Answer
$$
y= \int_{0}^{x} \sqrt {\sqrt {t}-1}) dt, \quad \quad 1 \leq x \leq 16
$$
The length of the given curve is
$$
\begin{aligned}
L &=\int_{1}^{16 } \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} dx\\
&=\frac{124}{5}
\end{aligned}
$$
Work Step by Step
$$
y= \int_{0}^{x} \sqrt {\sqrt {t}-1}) dt, \quad \quad 1 \leq x \leq 16
$$
$\Rightarrow$
$$
y^{\prime} =\frac{dy}{dx}=\sqrt {\sqrt {x}-1})
$$
$\Rightarrow$
$$
1+\left(\frac{d y}{d x}\right)^{2} =1+\left(\sqrt {\sqrt {x}-1}) \right)^{2} =\sqrt {x}.
$$
Let
$$
f(x)=\sqrt{1+\left(\frac{d y}{d x}\right)^{2}} =\sqrt{\sqrt {x}}
$$
Then the arc length is
$$
\begin{aligned}
L &=\int_{1}^{16} f(x ) dx \\
&=\int_{1}^{16 } \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} dx\\
&=\int_{1}^{16} \sqrt {\sqrt {x}} d x \\
&=\int_{1}^{16} x^{\frac{1}{4}} d x \\
&=\frac{4}{5} [x^{\frac{5}{4}} ]_{1}^{16} \\
&=\frac{4}{5} (32-1)\\
&=\frac{124}{5}
\end{aligned}
$$
the length of the given curve is $\frac{124}{5}$
