Question

Let \(\mathbf{F}(t)\) be continuous for \(a \leq t \leq b\) and let \(\mathbf{q}\) be a constant vector. Prove: a) \(\int_{a}^{b} \mathbf{q} \cdot \mathbf{F}(t) d t=\mathbf{q} \cdot \int_{a}^{b} \mathbf{F}(t) d t\) b) \(\int_{a}^{b} \mathbf{q} \times \mathbf{F}(t) d t=\mathbf{q} \times \int_{a}^{b} \mathbf{F}(t) d t\)

Short answer

Question: Prove the following properties of vector-valued integrals when \(\mathbf{q}\) is a constant vector: 1. \(\int_{a}^{b} \mathbf{q} \cdot \mathbf{F}(t) dt = \mathbf{q} \cdot \int_{a}^{b} \mathbf{F}(t) dt\) 2. \(\int_{a}^{b} \mathbf{q} \times \mathbf{F}(t) dt = \mathbf{q} \times \int_{a}^{b} \mathbf{F}(t) dt\)

Step-by-step solution

Prove the dot product property

To prove the dot product property, we start by writing the left-hand side of the equation (a): $$ \int_{a}^{b} \mathbf{q} \cdot \mathbf{F}(t) dt $$ Since \(\mathbf{q}\) is a constant vector, we can move it inside the integral and get: $$ \int_{a}^{b} \mathbf{q} \cdot \mathbf{F}(t) dt = \mathbf{q} \cdot \int_{a}^{b} \mathbf{F}(t) dt $$ This completes the proof for part (a).

Prove the cross product property

To prove the cross product property, we start by writing the left-hand side of the equation (b): $$ \int_{a}^{b} \mathbf{q} \times \mathbf{F}(t) dt $$ Since \(\mathbf{q}\) is a constant vector, we can move it inside the integral and get: $$ \int_{a}^{b} \mathbf{q} \times \mathbf{F}(t) dt = \mathbf{q} \times \int_{a}^{b} \mathbf{F}(t) dt $$ This completes the proof for part (b). Thus, we have proved both properties for vector-valued integrals involving dot and cross products when the vector \(\mathbf{q}\) is constant.

Key concepts

Vector Integration

In vector calculus, integration allows us to handle vector functions over an interval. When we integrate a vector function, the result is another vector. This operation extends the idea of integrating scalar functions to vectors, enabling complex calculations over paths or domains.

For any given vector function \( \mathbf{F}(t) \), representing a vector that changes with \( t \), vector integration involves calculating an integral over time or space.

The forward direction is to integrate component-wise:

• Break the vector into its components.
• Integrate each component separately as a regular scalar function.
• Combine the results back into a vector.

The key here is this component-wise integration, which maintains the orientation and dimensionality of the vector. Substitute the components back into the integral to find the full vector solution. This concept is essential in proving properties involving vector operations like the dot and cross product as seen in this problem.

Dot Product

The dot product is a fundamental vector operation, also known as the scalar product. It returns a scalar rather than a vector, signifying the magnitude of one vector in the direction of another.

Mathematically, for two vectors \( \mathbf{u} \) and \( \mathbf{v} \), the dot product is given by: \[ \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 \]

This operation has several key characteristics:

• Commutative: \( \mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u} \)
• Distributive over addition: \( \mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) = \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w} \)
• Scalar multiplier: If \( c \) is a constant, then \( c(\mathbf{u} \cdot \mathbf{v}) = (c\mathbf{u}) \cdot \mathbf{v} = \mathbf{u} \cdot (c\mathbf{v}) \)

Importantly, as shown in the exercise, if one of the vectors, like \( \mathbf{q} \), is a constant, it can be factored out of an integral, allowing simpler calculation of vector integrals involving the dot product.

Cross Product

The cross product is another essential vector operation, but unlike the dot product, it results in a vector. It is particularly valuable in physics for finding vectors perpendicular to two given vectors, like torque or angular momentum.

Given two vectors \( \mathbf{u} \) and \( \mathbf{v} \), the cross product is: \[ \mathbf{u} \times \mathbf{v} = (u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1) \]

Some important properties include:

• Anticommutative: \( \mathbf{u} \times \mathbf{v} = - (\mathbf{v} \times \mathbf{u}) \)
• Distributive over addition: \( \mathbf{u} \times (\mathbf{v} + \mathbf{w}) = \mathbf{u} \times \mathbf{v} + \mathbf{u} \times \mathbf{w} \)
• Zero vector: \( \mathbf{u} \times \mathbf{u} = \mathbf{0} \)

In the exercise, a constant vector \( \mathbf{q} \) can be factored out of the cross product integral, simplifying the calculation process and confirming the distributed nature of this vector operation.

Constant Vector Properties

Constant vectors play a vital role in vector integration because they simplify complex operations. A constant vector is a vector that does not change with respect to the variable of integration.

This characteristic allows us to move the constant vector in and out of dot and cross product integrals without affecting the operation's outcome. This property extensively simplifies the calculations involved in integrating vector functions.

Important features of constant vectors include:

• Scalability: A constant retains its magnitude across operations.
• Commutability: In expressions involving addition or multiplication, its position can often change without altering the result.
• Independence: Not influenced by the differential variable, can factor out of operations such as integrals.

In our exercise, these properties allow the constant vector \( \mathbf{q} \) to be factored out of both dot and cross product integrations, making the overall problem-solving approach more straightforward.