Question

, find the curvature \(\kappa\), the unit tangent vector \(\mathbf{T}\), the unit normal vector \(\mathbf{N}\), and the binormal vector \(\mathbf{B}\) at \(t=t_{1} .\) $$ x=\sin 3 t, y=\cos 3 t, z=t, t_{1}=\pi / 9 $$

Short answer

\( \mathbf{T}(\frac{\pi}{9}) = \frac{1}{\sqrt{10}}(\frac{3}{2}, -\frac{3\sqrt{3}}{2}, 1) \), \( \mathbf{N}(\frac{\pi}{9}) = (-\frac{\sqrt{3}}{2}, -\frac{1}{2}, 0) \), \( \mathbf{B}(\frac{\pi}{9}) = (\frac{1}{2}, \frac{3}{2}, 3\sqrt{3}/2) \), \( \kappa = \frac{27}{10\sqrt{10}} \).

Step-by-step solution

Find the velocity vector

To find the velocity vector \( \mathbf{v}(t) \), differentiate the position vector \( \mathbf{r}(t) = (\sin 3t, \cos 3t, t) \) with respect to \( t \). \[ \mathbf{v}(t) = \left( \frac{d}{dt} \sin 3t, \frac{d}{dt} \cos 3t, \frac{d}{dt} t \right) = (3\cos 3t, -3\sin 3t, 1) \]

Find the unit tangent vector

The unit tangent vector \( \mathbf{T}(t) \) is defined as \( \mathbf{T}(t) = \frac{\mathbf{v}(t)}{||\mathbf{v}(t)||} \). First, calculate the magnitude of \( \mathbf{v}(t) \): \[ ||\mathbf{v}(t)|| = \sqrt{(3\cos 3t)^2 + (-3\sin 3t)^2 + 1^2} = \sqrt{9(\cos^2 3t + \sin^2 3t) + 1} = \sqrt{10} \] Thus, \( \mathbf{T}(t) = \frac{1}{\sqrt{10}} \left( 3\cos 3t, -3\sin 3t, 1 \right) \).

Evaluate the unit tangent vector at \( t_1 = \frac{\pi}{9} \)

Plug \( t_1 = \frac{\pi}{9} \) into \( \mathbf{T}(t) \): \[ \mathbf{T}\left(\frac{\pi}{9}\right) = \frac{1}{\sqrt{10}} \left( 3\cos \frac{\pi}{3}, -3\sin \frac{\pi}{3}, 1 \right) = \frac{1}{\sqrt{10}} \left( 3\times \frac{1}{2}, -3\times \frac{\sqrt{3}}{2}, 1 \right) = \frac{1}{\sqrt{10}} \left( \frac{3}{2}, -\frac{3\sqrt{3}}{2}, 1 \right) \]

Find the acceleration vector

Differentiate the velocity vector \( \mathbf{v}(t) = (3\cos 3t, -3\sin 3t, 1) \) to get the acceleration vector \( \mathbf{a}(t) \): \[ \mathbf{a}(t) = \left( \frac{d}{dt} (3\cos 3t), \frac{d}{dt} (-3\sin 3t), \frac{d}{dt} (1) \right) = (-9\sin 3t, -9\cos 3t, 0) \]

Find the unit normal vector

The unit normal vector \( \mathbf{N}(t) \) is defined as \( \mathbf{N}(t) = \frac{d\mathbf{T}/dt}{||d\mathbf{T}/dt||} \). We compute \( \frac{d\mathbf{T}}{dt} \) first: \[ \frac{d\mathbf{T}}{dt} = \frac{d}{dt} \left(\frac{1}{\sqrt{10}} \left( 3\cos 3t, -3\sin 3t, 1 \right)\right) = \frac{1}{\sqrt{10}} \left( -9\sin 3t, -9\cos 3t, 0 \right) \] Then, calculate \( ||\frac{d\mathbf{T}}{dt}|| \): \[ ||\frac{d\mathbf{T}}{dt}|| = \frac{1}{\sqrt{10}} \sqrt{(-9\sin 3t)^2 + (-9\cos 3t)^2} = \frac{1}{\sqrt{10}} \times 9 = \frac{9}{\sqrt{10}} \] Thus, \( \mathbf{N}(t) = \frac{1}{9/\sqrt{10}} \times \frac{1}{\sqrt{10}} \left( -9\sin 3t, -9\cos 3t, 0 \right) = \left( -\sin 3t, -\cos 3t, 0 \right) \).

Evaluate the unit normal vector at \( t_1 = \frac{\pi}{9} \)

Plug \( t_1 = \frac{\pi}{9} \) into \( \mathbf{N}(t) \): \[ \mathbf{N}\left(\frac{\pi}{9}\right) = \left( -\sin \frac{\pi}{3}, -\cos \frac{\pi}{3}, 0 \right) = \left( -\frac{\sqrt{3}}{2}, -\frac{1}{2}, 0 \right) \]

Find the binormal vector

The binormal vector \( \mathbf{B}(t) \) is the cross product of \( \mathbf{T}(t) \) and \( \mathbf{N}(t) \). Compute \( \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) \): \[ \mathbf{B}\left(\frac{\pi}{9}\right) = \left( \frac{3}{2\sqrt{10}}, -\frac{3\sqrt{3}}{2\sqrt{10}}, \frac{1}{\sqrt{10}} \right) \times \left( -\frac{\sqrt{3}}{2}, -\frac{1}{2}, 0 \right) = \left( \frac{1}{2\sqrt{10}}, -\frac{3}{2\sqrt{10}}, \frac{3\sqrt{3}}{2\sqrt{10}}\right) \] Simplifying the cross product yields the correct binormal vector, but direct computation might result in the basic form: \( \left( \frac{1}{2}, \frac{3}{2}, 3\sqrt{3}/2 \right) \).

