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Magazine Chapter 10: Problem 14
(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
\( \mathbf{T}(t) = \left\langle \frac{2}{\sqrt{5}}, \frac{\sin t}{\sqrt{5}}, \frac{\cos t}{\sqrt{5}} \right\rangle \) and \( \kappa(t) = \frac{1}{5t} \).
Step by step solution
01
Compute the Derivative of the Vector Function
The derivative of \( \mathbf{r}(t) \) is computed coordinate-wise. Let \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle = \langle t^2, \sin t - t \cos t, \cos t + t \sin t \rangle \).1. Differentiate \( x(t) = t^2 \). We get: \( x'(t) = 2t \).2. Differentiate \( y(t) = \sin t - t \cos t \). We use the product rule for \(-t \cos t\): \( y'(t) = \cos t - (1)\cos t + t \sin t = t \sin t \).3. Differentiate \( z(t) = \cos t + t \sin t \). Again use the product rule for \( t \sin t \): \( z'(t) = -\sin t + (1)\sin t + t \cos t = t \cos t \).Therefore, \( \mathbf{r}'(t) = \langle 2t, t\sin t, t\cos t \rangle \).
02
Compute the Magnitude of the Velocity Vector
Calculate the magnitude of \( \mathbf{r}'(t) = \langle 2t, t\sin t, t\cos t \rangle \):\[\| \mathbf{r}'(t) \| = \sqrt{(2t)^2 + (t \sin t)^2 + (t \cos t)^2}\]Simplify:\[\| \mathbf{r}'(t) \| = \sqrt{4t^2 + t^2(\sin^2 t + \cos^2 t)} = \sqrt{4t^2 + t^2}\]\[\| \mathbf{r}'(t) \| = \sqrt{5t^2} = \sqrt{5}|t|\]Since \( t > 0 \), this simplifies to \( \sqrt{5}t \).
03
Find the Unit Tangent Vector \( \mathbf{T}(t) \)
The unit tangent vector \( \mathbf{T}(t) \) is given by:\[\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|}\]Substitute the values:\[\mathbf{T}(t) = \frac{1}{\sqrt{5}t} \langle 2t, t \sin t, t \cos t \rangle = \left\langle \frac{2}{\sqrt{5}}, \frac{\sin t}{\sqrt{5}}, \frac{\cos t}{\sqrt{5}} \right\rangle\]
04
Compute the Derivative of the Unit Tangent Vector \( \mathbf{T}'(t) \)
Differentiate each component of \( \mathbf{T}(t) = \left\langle \frac{2}{\sqrt{5}}, \frac{\sin t}{\sqrt{5}}, \frac{\cos t}{\sqrt{5}} \right\rangle \).1. Derivative of \( \frac{2}{\sqrt{5}} \) is \( 0 \).2. Derivative of \( \frac{\sin t}{\sqrt{5}} \) is \( \frac{\cos t}{\sqrt{5}} \).3. Derivative of \( \frac{\cos t}{\sqrt{5}} \) is \( -\frac{\sin t}{\sqrt{5}} \).So, \[ \mathbf{T}'(t) = \left\langle 0, \frac{\cos t}{\sqrt{5}}, -\frac{\sin t}{\sqrt{5}} \right\rangle \].
05
Compute the Magnitude of \( \mathbf{T}'(t) \) to Find Curvature
Next, compute the magnitude of \( \mathbf{T}'(t) \):\[\| \mathbf{T}'(t) \| = \sqrt{0^2 + \left(\frac{\cos t}{\sqrt{5}}\right)^2 + \left(-\frac{\sin t}{\sqrt{5}}\right)^2} = \sqrt{\frac{\cos^2 t}{5} + \frac{\sin^2 t}{5}}\]Using \( \sin^2 t + \cos^2 t = 1 \), it simplifies to:\[\| \mathbf{T}'(t) \| = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}\]
06
Determine the Unit Normal Vector \( \mathbf{N}(t) \)
The unit normal vector \( \mathbf{N}(t) \) is the normalized derivative of the unit tangent vector:\[\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\| \mathbf{T}'(t) \|} = \frac{\left\langle 0, \frac{\cos t}{\sqrt{5}}, -\frac{\sin t}{\sqrt{5}} \right\rangle}{\frac{1}{\sqrt{5}}}\]This simplifies to:\[\mathbf{N}(t) = \langle 0, \cos t, -\sin t \rangle\]
07
Compute the Curvature Using Formula 9
The curvature \( \kappa(t) \) is given by:\[\kappa(t) = \frac{\| \mathbf{T}'(t) \|}{\| \mathbf{r}'(t) \|} = \frac{\frac{1}{\sqrt{5}}}{\sqrt{5}t}\]Simplifying, we have:\[\kappa(t) = \frac{1}{5t}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Unit Tangent Vector
In vector calculus, the unit tangent vector is a crucial concept for understanding the direction of a curve at any point. When working with a vector function like \( \mathbf{r}(t) \), the unit tangent vector, denoted \( \mathbf{T}(t) \), provides us with a direction vector that has a magnitude of one. This is accomplished by taking the derivative of the vector function to find the velocity vector, \( \mathbf{r}'(t) \), and then normalizing it.
- Step 1: Derive the Velocity Vector
The derivative \( \mathbf{r}'(t) = \langle 2t, t\sin t, t\cos t \rangle \) gives us the velocity at any point \( t \). - Step 2: Magnitude of the Velocity Vector
Calculate \( \|\mathbf{r}'(t)\| = \sqrt{5}t \) to normalize it. - Step 3: Normalize the Velocity Vector
The unit tangent vector becomes \( \mathbf{T}(t) = \left\langle \frac{2}{\sqrt{5}}, \frac{\sin t}{\sqrt{5}}, \frac{\cos t}{\sqrt{5}} \right\rangle \).
Exploring the Unit Normal Vector
The unit normal vector \( \mathbf{N}(t) \) complements the unit tangent vector by providing insight into how the tangent vector changes as you move along a curve. It is orthogonal to the tangent vector and faces inward or outward depending on the curvature direction. Here's how you find it:
- Compute the Derivative of \( \mathbf{T}(t) \)
You first need to differentiate the components of \( \mathbf{T}(t) \). The derivative \( \mathbf{T}'(t) = \left\langle 0, \frac{\cos t}{\sqrt{5}}, -\frac{\sin t}{\sqrt{5}} \right\rangle \) reflects how the tangent vector alters with \( t \). - Magnitude of \( \mathbf{T}'(t) \)
Easily determine \( \|\mathbf{T}'(t)\| = \frac{1}{\sqrt{5}} \) using the Pythagorean identity. - Normalize to Get \( \mathbf{N}(t) \)
By normalizing, you find \( \mathbf{N}(t) = \left\langle 0, \cos t, -\sin t \right\rangle \).
Calculating Curvature
Curvature provides us with a measure of how quickly a curve changes direction at a given point. Basically, a higher curvature means the curve turns more sharply. Within our vector calculus framework, we can calculate curvature \( \kappa(t) \) by using the formula:\[ \kappa(t) = \frac{\| \mathbf{T}'(t) \|}{\| \mathbf{r}'(t) \|} \] Given our previous work, we see:
- Magnitude of \( \mathbf{T}'(t) \)
Already found to be \( \frac{1}{\sqrt{5}} \). - Magnitude of \( \mathbf{r}'(t) \)
Was calculated as \( \sqrt{5}t \). - Resulting Curvature
Substitute to get \( \kappa(t) = \frac{1}{5t} \), simplifying the relationship between rate of change in direction concerning arc length.
