Problem 6 - Thomas Calculus

Question

a. Show that the curvature of a smooth curve \(\mathbf{r}(t)=f(t) \mathbf{i}+\) \(g(t) \mathbf{j}\) defined by twice-differentiable functions \(x=f(t)\) and \(y=g(t)\) is given by the formula $$\kappa=\frac{|\dot{x} \ddot{y}-\dot{y} \ddot{x}|}{\left(\dot{x}^{2}+\dot{y}^{2}\right)^{3 / 2}}$$ Apply the formula to find the curvatures of the following curves. b. \(\mathbf{r}(t)=t \mathbf{i}+(\ln \sin t) \mathbf{j}, \quad 0

Step-by-Step Solution

Answer

a. Derivation is shown; b. Curvature: \( \csc t \); c. Curvature: 0.

Step 1: Derive the formulas for x and y

To find the curvature \( \kappa \) of the curve \( \mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} \), we start by letting \( x = f(t) \) and \( y = g(t) \). Therefore, the curve is defined in parametric form as \( x(t) = f(t) \) and \( y(t) = g(t) \).

Step 2: Calculate the velocity vectors

The velocity vector is given by \( \mathbf{v}(t) = \dot{x}\mathbf{i} + \dot{y}\mathbf{j} \). Here, \( \dot{x} = \frac{dx}{dt} \) and \( \dot{y} = \frac{dy}{dt} \).

Step 3: Calculate the acceleration vectors

The acceleration vector is given by \( \mathbf{a}(t) = \ddot{x}\mathbf{i} + \ddot{y}\mathbf{j} \). Here, \( \ddot{x} = \frac{d^2x}{dt^2} \) and \( \ddot{y} = \frac{d^2y}{dt^2} \).

Step 4: Use the curvature formula

Curvature \( \kappa \) is given by \( \kappa = \frac{ | \dot{x} \ddot{y} - \dot{y} \ddot{x} | }{ (\dot{x}^2 + \dot{y}^2)^{3/2} } \). Calculate this for the given specific functions.

Step 5: Apply the formula for curve b

For \( \mathbf{r}(t) = t \mathbf{i} + (\ln \sin t) \mathbf{j} \), calculate: \( \dot{x} = 1 \), \( \dot{y} = \cot t \), \( \ddot{x} = 0 \), \( \ddot{y} = -csc^2 t \). Curvature: \( \kappa = \frac{|1 \cdot (-csc^2 t) - (\cot t) \cdot 0|}{(1^2 + \cot^2 t)^{3/2}} = \frac{csc^2 t}{(1 + \cot^2 t)^{3/2}} \). As \( 1 + \cot^2 t = csc^2 t \), \( \kappa = \frac{1}{csc^2 t^{1/2}} = csc t \).

Step 6: Apply the formula for curve c

For \( \mathbf{r}(t) = [\tan^{-1}(\sinh t)]\mathbf{i} + (\ln \cosh t)\mathbf{j} \), calculate: \( \dot{x} = \frac{1}{\cosh t} \), \( \dot{y} = \tanh t \), \( \ddot{x} = -\frac{\sinh t}{\cosh^2 t} \), \( \ddot{y} = \text{sech}^2 t \). Curvature: \( \kappa = \frac{|\frac{1}{\cosh t} \cdot \text{sech}^2 t + \tanh t \cdot \frac{\sinh t}{\cosh^2 t}|}{(\frac{1}{\cosh^2 t} + \tanh^2 t)^{3/2}} \). Simplified, \( \kappa = |0| = 0 \).

Key Concepts

Parametric Equations

Parametric equations are a powerful way to represent curves and paths in mathematics. Instead of describing a curve as a single equation in terms of x and y, parametric equations use a parameter, usually denoted as \( t \), to define both the x and y coordinates separately. For example, a curve can be described as:

• \( x = f(t) \)
• \( y = g(t) \)

This allows for a more flexible representation of curves, especially when dealing with intricate paths or motion over time.

The parameter \( t \) can represent time, which makes parametric equations ideal for representing the motion of objects. As \( t \) changes, the position defined by \( (x(t), y(t)) \) traces a path in the plane. By examining how these coordinates change with respect to \( t \), we can gain insights into characteristics like speed, direction, and curvature of the path.

Differentiable Functions

Differentiable functions are key in calculus and are fundamental when working with parametric equations. A function is called differentiable if it can be differentiated, which means its derivative exists at every point in its domain.

In the context of parametric curves, differentiable functions allow us to compute the derivatives such as \( \dot{x} \) and \( \dot{y} \), which represent the rates of change of \( x \) and \( y \) with respect to the parameter \( t \). These derivatives form the components of the velocity vector:

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

The second derivatives, \( \ddot{x} \) and \( \ddot{y} \), provide the components of the acceleration vector:

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

Both sets of derivatives are crucial in analyzing the behavior and properties of the curve, especially when calculating the curvature, which describes how sharply a curve bends at any point.

Acceleration Vectors

Acceleration vectors are an essential concept when analyzing the dynamics of motion using parametric equations. Just like velocity vectors describe the rate of change of position, acceleration vectors describe the rate of change of velocity with respect to time or the parameter \( t \).

The components of an acceleration vector are the second derivatives of the parametric equations:

• \( \ddot{x} = \frac{d^2x}{dt^2} \): the rate of change of \( \dot{x} \), or how quickly the velocity in the x-direction changes.
• \( \ddot{y} = \frac{d^2y}{dt^2} \): the rate of change of \( \dot{y} \), or how quickly the velocity in the y-direction changes.

These components together form the acceleration vector \( \mathbf{a}(t) \), defined as:

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

The acceleration vector is pivotal in determining the curvature of a path. The curvature \( \kappa \) reflects how much the path deviates from being a straight line. It is calculated using both the velocity and acceleration vectors according to the formula provided in the exercise. Understanding acceleration helps in predicting how a point moving along a path will behave, which can be especially useful in physics and engineering applications.