Another way of deriving the equation of motion (5.16) is to use Lagrange's equations. Express the kinetic energy \(\frac{1}{2} m(\mathrm{~d} \boldsymbol{r} / \mathrm{d} t)^{2}\) in terms of \((x, y, z)\), and show that Lagrange's equations (3.44) reproduce (5.16) for the case where the force is conservative.

First, we'll express the given kinetic energy \(\frac{1}{2}m(\frac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t})^2\) in terms of \((x, y, z)\). We know that \(\boldsymbol{r} = x\boldsymbol{\hat{i}} + y\boldsymbol{\hat{j}} + z\boldsymbol{\hat{k}}\). Then, we can differentiate \(\boldsymbol{r}\) with respect to time and get the velocity vector \(\boldsymbol{v} = \frac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t} = \frac{\mathrm{d}x}{\mathrm{d}t}\boldsymbol{\hat{i}} + \frac{\mathrm{d}y}{\mathrm{d}t}\boldsymbol{\hat{j}} + \frac{\mathrm{d}z}{\mathrm{d}t}\boldsymbol{\hat{k}}\). Now, we can find the square of the velocity vector's magnitude: \((\boldsymbol{v})^2 = \left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}z}{\mathrm{d}t}\right)^2\) Then, the kinetic energy in terms of \((x, y, z)\) is given by \(T = \frac{1}{2}m\left(\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}z}{\mathrm{d}t}\right)^2\right)\)

Lagrange's equations are given by \(\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial T}{\partial \dot{q}_i}\right) - \frac{\partial T}{\partial q_i} = Q_i\), where \(q_i\) represents the generalized coordinates (in this case, \(x, y, z\)) and \(Q_i\) are the generalized forces. So, we need to calculate \(\frac{\partial T}{\partial \dot{x}}, \frac{\partial T}{\partial \dot{y}}, \frac{\partial T}{\partial \dot{z}}, \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z}\). Since \(T\) is dependent only on the velocities, we have \(\frac{\partial T}{\partial x} = \frac{\partial T}{\partial y} = \frac{\partial T}{\partial z} = 0\). And we can determine the partial derivatives with respect to the velocities as \(\frac{\partial T}{\partial \dot{x}} = m\dot{x}, \frac{\partial T}{\partial \dot{y}} = m\dot{y}, \frac{\partial T}{\partial \dot{z}} = m\dot{z}\). Now, let's write Lagrange's equations for each coordinate: 1. For \(x\): \(\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial T}{\partial \dot{x}}\right) - \frac{\partial T}{\partial x} = m\ddot{x} = Q_x\) 2. For \(y\): \(\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial T}{\partial \dot{y}}\right) - \frac{\partial T}{\partial y} = m\ddot{y} = Q_y\) 3. For \(z\): \(\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial T}{\partial \dot{z}}\right) - \frac{\partial T}{\partial z} = m\ddot{z} = Q_z\)

When the force is conservative, we can express the force as the negative gradient of a potential energy function \(U(x, y, z)\), i.e., \(\boldsymbol{F} = -\boldsymbol{\nabla}U\). Thus for each component of the force, we have \(Q_x = -\frac{\partial U}{\partial x}\), \(Q_y = -\frac{\partial U}{\partial y}\), and \(Q_z = -\frac{\partial U}{\partial z}\). Let's substitute these into the Lagrange's equations from Step 2: 1. \(m\ddot{x} = -\frac{\partial U}{\partial x}\) 2. \(m\ddot{y} = -\frac{\partial U}{\partial y}\) 3. \(m\ddot{z} = -\frac{\partial U}{\partial z}\) These equations are equivalent to equation (5.16) for conservative forces. Therefore, we have shown that Lagrange's equations can be used to derive the equation of motion.