The general solution of the surfaces which are perpendicular to the family of surfaces
z² = kxy, k ∈ $\mathbb{R}$
is

Question

The general solution of the surfaces which are perpendicular to the family of surfacesz2 = kxy, k &isin; $\mathbb{R}$is

Prepp Team · Accepted Answer

Orthogonal Surfaces Explained

We are asked to find the general solution of the surfaces that are perpendicular (orthogonal) to the given family of surfaces $z^2 = kxy$, where $k$ is a real constant.

A family of surfaces is defined by an equation of the form $f(x, y, z, k) = 0$. In this case, the given family is $f(x, y, z) = z^2 - kxy = 0$. The value of $k$ varies for different surfaces in the family.

The normal vector to a surface $f(x, y, z) = 0$ is given by its gradient $\nabla f$. For the given family, the gradient is:

$\nabla f = \left< \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right> = \left< \frac{\partial}{\partial x}(z^2 - kxy), \frac{\partial}{\partial y}(z^2 - kxy), \frac{\partial}{\partial z}(z^2 - kxy) \right> = \left< -ky, -kx, 2z \right>$

Let the orthogonal surfaces be represented by $g(x, y, z) = C$, where $C$ is a constant. The normal vector to these orthogonal surfaces is $\nabla g = \left< \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z} \right>$.

For the surfaces to be orthogonal, the normal vector of the given family ($\nabla f$) must be perpendicular to the normal vector of the orthogonal family ($\nabla g$). Their dot product must be zero:

$\nabla f \cdot \nabla g = 0$

$\left< -ky, -kx, 2z \right> \cdot \left< \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z} \right> = 0$

$-ky \frac{\partial g}{\partial x} - kx \frac{\partial g}{\partial y} + 2z \frac{\partial g}{\partial z} = 0$

This is a partial differential equation (PDE) for $g$. To solve it, we need to eliminate $k$. From the original equation of the family, $z^2 = kxy$, we can write $k = \frac{z^2}{xy}$ (assuming $x, y, z \neq 0$). Substituting this into the PDE:

$-\left(\frac{z^2}{xy}\right) y \frac{\partial g}{\partial x} - \left(\frac{z^2}{xy}\right) x \frac{\partial g}{\partial y} + 2z \frac{\partial g}{\partial z} = 0$

$-\frac{z^2}{x} \frac{\partial g}{\partial x} - \frac{z^2}{y} \frac{\partial g}{\partial y} + 2z \frac{\partial g}{\partial z} = 0$

Assuming $z \neq 0$, we can divide by $z$:

$-\frac{z}{x} \frac{\partial g}{\partial x} - \frac{z}{y} \frac{\partial g}{\partial y} + 2 \frac{\partial g}{\partial z} = 0$

Multiplying by $xy$ to clear denominators:

$-zy \frac{\partial g}{\partial x} - zx \frac{\partial g}{\partial y} + 2xy \frac{\partial g}{\partial z} = 0$

This is a first-order quasi-linear PDE of the form $P \frac{\partial g}{\partial x} + Q \frac{\partial g}{\partial y} + R \frac{\partial g}{\partial z} = 0$, where $P = -zy$, $Q = -zx$, and $R = 2xy$.

Solving the PDE using Characteristics

We solve this PDE using the method of characteristics. The characteristic equations are given by:

$\frac{dx}{P} = \frac{dy}{Q} = \frac{dz}{R}$

$\frac{dx}{-zy} = \frac{dy}{-zx} = \frac{dz}{2xy}$

Finding First Integrals

We need to find two independent first integrals from these equations.

First Integral

Consider the first two ratios:

$\frac{dx}{-zy} = \frac{dy}{-zx}$

Assuming $z \neq 0$, we can simplify:

$\frac{dx}{y} = \frac{dy}{x}$

$x \, dx = y \, dy$

Integrate both sides:

$\int x \, dx = \int y \, dy$

$\frac{x^2}{2} = \frac{y^2}{2} + C'$

$x^2 - y^2 = C_1$

So, $u = x^2 - y^2$ is a first integral.

Second Integral

Consider the first and third ratios:

$\frac{dx}{-zy} = \frac{dz}{2xy}$

Assuming $y \neq 0$, we can simplify:

$\frac{dx}{-z} = \frac{dz}{2x}$

$2x \, dx = -z \, dz$

$2x \, dx + z \, dz = 0$

Integrate both sides:

$\int 2x \, dx + \int z \, dz = C''$

$x^2 + \frac{z^2}{2} = C'''$

$2x^2 + z^2 = C_2$

So, $v = 2x^2 + z^2$ is another first integral.

These two integrals, $u = x^2 - y^2$ and $v = 2x^2 + z^2$, are independent.

General Solution of Orthogonal Surfaces

The general solution of the PDE $P \frac{\partial g}{\partial x} + Q \frac{\partial g}{\partial y} + R \frac{\partial g}{\partial z} = 0$ is given by $\varphi(u, v) = 0$, where $\varphi$ is an arbitrary differentiable function and $u, v$ are independent first integrals.

Therefore, the general solution of the surfaces orthogonal to $z^2 = kxy$ is:

$\varphi(x^2 - y^2, 2x^2 + z^2) = 0$

Here, $\varphi \in C^1(\mathbb{R}^2)$ indicates that $\varphi$ is a continuously differentiable function with two variables, mapping from $\mathbb{R}^2$ to $\mathbb{R}$.

Comparing this with the given options, we find that it matches option 3.

Prepp Team · Answer

&varphi; (x2 - y2 , x z) = 0, &varphi; &isin; C1  ( $\mathbb{R}$2 )

Prepp Team · Answer

&varphi; (x2 - y2 , x2 +  z2) = 0, &varphi; &isin; C1  ( $\mathbb{R}$2 )

Prepp Team · Answer

&varphi; (x2 + y2 , 3x2 -  z2) = 0, &varphi; &isin; C1  ( $\mathbb{R}$2 )