91Ó°ÊÓ

A tangent vector is a vector that is tangent to a curve at a given point, meaning it touches the curve at that point and has the same direction as the curve itself. Often, we can find tangent vectors by taking the derivative of the parametric equations describing a curve.
In the context of our exercise, we have a curve defined by the equation \(x^2 + y^2 = 4\), which can be expressed in parametric form using polar coordinates: \(x = 2\cos t\) and \(y = 2\sin t\).
To find the tangent vector to the curve at any point, we differentiate the parametric equation to get:

• \(C'(t) = \langle -2\sin t, 2\cos t \rangle\)

This tangent vector tells us the direction of the curve at any specific point. In the exercise, checking tangent vectors involves verifying if the vector from the field \(\mathbf{F}\) is parallel to the tangent vector of the curve. A vector field is tangent to a curve if at all points on the curve, the field vectors align with these tangent vectors.
In simpler terms, a field is tangent everywhere along the curve when its directions, at each point, match the curve’s local direction.

Normal vectors are perpendicular to a given curve at a point, meaning they point directly away from the curve. To determine when a vector field is normal to a curve, we analyze the dot product between vectors from the field and the curve’s tangent vectors.
In the exercise, this boils down to computing the dot product of the vector field \(\mathbf{F} = \langle x, y \rangle\) with the tangent vector \(C'(t) = \langle -2\sin t, 2\cos t \rangle\):

• \(\mathbf{F} \cdot C'(t) = \langle 2\cos t, 2\sin t \rangle \cdot \langle -2\sin t, 2\cos t \rangle\)
• The dot product simplifies to \(0\) along the curve if they are orthogonal.

In this solution, it turns out that no points satisfy this condition, making the vector field nowhere normal to the curve. Understanding normal vectors helps us comprehend how the curve behaves in relation to the surrounding space, as they tell us how to move perpendicularly off the curve.

Parametric curves are a powerful way to describe complicated shapes and paths by using parameters. Unlike regular equations that give direct relationships between \(x\) and \(y\), parametric equations describe \(x\) and \(y\) in terms of another variable, commonly denoted as \(t\).
In this exercise, the curve \(C\) is expressed parametrically with the relations:

• \(x = 2\cos t\)
• \(y = 2\sin t\)

Here, the parameter \(t\) can represent angles, making it especially useful in describing circles and ellipses.
The main advantage of parametric equations is their ability to describe not only the position but also the motion along the curve. By changing \(t\), you can track how a point moves, which lets us calculate tangent and normal vectors effortlessly, as seen in the solution.
Parametric curves transform calculus problems on motion and trajectory into more visual and solvable scenarios, granting a deeper insight into the curve's geometry.