Recall the aforementioned formula:

$\mathrm{Arc} \mathrm{Length} \= \int \sqrt{1\+{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}^2}\ⅆx$

The first step is to take the derivative of y with respect to x:

$\frac{\ⅆy}{\ⅆx}\= {\left(x-1\right)}^{\frac{1}{2}}\=\sqrt{x-1}$

Then substitute the derivative into the first formula to get:

$\mathrm{Arc} \mathrm{Length} \= {\int }_1^5\sqrt{1\+{\left(\sqrt{x-1}\right)}^2}\ⅆx$

$\mathrm{Arc} \mathrm{Length} \= {\int }_1^5\sqrt{1\+x-1}\ⅆx$

$\mathrm{Arc} \mathrm{Length} \= {\int }_1^5\sqrt{x}\ⅆx$

$\mathrm{Arc} \mathrm{Length} \= \frac{2}{3}{x}^{\frac{3}{2}}\genfrac{}{}{0ex}{}{}{\phantom{x\=a}}\|{{{{}_{}}_{}}_1{{{}^{}}^{}}^{}{}_{}}^5$

$\mathrm{Arc} \mathrm{Length} \= \frac{2}{3}\left(5^{\frac{3}{2}}-1^{\frac{3}{2}}\right)$

$\mathrm{Arc} \mathrm{Length} \=$6.79.

Recall the aforementioned formula:

$\mathrm{Arc} \mathrm{Length} \= \int \sqrt{{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)}^2\+{\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)}^2}\ⅆt$

To determine the arc length of the curve defined by these parametric equations, first take the derivative of x with respect to t:

$\frac{\ⅆx}{\ⅆ t}\= \sin \left(t\right)$

Next, take the derivative of y with respect to t:

$\frac{\ⅆy}{\ⅆ t}\= \cos \left(t\right)$

You can now substitute $\frac{\ⅆx}{\ⅆ t}$ and $\frac{\ⅆy}{\ⅆ t}$ into the first formula:

$\mathrm{Arc} \mathrm{Length} \= {\int }_0^{\frac{\mathrm{\π}}{2}}\sqrt{{\left(\sin \left(t\right)\right)}^2\+{\left(\cos \left(t\right)\right)}^2}\ⅆt$

$\mathrm{Arc} \mathrm{Length} \= {\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{{\sin }^2\left(t\right)\+{\cos }^2\left(t\right)}\ⅆt$.

Using the trig identity: ${\sin }^2\left(t\right)\+{\cos }^2\left(t\right)\=1$, you can simplify the above equation as:

$\mathrm{Arc} \mathrm{Length} \= {\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{1}\ⅆt$

$\mathrm{Arc} \mathrm{Length} \= {\int }_0^{\frac{\mathrm{\pi }}{2}}1 \ⅆt$

After integrating, this becomes:

$\mathrm{Arc} \mathrm{Length} \= t \|{{{{}_{}}_{}}_0}^{\frac{\mathrm{\π}}{2}}$

$\mathrm{Arc} \mathrm{Length} \= \frac{\mathrm{\π}}{2}-0$

$\mathrm{Arc} \mathrm{Length} \= \frac{\mathrm{\π}}{2}$.