# Find the arch length of the curve given by: x= (1/4)y^3 + (1/3y) ; 1 le y le 2 ?

May 2, 2018

the answer $L = {\int}_{1}^{2} \sqrt{1 + {\left(\frac{3}{4} \cdot {y}^{2} - \frac{1}{3 {y}^{2}}\right)}^{2}} \cdot \mathrm{dy} = \frac{23}{12} \left(u n i t e\right)$

#### Explanation:

we can determine the length of arc by using

$L = {\int}_{a}^{b} \sqrt{1 + {\left(f ' \left(y\right)\right)}^{2}} \cdot \mathrm{dy}$

$f \left(y\right) = \left(\frac{1}{4}\right) {y}^{3} + \left(\frac{1}{3 y}\right)$

$f ' \left(y\right) = \frac{3}{4} \cdot {y}^{2} - \frac{1}{3 {y}^{2}}$

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

$\int \sqrt{1 + {\left(\frac{3}{4} \cdot {y}^{2} - \frac{1}{3 {y}^{2}}\right)}^{2}} \cdot \mathrm{dy} = {y}^{3} / 4 - \frac{1}{3 \cdot y} = \frac{3 \cdot {y}^{4} - 4}{12 \cdot y}$

$L = {\int}_{1}^{2} \sqrt{1 + {\left(\frac{3}{4} \cdot {y}^{2} - \frac{1}{3 {y}^{2}}\right)}^{2}} \cdot \mathrm{dy} = \frac{23}{12} \left(u n i t e\right)$