# Derivative of scalar triple product

• V0ODO0CH1LD
In summary, the equation for the cross product is: σ'(t) . [σ'(t) x σ''(t)] + σ(t) . [σ''(t) x σ''(t)] = 0.
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

V0ODO0CH1LD said:

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

V0ODO0CH1LD said:
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.]

Thanks

## What is the definition of scalar triple product?

The scalar triple product is a mathematical operation that takes three vectors and returns a single scalar value. It is calculated by taking the dot product of one vector with the cross product of the other two vectors.

## What is the formula for calculating the derivative of scalar triple product?

The formula for the derivative of scalar triple product is d/dx (a · (b × c)) = (da/dx) · (b × c) + a · (db/dx × c) + a · (b × dc/dx), where a, b, and c are vectors and x is the variable with respect to which the derivative is taken.

## Why is the derivative of scalar triple product important?

The derivative of scalar triple product is important in vector calculus and physics as it allows us to calculate the rate of change of a scalar quantity with respect to a variable. It is also used in solving problems involving motion of particles and forces acting on them.

## What are some real-world applications of scalar triple product?

Scalar triple product has various applications in physics and engineering. It is used in calculating the moment of inertia in mechanics, calculating the work done by a force on a moving object, and determining the orientation of a plane in 3D space. It is also used in calculating the torque exerted by a force on a rigid body.

## Are there any limitations or assumptions when using scalar triple product?

One limitation of scalar triple product is that it can only be applied to three-dimensional vectors. Also, the vectors must be linearly independent for the cross product to exist. Additionally, the order of the vectors in the scalar triple product matters, as changing the order will result in a different scalar value. Lastly, the vectors must be continuous and differentiable for the derivative of scalar triple product to be calculated.

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