Problem 43 - Thomas Calculus

Question

Establish the following properties of integrable vector functions. a. The Constant Scalar Multiple Rule: $$\int_{a}^{b} k \mathbf{r}(t) d t=k \int_{a}^{b} \mathbf{r}(t) d t \quad( any scalar k)$$ The Rule for Negatives $$\int_{a}^{b}(-\mathbf{r}(t)) d t=-\int_{a}^{b} \mathbf{r}(t) d t$$ is obtained by taking \(k=-1\) b. The Sum and Difference Rules: $$\int_{a}^{b}\left(\mathbf{r}_{1}(t) \pm \mathbf{r}_{2}(t)\right) d t=\int_{a}^{b} \mathbf{r}_{1}(t) d t \pm \int_{a}^{b} \mathbf{r}_{2}(t) d t$$ c. The Constant Vector Multiple Rules: $$\int_{a}^{b} \mathbf{C} \cdot \mathbf{r}(t) d t=\mathbf{C} \cdot \int_{a}^{b} \mathbf{r}(t) d t \quad (any constant vector \mathbf{C} )$$ and $$\int_{a}^{b} \mathbf{C} \times \mathbf{r}(t) d t=\mathbf{C} \times \int_{a}^{b} \mathbf{r}(t) d t \quad( any constant vector \mathbf{C})$$

Step-by-Step Solution

Answer

The integrable vector function properties are verified by applying linearity and distributive rules of integration to vectors and scalar or vector constants.

Step 1: Understanding the Problem

We are tasked with verifying the rules of integration for vector functions. This involves proving the properties similar to scalar integration, but in the context of vector-valued functions. We will consider properties like constant scalar multiplication, the rules for negatives, sums and differences, and operations with constant vectors.

Step 2: Prove the Constant Scalar Multiple Rule

For this property, consider a vector function \( \mathbf{r}(t) \) and a scalar \( k \), both integrable over \([a, b]\). The constant scalar multiple rule states:\[ \int_{a}^{b} k \mathbf{r}(t) dt = k \int_{a}^{b} \mathbf{r}(t) dt \]This is proven by recognizing that the integration operation is linear, and thus constants can be factored out of the integral. Therefore, the scalar \( k \) can be pulled outside of the integral sign as follows:- Compute \( \int_{a}^{b} \mathbf{r}(t) dt = \mathbf{R}(b) - \mathbf{R}(a) \), where \( \mathbf{R}(t) \) is the antiderivative of \( \mathbf{r}(t) \).- Then \( \int_{a}^{b} k \mathbf{r}(t) dt = k( \mathbf{R}(b) - \mathbf{R}(a) ) = k \int_{a}^{b} \mathbf{r}(t) dt \). Hence, verified.

Step 3: Prove the Rule for Negatives

By setting \( k = -1 \) in the constant scalar multiple rule, we get:\[ \int_{a}^{b} (-\mathbf{r}(t)) dt = - \int_{a}^{b} \mathbf{r}(t) dt \]This follows directly from the fact that multiplying the function by -1 is equivalent to negating the integral's result due to linearity.

Step 4: Prove the Sum and Difference Rules

To prove the sum and difference rules for vector integration, consider two vector functions \( \mathbf{r}_1(t) \) and \( \mathbf{r}_2(t) \). The rule is:\[ \int_{a}^{b} (\mathbf{r}_{1}(t) \pm \mathbf{r}_{2}(t)) dt = \int_{a}^{b} \mathbf{r}_{1}(t) dt \pm \int_{a}^{b} \mathbf{r}_{2}(t) dt \]This property holds because integration is distributive over addition. To show this:- Let \( \mathbf{R}_1(t) \) and \( \mathbf{R}_2(t) \) be antiderivatives of \( \mathbf{r}_1(t) \) and \( \mathbf{r}_2(t) \), respectively.- Then \( \int_{a}^{b} (\mathbf{r}_1(t) \pm \mathbf{r}_2(t)) dt = (\mathbf{R}_1(b) \pm \mathbf{R}_2(b)) - (\mathbf{R}_1(a) \pm \mathbf{R}_2(a)) \).- Simplifying gives \( (\mathbf{R}_1(b) - \mathbf{R}_1(a)) \pm (\mathbf{R}_2(b) - \mathbf{R}_2(a)) \), which is \( \int_{a}^{b} \mathbf{r}_{1}(t) dt \pm \int_{a}^{b} \mathbf{r}_{2}(t) dt \). Thus, proven.

