Chapter 12: Problem 18
For the plane curves in Problems 17 through 21, find the unit tangent and normal vectors at the indicated point. \(x=t^{3}, y=t^{2}\) at \((-1,1)\)
Short Answer
Expert verified
Unit tangent vector is \( \left(\frac{3}{\sqrt{13}}, \frac{-2}{\sqrt{13}} \right) \), unit normal vector is \( \left(\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}} \right) \).
Step by step solution
01
Find the Derivatives
Given the parametric equations for the curve: \[ x = t^3 \]\[ y = t^2 \]First, find the derivatives with respect to \( t \):\[ \frac{dx}{dt} = 3t^2 \]\[ \frac{dy}{dt} = 2t \]
02
Determine the Value of Parameter t
To find the value of \( t \) at the point \((-1, 1)\), use the given parametric equations:\[ x = t^3 = -1 \] \[ y = t^2 = 1 \]Solving these, we get:- Since \( t^3 = -1 \), \( t = -1 \).- Since \( t^2 = 1 \), this is consistent for \( t=-1 \).
03
Evaluate Derivatives at t = -1
Substitute \( t = -1 \) into the derivatives:\[ \frac{dx}{dt}(-1) = 3(-1)^2 = 3 \]\[ \frac{dy}{dt}(-1) = 2(-1) = -2 \]
04
Calculate the Magnitude of the Tangent Vector
Find the magnitude of the tangent vector:\[ \| r'(t) \| = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \]
05
Compute the Unit Tangent Vector
The unit tangent vector \( T \) is the derivative vector divided by its magnitude:\[ T = \left(\frac{3}{\sqrt{13}}, \frac{-2}{\sqrt{13}} \right) \]
06
Derive the Unit Normal Vector
The unit normal vector \( N \) is obtained by rotating the unit tangent vector 90 degrees counterclockwise:\[ N = \left(\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}} \right) \]
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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
When exploring vector calculus, a crucial concept is the unit tangent vector. It helps us understand the direction of a curve at any specific point. The unit tangent vector, often represented by \( T \), is derived from the tangent vector of a curve.The tangent vector for a parametric curve defined by \( x = x(t) \) and \( y = y(t) \) is found by differentiating these equations with respect to the parameter \( t \). This gives us the velocities \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). Once we have these derivatives, the tangent vector can be formed as \( \langle \frac{dx}{dt}, \frac{dy}{dt} \rangle \). However, to convert this tangent vector into a unit tangent vector, which has a magnitude of 1, we need to normalize it.
- Calculate the magnitude of the tangent vector using the Pythagorean theorem: \( \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \).
- Divide each component of the tangent vector by this magnitude to get the unit tangent vector: \( T = \left( \frac{\frac{dx}{dt}}{\|r'(t)\|}, \frac{\frac{dy}{dt}}{\|r'(t)\|} \right) \).
Exploring the Unit Normal Vector
The unit normal vector, indicated as \( N \), plays a vital part in understanding the geometry of curves by pointing perpendicular to the tangent vector. It's derived by rotating the unit tangent vector 90 degrees.In 2D space, rotating a vector counterclockwise effectively swaps the components of the vector and introduces a sign change. Here's how:
- Given the unit tangent vector \( T = \left( a, b \right) \), the unit normal vector would be \( N = \left( -b, a \right) \).
- Normalize the resulting vector if necessary, even though it's already a unit vector from this operation.
Introduction to Parametric Equations
Parametric equations are a powerful way to define curves in a plane using a parameter \( t \). Instead of expressing one variable directly in terms of another (like \( y = f(x) \)), parametric equations use a third variable, \( t \), to define both \( x \) and \( y \).For example, consider \( x = t^3 \) and \( y = t^2 \). Here, the parameter \( t \) is used to trace out the curve in the \( xy \)-plane.
- By varying \( t \), you move along the curve.
- This method allows representation of more complex shapes, curves or paths which are difficult to define with traditional functions.