91Ó°ÊÓ

In vector calculus, derivatives are crucial for understanding how a vector function changes as its parameter varies. For a given vector function, like \( \mathbf{r}(t) = t\mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \), the derivative \( \mathbf{r}'(t) \) is computed by differentiating each component of the vector with respect to \( t \). This operation provides us with a new vector representing the rate of change of the original vector. It helps to visualize how the path defined by \( \mathbf{r}(t) \) evolves in space.

• The derivative of \( t \) is \( 1 \), giving \( \mathbf{i} \) as the new component from \( t\mathbf{i} \).
• The derivative of \( t^2 \) is \( 2t \), resulting in \( 2t\mathbf{j} \).
• The derivative of \( t^3 \) is \( 3t^2 \), which becomes the component \( 3t^2\mathbf{k} \).

From this, \( \mathbf{r}'(t) = \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \). By continuing this differentiation process, the second derivative \( \mathbf{r}''(t) \) describes the change of the first derivative, essentially showing how the rate of change itself is evolving. Here, \( \mathbf{r}''(t) = 0\mathbf{i} + 2\mathbf{j} + 6t\mathbf{k} \). This second derivative extends our understanding of the dynamics and curvature of the vector path.

The cross product is a fundamental operation in vector calculus, used to find a vector perpendicular to two given vectors in three-dimensional space. When handling derivative vectors like \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \), computing their cross product gives meaningful insights into the geometric properties of their relationship. The cross product of \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \) is calculated using a determinant:\[\mathbf{r}'(t) \times \mathbf{r}''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \1 & 2t & 3t^2 \0 & 2 & 6t \end{vmatrix}\]From this determinant, you compute each component of the resulting vector.

• The \( \mathbf{i} \)-component results from \( 12t^2 - 6t^2 = 6t^2 \).
• The \( \mathbf{j} \)-component is \( -6t \times 1 \), simplifying to \( -6t \).
• The \( \mathbf{k} \)-component is \( 2 \times 1 \), resulting in \( 2 \).

Thus, the cross product is \( 6t^2\mathbf{i} - 6t\mathbf{j} + 2\mathbf{k} \). This vector indicates the orientation and magnitude of the perpendicular axis to the plane formed by \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \).

The dot product is an operation that takes two vectors and returns a scalar, capturing the extent to which they align. In the problem, finding the dot product of the original vector \( \mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{j} + t^3\mathbf{k} \) and the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) = 6t^2\mathbf{i} - 6t\mathbf{j} + 2\mathbf{k} \) delivers a scalar value encapsulating this alignment. The dot product is computed by multiplying corresponding components and summing the results:

• The \( \mathbf{i} \)-components: \( t \times 6t^2 = 6t^3 \).
• The \( \mathbf{j} \)-components: \( t^2 \times -6t = -6t^3 \).
• The \( \mathbf{k} \)-components: \( t^3 \times 2 = 2t^3 \).

The sum of these products is \( 6t^3 - 6t^3 + 2t^3 = 2t^3 \). Finally, calculating the limit as \( t \to 1 \) gives \( 2(1)^3 = 2 \). This result, \( 2 \), reflects the precise combined interaction between the vector \( \mathbf{r}(t) \) and the orientation upon which the cross product's vector aligns at \( t = 1 \).