# Determining the Length of a Curve

## Key Questions

• We can find the arc length to be $\frac{1261}{240}$ by the integral
$L = {\int}_{1}^{2} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

Let us look at some details.

By taking the derivative,
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 {x}^{4}}{6} - \frac{3}{10 {x}^{4}}$

So, the integrand looks like:
sqrt{1+({dy}/{dx})^2}=sqrt{({5x^4)/6)^2+1/2+(3/{10x^4})^2
by completing the square
$= \sqrt{{\left(\frac{5 {x}^{4}}{6} + \frac{3}{10 {x}^{4}}\right)}^{2}} = \frac{5 {x}^{4}}{6} + \frac{3}{10 {x}^{4}}$

Now, we can evaluate the integral.
$L = {\int}_{1}^{2} \left(\frac{5 {x}^{4}}{6} + \frac{3}{10 {x}^{4}}\right) \mathrm{dx} = {\left[{x}^{5} / 6 - \frac{1}{10 {x}^{3}}\right]}_{1}^{2} = \frac{1261}{240}$

• It can be found by $L = {\int}_{0}^{4} \sqrt{1 + {\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)}^{2}} \mathrm{dy}$.

Let us evaluate the above definite integral.

By differentiating with respect to y,
$\frac{\mathrm{dx}}{\mathrm{dy}} = {\left(y - 1\right)}^{\frac{1}{2}}$

So, the integrand can be simplified as
$\sqrt{1 + {\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)}^{2}} = \sqrt{1 + {\left[{\left(y - 1\right)}^{\frac{1}{2}}\right]}^{2}} = \sqrt{y} = {y}^{\frac{1}{2}}$

Finally, we have
$L = \setminus {\int}_{0}^{4} {y}^{\frac{1}{2}} \mathrm{dy} = {\left[\frac{2}{3} {y}^{\frac{3}{2}}\right]}_{0}^{4} = \frac{2}{3} {\left(4\right)}^{\frac{3}{2}} - \frac{2}{3} {\left(0\right)}^{\frac{3}{2}} = \frac{16}{3}$

Hence, the arc length is $\frac{16}{3}$.

I hope that this helps.

• If you want to find the arc length of the graph of $y = f \left(x\right)$ from $x = a$ to $x = b$, then it can be found by
$L = {\int}_{a}^{b} \sqrt{1 + {\left[f ' \left(x\right)\right]}^{2}} \mathrm{dx}$