$\mathrm{\κ}$

$\=\frac{\|y″\|}{{\left(1\+{\left(y\prime \right)}^2\right)}^{3\/2}}$

$\=\frac{\frac{3}{4\sqrt{x}}}{{\left(1\+{\left(\frac{3}{2}\sqrt{x}\right)}^2\right)}^{3\/2}}$

$\=\frac{3}{4\sqrt{x} {\left(1\+\frac{9}{4}x\right)}^{3\/2}}$

$\=\frac{3}{\frac{4\sqrt{x} {\left(4\+9 x\right)}^{3\/2}}{4^{3\/2}}}$

$\=\frac{6}{\sqrt{x} {\left(4\+9 x\right)}^{3\/2}}$

Define the function $y\left(x\right)$ 

$y\left(x\right)\=x^{3\/2}$$\stackrel{\text{assign as function}}{\to }$$y$

Compute the curvature for $x\>0$

$\frac{y″\left(x\right)}{{\left(1\+{\left(y\prime \left(x\right)\right)}^2\right)}^{3\/2}}$ = $\frac{3}{4\sqrt{x}{\left(1\+\frac{9}{4}x\right)}^{3\/2}}$$\stackrel{\text{assuming positive}}{\to }$$\frac{6}{\sqrt{x}{\left(4\+9x\right)}^{3\/2}}$$$

Obtain $s\left(x\right)$, the arc-length function on the interval $\left[0\,x\right]$

${\int }_0^x\sqrt{1\+{\left(y\prime \left(t\right)\right)}^2}\mathit{\ⅆ}t$ = $-\frac{8}{27}\+\frac{1}{27}{\left(4\+9x\right)}^{3\/2}$$\stackrel{\text{assign to a name}}{\to }$$S$

Obtain $x\left(s\right)$, the inverse of the arc-length function

$X≔\mathrm{solve}\left(S\=s\,x\right)\left[1\right]$

$\frac{1}{9}{\left(27s\+8\right)}^{2\/3}-\frac{4}{9}$

Obtain $\mathrm{\θ}\left(s\right)\=\arctan \left(y\prime \left(s\right)\right)$, where the derivative is taken with respect to $x$

$\mathrm{\θ}≔\arctan \left(\mathrm{D}\left(y\right)\left(X\right)\right)$

$\arctan \left(\frac{3}{2}\sqrt{\frac{1}{9}{\left(27s\+8\right)}^{2\/3}-\frac{4}{9}}\right)$

Compute $\mathrm{\θ}\prime \left(s\right)$ and replace $s$ with $s\left(x\right)$

$\frac{\ⅆ}{\ⅆ s} \mathrm{\θ}$

$\frac{6}{\sqrt{\frac{1}{9}{\left(27s\+8\right)}^{2\/3}-\frac{4}{9}}\left(27s\+8\right)}$

$\stackrel{\text{evaluate at point}}{\to }$

$\frac{6}{\sqrt{\frac{1}{9}{\left({\left(4\+9x\right)}^{3\/2}\right)}^{2\/3}-\frac{4}{9}}{\left(4\+9x\right)}^{3\/2}}$

$\stackrel{\text{assuming positive}}{\to }$

$\frac{6}{\sqrt{x}{\left(4\+9x\right)}^{3\/2}}$