank you and sorry!

**tonio** · March 29th 2011, 03:19 AM

Originally Posted by Kimberly

Calculate the length of the curve:

$y=\frac{x^4}{4} + \frac{1}{8x^2}$ from $1\leq{x}\leq3$

I'll base myself on the formula: $\int_{a}^{b}\sqrt{1-(f'(x))^2}dx$

Oops! It is not minus but plus into the square root!

Step 1: Derive the function.

$y=\frac{x^4}{4} + \frac{1}{8x^2}$

$y' = x^3 - \frac{1}{4x^3}$

Step 2: Square the derivative.

$(x^3 - \frac{1}{4x^3})^2$

$x^6 + \frac{1}{16x^6} - \frac{1}{2}$

Step 3: Subtract the derivative from $1$

$1 - (x^6 + \frac{1}{16x^6} - \frac{1}{2})$

$\frac{3}{2} - \frac{1}{16x^6} - x^6$

This yields a monster of a integral...

$\int_{1}^{3}\sqrt{\frac{3}{2}-\frac{1}{16x^6}-x^6}dx$

And THAT I need help with...

EDIT: Sorry for the double post, Admin... my internet acted up. Please delete the other post. Thank you and sorry!

As noted above in blue, the derivative must go with plus and not with minus

within the square root, and then life looks way more beautiful:

$\displaystyle{\sqrt{1+f'(x)^2}=\sqrt{x^6+\frac{1}{ 2}+\frac{1}{16x^6}}=\sqrt{\left(x^3+\frac{1}{4x^3} \right)^2}=x^3+\frac{1}{4x^3}$ , and

thus the arc length wanted is obtained by an elementary definite integral:

$\displaystyle{\int\limits^3_1\sqrt{1+f'(x)^2}\,dx= \int\limits_1^3\left(x^3+\frac{1}{4x^3}\right)\,dx =$ ...etc.

Tonio

**Kimberly** · March 29th 2011, 04:37 AM

*blush*

Hehe... I knew that...

)

Thanks!

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