Решебник по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 94

\[\boxed{\mathbf{94}.}\]

\[T_{0} = 100{^\circ};\ \ T_{1} = 20{^\circ}:\]

\[\frac{\text{dT}}{\text{dt}} = - k(T - 20)\]

\[\frac{d(T - 20)}{\text{dt}} = - k(T - 20).\]

\[\int_{}^{}\frac{d(T - 20)}{\text{dt}} = \int_{}^{}{- k(T - 20)}\]

\[\int_{}^{}\frac{d(T - 20)}{\text{dt}} = - k\int_{}^{}{dt + C}\]

\[\ln|T - 20| = - kt + C\]

\[T - 20 = e^{- kt + C}\]

\[T = 20 + Ae^{- kt}.\]

\[T(0) = 20 + Ae^{- ok} = 100;\]

\[T(12) = 20 + Ae^{- 10k} = 60.\]

\[20 + Ae^{- ok} = 100\]

\[A = 80.\]

\[Подставим:\]

\[20 + Ae^{- 10k} = 60\]

\[20 + 80 \cdot e^{- 10k} = 60\]

\[80e^{- 10k} = 40\]

\[e^{- 10k} = \frac{1}{2}\]

\[e^{- k} = \left( \frac{1}{2} \right)^{\frac{1}{10}}.\]

\[Время,\ через\ которое\ самовар\ \]

\[остынет\ до\ 30{^\circ}:\]

\[T(t) = 20 + 80 \cdot e^{- tk} = 30\]

\[80e^{- tk} = 10\]

\[80 \cdot \left( \frac{1}{2}^{0,1} \right)^{t} = 10\]

\[\left( \frac{1}{2}^{0,1} \right)^{t} = \frac{1}{8}\]

\[\left( \frac{1}{2}^{0,1} \right)^{t} = \left( \frac{1}{2} \right)^{3}\]

\[0,1t = 3\]

\[t = 30\ (с).\]

\[Ответ:через\ 30\ с.\]