Question

Find the differential equation of the space curve in which the two families of surfaces ax2+by2+cz2=u and a2x2+b2y2+c2z2=v intersect .

Answer 1

Answer on Question #40388 – Math - Differential Calculus

Find the differential equation of the space curve in which the two families of surfaces $ax^2 + by^2 + cz^2 = u$ and $a^2x^2 + b^2y^2 + c^2z^2 = v$ intersect.

Solution.

We have the two families of surfaces:

$$u(x, y, z) = ax^2 + by^2 + cz^2 = 0, \quad v(x, y, z) = a^2x^2 + b^2y^2 + c^2z^2 = 0$$

Let's find the differential equations of the space curve. Form the matrix of partial derivatives:

$$\left( \begin{array}{ccc} u_x' & u_y' & u_z' \\ v_x' & v_y' & v_z' \end{array} \right) = \left( \begin{array}{ccc} 2ax & 2by & 2cz \\ 2a^2x & 2b^2y & 2c^2z \end{array} \right).$$

Then find determinants of the matrix:

$$\left| \begin{array}{cc} u_x' & u_y' \\ v_x' & v_y' \end{array} \right| = \left| \begin{array}{cc} 2ax & 2by \\ 2a^2x & 2b^2y \end{array} \right| = 4ab^2xy - 4a^2bxy = 4abxy(b - a) = f_1$$ $$\left| \begin{array}{cc} u_y' & u_z' \\ v_y' & v_z' \end{array} \right| = \left| \begin{array}{cc} 2by & 2cz \\ 2b^2y & 2c^2z \end{array} \right| = 4bc^2yz - 4b^2cyz = 4bcyz(c - b) = f_2$$ $$\left| \begin{array}{cc} u_z' & u_x' \\ v_z' & v_x' \end{array} \right| = \left| \begin{array}{cc} 2cz & 2ax \\ 2c^2z & 2a^2x \end{array} \right| = 4ca^2zx - 4c^2azx = 4cazx(a - c) = f_3$$

So, we have the differential equations:

$$\left\{ \begin{array}{l} 4abxy(b - a) = f_1 \\ 4bcyz(c - b) = f_2 \\ 4cazx(a - c) = f_3 \end{array} \right.$$

that completely describe the space curve.