Chapter 13: Problem 4
Find \(\mathbf{T}, \mathbf{N},\) and \(\kappa\) for the plane curves. \(\mathbf{r}(t)=(\cos t+t \sin t) \mathbf{i}+(\sin t-t \cos t) \mathbf{j}, \quad t>0\)
Short Answer
Expert verified
\( \mathbf{T} = (- \cos t, \sin t), \mathbf{N} = (\sin t, \cos t), \kappa = \frac{1}{|t|} \)."
Step by step solution
01
Find the Derivative of the Position Vector
To find the tangent vector \( \mathbf{T} \), compute the derivative of the position vector \( \mathbf{r}(t) \). This derivative is \( \mathbf{r}'(t) = \big( \frac{d}{dt}(\cos t + t \sin t), \frac{d}{dt}(\sin t - t \cos t) \big) \). Calculating these derivatives, we get \( \mathbf{r}'(t) = (-t \cos t, t \sin t) \).
02
Normalize the Tangent Vector
The unit tangent vector \( \mathbf{T} \) is the normalized derivative of the position vector. Calculate the magnitude of \( \mathbf{r}'(t) \): \( |\mathbf{r}'(t)| = \sqrt{(-t \cos t)^2 + (t \sin t)^2} = \sqrt{t^2} = |t| \). The unit tangent vector \( \mathbf{T} = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} = \left( - \cos t, \sin t \right) \).
03
Find the Derivative of the Tangent Vector
Compute the derivative of the unit tangent vector \( \mathbf{T}(t) = ( - \cos t, \sin t) \) to find the normal vector. This derivative is \( \mathbf{T}'(t) = ( \sin t, \cos t) \).
04
Normalize the Normal Vector
The principal unit normal vector \( \mathbf{N} \) is the normalized derivative of the tangent vector. The magnitude of \( \mathbf{T}'(t) \) is \( |\mathbf{T}'(t)| = \sqrt{(\sin t)^2 + (\cos t)^2} = 1 \). So, \( \mathbf{N} = \mathbf{T}'(t) = ( \sin t, \cos t) \).
05
Compute the Curvature \(\kappa\)
The curvature \( \kappa \) is given by \( \kappa = \frac{| \mathbf{T}'(t) |}{| \mathbf{r}'(t) |} = \frac{1}{|t|} \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Tangent Vector
When you think about a tangent vector, picture it as the arrow that just touches the curve at a specific point, pointing in the direction the curve is headed. This vector tells you the immediate direction in which the curve is moving.
- **Find the Derivative:** The first step in finding the tangent vector is to take the derivative of the position vector, \( \mathbf{r}(t) \). This gives you \( \mathbf{r}'(t) \), which is essentially the velocity of the object moving along the curve.- **Normalization:** To convert this tangent vector into a unit tangent vector, \( \mathbf{T} \), you normalize it by dividing by its magnitude. This step helps us understand the direction of the tangent without being concerned about its length.For the given curve, \( \mathbf{r}'(t) = (-t \cos t, t \sin t) \), and once normalized, we have \( \mathbf{T} = (-\cos t, \sin t) \). This means our tangent vector points exactly in the direction defined by the angle \( t \).
- **Find the Derivative:** The first step in finding the tangent vector is to take the derivative of the position vector, \( \mathbf{r}(t) \). This gives you \( \mathbf{r}'(t) \), which is essentially the velocity of the object moving along the curve.- **Normalization:** To convert this tangent vector into a unit tangent vector, \( \mathbf{T} \), you normalize it by dividing by its magnitude. This step helps us understand the direction of the tangent without being concerned about its length.For the given curve, \( \mathbf{r}'(t) = (-t \cos t, t \sin t) \), and once normalized, we have \( \mathbf{T} = (-\cos t, \sin t) \). This means our tangent vector points exactly in the direction defined by the angle \( t \).
Normal Vector
A normal vector is perpendicular to the tangent vector and points towards the inside of the curve's bend. It's like the curve's imaginary 'compass needle,' always at a right angle to where you're heading instantaneously.
- **Differentiate the Tangent Vector:** The normal vector originates from taking the derivative of the unit tangent vector \( \mathbf{T}(t) \). This derivative gives you the change in direction of the tangent vector.- **Normalization Again:** Like the tangent vector, the normal vector \( \mathbf{N} \) also needs to be a unit vector, ensuring it has a length of one unit.For our curve, taking the derivative yields \( \mathbf{T}'(t) = (\sin t, \cos t) \), and it happens to already be a unit vector: \( \mathbf{N} = (\sin t, \cos t) \). This normal vector gives us an insight into the orientation and the curvature of our path.
- **Differentiate the Tangent Vector:** The normal vector originates from taking the derivative of the unit tangent vector \( \mathbf{T}(t) \). This derivative gives you the change in direction of the tangent vector.- **Normalization Again:** Like the tangent vector, the normal vector \( \mathbf{N} \) also needs to be a unit vector, ensuring it has a length of one unit.For our curve, taking the derivative yields \( \mathbf{T}'(t) = (\sin t, \cos t) \), and it happens to already be a unit vector: \( \mathbf{N} = (\sin t, \cos t) \). This normal vector gives us an insight into the orientation and the curvature of our path.
Curvature
Curvature is like the measure of how sharply a curve bends at a given point. It’s akin to asking, 'How tight is the curve here?' Higher curvature means a tighter bend and vice versa.
- **Formula for Curvature:** For curves parameterized by \( t \), curvature \( \kappa \) is calculated using the derivative of the tangent vector and the derivative of the position vector.- **Calculation:** In our example, we find \( \kappa \) using the formula \( \kappa = \frac{| \mathbf{T}'(t) |}{| \mathbf{r}'(t) |} \), which gives us \( \kappa = \frac{1}{|t|} \).This concise measure allows you to understand how much the path turns at any point \( t \). Curvature becomes extremely useful in fields like physics and engineering, where knowing how a path bends can be vital to design.
- **Formula for Curvature:** For curves parameterized by \( t \), curvature \( \kappa \) is calculated using the derivative of the tangent vector and the derivative of the position vector.- **Calculation:** In our example, we find \( \kappa \) using the formula \( \kappa = \frac{| \mathbf{T}'(t) |}{| \mathbf{r}'(t) |} \), which gives us \( \kappa = \frac{1}{|t|} \).This concise measure allows you to understand how much the path turns at any point \( t \). Curvature becomes extremely useful in fields like physics and engineering, where knowing how a path bends can be vital to design.