Question

A curve \(C\) is described along with 2 points on \(C\). (a) Using a sketch, determine at which of these points the curvature is greater. (b) Find the curvature \(\kappa\) of \(C\), and evaluate \(\kappa\) at each of the 2 given points. \(C\) is defined by \(\vec{r}(t)=\langle 4 t+2,3 t-1,2 t+5\rangle ;\) points given at \(t=0\) and \(t=1\).

Short answer

The curvature at both points \( t = 0 \) and \( t = 1 \) is zero; the curve is a straight line.

Step-by-step solution

Understanding the Problem

We have a vector function \( \vec{r}(t) = \langle 4t+2, 3t-1, 2t+5 \rangle \) that defines a curve \( C \). We need to find and compare the curvature \( \kappa \) at the points corresponding to \( t = 0 \) and \( t = 1 \).

Formula for Curvature

The curvature \( \kappa \) of a curve given by a position vector \( \vec{r}(t) \) is defined as \( \kappa = \frac{\| \vec{v}(t) \times \vec{a}(t) \|}{\| \vec{v}(t) \|^3} \), where \( \vec{v}(t) \) is the velocity vector \( \frac{d\vec{r}}{dt} \) and \( \vec{a}(t) \) is the acceleration vector \( \frac{d^2\vec{r}}{dt^2} \).

Compute Velocity and Acceleration

Calculate the first derivative (velocity) and second derivative (acceleration) of \( \vec{r}(t) \):- \( \vec{v}(t) = \frac{d}{dt}\langle 4t+2,3t-1,2t+5\rangle = \langle 4, 3, 2 \rangle \)- \( \vec{a}(t) = \frac{d^2}{dt^2}\langle 4t+2,3t-1,2t+5\rangle = \langle 0, 0, 0 \rangle \)

Compute the Cross Product

Find the cross product \( \vec{v}(t) \times \vec{a}(t) \):\[\vec{v}(t) \times \vec{a}(t) = \langle 4, 3, 2 \rangle \times \langle 0, 0, 0 \rangle = \langle 0, 0, 0 \rangle\]Since the cross product is zero, the vectors are parallel or \( \vec{a}(t) \) is zero.

Calculate Magnitude and Curvature

The magnitude of \( \vec{v}(t) \) is:\[\| \vec{v}(t) \| = \sqrt{4^2 + 3^2 + 2^2} = \sqrt{29}\]The curvature \( \kappa = \frac{\| \vec{v}(t) \times \vec{a}(t) \|}{\| \vec{v}(t) \|^3} = \frac{0}{(\sqrt{29})^3} = 0 \).This shows that the curve is a straight line with zero curvature at both points \( t=0 \) and \( t=1 \).

Conclusion and Answer

Since the curvature is 0 at both points, there is no difference in curvature. The curve is a straight line indicating a constant velocity and no acceleration.

Key concepts

Vector Calculus

Vector calculus is an essential branch of mathematics that deals with vector fields and operations on vector spaces. In simpler terms, it is the calculus of vector fields. It extends techniques from calculus, like differentiation and integration, to vector functions. Vector functions are expressions like \( \vec{r}(t) = \langle 4t+2, 3t-1, 2t+5 \rangle \), which describe a path or curve in a multi-dimensional space.

Vector calculus is particularly important when we need to describe and analyze the motion of objects in physics or engineering. It helps us comprehend the orientation, direction, and rate of change of vector quantities, making it useful in solving problems related to motion in three-dimensional space.

• Vectors represent quantities that have both magnitude and direction.
• Differentiating vectors allows us to find velocity and acceleration.
• Integration of vectors can help calculate pathways or fluxes.

Its application extends to physical phenomena such as electromagnetism, fluid flow, and the motion of celestial bodies, making it indispensable in the fields of physics and engineering.

Curvature Formula

The curvature formula is instrumental in understanding the shape or bend of a curve at a specific point. In vector calculus, the curvature \( \kappa \) indicates how sharply a curve turns at a particular point. The formula for the curvature of a parametric vector function \( \vec{r}(t) \) is given by \( \kappa = \frac{\| \vec{v}(t) \times \vec{a}(t) \|}{\| \vec{v}(t) \|^3} \), where:

• \( \vec{v}(t) = \frac{d}{dt} \vec{r}(t) \) is the velocity vector.
• \( \vec{a}(t) = \frac{d^2}{dt^2} \vec{r}(t) \) is the acceleration vector.

The numerator, \( \| \vec{v}(t) \times \vec{a}(t) \| \), is the magnitude of the cross product of the velocity and acceleration vectors. This represents the amount of rotation or "twisting" of the curve at the point.
The denominator, \( \| \vec{v}(t) \|^3 \), is the cubed magnitude of the velocity vector, normalizing the measure of curvature considering the speed of traversal along the curve.

When the curvature is zero, as seen in this exercise, it suggests the trajectory is a straight line with no bending. The path remains uniform, indicating no change in orientation, exemplified by constant velocity and zero acceleration.

Velocity and Acceleration Vectors

Velocity and acceleration vectors are fundamental in understanding motion in vector calculus. These vectors help us explore how the position of an object changes over time:

- **Velocity vector**, \( \vec{v}(t) \): This is the first derivative of the position vector \( \vec{r}(t) \) with respect to time \( t \). It represents the rate of change of position and describes the speed and direction of movement. In our problem, \( \vec{v}(t) = \langle 4, 3, 2 \rangle \), signifies constant speed in a straight line, as its components don't depend on \( t \).

- **Acceleration vector**, \( \vec{a}(t) \): This is the second derivative of the position vector, illustrating the rate of change of velocity. It tells us how quickly the velocity changes over time. Here, \( \vec{a}(t) = \langle 0, 0, 0 \rangle \), meaning there is no change in speed or direction of the path.

A constant velocity with zero acceleration suggests that the object is moving in a straight line without speeding up or slowing down. By comprehending these vectors, we can delve into intricate systems like orbits, trajectories, and flows in diverse physical settings.