1. **Problem:** Prove that $$\frac{d}{dt} e_{\rho} = \dot{\phi} e_{\phi}, \quad \frac{d}{dt} e_{\phi} = -\dot{\phi} e_{\rho}$$ where dots denote differentiation with respect to time $t$. 2. **Understanding the unit vectors:** In cylindrical coordinates, the unit vectors $e_{\rho}$ and $e_{\phi}$ depend on the angle $\phi$ and change direction as $\phi$ changes with time. 3. **Express $e_{\rho}$ and $e_{\phi}$ in Cartesian coordinates:** $$e_{\rho} = \cos \phi \, \mathbf{i} + \sin \phi \, \mathbf{j}$$ $$e_{\phi} = -\sin \phi \, \mathbf{i} + \cos \phi \, \mathbf{j}$$ 4. **Differentiate $e_{\rho}$ with respect to time:** $$\frac{d}{dt} e_{\rho} = \frac{d}{dt} (\cos \phi \, \mathbf{i} + \sin \phi \, \mathbf{j}) = -\sin \phi \, \dot{\phi} \, \mathbf{i} + \cos \phi \, \dot{\phi} \, \mathbf{j} = \dot{\phi} (-\sin \phi \, \mathbf{i} + \cos \phi \, \mathbf{j}) = \dot{\phi} e_{\phi}$$ 5. **Differentiate $e_{\phi}$ with respect to time:** $$\frac{d}{dt} e_{\phi} = \frac{d}{dt} (-\sin \phi \, \mathbf{i} + \cos \phi \, \mathbf{j}) = -\cos \phi \, \dot{\phi} \, \mathbf{i} - \sin \phi \, \dot{\phi} \, \mathbf{j} = -\dot{\phi} (\cos \phi \, \mathbf{i} + \sin \phi \, \mathbf{j}) = -\dot{\phi} e_{\rho}$$ 6. **Conclusion:** We have shown that $$\frac{d}{dt} e_{\rho} = \dot{\phi} e_{\phi}, \quad \frac{d}{dt} e_{\phi} = -\dot{\phi} e_{\rho}$$ which completes the proof.