In multivariable calculus, the terms path and curve have distinct but related meanings:

1. Curve:

   • A curve is generally a geometric object, described as a set of points in space that smoothly connect without sharp angles or breaks. In mathematical terms, it is a continuous image of an interval.
   • It is often represented as the trace of a function $( \mathbf{r}(t) = (x(t), y(t), z(t)) )$ , where $( t )$ is a parameter typically representing time or arc length, and $( \mathbf{r}(t) )$ describes the position of a point in space.

2. Path:

   • A path includes the notion of movement along a curve, described by a parameterized function $( \mathbf{r}(t) )$, where $( t )$ varies over an interval $([a, b])$ . A path not only defines the shape of the curve but also gives a direction and often a speed of traversal.
   • The parameter $( t )$ indicates how a particle moves along the curve, defining directionality and possibly how fast the particle moves (if the speed varies).

Mathematical Definition

• A curve is a set $( C = { \mathbf{r}(t) : t \in [a, b] } )$ for a continuous function $( \mathbf{r}: [a, b] \to \mathbb{R}^n )$ .
• A path is a function $( \mathbf{r}(t) = (x(t), y(t), z(t), \dots) )$ that is continuous and differentiable (in most cases), mapping a parameter $( t )$ to points in $( \mathbb{R}^n )$ .

Examples:

• Path with parameterization: Suppose $( \mathbf{r}(t) = (\cos t, \sin t) )$ , $( t \in [0, 2\pi] )$ . This defines a circular path in the $( xy )$ -plane.

• Curve from a path: The curve is the geometric locus $( C = { (\cos t, \sin t) : t \in [0, 2\pi] } )$ , which represents the unit circle in the plane.

Differences:

• A curve is often considered as the "shape" (the set of points), whereas a path provides a "journey" along the shape, including orientation and velocity. For example, $( \mathbf{r}_1(t) = (\cos t, \sin t) )$ and $( \mathbf{r}_2(t) = (\cos(-t), \sin(-t)) )$ describe the same curve (circle), but opposite paths (directions).

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Velocity and Tangency

The velocity vector on a path describes the direction and rate of change of position as a particle moves along the path. It is derived from the parameterized representation of the path and is defined as the derivative of the position vector $(\mathbf{r}(t))$ with respect to the parameter $(t)$ .

Formula for the Velocity Vector

If a path is parameterized by a function $(\mathbf{r}(t) = (x(t), y(t), z(t)))$ , the velocity vector is given by:

\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \left(\frac{dx(t)}{dt}, \frac{dy(t)}{dt}, \frac{dz(t)}{dt}\right).

Interpretation

1. Direction: The velocity vector points in the direction of instantaneous motion along the path.
2. Magnitude: The magnitude of the velocity vector, $(|\mathbf{v}(t)|)$ , represents the speed of motion at time $(t)$ .

Example

Suppose the path is parameterized as:

\mathbf{r}(t) = (\cos t, \sin t, t), \quad t \in [0, 2\pi].

The velocity vector is:

\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \left(-\sin t, \cos t, 1\right).

• At $(t = 0), (\mathbf{v}(0) = (0, 1, 1))$ .
• At $(t = \pi/2), (\mathbf{v}(\pi/2) = (-1, 0, 1))$ .

Geometric Representation

The velocity vector at any point on the path is a tangent vector to the curve at that point, pointing in the direction of motion. Its length corresponds to the speed.

To find the tangent vector and the equation of the tangent line at a specific point on a parameterized path, follow these steps:

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1. Tangent Vector

The tangent vector is the velocity vector at a given point.

• Suppose the path is parameterized as $(\mathbf{r}(t) = (x(t), y(t), z(t)))$ , and you want the tangent vector at $(t = t_0)$ .
• Compute the derivative of $(\mathbf{r}(t))$ :

  \mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \left(\frac{dx(t)}{dt}, \frac{dy(t)}{dt}, \frac{dz(t)}{dt}\right).

• Evaluate $(\mathbf{v}(t)) at (t = t_0)$ :

  \mathbf{v}(t_0) = \left(\frac{dx(t)}{dt}\bigg|_{t=t_0}, \frac{dy(t)}{dt}\bigg|_{t=t_0}, \frac{dz(t)}{dt}\bigg|_{t=t_0}\right).

This gives the tangent vector at $(t = t_0)$ .

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2. Equation of the Tangent Line

The tangent line is a line passing through the point $(\mathbf{r}(t_0))$ in the direction of the tangent vector $(\mathbf{v}(t_0))$.

1. Point on the Line: The point $(\mathbf{r}(t_0) = (x(t_0), y(t_0), z(t_0)))$ is on the tangent line.

2. Direction of the Line: The tangent vector $(\mathbf{v}(t_0))$ gives the direction.

3. Parametric Equation: The tangent line can be written as:

   \mathbf{l}(s) = \mathbf{r}(t_0) + s \mathbf{v}(t_0),

where $(s)$ is a real parameter.

In coordinates, this is:

   \mathbf{l}(s) = \big(x(t_0) + s v_x, \; y(t_0) + s v_y, \; z(t_0) + s v_z\big),

where $(\mathbf{v}(t_0) = (v_x, v_y, v_z))$.

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Example

Let $(\mathbf{r}(t) = (\cos t, \sin t, t))$ , and find the tangent vector and tangent line at $(t = \pi/4)$ .

1. Compute the Derivative:

   \mathbf{v}(t) = \frac{d}{dt} (\cos t, \sin t, t) = (-\sin t, \cos t, 1).

2. Evaluate at $(t = \pi/4)$ :

   \mathbf{v}(\pi/4) = \left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, 1\right).

3. Point on the Curve:

   \mathbf{r}(\pi/4) = \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, \frac{\pi}{4}\right).

4. Tangent Line: Parametric equation:

   \mathbf{l}(s) = \mathbf{r}(\pi/4) + s \mathbf{v}(\pi/4),

which expands to:

   \mathbf{l}(s) = \left(\frac{\sqrt{2}}{2} - s\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} + s\frac{\sqrt{2}}{2}, \frac{\pi}{4} + s\right).

Visualization

The tangent vector is the arrow pointing in the direction of motion, while the tangent line is the infinite extension of that arrow through the point.