1. **State the problem:** Find the differential equations of the space curves formed by the intersection of the two families of surfaces given by: $$u = x^2 + y^2 + z^2 = c_1$$ $$v = x + z = c_2$$ where $c_1$ and $c_2$ are constants. 2. **Understand the surfaces:** - The first family $u = c_1$ represents spheres centered at the origin with radius $\sqrt{c_1}$. - The second family $v = c_2$ represents planes with equation $x + z = c_2$. 3. **Find gradients:** The curves of intersection satisfy both surfaces simultaneously. The tangent vector to the curve is perpendicular to both gradients $\nabla u$ and $\nabla v$. Calculate gradients: $$\nabla u = \left(\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial u}{\partial z}\right) = (2x, 2y, 2z)$$ $$\nabla v = \left(\frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}, \frac{\partial v}{\partial z}\right) = (1, 0, 1)$$ 4. **Find direction of tangent vector:** The tangent vector $\mathbf{T}$ is perpendicular to both gradients, so: $$\mathbf{T} = \nabla u \times \nabla v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2x & 2y & 2z \\ 1 & 0 & 1 \end{vmatrix}$$ Calculate the cross product: $$\mathbf{T} = \mathbf{i}(2y \cdot 1 - 2z \cdot 0) - \mathbf{j}(2x \cdot 1 - 2z \cdot 1) + \mathbf{k}(2x \cdot 0 - 2y \cdot 1)$$ $$= (2y)\mathbf{i} - (2x - 2z)\mathbf{j} - (2y)\mathbf{k}$$ So, $$\mathbf{T} = (2y, -(2x - 2z), -2y) = (2y, 2z - 2x, -2y)$$ 5. **Write the system of differential equations:** The tangent vector components correspond to derivatives with respect to a parameter $t$ along the curve: $$\frac{dx}{dt} = 2y$$ $$\frac{dy}{dt} = 2z - 2x$$ $$\frac{dz}{dt} = -2y$$ 6. **Simplify by dividing by 2:** $$\frac{dx}{dt} = 2y \Rightarrow \frac{dx}{dt} = 2y$$ Divide both sides by 2: $$\frac{\cancel{2}y}{\cancel{2}} = y$$ Similarly for others: $$\frac{dx}{dt} = 2y \Rightarrow \frac{dx}{dt} = 2y$$ $$\frac{dy}{dt} = 2z - 2x \Rightarrow \frac{dy}{dt} = 2(z - x)$$ $$\frac{dz}{dt} = -2y$$ Dividing all by 2: $$\frac{dx}{dt} = y$$ $$\frac{dy}{dt} = z - x$$ $$\frac{dz}{dt} = -y$$ 7. **Final differential equations of the space curves:** $$\boxed{\frac{dx}{dt} = y, \quad \frac{dy}{dt} = z - x, \quad \frac{dz}{dt} = -y}$$ These equations describe the tangent direction of the curves formed by the intersection of the given surfaces.