# Derivative of scalar triple product

1. Jun 24, 2014

### V0ODO0CH1LD

1. The problem statement, all variables and given/known data

If u(t) = σ(t) . [σ'(t) x σ''(t)], show that u'(t) = σ(t) . [σ'(t) x σ'''(t)].

2. Relevant equations

The rules for differentiating dot products and cross products, respectively, are:

d/dt f(t) . g(t) = f'(t) . g(t) + f(t) . g'(t)

d/dt f(t) x g(t) = f'(t) x g(t) + f(t) x g'(t)

3. The attempt at a solution

So I applied each individual rule to σ(t) . [σ'(t) x σ''(t)] to get

σ'(t) . [σ'(t) x σ''(t)] + σ(t) . [σ''(t) x σ''(t) + σ'(t) x σ'''(t)]

which I can expand into

σ'(t) . [σ'(t) x σ''(t)] + σ(t) . [σ''(t) x σ''(t)] + σ(t) . [σ'(t) x σ'''(t)]

meaning σ'(t) . [σ'(t) x σ''(t)] + σ(t) . [σ''(t) x σ''(t)] should equal zero, but I don't see why it would.

Thanks

2. Jun 24, 2014

### pasmith

Basic facts about the cross product:

For any vector $\mathbf{a}$, we have $\mathbf{a} \times \mathbf{a} = \mathbf{0}$.

For any vectors $\mathbf{a}$ and $\mathbf{b}$, the vector $\mathbf{a} \times \mathbf{b}$ is orthogonal to both $\mathbf{a}$ and $\mathbf{b}$, so $$\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = 0 = \mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}).$$

3. Jun 24, 2014

### moontiger

The cross product of any vector with itself is zero, so the second term is zero.

The first term is zero because the cross product σ'(t) x σ''(t) is orthogonal to σ'(t). The dot product of a vector with a vector orthogonal to itself is zero.

Hope that helps.

[Edit: Oops...I was posting at the same time as pasmith. What pasmith said.]

4. Jun 24, 2014

Thanks