# How do you find the lengths of the curve x=(y^4+3)/(6y) for 3<=y<=8?

May 19, 2018

Use the arc length formula.

#### Explanation:

$x = \frac{{y}^{4} + 3}{6 y} = {y}^{3} / 6 + \frac{1}{2 y}$

$x ' = \frac{1}{2} \left({y}^{2} - \frac{1}{y} ^ 2\right)$

Arc length is given by:

$L = {\int}_{3}^{8} \sqrt{1 + \frac{1}{4} {\left({y}^{2} - \frac{1}{y} ^ 2\right)}^{2}} \mathrm{dy}$

Expand the square:

$L = {\int}_{3}^{8} \sqrt{1 + \frac{1}{4} \left({y}^{4} - 2 + \frac{1}{y} ^ 4\right)} \mathrm{dy}$

Combine terms:

$L = \frac{1}{2} {\int}_{3}^{8} \sqrt{{y}^{4} + 2 + \frac{1}{y} ^ 4} \mathrm{dy}$

Factorize:

$L = \frac{1}{2} {\int}_{3}^{8} \sqrt{{\left({y}^{2} + \frac{1}{y} ^ 2\right)}^{2}} \mathrm{dy}$

Hence

$L = \frac{1}{2} {\int}_{3}^{8} \left({y}^{2} + \frac{1}{y} ^ 2\right) \mathrm{dy}$

Integrate term by term:

$L = \frac{1}{2} {\left[{y}^{3} / 3 - \frac{1}{y}\right]}_{3}^{8}$

Insert the limits of integration:

$L = \frac{1}{6} \left({8}^{3} - {3}^{3}\right) - \frac{1}{2} \left(\frac{1}{8} - \frac{1}{3}\right)$

Simplify:

$L = \frac{485}{6} + \frac{5}{48} = \frac{1295}{6}$