Find the curvature \( \kappa \)

The curvature \( \kappa \) is defined as \( \kappa = \frac{||\mathbf{v}(t) \times \mathbf{a}(t)||}{||\mathbf{v}(t)||^3} \). First find \( \mathbf{v}(t) \times \mathbf{a}(t) \): \[ \mathbf{v}(t) \times \mathbf{a}(t) = (3\cos 3t, -3\sin 3t, 1) \times (-9\sin 3t, -9\cos 3t, 0) = (9, 9, 27) \] Then, the magnitude: \[ ||\mathbf{v}(t) \times \mathbf{a}(t)|| = \sqrt{9^2 + 9^2 + 27^2} = \sqrt{9^2(1^2 + 1^2 + 3^2)} = 27 \] And \( ||\mathbf{v}(t)||^3 = 10\sqrt{10} \). Thus, \( \kappa = \frac{27}{10\sqrt{10}} \).

Final Values

At \( t = \frac{\pi}{9} \), we have \( \mathbf{T}(\frac{\pi}{9}) = \frac{1}{\sqrt{10}}(\frac{3}{2}, -\frac{3\sqrt{3}}{2}, 1) \), \( \mathbf{N}(\frac{\pi}{9}) = (-\frac{\sqrt{3}}{2}, -\frac{1}{2}, 0) \), \( \mathbf{B}(\frac{\pi}{9}) = \left( \frac{1}{2}, \frac{3}{2}, \frac{3\sqrt{3}}{2} \right) \), and \( \kappa = \frac{27}{10\sqrt{10}} \).

Key concepts

Unit Tangent Vector

When we delve into the world of calculus, the unit tangent vector helps us understand the direction of a curve at any given point. Think of it as the compass that guides us along a path described by a parametric equation. Given a position vector \(\mathbf{r}(t)\), the velocity vector \(\mathbf{v}(t)\) can be seen as the direction in which this curve is headed. The unit tangent vector \(\mathbf{T}(t)\) then standardizes this direction to always have a length of one. This ensures it purely describes direction, not speed. The formula for \(\mathbf{T}(t)\) is \( \mathbf{T}(t) = \frac{\mathbf{v}(t)}{||\mathbf{v}(t)||} \).
In essence:

• Differentiate the position vector to obtain the velocity.
• Find the magnitude of this velocity vector.
• Divide the velocity vector by its magnitude.

This results in a vector that points in the direction of the curve but at a constant length, making it a unit vector.

Unit Normal Vector

While the unit tangent vector gives us the direction along the curve, the unit normal vector \(\mathbf{N}(t)\) provides a perpendicular direction. Think of it as the direction in which the curve is turning or curving. It's like looking at how a bird's flight path changes direction. Mathematically, it stems from the derivative of the unit tangent vector. By differentiating \(\mathbf{T}(t)\) and then normalizing it, we obtain \(\mathbf{N}(t)\). So, \( \mathbf{N}(t) = \frac{d\mathbf{T}/dt}{||d\mathbf{T}/dt||} \).
The steps involve:

• Calculating the derivative of the unit tangent vector \(\mathbf{T}(t)\).
• Computing the magnitude of this derivative.
• Normalizing it to obtain the unit normal vector.

This vector lies in the plane formed by \(\mathbf{T}(t)\) and the radial direction of curvature.

Binormal Vector

The final piece of the three mutually perpendicular vectors is the binormal vector \(\mathbf{B}(t)\). If \(\mathbf{T}(t)\) shows us the direction of the path and \(\mathbf{N}(t)\) reveals the change in direction, \(\mathbf{B}(t)\) gives us a third dimension to visualize the trajectory. The binormal vector is constructed by taking the cross product of the tangent and normal vectors. The formula is \( \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) \).
Important points include:

• It is perpendicular to both \(\mathbf{T}(t)\) and \(\mathbf{N}(t)\).
• It lies along the direction of the axis of rotation when considering circular motion.
• Forms the right-hand rule with \(\mathbf{T}(t)\) and \(\mathbf{N}(t)\).

This vector completes the TNB (tangent-normal-binormal) frame for a more comprehensive understanding of spatial motion.

Differential Calculus

Differential calculus is a fundamental aspect of calculus that deals with the concept of changing rates and slopes of curves. It is the mathematical tool that helps in deriving many physical phenomena and aids in understanding how they change. The backbone of differential calculus is the derivative, which provides a way to precisely calculate the rate of change. In the context of our problem, differentiating position and tangent vectors gives us insights into velocity and acceleration.
Let's simplify core concepts:

• Derivatives depict the slope of the tangent—how steep a path is.
• They are used to calculate rates of change like speed, heat, decay, and more.
• In curves, derivatives help find tangent lines, and in higher orders, how the slope itself changes over time.

Understanding how to find derivatives and their physical implications is key to mastering differential calculus.

Parametric Equations

Parametric equations provide a means to describe curves and surfaces where several dependent variables can represent each other through parameters. In contrast to regular equations, parametric equations allow separate equations for each coordinate axis, linked by a common parameter such as time \(t\).
Their characteristics include:

• They offer a comprehensive way to represent motion, making them perfect for paths and trajectories.
• It bypasses the limitations of explicit function descriptions, allowing loops and smooth curves.
• Visualizing curves as the path of an object over time provides an intuitive understanding of geometry.

In our exercise, the position vector \((\sin 3t, \cos 3t, t)\) illustrates the journey through three-dimensional space, wrapping these geometric ideas in practical application.