How do I find the arc length of
$x=\sqrt{y-y^2}+\sin^{-1}\sqrt{y}$
From $y=0 \;to\;x= {3\over 4}$
The answer is =$\sqrt3$
. But I got the incorrect answer. Anyone explain it step by step. Thanks in advance.
$\displaystyle L=\int\limits _c^d \sqrt {1+\left ( \dfrac {\mathrm dx}{\mathrm dy}\right)^2}\,\mathrm dy\\\dfrac{\mathrm dx}{\mathrm dx}=\dfrac {1-2y}{2\sqrt {y-y^2}}+\dfrac {1}{2\sqrt y \;\sqrt {1-y}}\\=\dfrac {2-2y}{2\sqrt y \;\sqrt {1-y}}\\=\sqrt {\dfrac {1-y}{y}}\\\Rightarrow 1+\Big(\dfrac{\mathrm dx}{\mathrm dy}\Big)^2=1+\dfrac{1-y}{y}=\dfrac {1}{y}$
$\therefore\;L=\displaystyle\int\limits _0^{3/4} \sqrt {\dfrac {1}{y}}\,\mathrm dy\\=2\Bigg[\sqrt y\,\Bigg]_0^{3/4}=2\left[\,\dfrac{\sqrt3}{\sqrt4}-0\right]\\=\sqrt{3}$
Here is the similar question with the complete solution:
Arc length