Problem 46 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The limit of the cross product is the cross product of the limits: \(\mathbf{A} \times \mathbf{B}\).

Step 1: Express the Cross Product Using Determinant Formula

The cross product of two vector functions \(\mathbf{r}_1(t)\) and \(\mathbf{r}_2(t)\) can be written as a determinant:\[\mathbf{r}_1(t) \times \mathbf{r}_2(t) = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \f_1(t) & f_2(t) & f_3(t) \g_1(t) & g_2(t) & g_3(t) \end{vmatrix}\] Evaluating the determinant, the cross product is:\[(f_2(t)g_3(t) - f_3(t)g_2(t))\mathbf{i} - (f_1(t)g_3(t) - f_3(t)g_1(t))\mathbf{j} + (f_1(t)g_2(t) - f_2(t)g_1(t))\mathbf{k}\]

Step 2: Apply the Limit to Each Component

To find \(\lim _{t \rightarrow t_0} \left(\mathbf{r}_{1}(t) \times \mathbf{r}_{2}(t)\right)\), apply the limit to each component of the cross product separately, using the limit product rule for scalar functions:\[\lim _{t \rightarrow t_0} \left(f_2(t)g_3(t) - f_3(t)g_2(t)\right) \mathbf{i}, - \lim _{t \rightarrow t_0} \left(f_1(t)g_3(t) - f_3(t)g_1(t)\right) \mathbf{j}, \lim _{t \rightarrow t_0} \left(f_1(t)g_2(t) - f_2(t)g_1(t)\right) \mathbf{k}\]

Step 3: Use the Limit Product Rule

The rule states that \(\lim _{t \rightarrow t_{0}} (f(t)g(t)) = \left(\lim _{t \rightarrow t_{0}} f(t)\right)\left(\lim _{t \rightarrow t_{0}} g(t)\right)\) if these limits exist. Apply this to each product term:\[\lim _{t \rightarrow t_0} f_2(t)g_3(t) = A_2B_3, \quad \lim _{t \rightarrow t_0} f_3(t)g_2(t) = A_3B_2\]Substituting, the first component becomes \((A_2B_3 - A_3B_2)\mathbf{i}\). This process repeats analogously for the other components.

Step 4: Assemble the Final Cross Product

After finding limits of individual components, assemble them into the final vector:\[\lim _{t \rightarrow t_0} \left(\mathbf{r}_{1}(t) \times \mathbf{r}_{2}(t)\right) = (A_2B_3 - A_3B_2)\mathbf{i} - (A_1B_3 - A_3B_1)\mathbf{j} + (A_1B_2 - A_2B_1)\mathbf{k}\]This matches \(\mathbf{A} \times \mathbf{B}\).

Step 5: Confirm the Result

Summarize the outcome: The limit of the cross product is exactly the cross product of the limits, confirming that: \[\lim _{t \rightarrow t_0} \left(\mathbf{r}_{1}(t) \times \mathbf{r}_{2}(t)\right) = \mathbf{A} \times \mathbf{B}\]

Key Concepts

Cross Product

The cross product is a fundamental operation in vector calculus used to find a vector perpendicular to two given vectors in three-dimensional space. Imagine you have two vectors, say \( \mathbf{r}_1(t) \) and \( \mathbf{r}_2(t) \), and let's compute their cross product. It's crucial because it gives insights into the orientation and area of parallelograms defined by these vectors.

Given two vectors:

• \( \mathbf{r}_1(t) = f_1(t) \mathbf{i} + f_2(t) \mathbf{j} + f_3(t) \mathbf{k} \)
• \( \mathbf{r}_2(t) = g_1(t) \mathbf{i} + g_2(t) \mathbf{j} + g_3(t) \mathbf{k} \)

you can express their cross product in determinant form, a handy tool for simplifying calculations. It involves arranging vector components in a matrix and solving the determinant to produce a new vector:\[\mathbf{r}_1(t) \times \mathbf{r}_2(t) = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \ f_1(t) & f_2(t) & f_3(t) \ g_1(t) & g_2(t) & g_3(t)\end{vmatrix}\]

This results in a new vector that conveys much about the initial two vectors, especially their spatial relationship.

Determinant Formula

The determinant formula plays a key role in solving cross products of vectors. It's like a compact way to code the transformation required to find the perpendicularity and direction indicated by the cross product.

For vectors \( \mathbf{r}_1(t) \) and \( \mathbf{r}_2(t) \), the determinant is computed by first setting up a 3x3 matrix using the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) and the components of each vector. Calculate the determinant by expanding along the first row. This is done by multiplying across diagonals in a specified order, followed by the subtraction of the products on alternate diagonals:

• \((f_2(t)g_3(t) - f_3(t)g_2(t)) \mathbf{i}\)
• \((-1)(f_1(t)g_3(t) - f_3(t)g_1(t)) \mathbf{j}\)
• \((f_1(t)g_2(t) - f_2(t)g_1(t)) \mathbf{k}\)

This formulation allows us to derive a vector pointing in the correct direction, reflecting both magnitude and orientation determined by the vector components.

Grasping the formula's structure is essential, as it not only aids in cross product evaluations but also in understanding how various linear transformations affect vector spaces.

Limit Product Rule

The Limit Product Rule is vital in calculus, especially when determining the limits of products. It states that:\[\lim_{t \to t_0} (f(t) \cdot g(t)) = \left(\lim_{t \to t_0} f(t)\right) \cdot \left(\lim_{t \to t_0} g(t)\right)\]if both limits exist.

In our scenario, we've got functions \( f_i(t) \) and \( g_i(t) \) making up the vector components of \( \mathbf{r}_1(t) \) and \( \mathbf{r}_2(t) \). Applying this rule to each scalar product term in the cross product helps to find the limit of each resulting component separately.

• Evaluate the limits of individual terms like \( f_2(t)g_3(t) \) or \( f_3(t)g_2(t) \) as \( t \to t_0 \).
• Multiply the component limits to obtain the final limit vector components, such as \( \lim_{t \to t_0} (f_2(t)g_3(t) - f_3(t)g_2(t)) \mathbf{i} \).

This rule confirms that the limit of the vector product is indeed the product of the limits, making it a powerful and proofworthy method to handle complex calculus problems involving vector limits.

Vector Calculus

Vector calculus extends basic calculus through vectors and covers operations like differentiation and integration in multiple dimensions. Unlike simple scalar calculus, vector calculus handles quantities with magnitude and direction, making it suitable for fields like physics and engineering.

When studying limits of vector functions, one essential concept is maintaining vector integrity through transformations and operations like cross products. In vector calculus:

• Vectors are not limited to numerical values but have direction and dimensions.
• Operations maintain not only numerical result but vector-oriented results such as direction and space alignment.
• Such transformation helps in assessing physical phenomena, like the orientation of forces or the flow of currents.

Tools like the cross product, determinants, and limit rules allow analysts to gauge predictions about vector behavior under varying conditions. This makes vector calculus rich and deeply integrated into applied sciences.