Задача 61688 Вычислить [m]d^2y/dx^2[/m]...
Условие
Решение
[m]y``_{x}=(\frac{y`_{t}}{x`_{t}})`_{x}=\frac{(\frac{y`_{t}}{x`_{t}})`_{t}}{x`_{t}}=\frac{y``_{tt}\cdot x`_{t}-y`_{t}\cdot x``_{tt}}{(x`_{t})^3}[/m]
[m]\left\{\begin {matrix}x`_{t}=(\sqrt{1-t^2})`_{t}\\y`_{t}=(\frac{t}{\sqrt{1-t^2}})`_{t}\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}x`_{t}=\frac{1}{2\sqrt{1-t^2}}\cdot (1-t^2)`_{t}\\y`_{t}=\frac{(t)`\cdot \sqrt{1-t^2}-t\cdot (\sqrt{1-t^2})`}{(\sqrt{1-t^2})^2}\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}x`_{t}=\frac{-t}{\sqrt{1-t^2}}\\y`_{t}=\frac{\sqrt{1-t^2}-t\cdot \frac{-t}{\sqrt{1-t^2}}}{1-t^2}\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}x`_{t}=\frac{-t}{\sqrt{1-t^2}}\\y`_{t}=\frac{1-t^2+t^2}{\sqrt{1-t^2}\cdot (1-t^2)}\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}x`_{t}=\frac{-t}{\sqrt{1-t^2}}\\y`_{t}=\frac{1}{\sqrt{1-t^2}\cdot (1-t^2)}\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}x``_{t}=(\frac{-t}{\sqrt{1-t^2}})`_{t}\\y``_{t}=(\frac{1}{\sqrt{1-t^2}\cdot (1-t^2)})`_{t}\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}x``_{t}=\frac{-1}{\sqrt{1-t^2}\cdot (1-t^2)}\\y``_{t}=( (1-t^2)^{-\frac{3}{2}})`_{t}\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}x``_{t}=\frac{-1}{\sqrt{1-t^2}\cdot (1-t^2)}\\y``_{t}=-\frac{3}{2}( (1-t^2)^{-\frac{3}{2}-1}\cdot (1-t^2)`_{t}\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}x``_{t}=\frac{-1}{\sqrt{1-t^2}\cdot (1-t^2)}\\y``_{t}=-\frac{3t}{2\sqrt{1-t^2}\cdot (1-t^2)^2}\end {matrix}\right.[/m]
и подставляем в формулу...