Problem 47 If $\mathbf{u}, \mathbf{v},$ a... [FREE SOLUTION]

Chapter 15: Problem 47

If $\mathbf{u}, \mathbf{v},$ and $\mathbf{w}$ are differentiable, prove that $$ \begin{aligned} D_{+}[\mathbf{u}(t) \cdot \mathbf{v}(t) \times \mathbf{w}(t)]=&\left[\mathbf{u}^{\prime}(t)+\mathbf{v}(t) \times \mathbf{w}(t)\right] \\ &+\left[\mathbf{u}(t) \cdot \mathbf{v}^{\prime}(t) \times \mathbf{w}(t)\right] \\\ &+\left[\mathbf{u}(t) \cdot \mathbf{v}(t) \times \mathbf{w}^{\prime}(t)\right] . \end{aligned} $$

Short Answer

Expert verified

The derivative of the scalar triple product is the sum of products derived by differentiating one vector at a time.

Step by step solution

Recall the Derivative of a Cross Product

The derivative of a cross product of two vectors $ \mathbf{a}(t) $ and $ \mathbf{b}(t) $, where each is differentiable, is given by: \[\frac{d}{dt} [\mathbf{a}(t) \times \mathbf{b}(t)] = \mathbf{a}'(t) \times \mathbf{b}(t) + \mathbf{a}(t) \times \mathbf{b}'(t).\]In this exercise, we have three vectors $ \mathbf{u}(t), \mathbf{v}(t), $ and $ \mathbf{w}(t) $. We are working toward expressing the derivative of their scalar triple product.

Apply Product Rule to Scalar Triple Product

Remember that the scalar triple product $ \mathbf{u}(t) \cdot (\mathbf{v}(t) \times \mathbf{w}(t)) $ can be considered as a function of three vectors.For differentiating with respect to time $ t $, we can apply the product rule as follows: \[D_{+}[\mathbf{u}(t) \cdot (\mathbf{v}(t) \times \mathbf{w}(t))] = \mathbf{u}'(t) \cdot (\mathbf{v}(t) \times \mathbf{w}(t)) + \mathbf{u}(t) \cdot \frac{d}{dt} [\mathbf{v}(t) \times \mathbf{w}(t)].\]

Apply Derivative of a Cross Product

According to Step 1, differentiate the cross product part separately:\[\frac{d}{dt} [\mathbf{v}(t) \times \mathbf{w}(t)] = \mathbf{v}'(t) \times \mathbf{w}(t) + \mathbf{v}(t) \times \mathbf{w}'(t).\]Substitute this back into the expression derived in Step 2:

Substitute Back and Simplify

Now substitute the result from Step 3 back into the expression from Step 2:\[D_{+}[\mathbf{u}(t) \cdot (\mathbf{v}(t) \times \mathbf{w}(t))] = \mathbf{u}'(t) \cdot (\mathbf{v}(t) \times \mathbf{w}(t))+ \mathbf{u}(t) \cdot (\mathbf{v}'(t) \times \mathbf{w}(t))+ \mathbf{u}(t) \cdot (\mathbf{v}(t) \times \mathbf{w}'(t)).\]This confirms that each term in the derivative corresponds to differentials of $ \mathbf{u}, \mathbf{v}, $ and $ \mathbf{w} $.

Conclude the Derivative Expression

The differentiated expression correctly breaks into the expected linear combination of components:\[D_{+} [\mathbf{u}(t) \cdot \mathbf{v}(t) \times \mathbf{w}(t)] = \mathbf{u}'(t) \cdot (\mathbf{v}(t) \times \mathbf{w}(t)) + \mathbf{u}(t) \cdot \mathbf{v}'(t) \times \mathbf{w}(t) + \mathbf{u}(t) \cdot \mathbf{v}(t) \times \mathbf{w}'(t).\]That illustrates the complete expression for the derivative.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Scalar Triple Product

The scalar triple product is a significant concept in vector calculus. It involves three vectors and is denoted as $ \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) $. This product gives a scalar value, unlike the cross product, which results in a vector. The scalar triple product is often used to determine the volume of a parallelepiped formed by three vectors. Here's why it's useful:

If the scalar triple product is zero, the vectors are coplanar.
The absolute value of the scalar triple product equals the volume of the parallelepiped.

Understanding the scalar triple product's geometric and algebraic properties helps in solving many physics and engineering problems where spatial relationships are important.

Cross Product

The cross product, symbolized by $ \times $, is an operation performed on two vectors in three-dimensional space, resulting in a vector that is perpendicular to the plane containing the original vectors. It is a key tool in vector calculus and provides both magnitude and direction. Here are some essential properties:

The magnitude of the cross product $ \mathbf{a} \times \mathbf{b} $ is equal to the area of the parallelogram that the vectors span: $ |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}| \sin(\theta) $.
The direction is given by the right-hand rule, meaning the resulting vector points perpendicularly outward from the surface formed by $ \mathbf{a} $ and $ \mathbf{b} $.
The cross product is anti-commutative, so $ \mathbf{a} \times \mathbf{b} = - (\mathbf{b} \times \mathbf{a}) $.

Knowing how to compute and interpret the cross product is vital for understanding forces and rotations in physics.

Product Rule

The product rule is a fundamental principle in calculus used for finding the derivative of the product of two functions. In the context of vector calculus, it extends to products involving vector functions. For the scalar triple product, applying the product rule allows us to effectively break down complex derivatives into manageable parts. Here's how it works:

For two differentiable functions $ f(t) $ and $ g(t) $, the product rule states: $ (fg)' = f'g + fg' $.
This rule can be extended to functions involving more than two factors, like in the scalar triple product difference, where each element must be treated using respective derivatives.

By using the product rule, one can find derivatives efficiently, especially in terms of systems involving multiple interacting vector quantities.

Differentiable Functions

Differentiability is an essential concept in calculus, indicating that a function has a derivative at each point in its domain. For vector-valued functions, being differentiable implies each component of the function can be individually derived. Here's what differentiability means in practical terms:

A function $ f(t) $ is differentiable at $ t $ if it is continuous at that point, and if the derivative $ f'(t) $ exists.
For vector functions $ \mathbf{u}(t) $, $ \mathbf{v}(t) $, and $ \mathbf{w}(t) $, being differentiable means $ \mathbf{u}'(t) $, $ \mathbf{v}'(t) $, and $ \mathbf{w}'(t) $ can be calculated, thus ensuring the accurate application of calculus techniques like derivative and product rule.

Differentiability provides the mathematical framework needed to analyze how functions change, critical for applications in physics and engineering where changes in quantities are examined over time.

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Short Answer

Step by step solution

Recall the Derivative of a Cross Product

Apply Product Rule to Scalar Triple Product

Apply Derivative of a Cross Product

Substitute Back and Simplify

Conclude the Derivative Expression

Key Concepts

Scalar Triple Product

Cross Product

Product Rule

Differentiable Functions

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