1. **Problem statement:** Find the differential equations of the space curves formed by the intersection of the two families of surfaces given by $$u = x^2 + y^2 + z^2 = c_1$$ and $$v = x + z = c_2$$ where $c_1$ and $c_2$ are constants. 2. **Step 1: Understand the problem** The space curves are the intersection of surfaces $u = c_1$ and $v = c_2$. To find the differential equations of these curves, we use the fact that along the curve both $u$ and $v$ are constant. 3. **Step 2: Use total differentials** Since $u$ and $v$ are constant on the curve, their total differentials must be zero: $$du = 0, \quad dv = 0$$ Calculate $du$ and $dv$: $$du = \frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy + \frac{\partial u}{\partial z} dz$$ $$dv = \frac{\partial v}{\partial x} dx + \frac{\partial v}{\partial y} dy + \frac{\partial v}{\partial z} dz$$ 4. **Step 3: Compute partial derivatives** For $u = x^2 + y^2 + z^2$: $$\frac{\partial u}{\partial x} = 2x, \quad \frac{\partial u}{\partial y} = 2y, \quad \frac{\partial u}{\partial z} = 2z$$ For $v = x + z$: $$\frac{\partial v}{\partial x} = 1, \quad \frac{\partial v}{\partial y} = 0, \quad \frac{\partial v}{\partial z} = 1$$ 5. **Step 4: Write the differential equations** From $du=0$: $$2x dx + 2y dy + 2z dz = 0$$ From $dv=0$: $$dx + 0 \cdot dy + dz = 0 \implies dx + dz = 0$$ 6. **Step 5: Simplify and find relations** From $dv=0$: $$dz = -dx$$ Substitute into $du=0$: $$2x dx + 2y dy + 2z dz = 0 \implies 2x dx + 2y dy + 2z (-dx) = 0$$ $$2x dx + 2y dy - 2z dx = 0$$ $$2(x - z) dx + 2y dy = 0$$ Divide both sides by 2: $$(x - z) dx + y dy = 0$$ 7. **Step 6: Write the system of differential equations for the space curves** $$\begin{cases} dz = -dx \\ (x - z) dx + y dy = 0 \end{cases}$$ **Final answer:** The differential equations of the space curves formed by the intersection of the surfaces are $$dz = -dx$$ and $$(x - z) dx + y dy = 0$$