Problem 32 - Thomas Calculus

Question

Limits of cross products of vector functions Suppose that \(\mathbf{r}_{1}(t)=f_{1}(t) \mathbf{i}+f_{2}(t) \mathbf{j}+f_{3}(t) \mathbf{k}, \mathbf{r}_{2}(t)=g_{1}(t) \mathbf{i}+g_{2}(t) \mathbf{j}+g_{3}(t) \mathbf{k},\) \(\lim _{t \rightarrow t_{0}} \mathbf{r}_{1}(t)=\mathbf{A},\) and \(\lim _{t \rightarrow t_{0}} \mathbf{r}_{2}(t)=\mathbf{B} .\) Use the determinant for- mula for cross products and the Limit Product Rule for scalar functions to show that \begin{equation} \lim _{t \rightarrow t_{0}}\left(\mathbf{r}_{1}(t) \times \mathbf{r}_{2}(t)\right)=\mathbf{A} \times \mathbf{B} \end{equation}

Step-by-Step Solution

Answer

The limit of the cross product is equal to the cross product of the limits: \( \lim_{t \to t_0} \left( \mathbf{r}_1(t) \times \mathbf{r}_2(t) \right) = \mathbf{A} \times \mathbf{B} \).

Step 1: Express Cross Product as a Determinant

The cross product of two vectors \( \mathbf{r}_1(t) \) and \( \mathbf{r}_2(t) \) can be represented as a determinant:\[\mathbf{r}_1(t) \times \mathbf{r}_2(t) = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \f_1(t) & f_2(t) & f_3(t) \g_1(t) & g_2(t) & g_3(t) \\end{vmatrix}\]This expands into the following expression:\[\mathbf{r}_1(t) \times \mathbf{r}_2(t) = \left(f_2(t)g_3(t) - f_3(t)g_2(t)\right)\mathbf{i} - \left(f_1(t)g_3(t) - f_3(t)g_1(t)\right)\mathbf{j} + \left(f_1(t)g_2(t) - f_2(t)g_1(t)\right)\mathbf{k}\]

Step 2: Apply the Limit to Each Component of the Cross Product

Applying the limit to the components as \( t \to t_0 \):\[\lim_{t \to t_0}\left(f_2(t)g_3(t) - f_3(t)g_2(t)\right)\mathbf{i}, \lim_{t \to t_0}\left(f_3(t)g_1(t) - f_1(t)g_3(t)\right)\mathbf{j}, \lim_{t \to t_0}\left(f_1(t)g_2(t) - f_2(t)g_1(t)\right)\mathbf{k}\]

Step 3: Use the Limit Product Rule

By the Limit Product Rule, \( \lim_{t \to t_0} \left( f_i(t) \cdot g_j(t) \right) = \lim_{t \to t_0} f_i(t) \, \lim_{t \to t_0} g_j(t) \): - \( \lim_{t \to t_0} f_2(t) g_3(t) = A_2 B_3 \)- \( \lim_{t \to t_0} f_3(t) g_2(t) = A_3 B_2 \) (and apply similarly to other paired components).

Step 4: Simplify Expression Using Limits

Using the limits found in Step 3, the components of the limit of the cross product are:\[\begin{align*}\lim_{t \to t_0}(\mathbf{r}_1(t) \times \mathbf{r}_2(t)) & = \left( A_2 B_3 - A_3 B_2 \right)\mathbf{i} - \left( A_1 B_3 - A_3 B_1 \right)\mathbf{j} + \left( A_1 B_2 - A_2 B_1 \right)\mathbf{k} \& = \mathbf{A} \times \mathbf{B}\end{align*}\]

Step 5: Conclusion

Therefore, we have shown that: \[ \lim_{t \to t_0} \left( \mathbf{r}_1(t) \times \mathbf{r}_2(t) \right) = \mathbf{A} \times \mathbf{B} \]The limit of the cross product of the vector functions is equal to the cross product of their respective limits.

Key Concepts

Cross Product

The cross product is a vital operation in vector calculus, primarily used with three-dimensional vectors. It combines two vectors to produce another vector that is perpendicular to the original pair. Unlike the dot product, the result of a cross product is a vector rather than a scalar. Consider vectors \( \mathbf{u} \) and \( \mathbf{v} \). The cross product \( \mathbf{u} \times \mathbf{v} \) is calculated using the determinant of a matrix built from the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \), along with the components of \( \mathbf{u} \) and \( \mathbf{v} \). The resulting vector illustrates the area and orientation of the parallelogram defined by \( \mathbf{u} \) and \( \mathbf{v} \). Specifically, for vectors \( \mathbf{r}_1(t) \) and \( \mathbf{r}_2(t) \), the cross product unfolds into:

• \( (f_2(t)g_3(t) - f_3(t)g_2(t)) \mathbf{i} \)
• \( -(f_1(t)g_3(t) - f_3(t)g_1(t)) \mathbf{j} \)
• \( (f_1(t)g_2(t) - f_2(t)g_1(t)) \mathbf{k} \)

This formula configures the physical intuition behind torque and angular momentum.

Limits of Vector Functions

Understanding limits of vector functions involves approaching a particular value, \( t_0 \), for which the vector function \( \mathbf{r}(t) \) becomes consistently close to a vector \( \mathbf{A} \). This concept is analogous to limits of scalar functions but takes place in a vector space. With vector functions \( \mathbf{r}_1(t) \) and \( \mathbf{r}_2(t) \), their limits at \( t \rightarrow t_0 \) give vectors \( \mathbf{A} \) and \( \mathbf{B} \). Essentially breaking down into a sequence of coordinate-wise limits for each vector component. When evaluating limits, the operations indicate that we can take the limit of each component separately, leading naturally to the limit of the entire vector function as the collection of these limits. This property simplifies the process and ensures that even complex vector expressions remain manageable when approaching a limit.

Determinant Formula

In vector calculus, the determinant plays a pivotal role when analyzing the cross product. It organizes three-dimensional vectors and helps compute their product using a 3x3 matrix. This matrix has the unit basis vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) as the first row, and the components of the two vectors as the subsequent rows. Computing the determinant involves:

• Multiplying and summing certain components to form the resulting cross vector.
• Creating a structured approach to keep track of multiple dimensions simultaneously.

The formula for the determinant allows algebraic expressions of the cross product to transform into a vector defined with \( i, j, k \) components. This approach not only organizes the computation but also provides geometric insights into the nature of these operations.

Limit Product Rule

The Limit Product Rule is essential when operating with limits in calculus. This principle states that the limit of the product of functions equals the product of their individual limits, assuming each limit independently exists. Applied to vector functions, it simplifies the evaluation by separating complex products into simpler, manageable parts. To illustrate, consider the product \( f(t)g(t) \), with limits at a point \( t_0 \). This rule allows us to express it as:

• \( \lim_{t \to t_0}f(t)g(t) = (\lim_{t \to t_0}f(t)) (\lim_{t \to t_0}g(t)) \)

This simplification step was pivotal in computing the cross product limit from the original exercise. It confirms that the limit of the whole expression equals the limit of the cross product of its components, ensuring the consistency and accuracy necessary for such vector operations.