# Find the differential equations of the space curve in which the two families of surfaces u=x^2-y^2=c_1 and v=y^2-z^2=c_2 intersect?

Jul 4, 2018

• $\frac{\mathrm{dx}}{y z} = \frac{\mathrm{dy}}{x z} = \frac{\mathrm{dz}}{x y}$

#### Explanation:

Because $u \left(\boldsymbol{x}\right)$ and $v \left(\boldsymbol{x}\right)$ are level surfaces: $\mathrm{du} = \mathrm{dv} = 0$

• $\mathrm{du} = {\underbrace{2 x}}_{{u}_{x}} \mathrm{dx} - {\underbrace{2 y}}_{{u}_{y}} \mathrm{dy} = 0 q \quad \implies x \mathrm{dx} = y \mathrm{dy} q \quad \mathbb{A}$

Likewise;

• $\mathrm{dv} = 2 y \mathrm{dy} - 2 z \mathrm{dz} = 0 q \quad \implies y \mathrm{dy} = z \mathrm{dz} q \quad \mathbb{B}$

$\mathbb{A} \text{ & } \mathbb{B} \implies x \mathrm{dx} = y \mathrm{dy} = z \mathrm{dz}$

$\therefore \frac{x \mathrm{dx}}{x y z} = \frac{y \mathrm{dy}}{x y z} = \frac{z \mathrm{dz}}{x y z}$

• $\frac{\mathrm{dx}}{y z} = \frac{\mathrm{dy}}{x z} = \frac{\mathrm{dz}}{x y}$