Derivative of scalar triple product

[V0ODO0CH1LD]

Homework Statement

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

Homework 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)

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

 

[pasmith]

Homework Statement

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

Homework 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)

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.

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 $$\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = 0 = \mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}).$$

 

[moontiger]

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

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.]

 

[V0ODO0CH1LD]

Thanks