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Доказать, что касательные плоскости к конусу $z = xf(y/x)$ проходят через его вершину

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Find derivatives using the quotient rule II

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Q. Доказать, что касательные плоскости к конусу

z = xf(y/x)

проходят через его вершину

Find Tangent Plane Equation: Let's first find the equation of the tangent plane to the cone at a point $(x_0, y_0, z_0)$ where $z_0 = x_0f(y_0/x_0)$ . To find the equation of the tangent plane, we need to calculate the partial derivatives of $z$ with respect to $x$ and $y$ at the point $(x_0, y_0)$ .
Calculate Partial Derivatives: The partial derivative of $z$ with respect to $x$ is given by: $\newline$ $\frac{\partial z}{\partial x} = f\left(\frac{y}{x}\right) + x f'\left(\frac{y}{x}\right)\left(-\frac{y}{x^2}\right) = f\left(\frac{y}{x}\right) - \left(\frac{y}{x}\right)f'\left(\frac{y}{x}\right)$ $\newline$ We evaluate this derivative at the point $(x_0, y_0)$ .
Equation of Tangent Plane: The partial derivative of $z$ with respect to $y$ is given by: $\frac{\partial z}{\partial y} = x(\frac{1}{x})f'(\frac{y}{x}) = f'(\frac{y}{x})$ We evaluate this derivative at the point $(x_0, y_0)$ .
Check Vertex Satisfaction: The equation of the tangent plane at the point $(x_0, y_0, z_0)$ is then: $\newline$ $z - z_0 = \left(\frac{\partial z}{\partial x}\right)\bigg|_{(x_0, y_0)}(x - x_0) + \left(\frac{\partial z}{\partial y}\right)\bigg|_{(x_0, y_0)}(y - y_0)$ $\newline$ Substituting the partial derivatives we found, we get: $\newline$ $z - x_0f\left(\frac{y_0}{x_0}\right) = \left[f\left(\frac{y_0}{x_0}\right) - \left(\frac{y_0}{x_0}\right)f'\left(\frac{y_0}{x_0}\right)\right](x - x_0) + f'\left(\frac{y_0}{x_0}\right)(y - y_0)$
Simplify Equation: Now, we need to check if the vertex of the cone, which is the origin $(0, 0, 0)$ , satisfies the equation of the tangent plane. If we substitute $x = 0$ , $y = 0$ , and $z = 0$ into the equation of the tangent plane, we should get a true statement if the tangent plane passes through the vertex.
Conclusion: Substituting $x = 0$ , $y = 0$ , and $z = 0$ into the equation of the tangent plane, we get: $\newline$ $0 - x_0f(y_0/x_0) = [f(y_0/x_0) - (y_0/x_0)f'(y_0/x_0)](0 - x_0) + f'(y_0/x_0)(0 - y_0)$ $\newline$ Simplifying, we get: $\newline$ $-x_0f(y_0/x_0) = -x_0[f(y_0/x_0) - (y_0/x_0)f'(y_0/x_0)] - y_0f'(y_0/x_0)$
Conclusion: Substituting $x = 0$ , $y = 0$ , and $z = 0$ into the equation of the tangent plane, we get: $\newline$ $0 - x_0f(y_0/x_0) = [f(y_0/x_0) - (y_0/x_0)f'(y_0/x_0)](0 - x_0) + f'(y_0/x_0)(0 - y_0)$ $\newline$ Simplifying, we get: $\newline$ $-x_0f(y_0/x_0) = -x_0[f(y_0/x_0) - (y_0/x_0)f'(y_0/x_0)] - y_0f'(y_0/x_0)$ Notice that the terms on the right side of the equation can be grouped to cancel each other out: $\newline$ $-x_0f(y_0/x_0) = -x_0f(y_0/x_0) + x_0(y_0/x_0)f'(y_0/x_0) - y_0f'(y_0/x_0)$ $\newline$ The terms $x_0(y_0/x_0)f'(y_0/x_0)$ and $-y_0f'(y_0/x_0)$ cancel each other out, leaving us with: $\newline$ $-x_0f(y_0/x_0) = -x_0f(y_0/x_0)$
Conclusion: Substituting $x = 0$ , $y = 0$ , and $z = 0$ into the equation of the tangent plane, we get: $\newline$ $0 - x_0f(y_0/x_0) = [f(y_0/x_0) - (y_0/x_0)f'(y_0/x_0)](0 - x_0) + f'(y_0/x_0)(0 - y_0)$ $\newline$ Simplifying, we get: $\newline$ $-x_0f(y_0/x_0) = -x_0[f(y_0/x_0) - (y_0/x_0)f'(y_0/x_0)] - y_0f'(y_0/x_0)$ Notice that the terms on the right side of the equation can be grouped to cancel each other out: $\newline$ $-x_0f(y_0/x_0) = -x_0f(y_0/x_0) + x_0(y_0/x_0)f'(y_0/x_0) - y_0f'(y_0/x_0)$ $\newline$ The terms $x_0(y_0/x_0)f'(y_0/x_0)$ and $-y_0f'(y_0/x_0)$ cancel each other out, leaving us with: $\newline$ $-x_0f(y_0/x_0) = -x_0f(y_0/x_0)$ Since both sides of the equation are equal, we have shown that the origin $(0, 0, 0)$ satisfies the equation of the tangent plane. Therefore, the tangent plane to the cone $y = 0$ $0$ at any point $y = 0$ $1$ passes through the vertex of the cone.

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