To prove that the total energy function $E(t)$ of the particle is constant for all time $t$, we need to show that the derivative of $E(t)$ with respect to $t$ is zero, i.e., $\frac{dE(t)}{dt} = 0$.

Step 1: Define the energy function $E(t)$

The total energy function $E(t)$ is given by: $E(t) = V(\vec{r}(t)) + \frac{m}{2} || \vec{r}'(t) ||^2$ Here:

• $V(\vec{r}(t))$ is the potential energy, which depends on the position vector $\vec{r}(t) = (x(t), y(t), z(t))$.
• $\frac{m}{2} || \vec{r}'(t) ||^2$ is the kinetic energy, where $\vec{r}'(t)$ is the velocity vector and $m$ is the mass of the particle.

Step 2: Differentiate $E(t)$ with respect to $t$

We need to differentiate the total energy function with respect to time $t$. Using the chain rule for differentiation, we get: $\frac{dE(t)}{dt} = \frac{d}{dt} \left( V(\vec{r}(t)) \right) + \frac{d}{dt} \left( \frac{m}{2} || \vec{r}'(t) ||^2 \right)$

2.1: Differentiate the potential energy term $V(\vec{r}(t))$

By the chain rule: $\frac{d}{dt} \left( V(\vec{r}(t)) \right) = \nabla V(\vec{r}(t)) \cdot \vec{r}'(t)$ where $\nabla V(\vec{r}(t))$ is the gradient of the potential function $V$, and $\vec{r}'(t)$ is the velocity vector.

2.2: Differentiate the kinetic energy term $\frac{m}{2} || \vec{r}'(t) ||^2$

The kinetic energy is $\frac{m}{2} || \vec{r}'(t) ||^2$, where $|| \vec{r}'(t) ||^2 = \vec{r}'(t) \cdot \vec{r}'(t)$. Differentiating this term with respect to $t$ gives: $\frac{d}{dt} \left( \frac{m}{2} || \vec{r}'(t) ||^2 \right) = m \vec{r}'(t) \cdot \vec{r}''(t)$ where $\vec{r}''(t)$ is the acceleration vector.

Step 3: Use Newton's Second Law

According to Newton's Second Law, the force acting on the particle is: $m \vec{r}''(t) = \vec{F}(\vec{r}(t))$ where $\vec{F}(\vec{r}(t)) = -\nabla V(\vec{r}(t))$. Substituting this into the equation for the kinetic energy derivative, we get: $m \vec{r}'(t) \cdot \vec{r}''(t) = \vec{r}'(t) \cdot \vec{F}(\vec{r}(t)) = \vec{r}'(t) \cdot \left( -\nabla V(\vec{r}(t)) \right)$ Thus, we have: $m \vec{r}'(t) \cdot \vec{r}''(t) = - \vec{r}'(t) \cdot \nabla V(\vec{r}(t))$

Step 4: Combine the results

Now, substitute the results from Steps 2.1 and 2.2 into the expression for $\frac{dE(t)}{dt}$: $\frac{dE(t)}{dt} = \nabla V(\vec{r}(t)) \cdot \vec{r}'(t) + m \vec{r}'(t) \cdot \vec{r}''(t)$ Using the fact that $m \vec{r}''(t) = - \nabla V(\vec{r}(t))$, this becomes: $\frac{dE(t)}{dt} = \nabla V(\vec{r}(t)) \cdot \vec{r}'(t) - \nabla V(\vec{r}(t)) \cdot \vec{r}'(t) = 0$

Step 5: Conclusion

Since $\frac{dE(t)}{dt} = 0$, the total energy $E(t)$ is constant for all $t$.

Thus, we have proven that the total energy of the particle remains constant over time.

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Would you like more details or further explanation?

Here are five related questions:

1. What is the physical interpretation of a potential function in a force field?
2. How can you determine the force from a given potential function?
3. What is the significance of energy conservation in a closed system?
4. How does the gradient of a scalar field relate to the direction of the force?
5. What are the conditions under which Newton's laws hold in this scenario?

Tip: The potential energy function $V(\vec{r}(t))$ often simplifies the process of finding the force in conservative systems.