Функции $y$ и $z$ независимой переменной $x$ заданы системой уравнений $x^2 + y^2 - z^2 = 0$ , $x^2 + 2y^2 + 3z^2 = 1$ . Найти $dy$ , $dz$ , $d^{2}y$ , $d^{2}z$ .

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Algebra 1

Transformations of quadratic functions

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Q. Функции

y

z

независимой переменной

x

заданы системой уравнений

x^2 + y^2 - z^2 = 0

x^2 + 2y^2 + 3z^2 = 1

. Найти

dy

dz

d^{2}y

d^{2}z

Differentiate Equations: Differentiate both equations with respect to $x$ to find $\frac{dy}{dx}$ and $\frac{dz}{dx}$ . For the first equation: $2x + 2y\frac{dy}{dx} - 2z\frac{dz}{dx} = 0$ . For the second equation: $2x + 4y\frac{dy}{dx} + 6z\frac{dz}{dx} = 0$ .
Solve Linear Equations: Solve the system of linear equations for $\frac{dy}{dx}$ and $\frac{dz}{dx}$ . Let's call $\frac{dy}{dx} = p$ and $\frac{dz}{dx} = q$ for simplicity. We have: $2x + 2yp - 2zq = 0$ , ( $1$ ) $2x + 4yp + 6zq = 0$ . ( $2$ )
Eliminate Variable: Multiply equation ( $1$ ) by $2$ and subtract from equation ( $2$ ) to eliminate $x$ . $\newline$ $4x + 4yp - 4zq = 0, \quad (3)$ $\newline$ $2x + 4yp + 6zq = 0. \quad (4)$ $\newline$ Subtracting $(3)$ from $(4)$ gives: $\newline$ $2x + 4yp + 6zq - (4x + 4yp - 4zq) = 0 - 0,$ $\newline$ $2zq + 10zq = 0,$ $\newline$ $12zq = 0,$ $\newline$ $q = \frac{dz}{dx} = 0.$
Substitute and Solve: Substitute $q = 0$ back into equation $(1)$ to find $p$ . $\newline$ $2x + 2yp - 2z(0) = 0,$ $\newline$ $2x + 2yp = 0,$ $\newline$ $p = \frac{dy}{dx} = -\frac{x}{y}.$
Find Second Derivatives: Differentiate $p$ and $q$ with respect to $x$ to find $d^{2}y$ and $d^{2}z$ . Since $q = \frac{dz}{dx} = 0$ , $d^{2}z = \frac{d(\frac{dz}{dx})}{dx} = 0$ . For $p = \frac{dy}{dx} = -\frac{x}{y}$ , use the quotient rule: $\frac{d(\frac{u}{v})}{dx} = \frac{v(\frac{du}{dx}) - u(\frac{dv}{dx})}{v^{2}}$ . $\frac{d(-\frac{x}{y})}{dx} = \frac{y(-1) - (-x)(\frac{dy}{dx})}{y^{2}}$ , $q$ $0$ , $q$ $1$ .