Question

50 баллов!! Ребят, всем привет, очень срочно нужно 3 задание.

Answer 1

$y=\sqrt{2x-x^2}\; ,\; \; x=\frac{1}{4}\; ,\; x=1\\\\\\1+(y')^2=1+\Big (\frac{1}{2\sqrt{2x-x^2}}\cdot (2-2x)\Big )^2=1+\frac{(1-x)^2}{2x-x^2}=\\\\=\frac{2x-x^2+1-2x+x^2}{2x-x^2}=\frac{1}{2x-x^2}=-\frac{1}{x^2-2x}=-\frac{1}{(x-1)^2-1}=\frac{1}{1-(x-1)^2}\\\\\\l=\int\limits^a_b \sqrt{1+(y')^2} \, dx=\int\limits^1_{\frac{1}{4}} \frac{dx}{\sqrt{1-(x-1)^2}}=\int\limits^1_{\frac{1}{4}}\, \frac{d(x-1)}{\sqrt{1-(x-1)^2}}=\\\\=arcsin(x-1)\Big |_{\frac{1}{4}}^1=arcsin0-arcsin(-\frac{3}{4})=arcsin\frac{3}{4}$

$P.S.\; \; y=\sqrt{2x-x^2}\; \; \to \; \; y^2=2x-x^2\; ,\\\\x^2-2x+y^2=0\\\\(x-1)^2+y^2=1\; \; -\; \; okryznost\; centr\; (1,0)\; ,\; R=1\; .$