We are tasked with finding the time derivative of the expression $u(t) \cdot (v(t) \times w(t))$, where $u(t)$, $v(t)$, and $w(t)$ are time-dependent vector functions.

We will use the product rule for derivatives and vector identities to approach this. Here's the step-by-step breakdown:

Expression:

$\frac{d}{dt} \left( u(t) \cdot (v(t) \times w(t)) \right)$

Apply the Product Rule:

We treat this as a product of two functions: $u(t)$ and $v(t) \times w(t)$. The product rule states: $\frac{d}{dt} (f \cdot g) = \frac{d}{dt} f \cdot g + f \cdot \frac{d}{dt} g$ So, we have: $\frac{d}{dt} \left( u(t) \cdot (v(t) \times w(t)) \right) = \frac{d}{dt} u(t) \cdot (v(t) \times w(t)) + u(t) \cdot \frac{d}{dt} (v(t) \times w(t))$

Step 1: Derivative of $u(t)$:

$\frac{d}{dt} u(t) \cdot (v(t) \times w(t)) = \dot{u}(t) \cdot (v(t) \times w(t))$ where $\dot{u}(t)$ is the time derivative of $u(t)$.

Step 2: Derivative of $v(t) \times w(t)$:

We use the product rule for the cross product: $\frac{d}{dt} (v(t) \times w(t)) = \dot{v}(t) \times w(t) + v(t) \times \dot{w}(t)$ where $\dot{v}(t)$ and $\dot{w}(t)$ are the time derivatives of $v(t)$ and $w(t)$, respectively.

Substitute this back into the main equation:

$\frac{d}{dt} \left( u(t) \cdot (v(t) \times w(t)) \right) = \dot{u}(t) \cdot (v(t) \times w(t)) + u(t) \cdot (\dot{v}(t) \times w(t) + v(t) \times \dot{w}(t))$

Final Expression:

$\frac{d}{dt} \left( u(t) \cdot (v(t) \times w(t)) \right) = \dot{u}(t) \cdot (v(t) \times w(t)) + u(t) \cdot (\dot{v}(t) \times w(t)) + u(t) \cdot (v(t) \times \dot{w}(t))$

This is the desired expression for the time derivative of $u(t) \cdot (v(t) \times w(t))$.

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Would you like more details on any step? Here are 5 related questions for further exploration:

1. How does the product rule apply to scalar and vector functions?
2. Can you explain the product rule for cross products in more depth?
3. How would this derivative change if any of the vectors were constant?
4. What physical examples use this type of derivative involving vector products?
5. How does this result relate to angular momentum in physics?

Tip: Always remember that when differentiating products of vectors, both the scalar and vector product rules need to be applied carefully!