We are given that:

$u(t) = \mathbf{r}(t) \cdot [\mathbf{r}'(t) \times \mathbf{r}''(t)]$

We need to show that:

$u'(t) = \mathbf{r}(t) \cdot [\mathbf{r}'(t) \times \mathbf{r}'''(t)]$

Step-by-Step Solution:

1. Differentiate $u(t)$ using the product rule:

   Since $u(t)$ is the dot product of $\mathbf{r}(t)$ with a vector, we apply the product rule for differentiation. Let:

   $\mathbf{v}(t) = \mathbf{r}'(t) \times \mathbf{r}''(t)$

   Then $u(t) = \mathbf{r}(t) \cdot \mathbf{v}(t)$, and differentiating this with respect to $t$, we get:

   $u'(t) = \frac{d}{dt} \left( \mathbf{r}(t) \cdot \mathbf{v}(t) \right)$

   Using the product rule for derivatives:

   $u'(t) = \mathbf{r}'(t) \cdot \mathbf{v}(t) + \mathbf{r}(t) \cdot \mathbf{v}'(t)$

2. Simplify $\mathbf{v}(t) = \mathbf{r}'(t) \times \mathbf{r}''(t)$:

   Now, compute $\mathbf{v}'(t)$:

   $\mathbf{v}'(t) = \frac{d}{dt} \left( \mathbf{r}'(t) \times \mathbf{r}''(t) \right)$

   Using the product rule for the cross product:

   $\mathbf{v}'(t) = \mathbf{r}''(t) \times \mathbf{r}''(t) + \mathbf{r}'(t) \times \mathbf{r}'''(t)$

   The term $\mathbf{r}''(t) \times \mathbf{r}''(t)$ is zero because the cross product of any vector with itself is zero. Thus:

   $\mathbf{v}'(t) = \mathbf{r}'(t) \times \mathbf{r}'''(t)$

3. Substitute back into the derivative of $u(t)$:

   Now substitute $\mathbf{v}(t) = \mathbf{r}'(t) \times \mathbf{r}''(t)$ and $\mathbf{v}'(t) = \mathbf{r}'(t) \times \mathbf{r}'''(t)$ into the expression for $u'(t)$:

   $u'(t) = \mathbf{r}'(t) \cdot (\mathbf{r}'(t) \times \mathbf{r}''(t)) + \mathbf{r}(t) \cdot (\mathbf{r}'(t) \times \mathbf{r}'''(t))$

4. Simplify the first term:

   The first term $\mathbf{r}'(t) \cdot (\mathbf{r}'(t) \times \mathbf{r}''(t))$ is zero because the dot product of any vector with the cross product of itself and another vector is always zero. Hence, the term vanishes:

   $\mathbf{r}'(t) \cdot (\mathbf{r}'(t) \times \mathbf{r}''(t)) = 0$

5. Final result:

   Therefore, we are left with:

   $u'(t) = \mathbf{r}(t) \cdot (\mathbf{r}'(t) \times \mathbf{r}'''(t))$

This is exactly what we were asked to show.

Conclusion:

$u'(t) = \mathbf{r}(t) \cdot [\mathbf{r}'(t) \times \mathbf{r}'''(t)]$

Would you like further clarification or details on any step?

Related Questions:

1. How is the cross product related to the differentiation of vector functions?
2. What is the geometric interpretation of $\mathbf{r}'(t) \times \mathbf{r}''(t)$?
3. How does the dot product influence the result when differentiating?
4. Can this method be extended to higher derivatives?
5. What are some applications of these types of vector derivatives in physics?

Tip:

Always check if any terms in your differentiation vanish, as it simplifies the process greatly!