Step 5: Prove the Constant Vector Multiple Rules for Dot Product

Given a constant vector \( \mathbf{C} \) and a vector function \( \mathbf{r}(t) \), to show:\[ \int_{a}^{b} \mathbf{C} \cdot \mathbf{r}(t) dt = \mathbf{C} \cdot \int_{a}^{b} \mathbf{r}(t) dt \]This result can be proved by noting that the dot product is distributive over addition and maintains linearity. If \( \mathbf{R}(t) \) is an antiderivative of \( \mathbf{r}(t) \):- Calculate \( \int_{a}^{b} \mathbf{C} \cdot \mathbf{r}(t) dt = \mathbf{C} \cdot (\mathbf{R}(b) - \mathbf{R}(a)) \).- This simplifies to \( \mathbf{C} \cdot \int_{a}^{b} \mathbf{r}(t) dt \), confirming the property.

Step 6: Prove the Constant Vector Multiple Rules for Cross Product

For another constant vector \( \mathbf{C} \), show:\[ \int_{a}^{b} \mathbf{C} \times \mathbf{r}(t) dt = \mathbf{C} \times \int_{a}^{b} \mathbf{r}(t) dt \]By the properties of the cross product, which is linear as well:- Find \( \int_{a}^{b} \mathbf{C} \times \mathbf{r}(t) dt = \mathbf{C} \times (\mathbf{R}(b) - \mathbf{R}(a)) \).- This results in \( \mathbf{C} \times \int_{a}^{b} \mathbf{r}(t) dt \), affirming the property.

Key Concepts

Integration of Vector Functions

When dealing with vector calculus, sometimes we need to integrate vector functions. Imagine a vector function as a curve in space, where each point is a vector. To integrate a vector function, you calculate the antiderivative of each component of the vector.

If you have a vector function \( \mathbf{r}(t) = \langle f_1(t), f_2(t), f_3(t) \rangle \), its integral over an interval \([a, b]\) is another vector whose components are the integrals of \( f_1(t) \), \( f_2(t) \), and \( f_3(t) \) from \( a \) to \( b \). Formally, this means:

• \( \int_{a}^{b} \mathbf{r}(t) dt = \langle \int_{a}^{b} f_1(t) dt, \int_{a}^{b} f_2(t) dt, \int_{a}^{b} f_3(t) dt \rangle \).

This process allows the calculation of vector areas, moments, and other vector quantities which are useful in physics and engineering.

Integration Properties

Much like scalar integration, vector integration has properties that make calculations easier. These are linearity, subtraction, and addition. These properties simplify the integration process for vector functions.

**Linearity**: It allows you to factor out constants from the integral. For example, for a scalar \( k \),

• \( \int_{a}^{b} k \mathbf{r}(t) dt = k \int_{a}^{b} \mathbf{r}(t) dt \).

**Subtraction and Addition**: If you have two vector functions \( \mathbf{r}_1(t) \) and \( \mathbf{r}_2(t) \), these rules apply:

• \( \int_{a}^{b} (\mathbf{r}_1(t) \pm \mathbf{r}_2(t)) dt = \int_{a}^{b} \mathbf{r}_1(t) dt \pm \int_{a}^{b} \mathbf{r}_2(t) dt \).

These integration properties are essential as they ensure the operations are consistent with the geometry of the vector spaces involved.

Linear Algebra

Linear algebra forms the backbone of vector calculus. It provides tools such as matrices and vectors that help in performing complex integrations.

Matrix transformations are akin to transformations in geometry, allowing shifts, rotations, and scaling of vector spaces. When integrating vector functions, linear algebra concepts help manage these transformations during integration.

Understanding how vectors add or scale, how matrices operate on vectors, and how these interactions affect integrals is crucial. It means knowing how to work with and visualize vectors in multi-dimensional spaces.

Dot Product

The dot product is an operation between two vectors that results in a scalar. This operation is crucial because it helps find angles, lengths, and projections of vectors.

During integration, when you integrate the dot product of one constant vector \( \mathbf{C} \) and another function, you simplify calculations. The property states that you can "factor out" the constant vector:

• \( \int_{a}^{b} \mathbf{C} \cdot \mathbf{r}(t) dt = \mathbf{C} \cdot \int_{a}^{b} \mathbf{r}(t) dt \).

This is helpful for simplifying equations and can make solving some physics and engineering problems more straightforward.

Cross Product

The cross product operation produces a vector that is perpendicular to two given vectors. It is commonly used in physics to determine torque or angular momentum.

Like the dot product, the cross product can also be integrated with simplifications. For a constant vector \( \mathbf{C} \), the property allows you to separate it from the integral:

• \( \int_{a}^{b} \mathbf{C} \times \mathbf{r}(t) dt = \mathbf{C} \times \int_{a}^{b} \mathbf{r}(t) dt \).

Applying this property means focusing on the orientation and magnitude changes in multi-dimensional spaces and greatly simplifies complex vector integration calculations.