Question

Relationship between \(\mathbf{T}, \mathbf{N},\) and a Show that if an object accelerates in the sense that \(d^{2} s / d t^{2}>0\) and \(\kappa \neq 0,\) then the acceleration vector lies between \(\mathbf{T}\) and \(\mathbf{N}\) in the plane of \(\mathbf{T}\) and \(\mathbf{N}\). If an object decelerates in the sense that \(d^{2} s / d t^{2}<0,\) then the acceleration vector lies in the plane of \(\mathbf{T}\) and \(\mathbf{N},\) but not between \(\mathbf{T}\) and \(\mathbf{N}\)

Short answer

Short Answer: In the case of positive acceleration, the acceleration vector lies between the Tangent vector (T) and the Normal vector (N) in the plane of T and N, since both the components of the acceleration vector are positive. In the case of negative acceleration, the acceleration vector still lies in the plane of T and N, but not between T and N, because the component along the T vector becomes negative while the component along the N vector remains positive.

Step-by-step solution

Case 1: Positive Acceleration

If an object accelerates in the sense that \(d^{2} s / d t^{2}>0\) and \(\kappa \neq 0,\) then the acceleration vector lies between \(\mathbf{T}\) and \(\mathbf{N}\) in the plane of \(\mathbf{T}\) and \(\mathbf{N}\). To prove this, we can take a look at the acceleration vector formula: \(\mathbf{a} = \frac{d^2 s}{dt^2} \mathbf{T} + (\frac{ds}{dt})^2 \kappa \mathbf{N}\) Since both \(\frac{d^2 s}{dt^2} >0\) and \(\kappa \neq 0\), this indicates that both the components of the acceleration vector are positive. As a result, the acceleration vector will be a linear combination of the T and N vectors in the positive direction, which means that the acceleration vector will lie between T and N in the plane of T and N.

Case 2: Negative Acceleration

If an object decelerates in the sense that \(d^{2} s / d t^{2}<0,\) then the acceleration vector lies in the plane of \(\mathbf{T}\) and \(\mathbf{N},\) but not between \(\mathbf{T}\) and \(\mathbf{N}\). Again, we take a look at the acceleration vector formula: \(\mathbf{a} = \frac{d^2 s}{dt^2} \mathbf{T} + (\frac{ds}{dt})^2 \kappa \mathbf{N}\) With \(\frac{d^2 s}{dt^2} <0\) and \(\kappa \neq 0\), the component along the T vector becomes negative while the component along the N vector remains positive. Consequently, the acceleration vector will still lie in the plane of T and N, but in the direction opposite of the T vector. This means that the acceleration vector will not lie between T and N as it does in the case of positive acceleration. With the analysis of these two cases, we have proved the given statements in the exercise.

Key concepts

Tangent Vector

The tangent vector, often denoted as \( \mathbf{T} \), is a fundamental concept in vector calculus and physics. It represents the direction of a curve at a given point and is tangent to the path an object is following.
This vector is essential when considering motion, as it shows the instantaneous direction of travel.

• Characteristics: The tangent vector has a magnitude of one and is always oriented in the direction of motion.
• Calculation: For a parameterized curve \( \mathbf{r}(t) \), the tangent vector is obtained by differentiating the position vector \( \mathbf{r} \) with respect to \( t \) and normalizing it, i.e., \( \mathbf{T} = \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|} \).

Understanding the tangent vector helps us describe how quickly and in what direction an object is moving along a path, which is crucial when analyzing acceleration vectors.

Normal Vector

The normal vector, denoted as \( \mathbf{N} \), plays an important role alongside the tangent vector in describing motion along a curve. Whereas the tangent vector represents the direction of motion, the normal vector indicates the direction in which the path is curving.

• Meaning: The normal vector is perpendicular to the tangent vector at any given point on the curve.
• Rotation and Curvature: This vector shows where the curve is bending and is crucial for calculating curvature.
• Calculation: From the tangent vector, the normal vector is obtained by differentiating \( \mathbf{T} \) with respect to the parameter \( t \) and normalizing the result, i.e., \( \mathbf{N} = \frac{\mathbf{T}'(t)}{\| \mathbf{T}'(t) \|} \).

The normal vector is vital for understanding how the acceleration vector can be decomposed into components: one aligned with the tangent direction and another perpendicular to it, indicating changes in direction rather than speed.

Curvature

Curvature, represented by the symbol \( \kappa \), quantifies how sharply a curve bends at a particular point. It gives us a measure of the rate at which the direction of the tangent vector is changing.

• Definition: Mathematically, curvature is defined as the magnitude of the rate of change of the tangent vector with respect to arc length, \( s \): \( \kappa = \left\| \frac{d \mathbf{T}}{ds} \right\| \).
• Geometric Interpretation: Curvature can be thought of as the inverse of the radius of the osculating circle at a point on the curve. A higher curvature indicates a tighter bend.
• Role in Acceleration: In the analysis of acceleration vectors, curvature signifies the strength of the component of acceleration perpendicular to the tangent vector, essentially pointing in the direction of the normal vector \( \mathbf{N} \).

Understanding curvature is essential for comprehending how the overall acceleration vector behaves as it combines directional changes with speed variations on a curve.