91Ó°ÊÓ

In mathematics, a parametric curve is a curve that is expressed through a set of functions, each representing a component of a point in space. For the given exercise, the vector equation \( \mathbf{r}(t) = t \mathbf{i} + \sin t \mathbf{j} + \cos t \mathbf{k} \) is an example of a parametric curve in three-dimensional space. Here, the parameter \( t \) acts as a controller, moving a point along the curve as \( t \) changes. Parametric equations are powerful when dealing with curves because they allow us to describe complex geometric shapes in a straightforward way.

• The component \( t \mathbf{i} \) suggests movement along the x-axis
• \( \sin t \mathbf{j} \) implies an oscillation along the y-axis
• \( \cos t \mathbf{k} \) indicates another oscillation along the z-axis

The elegance of parametric curves is that they offer a unique blend of simplicity and flexibility in modeling spatial relationships.

Calculating the derivative of a vector function is essential in understanding various properties of curves, such as their length, orientation, and more. For the given vector function \( \mathbf{r}(t) = t\mathbf{i} + \sin t\mathbf{j} + \cos t \mathbf{k} \), the derivative \( \mathbf{r}'(t) \) is determined by differentiating each component with respect to \( t \). This yields:
\[ \mathbf{r}'(t) = \frac{d}{dt}(t) \mathbf{i} + \frac{d}{dt}(\sin t) \mathbf{j} + \frac{d}{dt}(\cos t) \mathbf{k} = \mathbf{i} + \cos t \mathbf{j} - \sin t \mathbf{k} \]
The derivative \( \mathbf{r}'(t) \) indicates the velocity of a point as it travels along the curve. In simpler terms, it reveals how fast and in what direction the point is moving at any given moment.

• The term \( \mathbf{i} \) shows constant motion along the x-axis
• The \( \cos t \mathbf{j} \) indicates varying changes in velocity along the y-axis
• \(-\sin t \mathbf{k} \) represents variable velocity along the z-axis

These derivatives are key for further calculations like the curve's length.

The length of a curve in three-dimensional space involves the integration of the magnitude of its derivative. This tells us how far a point moves along a curve. In the exercise, the length is derived by integrating the magnitude of \( \mathbf{r}'(t) \). We begin by finding the magnitude:
\[ \| \mathbf{r}'(t) \| = \sqrt{(1)^2 + (\cos t)^2 + (-\sin t)^2} \]
Utilizing the trigonometric identity \( \cos^2 t + \sin^2 t = 1 \), simplifies the expression to:
\[ \| \mathbf{r}'(t) \| = \sqrt{2} \]
With the magnitude \( \sqrt{2} \), the curve length is obtained by integrating with respect to \( t \):
\[ L = \int_{0}^{2} \sqrt{2} \, dt \]
This simplifies to \( L = 2\sqrt{2} \), which is the total length of the curve over the given interval from \( t = 0 \) to \( t = 2 \).

In this context, integration is used to find the total length of the curve. Particularly, it involves integrating a constant, which can simplify calculations. The integral in our exercise:
\[ L = \int_{0}^{2} \sqrt{2} \, dt \]
Given \( \sqrt{2} \) is a constant, it can be taken out of the integral, transforming the problem into a straightforward evaluation:
\[ \sqrt{2} \cdot \int_{0}^{2} \, dt \]
This integral results in:
\[ \sqrt{2} \cdot (2 - 0) = 2\sqrt{2} \]
Thus providing the length of the curve as \( 2\sqrt{2} \). Integration thus becomes a critical tool in calculating various properties like areas, volumes, and in this case, curve length. It helps us in accumulating values that vary according to a certain rule or function, offering deeper insights into geometric shapes and their dimensions.