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A parametric curve is a way of representing a curve using a set of equations. These equations express the coordinates of the points on the curve as functions of a variable, commonly denoted as \( t \). Each value of \( t \) corresponds to a specific point on the curve. For example, with the vector equation \( \mathbf{r}(t) = (1 + \cos t) \mathbf{i} + (2 + \sin t) \mathbf{j} \),

• \( x(t) = 1 + \cos t \) represents the horizontal component, and
• \( y(t) = 2 + \sin t \) represents the vertical component.

By varying \( t \), you trace a path called a curve in the plane. In this case, the curve traces a circle because both \( x \) and \( y \) components are based on trigonometric functions of \( t \). As \( t \) moves from \( 0 \) to \( 2\pi \), it completes one full revolution, sketching a circle in the plane.

A plane curve is simply a curve that lies on a two-dimensional plane. It can be defined by a function \( y = f(x) \) or by parametric equations of the form \( x = f(t) \) and \( y = g(t) \). Plane curves can take on various shapes, such as lines, circles, or more complex patterns. In the example of \( \mathbf{r}(t) = (1 + \cos t) \mathbf{i} + (2 + \sin t) \mathbf{j} \), the plane curve drawn is a circle. This particular plane curve is centered at (1, 2) with a radius of one unit. The circle lies entirely on the plane defined by the coordinate axes \( x \) and \( y \). Understanding plane curves helps in recognizing how certain algebraic expressions can manifest as geometric figures in two-dimensional space.

The tangent vector provides vital information about the direction and slope of a curve at a particular point. If you envision a curve as a road, the tangent vector tells us the direction the road is heading at any specific moment. To find the tangent vector of a parametric curve \( \mathbf{r}(t) \), you calculate its derivative, noted as \( \mathbf{r}'(t) \). This derivative,

• \( x'(t) = \frac{d}{dt}(1 + \cos t) = -\sin t \)
• \( y'(t) = \frac{d}{dt}(2 + \sin t) = \cos t \)

forms the vector \( \mathbf{r}'(t) = -\sin t \mathbf{i} + \cos t \mathbf{j} \). At \( t = \pi/6 \), it becomes \( -1/2 \mathbf{i} + \sqrt{3}/2 \mathbf{j} \). This vector is tangent to the circle at the point \((1 + \sqrt{3}/2, 2.5)\) and points in the direction of the curve's motion.

The derivative of a vector function, like the one considered in this task, gives a vector that represents the rate of change of the original vector function with respect to \( t \). It provides insights into how the vector changes as you move along the curve.Taking the derivative of each component of \( \mathbf{r}(t) \) gives us:

• \( \frac{dx}{dt} = \frac{d}{dt}(1 + \cos t) = -\sin t \)
• \( \frac{dy}{dt} = \frac{d}{dt}(2 + \sin t) = \cos t \)

Thus, the derivative \( \mathbf{r}'(t) = -\sin t \mathbf{i} + \cos t \mathbf{j} \) is the tangent vector, as previously discussed.The significance of finding the derivative lies in its ability to provide instantaneous information about the slope and direction of the curve at any point \( t \). This is crucial for analyzing and sketching curves, as well as understanding their behavior in a plane.