Problem 34 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the curvature \(\kappa\), the unit tangent vector \(\mathbf{T}\), the unit normal vector \(\mathbf{N}\), and the binormal vector \(\mathbf{B}\) at \(t=t_{1}\). \(x=\ln t, y=3 t, z=t^{2}, t_{1}=2\)

Step-by-Step Solution

Answer

The unit tangent vector \( \mathbf{T}(2) = \frac{1}{5.025}(0.5, 3, 4) \).

Step 1: Parametrize the curve

The curve is given by the vector function \( \mathbf{r}(t) = ( \ln t, 3t, t^2 ) \). Here, \( x = \ln t \), \( y = 3t \), \( z = t^2 \) are the components of the curve.

Step 2: Find the derivative \( \mathbf{r}'(t) \)

Differentiate each component of \( \mathbf{r}(t) \): \[ \mathbf{r}'(t) = \left( \frac{1}{t}, 3, 2t \right) \] This is the velocity vector at any parameter \( t \).

Step 3: Evaluate \( \mathbf{r}'(t_1) \)

Substitute \( t_1 = 2 \) to find the velocity vector at \( t = 2 \):\[ \mathbf{r}'(2) = \left( \frac{1}{2}, 3, 4 \right) \].

Step 4: Find the unit tangent vector \( \mathbf{T}(t) \)

Calculate the magnitude of \( \mathbf{r}'(t_1) \):\[ |\mathbf{r}'(2)| = \sqrt{\left(\frac{1}{2}\right)^2 + 3^2 + 4^2} = \sqrt{\frac{1}{4} + 9 + 16} = \sqrt{25.25} = 5.025 \] The unit tangent vector is: \[ \mathbf{T}(2) = \frac{1}{5.025}\left( \frac{1}{2}, 3, 4 \right) \].

Step 5: Find the derivative \( \mathbf{T}'(t) \)

Differentiate \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} \) with respect to \( t \). Substitute \( t_1 = 2 \) to find \( \mathbf{T}'(2) \).

Step 6: Find the unit normal vector \( \mathbf{N}(2) \)

The unit normal vector \( \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|} \). Calculate \( \mathbf{T}'(2) \) and its magnitude, then normalize it.

Step 7: Calculate the binormal vector \( \mathbf{B}(2) \)

The binormal vector is given by \( \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) \). Calculate this cross product at \( t_1 = 2 \).

Step 8: Calculate the curvature \( \kappa(t) \)

Curvature is given by the formula \[ \kappa(t) = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3} \].Derivative \( \mathbf{r}''(t) \) can be computed as: \[ \mathbf{r}''(t) = \left( -\frac{1}{t^2}, 0, 2 \right) \] Substitute \( t_1 = 2 \) into the formula to find the curvature at \( t = 2 \).

Key Concepts

Unit Tangent Vector

The unit tangent vector, often denoted as \( \mathbf{T}(t) \), provides essential information about the direction of a curve at any given point. Imagine walking along a twisting path; the direction you're facing at any point is the tangent vector. The formula to calculate the unit tangent vector is: - \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} \) - \( \mathbf{r}'(t) \) is the derivative of the position vector \( \mathbf{r}(t) \). It's like the velocity, showing how fast and in what direction you're moving on the curve.- \(|\mathbf{r}'(t)|\) is the magnitude—or length—of the velocity vector. It's the speed at which you are moving along the curve.
When you divide the velocity vector by its magnitude, you're left with a vector of length one, which only indicates direction—not speed.
The resulting \( \mathbf{T}(t) \) remains normal to the curve, ensuring it always points just along the path, without veering off to the sides.

Unit Normal Vector

The unit normal vector, \( \mathbf{N}(t) \), helps describe how the curve bends or turns in space. It's like the guide rails of a curvy train track, showing where and how sharply the track is turning.To find the unit normal vector: 1. Differentiate the unit tangent vector \( \mathbf{T}(t) \) to get \( \mathbf{T}'(t) \). This new derivative vector shows the change in direction along the path. 2. Calculate the magnitude of \( \mathbf{T}'(t) \). 3. Normalize by dividing \( \mathbf{T}'(t) \) by its magnitude: \[ \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|} \] With this normalization, \( \mathbf{N}(t) \) will have a length of one and direction perpendicular to the tangent vector, pointing towards the center of curvature. Think of \( \mathbf{N}(t) \) as always being 'normal' or at right angles to \( \mathbf{T}(t) \), reinforcing the curve's changing direction.

Binormal Vector

The binormal vector, denoted as \( \mathbf{B}(t) \), completes the orthogonal triad of vectors used to describe the geometry of the curve in space. It's the vector you get when you cross the unit tangent vector with the unit normal vector.To compute the binormal vector:

• Perform the cross product between the unit tangent vector \( \mathbf{T}(t) \) and the unit normal vector \( \mathbf{N}(t) \):

\[ \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) \] The resulting \( \mathbf{B}(t) \) will be perpendicular to both \( \mathbf{T}(t) \) and \( \mathbf{N}(t) \). This means it points outward, away from the plane formed by \( \mathbf{T}(t) \) and \( \mathbf{N}(t) \), offering a sense of the curve's twist or spiraling nature, like how a rollercoaster climbs or falls in 3D space.This trio—\( \mathbf{T}(t) \), \( \mathbf{N}(t) \), and \( \mathbf{B}(t) \)—are fundamental in understanding the three-dimensional orientation of curves, often used in applications from computer graphics to physics.