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Chapter 11: Problem 13

Find the unit tangent vector \(\mathbf{T}\) and the curvature \(\kappa\) for the following parameterized curves. $$\mathbf{r}(t)=\langle 2 t, 4 \sin t, 4 \cos t\rangle$$

Short Answer

Expert verified
Question: Find the unit tangent vector T(t) and the curvature 魏 for the following parameterized curve: \(\mathbf{r}(t) = \langle 2t, 4\sin t, 4\cos t \rangle\) Answer: The unit tangent vector T(t) is given by: \(\mathbf{T}(t) = \langle \frac{2}{\sqrt{20}}, \frac{4 \cos t}{\sqrt{20}}, \frac{-4 \sin t}{\sqrt{20}} \rangle\) And the curvature 魏 is given by: \(\kappa = \frac{4}{5 \sqrt{20}}\)

Step by step solution

01

Find the first and second derivatives

We need to find \(\mathbf{r'}(t)\) and \(\mathbf{r''}(t)\). $$\mathbf{r}(t) = \langle 2 t, 4 \sin t, 4 \cos t\rangle$$ Taking the first derivative: $$\mathbf{r'}(t) = \langle 2, 4 \cos t, -4 \sin t\rangle$$ Taking the second derivative: $$\mathbf{r''}(t) = \langle 0, -4 \sin t, -4 \cos t\rangle$$
02

Calculate the unit tangent vector \(\mathbf{T}\)

We need to normalize the first derivative to obtain the unit tangent vector \(\mathbf{T}(t)\). First, compute the magnitude of \(\mathbf{r'}(t)\): $$||\mathbf{r'}(t)|| = \sqrt{(2)^2 + (4 \cos t)^2 + (-4 \sin t)^2} = \sqrt{4 + 16(\cos^2 t + \sin^2 t)} = \sqrt{4 + 16} = \sqrt{20}$$ Now, normalize \(\mathbf{r'}(t)\): $$\mathbf{T}(t) = \frac{\mathbf{r'}(t)}{||\mathbf{r'}(t)||} = \langle \frac{2}{\sqrt{20}}, \frac{4 \cos t}{\sqrt{20}}, \frac{-4 \sin t}{\sqrt{20}} \rangle$$
03

Calculate the curvature \(\kappa\)

Using the formula for curvature \(\kappa = \frac{||\mathbf{r'}(t) \times \mathbf{r''}(t)||}{||\mathbf{r'}(t)||^3}\), we will compute the cross product and the curvature. First, calculate \(\mathbf{r'}(t) \times \mathbf{r''}(t)\): $$\mathbf{r'}(t) \times \mathbf{r''}(t) = \langle 2, 4 \cos t, -4 \sin t\rangle \times \langle 0, -4 \sin t, -4 \cos t\rangle = \langle -16 \cos^2 t - 16 \sin^2 t, 0, 2 \rangle$$ Compute the magnitude: $$||\mathbf{r'}(t) \times \mathbf{r''}(t)|| = \sqrt{(-16 \cos^2 t - 16 \sin^2 t)^2 + 0^2 + (2)^2} = \sqrt{(16\cos^2t + 16\sin^2t)^2 + 4} = \sqrt{256} = 16$$ Now, we will find the curvature \(\kappa\): $$\kappa = \frac{||\mathbf{r'}(t) \times \mathbf{r''}(t)||}{||\mathbf{r'}(t)||^3} = \frac{16}{(\sqrt{20})^3} = \frac{16}{20 \sqrt{20}} = \frac{4}{5 \sqrt{20}}$$ So, the unit tangent vector is given by: $$\mathbf{T}(t) = \langle \frac{2}{\sqrt{20}}, \frac{4 \cos t}{\sqrt{20}}, \frac{-4 \sin t}{\sqrt{20}} \rangle$$ And the curvature is given by: $$\kappa = \frac{4}{5 \sqrt{20}}$$

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Key Concepts

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

Unit Tangent Vector and its Importance
Imagine walking along a path. At every point, you might ask, "Which way am I headed?" This is precisely what the **unit tangent vector** describes for parameterized curves. The unit tangent vector, denoted as \( \mathbf{T}(t) \), represents the direction in which a curve moves at any given point. It's a normalized vector, meaning it has a length of one, ensuring that it only provides direction, not magnitude.
To find \( \mathbf{T}(t) \), we start with the first derivative of the curve, \( \mathbf{r'}(t) \), which gives the rate of change or the instantaneous direction of the curve.
  • First, calculate \( \mathbf{r'}(t) \).
  • Compute its magnitude, \(||\mathbf{r'}(t)||\).
  • Normalize by dividing each component of \( \mathbf{r'}(t) \) by its magnitude: \( \mathbf{T}(t) = \frac{\mathbf{r'}(t)}{||\mathbf{r'}(t)||} \).
For our specific example with \( \mathbf{r}(t)=\langle 2t, 4 \sin t, 4 \cos t \rangle \), the unit tangent vector \( \mathbf{T}(t) \) is calculated as \( \langle \frac{2}{\sqrt{20}}, \frac{4 \cos t}{\sqrt{20}}, \frac{-4 \sin t}{\sqrt{20}} \rangle \). This vector efficiently conveys only the curve's direction without being influenced by the curve's speed.
Understanding Parameterized Curves
Understanding parameterized curves is like unraveling how shapes form as you alter a variable鈥攊n this case, \( t \). They give you the freedom to trace a path as a function of some parameter. Here, \( \mathbf{r}(t)=\langle 2t, 4 \sin t, 4 \cos t \rangle \) is a parameterized curve where each function within the brackets serves a distinct role.
  • \( 2t \): Controls how fast and in which direction the path moves along the x-axis as \( t \) changes.
  • \( 4 \sin t \) and \( 4 \cos t \): These trigonometric functions describe circular motion on the y and z axes, forming a helix due to the \( 2t \) term.
The beauty of parameterized curves is that they allow the tracing of complex paths, combining linear and rotational motions seamlessly. This is what makes them ideal when modeling real-life scenarios, like the trajectory of a thrown ball or the path of an orbiting satellite. By changing the parameter \( t \), you can "walk" along the curve and capture every nuance of the shape.
Role and Computation of Derivatives
Derivatives are the champions of calculus, acting like detectives that unravel the detailed behavior of functions. For parameterized curves, derivatives help in understanding how the curve evolves as we change the parameter \( t \).
In the context of our curve \( \mathbf{r}(t)=\langle 2t, 4 \sin t, 4 \cos t \rangle \), we compute its derivative to find \( \mathbf{r'}(t) \). Here's why derivatives are essential:
  • First Derivative \( \mathbf{r'}(t) \): Tells you the velocity vector of the point on the curve, indicating the rate and direction of change.
  • Second Derivative \( \mathbf{r''}(t) \): Offers insights into the acceleration, demonstrating how the curve's direction itself is changing.
Using derivatives:
  • \( d/dt \langle 2t, 4 \sin t, 4 \cos t \rangle = \langle 2, 4 \cos t, -4 \sin t \rangle \)
  • Further derivation gives \( \mathbf{r''}(t)= \langle 0, -4 \sin t, -4 \cos t \rangle \)
These calculations allow us to go deeper into the curve's nature, precisely mapping how it shifts through space and time, thus arming us with the ability to describe the curve's behavior fully. Thus, whether you're studying physical phenomena or purely abstract shapes, derivatives make the hidden dynamics of parameterized curves visible.

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Most popular questions from this chapter

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