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 \( \kappa \) is zero at both \( t=0 \) and \( t=1 \). The curvature is not greater at any point because it is zero throughout.

Step-by-step solution

Parametric Derivatives

First, we need to find the first and second derivatives of the vector function \( \vec{r}(t) = \langle 4t+2, 3t-1, 2t+5 \rangle \). The first derivative, \( \vec{r}'(t) \), is obtained by differentiating each component with respect to \( t \):\[ \vec{r}'(t) = \langle 4, 3, 2 \rangle \]. The second derivative, \( \vec{r}''(t) \), is obtained by differentiating each component of \( \vec{r}'(t) \): \[ \vec{r}''(t) = \langle 0, 0, 0 \rangle \].

Curvature Formula

The curvature \( \kappa \) of a space curve defined by a vector function \( \vec{r}(t) \) is given by:\[ \kappa = \frac{|| \vec{r}'(t) \times \vec{r}''(t) ||}{||\vec{r}'(t)||^3} \]. In our case, we have \( \vec{r}'(t) = \langle 4, 3, 2 \rangle \) and \( \vec{r}''(t) = \langle 0, 0, 0 \rangle \).

Cross Product Calculation

The cross product \( \vec{r}'(t) \times \vec{r}''(t) \) of our derivatives is:\[ \vec{r}'(t) \times \vec{r}''(t) = \langle 4, 3, 2 \rangle \times \langle 0, 0, 0 \rangle = \langle 0, 0, 0 \rangle \].

Magnitude Calculations

Now calculate the magnitudes:1. \( ||\vec{r}'(t) \times \vec{r}''(t)|| = ||\langle 0, 0, 0 \rangle|| = 0. \)2. \( ||\vec{r}'(t)|| = ||\langle 4, 3, 2 \rangle|| = \sqrt{4^2 + 3^2 + 2^2} = \sqrt{29}. \)

Curvature Evaluation

Using the previously obtained values:\[ \kappa = \frac{0}{(\sqrt{29})^3} = 0. \]Hence, at any value of \( t \), the curvature of \( C \) is zero.

Analysis and Evaluation at Points

Since \( \kappa = 0 \) for any value of \( t \), it means there is no curvature at either of the two points, \( t=0 \) and \( t=1 \). The curve \( C \) is a straight line and has constant zero curvature everywhere.

Key concepts

Parametric Equations

Parametric equations are a way to describe mathematical objects using parameters. Unlike traditional linear equations that relate variables directly, parametric equations define each coordinate as a separate function of a common parameter.

In the context of curves in three-dimensional space, parametric equations help us understand how an object moves through space over time. For example, the equation \( \vec{r}(t) = \langle 4t+2, 3t-1, 2t+5 \rangle \) describes a straight line in space as \( t \) changes. Here, \( t \) is the parameter, allowing the representation of each x, y, and z coordinate as separate functions:

• x-coordinate: \( 4t + 2 \)
• y-coordinate: \( 3t - 1 \)
• z-coordinate: \( 2t + 5 \)

This approach is especially useful in calculus when analyzing the dynamics of curves and trajectories.

Vector Functions

Vector functions are powerful tools in calculus, involving functions that output vectors instead of scalar values. They are widely used to depict curves and surfaces in space.

A vector function, like \( \vec{r}(t) = \langle 4t+2, 3t-1, 2t+5 \rangle \), provides a compact way to represent the motion of a particle along the curve. It captures the essence of the curve by encapsulating all three coordinates (x, y, and z) into one unified expression.

Calculating derivatives of vector functions is key to understanding changing behavior on curves. For each component of the vector, you differentiate independently with respect to the parameter \( t \). This results in the first derivative \( \vec{r}'(t) = \langle 4, 3, 2 \rangle \), giving the direction of the curve at any point in space. The second derivative, \( \vec{r}''(t) = \langle 0, 0, 0 \rangle \), indicates that the curve is indeed straight, showing constant direction.

Cross Product

The cross product is an essential operation in vector calculus, used to find a vector orthogonal to two given vectors in three-dimensional space. It's crucial in understanding geometric concepts like area, volume, and orientation.

In curvature analysis, the cross product of the first and second derivatives of a curve, \( \vec{r}'(t) \times \vec{r}''(t) \), is applied for determining if the curve changes direction.

For our exercise, evaluating the cross product of \( \vec{r}'(t) = \langle 4, 3, 2 \rangle \) and \( \vec{r}''(t) = \langle 0, 0, 0 \rangle \), we find:

• The cross product is \( \langle 0, 0, 0 \rangle \), indicating no orthogonal vector can be formed, as expected from a linear path.

This shows that the curve is a straight line, with the cross product confirming no deviation in direction.

Derivative Calculations

Derivative calculations play a pivotal role in examining the nature of curves by providing insights on rates of change. In the context of curvature, derivatives are used to understand how a curve bends in space.

The first derivative, \( \vec{r}'(t) \), indicates the tangent vector at any point on the curve, showing how the position changes with respect to the parameter \( t \). The magnitude of this derivative, \( ||\vec{r}'(t)|| \), offers the speed of movement along the curve.

The second derivative, \( \vec{r}''(t) \), helps us explore the acceleration and how the direction of the curve's tangent changes. If \( \vec{r}''(t) \) is zero, as in this exercise, it implies that the curve is straight with no acceleration, hence a curvature of zero.

• Key calculations include:
• \( ||\vec{r}'(t)|| = \sqrt{4^2 + 3^2 + 2^2} = \sqrt{29} \), determining the trajectory's speed.
• \( \kappa = \frac{||\vec{r}'(t) \times \vec{r}''(t)||}{||\vec{r}'(t)||^3} = 0\), indicating no curvature.

In summary, derivative calculations confirm the linearity of the curve in this specific example.