Problem 69 - Calculus 3

Question

Evaluate \(\frac{d}{d t}\left[\mathbf{u}(t) \times \mathbf{u}^{\prime}(t)\right]\) given \(\mathbf{u}(t)=t^{2} \mathbf{i}-2 t \mathbf{j}+\mathbf{k}\)

Step-by-Step Solution

Answer

The derivative is the zero vector, \( \mathbf{0} \).

Step 1: Understand the Function Elements

Identify that \( \mathbf{u}(t) = t^2 \mathbf{i} - 2t \mathbf{j} + \mathbf{k} \) and its derivative \( \mathbf{u}'(t) \) is needed. Recognize that \( \mathbf{u}(t) \) is a vector-valued function consisting of time-dependent components.

Step 2: Find the Derivative of the Vector Function

Compute \( \mathbf{u}'(t) \), the derivative of \( \mathbf{u}(t) \). Use differentiation rules for each component: \( \mathbf{u}'(t) = \frac{d}{dt}(t^2 \mathbf{i} - 2t \mathbf{j} + \mathbf{k}) = 2t \mathbf{i} - 2 \mathbf{j} \).

Step 3: Use the Product Rule for Vector Cross Product

To differentiate a cross product \( \mathbf{a}(t) \times \mathbf{b}(t) \), use the product rule: \[ \frac{d}{dt}[\mathbf{a}(t) \times \mathbf{b}(t)] = \mathbf{a}'(t) \times \mathbf{b}(t) + \mathbf{a}(t) \times \mathbf{b}'(t) \]. Here, \( \mathbf{a}(t) = \mathbf{u}(t) \) and \( \mathbf{b}(t) = \mathbf{u}'(t) \).

Step 4: Differentiate \( \mathbf{u}(t) \) and \( \mathbf{u}'(t) \)

Differentiate to find \( \mathbf{u}'(t) \) again (to be rigorous): it's \( 2t \mathbf{i} - 2 \mathbf{j} \). Now differentiate \( \mathbf{u}''(t) \) which is the zero vector \( \mathbf{0} \), as the components were constant linear terms in \( \mathbf{u}'(t) \).

Step 5: Compute Each Part of the Product Rule

Calculate \( \mathbf{u}''(t) \times \mathbf{u}'(t) = \mathbf{0} \times (2t\mathbf{i} - 2\mathbf{j}) = \mathbf{0} \). Next, calculate \( \mathbf{u}(t) \times \mathbf{u}''(t) = (t^2 \mathbf{i} - 2t \mathbf{j} + \mathbf{k}) \times \mathbf{0} = \mathbf{0} \).

Step 6: Combine Results Using the Product Rule

The derivative \( \frac{d}{dt} [\mathbf{u}(t) \times \mathbf{u}'(t)] = \mathbf{0} + \mathbf{0} = \mathbf{0} \).

Step 7: Conclusion

The rate of change of the cross product of \( \mathbf{u}(t) \) and its derivative is the zero vector, \( \mathbf{0} \).

Key Concepts

Vector Cross Product

In vector calculus, the cross product (or vector product) is a binary operation on two vectors in three-dimensional space. The result of a cross product is another vector that is perpendicular to the plane formed by the two original vectors. The magnitude of the cross product is equal to the area of the parallelogram that the two vectors span. This operation is denoted as \( \mathbf{a} \times \mathbf{b} \).
The cross product has some interesting properties:

• Non-commutative: \( \mathbf{a} \times \mathbf{b} eq \mathbf{b} \times \mathbf{a} \)
• Anti-commutative: \( \mathbf{a} \times \mathbf{b} = - ( \mathbf{b} \times \mathbf{a} ) \)
• Distributive over vector addition: \( \mathbf{a} \times ( \mathbf{b} + \mathbf{c} ) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c} \)
• Orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \): The result vector is perpendicular to both operands.

The cross product is useful in physics and engineering, especially for calculations involving rotation and forces.

Derivative of Vector-Valued Function

Vector-valued functions are functions that take real numbers as input and output vectors. The differentiation of such functions involves differentiating each component with respect to the variable. For example, if a vector-valued function \( \mathbf{u}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k} \), its derivative \( \mathbf{u}'(t) \) is computed as:
\[ \mathbf{u}'(t) = \frac{d}{dt}(f(t))\mathbf{i} + \frac{d}{dt}(g(t))\mathbf{j} + \frac{d}{dt}(h(t))\mathbf{k} \]
In our original exercise, the function \( \mathbf{u}(t) = t^2 \mathbf{i} - 2t \mathbf{j} + \mathbf{k} \) was differentiated to yield \( \mathbf{u}'(t) = 2t \mathbf{i} - 2\mathbf{j} \). Each component is treated individually:

• The derivative of \( t^2 \) is \( 2t \), giving the first component.
• \( -2t \) becomes \( -2 \) in the second component.
• The derivative of 1 with respect to \( t \) is 0, hence the third component, \( \mathbf{k} \), vanishes.

Understanding how to differentiate each component properly is key when working with vector-valued functions.

Product Rule for Derivatives

When dealing with the derivative of the product of two functions, the well-known product rule is applied. This rule can also be extended to the cross product of two vector-valued functions. For scalar functions, if you have products \( f(t) \times g(t) \), the derivative is:
\[ \frac{d}{dt}[f(t)g(t)] = f'(t)g(t) + f(t)g'(t) \]
For vectors \( \mathbf{a}(t) \) and \( \mathbf{b}(t) \), the derivative of their cross product is given by:

• \( \frac{d}{dt}[\mathbf{a}(t) \times \mathbf{b}(t)] = \mathbf{a}'(t) \times \mathbf{b}(t) + \mathbf{a}(t) \times \mathbf{b}'(t) \)

This formula ensures both \( \mathbf{a}(t) \) and \( \mathbf{b}(t) \) are treated correctly in terms of their time dependence. In our exercise, this rule ensures that both the vector \( \mathbf{u}(t) \) and its derivative \( \mathbf{u}'(t) \) are differentiated and crossed properly. Each term in the expanded form is evaluated independently, and their sum results in the derivative of the cross product.