\[\begin{array}{l}
{\text{Точка }}P\left( {x\left( t \right),y\left( t \right)} \right){\text{ движется по кривой }}y = {x^{\frac{3}{2}}}{\text{ со скоростью }}\left| {v\left( t \right)} \right| = 1. \hfill \\
{\text{Найдите функцию }}x\left( t \right){\text{, если }}x\left( 0 \right) = 0. \hfill \\
\end{array}\]
\[\begin{array}{l}
\int\limits_0^t {\sqrt {{{\left( {x{{\left( t \right)}^\prime }} \right)}^2} + {{\left( {{{\left( {h\left( {x\left( t \right)} \right)} \right)}^\prime }} \right)}^2}} dt} = g\left( t \right) \Leftrightarrow \left[ \begin{array}{l}
{\text{дифференцируем}} \hfill \\
{\text{равенство по }}t \hfill \\
\end{array} \right] \Leftrightarrow \hfill \\
\sqrt {{{\left( {x{{\left( t \right)}^\prime }} \right)}^2} + {{\left( {{{\left( {h\left( {x\left( t \right)} \right)} \right)}^\prime }} \right)}^2}} = g'\left( t \right) \Leftrightarrow \hfill \\
{\left( {x{{\left( t \right)}^\prime }} \right)^2} + {\left( {{{\left( {h\left( {x\left( t \right)} \right)} \right)}^\prime }} \right)^2} = g'\left( t \right) \Leftrightarrow {\left( {x'} \right)^2} + \frac{9}{4}x \cdot {\left( {x'} \right)^2} = 1 \Leftrightarrow \hfill \\
x\left( t \right) = \frac{{{{\left( {27t + 8} \right)}^{\frac{2}{3}}} - 4}}{9}. \hfill \\
\end{array}\]
\[\begin{array}{l}
{\left( {x'} \right)^2} + \frac{9}{4}x \cdot {\left( {x'} \right)^2} = 1 \Leftrightarrow x' = \sqrt {\frac{4}{{4 + 9x}}} \Leftrightarrow {\left( {4 + 9x} \right)^\prime } = \pm \frac{{18}}{{\sqrt {4 + 9x} }} \hfill \\
{\left( {{y^2}} \right)^\prime } = \pm \frac{{18}}{y} \Leftrightarrow yy' = \pm \frac{9}{y} \Leftrightarrow {y^2}dy = \pm 9dx \Leftrightarrow {y^3} = \pm 27x + C \Leftrightarrow \hfill \\
{\left( {\sqrt {4 + 9x} } \right)^3} = \pm 27x + C \hfill \\
\end{array}\]
\[x\left( t \right) = \frac{{{{\left( {27t + 8} \right)}^{\frac{2}{3}}} - 4}}{9}\]