91Ó°ÊÓ

Understanding the concept of a vector function is key to exploring parametric equations. A vector function, like \( \mathbf{r}(t) = (t^{2} + 1) \mathbf{i} + t \mathbf{j} + 3 \mathbf{k} \), assigns a vector to every value of \( t \). This vector represents a point in three-dimensional space, where each component corresponds to a spatial direction: \( x, y, \) and \( z \).

• The first component, \( (t^{2} + 1) \), affects the x-direction.
• The second component, \( t \), affects the y-direction.
• The third component, \( 3 \), affects the z-direction.

This structure allows vector functions to describe complex paths through space by changing \( t \). In this case, a unique curve is formed as \( t \) varies over all real numbers. The role of \( t \) is to parameterize this path, providing a means to calculate each point along the curve precisely.

Curve sketching with parametric equations involves translating the algebraic expressions of a vector function into a visual representation. This can help us understand how the curve behaves in three dimensions. Given \( \mathbf{r}(t) = (t^{2} + 1) \mathbf{i} + t \mathbf{j} + 3 \mathbf{k} \), we need to consider each parametric equation:

• \( x = t^2 + 1 \)
• \( y = t \)
• \( z = 3 \)

The task is to visualize this information on a graph. The curve appears as a projection onto the YZ-plane, which simplifies to a standard parabola \( x = y^2 + 1 \) in the plane \( z = 3 \). To sketch the curve, plot this parabola parallel to and above the x-axis at \( z = 3 \). Remember to indicate the orientation by noting how \( y \) (or \( t \)) progresses from negative to positive values.

Parametric equations allow us to describe a curve in three-dimensional space explicitly. This means specifying each coordinate \( x, y, \) and \( z \) as a function of one or more parameters. In our scenario, the parameter \( t \) informs us of the position of each point on the curve by:

• \( x = t^2 + 1 \)
• \( y = t \)
• \( z = 3 \)

Here, \( z \) remains constant, which means this curve lies flat on the width defined by \( z = 3 \). Parametric equations like these offer more flexibility and detail in plotting compared to standard Cartesian equations by allowing the curve's path and orientation to be explicitly managed through the variation of \( t \). Analyzing how \( x, y, \) and \( z \) change lets us predict and sketch the curve’s path with precision.

Understanding planes in three dimensions is crucial when dealing with parametric surfaces or curves like the one in this exercise. A plane is essentially a flat, two-dimensional surface extending infinitely in three-dimensional space. In the context of our problem, we have a specific plane given by \( z = 3 \). This plane is parallel to the xy-plane but is exactly 3 units above it.

• Every point on this plane has a z-coordinate of 3.
• Curves confined to this plane will not move in the z-direction.

When sketching the curve, understanding the plane relationship is vital to accurately place the curve in space. Since our curve is \( x = y^2 + 1 \) on the \( z = 3 \) plane, we know it will look like a simple parabola located on this horizontal plane, letting us further notice its connection to axes.