Problem 6 - Thomas Calculus

Question

A formula for the curvature of a parametrized plane curve $$ \begin{array}{l}{\text { a. Show that the curvature of a smooth curve } \mathbf{r}(t)=f(t) \mathbf{i}+} \\ {g(t) \mathbf{j} \text { defined by twice- differentiable functions } x=f(t) \text { and }} \\ {y=g(t) \text { is given by the formula }}\end{array} $$ $$ \kappa=\frac{|\dot{x} \ddot{y}-\dot{y} \ddot{x}|}{\left(\dot{x}^{2}+\dot{y}^{2}\right)^{3 / 2}} $$ The dots in the formula denote differentiation with respect to \(t\) one derivative for each dot. Apply the formula to find the curvatures of the following curves. $$ \begin{array}{l}{\text { b. } \mathbf{r}(t)=t \mathbf{i}+(\ln \sin t) \mathbf{j}, \quad 0< t <\pi} \\ {\text { c. } \mathbf{r}(t)=\left[\tan ^{-1}(\sinh t)\right] \mathbf{i}+(\ln \cosh t) \mathbf{j}}\end{array} $$

Step-by-Step Solution

Answer

Curvature for b: \( \kappa = \sin t \). Curvature for c: \( \kappa = \frac{1}{\cosh^3 t} \).

Step 1: Understanding the Problem

We are given two tasks: first, to demonstrate how the curvature of a parametrized plane curve is represented by a given formula, and second, to use this formula to calculate the curvature of specific curves. The curvature formula is given as: \[ \kappa=\frac{|\dot{x} \ddot{y}-\dot{y} \ddot{x}|}{\left(\dot{x}^{2}+\dot{y}^{2}\right)^{3 / 2}} \] where \( x = f(t) \) and \( y = g(t) \).

Step 2: Derive the Curvature Formula

Consider the curve \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} \). The velocity vector is \( \mathbf{v}(t) = \dot{x}(t) \mathbf{i} + \dot{y}(t) \mathbf{j} \) and the acceleration vector is \( \mathbf{a}(t) = \ddot{x}(t) \mathbf{i} + \ddot{y}(t) \mathbf{j} \). The curvature \( \kappa \) is defined as \( \|\frac{d\mathbf{T}}{ds}\| \), where \( \mathbf{T} \) is the unit tangent vector, and can be simplified to the given formula using arc length \( ds = \sqrt{\dot{x}^2 + \dot{y}^2} \, dt \).

Step 3: Compute the Derivatives for Curve b

For \( \mathbf{r}(t) = t \mathbf{i} + (\ln \sin t) \mathbf{j} \):\[\dot{x}(t) = 1, \quad \dot{y}(t) = \cot t\]\[\ddot{x}(t) = 0, \quad \ddot{y}(t) = -\csc^2 t\]

Step 4: Calculate Curvature for Curve b

Using the formula, substitute the derivatives: \[ \kappa = \frac{|1 \cdot (-\csc^2 t) - \cot t \cdot 0|}{(1^2 + \cot^2 t)^{3/2}} = \frac{\csc^2 t}{(1 + \cot^2 t)^{3/2}} \] Simplify using \(1 + \cot^2 t = \csc^2 t\), resulting in \( \kappa = \sin t \).

Step 5: Compute the Derivatives for Curve c

For \( \mathbf{r}(t) = [\tan^{-1}(\sinh t)] \mathbf{i} + (\ln \cosh t) \mathbf{j} \):\[ \dot{x}(t) = \frac{1}{\cosh t}, \quad \dot{y}(t) = \tanh t \] \[ \ddot{x}(t) = -\frac{\sinh t}{\cosh^2 t}, \quad \ddot{y}(t) = \text{sech}^2 t \]

Step 6: Calculate Curvature for Curve c

Substitute derivatives into the formula: \[ \kappa = \frac{\left| \frac{1}{\cosh t} \cdot \text{sech}^2 t - \tanh t \cdot \left(-\frac{\sinh t}{\cosh^2 t}\right) \right|}{\left( \frac{1}{\cosh^2 t} + \tanh^2 t \right)^{3/2}} \] Simplify the expression to find \( \kappa = \frac{1}{\cosh^3 t} \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves calculating the rate at which a function is changing at any given point. It is crucial in understanding the behavior of functions and curves. When we talk about a parametrized curve like \( \mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} \), we are interested in how the curve changes as the parameter \( t \) varies.

In the context of this exercise, differentiation helps us find the first and second derivatives of the functions \( x(t) \) and \( y(t) \). These derivatives are denoted by \( \dot{x}(t) \) and \( \ddot{x}(t) \) (similarly for \( y(t) \)).

• \( \dot{x}(t) \) is the derivative of \( x(t) \) with respect to \( t \) and represents the rate of change of \( x(t) \).
• \( \ddot{x}(t) \) is the second derivative of \( x(t) \) and indicates how the rate of change of \( x(t) \) is changing over time.

By computing these derivatives, we can understand how the curve bends and utilizes these values in the curvature formula \( \kappa \). Differentiation therefore underpins the process of determining curvature by quantifying changes in the curve's trajectory.

Velocity and Acceleration Vectors

In analyzing motion along a curve, velocity and acceleration vectors offer insights into how an object is traveling. The velocity vector describes the speed and direction of movement, while the acceleration vector captures changes in the velocity.

For a parametrized curve \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} \), the velocity vector is given by:

• \( \mathbf{v}(t) = \dot{x}(t) \mathbf{i} + \dot{y}(t) \mathbf{j} \)

This vector indicates the direction in which the curve is "traveling" at each point \( t \). Similarly, the acceleration vector, which represents changes in this velocity, is computed as:

• \( \mathbf{a}(t) = \ddot{x}(t) \mathbf{i} + \ddot{y}(t) \mathbf{j} \)

Understanding these vectors allows us to compute the curvature \( \kappa \) of the curve using their derivatives. In effect, the relationship between the velocity and acceleration vectors plays a key role in finding how sharply a curve bends at any point.

Arc Length Calculation

Arc length is the measure of the distance along a curve between two points. In calculus, calculating arc length often requires the use of integrals and derivatives. For a parametrized curve, we can simplify the process using its derivative.

To find the arc length \( ds \) of a curve \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} \), we use the formula:
\[ ds = \sqrt{\dot{x}^2 + \dot{y}^2} \, dt \]
This shows that the infinitesimal arc length, \( ds \), depends on the rate of change in both \( x \) and \( y \) as given by their derivatives.

• The term \( \sqrt{\dot{x}^2 + \dot{y}^2} \) represents the magnitude of the velocity vector, essentially telling us how fast we are moving along the curve.

In the context of curvature, this arc length element is crucial because it is used to normalize the unit tangent vector and thus compute the curvature \( \kappa \). Understanding the concept of arc length helps to bridge the idea between the physical distance along the curve and how we mathematically handle the curvature over that length.