№1841
\[\begin{array}{l}
{\text{Пусть }}\kappa \left( t \right) = \frac{{x'y'' - x''y'}}{{{{\left( {{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2}} \right)}^{\frac{3}{2}}}}}{\text{ - кривизна регулярной кривой }}\gamma , \hfill \\
I = \int\limits_\gamma {\kappa \left( t \right)d\gamma } = \int\limits_a^b {\kappa \left( t \right)\sqrt {{{\left( {x'\left( t \right)} \right)}^2} + {{\left( {y'\left( t \right)} \right)}^2}} dt} , \hfill \\
{l_1}{\text{ - касательная к }}\gamma {\text{ при }}t = a,{\text{ }}{l_2}{\text{ - касательная к }}\gamma {\text{ при }}t = b. \hfill \\
{\text{Докажите}}{\text{, что }}\left| I \right|{\text{ равен углу между }}{l_1}{\text{ и }}{l_2}. \hfill \\
\end{array}\]
Your solution
smart: 5
y = f\left( x \right) \Rightarrow \kappa \left( x \right) = \frac{{y''}}{{{{\left( {1 + {{\left( {y'} \right)}^2}} \right)}^{\frac{3}{2}}}}} \Rightarrow \hfill \\
\int\limits_a^b {\kappa \left( x \right)ds} = \int\limits_a^b {\frac{{y''}}{{{{\left( {1 + {{\left( {y'} \right)}^2}} \right)}^{\frac{3}{2}}}}}\sqrt {1 + {{\left( {y'} \right)}^2}} dx} = \hfill \\
\int\limits_a^b {\frac{{y''}}{{1 + {{\left( {y'} \right)}^2}}}dx} = \int\limits_a^b {\frac{{dy'}}{{1 + {{\left( {y'} \right)}^2}}}} = \operatorname{arctg} \left( {y'\left( b \right)} \right) - \operatorname{arctg} \left( {y'\left( a \right)} \right). \hfill \\
\end{array} \]