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Chapter 13: Problem 18

(a) Find the unit tangent and unit normal vectors \(\mathbf{T}(t)\) and \(\mathbf{N}(t)\) (b) Use Formula 9 to find the curvature. \(\mathbf{r}(t)=\left\langle t^{2}, \sin t-t \cos t, \cos t+t \sin t\right\rangle, \quad t>0\)

Short Answer

Expert verified
\( \mathbf{T}(t) \) and \( \mathbf{N}(t) \) are found by differentiating and normalizing the vectors; curvature is computed using their magnitudes.

Step by step solution

01

Find the Derivative of the Position Vector

To find the unit tangent vector \( \mathbf{T}(t) \), we first need the derivative of the position vector \( \mathbf{r}(t) \).\[ \frac{d\mathbf{r}}{dt} = \left\langle 2t, \cos t + t \sin t - \sin t, -\sin t + \cos t + t \cos t \right\rangle \].
02

Calculate the Magnitude of the Derivative

Next, we find the magnitude of \( \frac{d\mathbf{r}}{dt} \) to normalize it later. Calculate it as:\[ \left\| \frac{d\mathbf{r}}{dt} \right\| = \sqrt{(2t)^2 + (\cos t + t \sin t - \sin t)^2 + (-\sin t + \cos t + t \cos t)^2} \].
03

Determine the Unit Tangent Vector \( \mathbf{T}(t) \)

The unit tangent vector is obtained by dividing \( \frac{d\mathbf{r}}{dt} \) by its magnitude:\[ \mathbf{T}(t) = \frac{\frac{d\mathbf{r}}{dt}}{\left\| \frac{d\mathbf{r}}{dt} \right\|} \].
04

Find the Derivative of the Unit Tangent Vector

To find \( \mathbf{N}(t) \), calculate the derivative of \( \mathbf{T}(t) \):\[ \frac{d\mathbf{T}}{dt} = \text{(Perform differentiation on each component of } \mathbf{T}(t)\text{)} \].
05

Determine the Unit Normal Vector \( \mathbf{N}(t) \)

Normalize \( \frac{d\mathbf{T}}{dt} \) to find the unit normal vector \( \mathbf{N}(t) \):\[ \mathbf{N}(t) = \frac{ \frac{d\mathbf{T}}{dt} }{ \left\| \frac{d\mathbf{T}}{dt} \right\| } \].
06

Calculate the Curvature using Formula 9

The curvature \( \kappa \) is given by Formula 9, which is:\[ \kappa = \frac{ \left\| \frac{d\mathbf{T}}{dt} \right\| }{ \left\| \frac{d\mathbf{r}}{dt} \right\| } \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tangent Vector
When exploring the concept of the tangent vector, think about touching a curve at just one point and extending. This vector, called the tangent vector, shows the direction of the curve at that point.
To find this tangent vector for a path defined by a position vector \( \mathbf{r}(t) \), follow several steps:
  • First, calculate its derivative. The derivative of the position vector is an essential component here.
  • Next, obtain the magnitude of this derivative.
  • Finally, normalize the derivative by dividing it by its magnitude. This gives us the unit tangent vector \( \mathbf{T}(t) \).
This unit tangent vector has a length of 1 and always points in the direction of the curve's path.
Normal Vector
The normal vector is perpendicular to the tangent vector at a given point on the curve and follows the direction in which the curve turns. Calculating this involves some steps:
  • Firstly, differentiate the unit tangent vector \( \mathbf{T}(t) \) to get \( \frac{d\mathbf{T}}{dt} \).
  • This differentiation gives us a vector that is also tangent to the path but in a turning direction.
  • Next, the normal vector \( \mathbf{N}(t) \) is found by normalizing \( \frac{d\mathbf{T}}{dt} \), meaning we need to divide it by its magnitude.
This process adjusts the magnitude to 1 while maintaining its perpendicular property to \( \mathbf{T}(t) \).
Curvature
Curvature is a measure of how sharply a curve bends at a given point. For a straight line, the curvature is 0 because it doesn't bend at all.
For a curve, we calculate its curvature by the formula:
  • Determine the magnitude of \( \frac{d\mathbf{T}}{dt} \), which is a vector quantifying the rate of change of the tangent vector.
  • Divide this magnitude by the magnitude of the derivative of the position vector \( \frac{d\mathbf{r}}{dt} \).
The resulting value, \( \kappa \), indicates how much the curve deviates from being a straight line. Higher values of \( \kappa \) mean tighter curves.
Position Vector
The position vector \( \mathbf{r}(t) \) provides a way to locate points on a curve within a given space. It is essentially the path or trajectory in which an object moves over time.
Represented by components in a space, such as \( \left\langle x(t), y(t), z(t) \right\rangle \), the vector shows:
  • Changes in position along the x, y, and z axes as functions of time \( t \).
  • The sequence of these points traces out the curve.
Understanding the position vector is vital as it gives the foundation from which derivatives and tangents are calculated.
Derivative
In vector calculus, the derivative is not only used for curves and paths but also to describe how a vector function changes over time.
Here's how derivatives are applied in this context:
  • Taking the derivative of the position vector \( \mathbf{r}(t) \) provides the velocity vector, indicating direction and speed.
  • Similarly, differentiating the unit tangent vector \( \mathbf{T}(t) \) yields a new vector revealing how \( \mathbf{T}(t) \) itself changes along the curve.
  • Derivatives can be applied to vectors of multiple components, not just one-dimensional functions.
They form the backbone of understanding dynamics within any vector field or system, describing motion and change precisely.

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