Problem 6 - Thomas Calculus

Question

A formula for the curvature of a parametrized plane curve $$ \begin{array}{c}{\text { a. Show that the curvature of a smooth curve } \mathbf{r}(t)=f(t) \mathbf{i}+} \\ {g(t) \mathbf{j} \text { defined by twice- differentiable functions } x=f(t) \text { and }} \\ {y=g(t) \text { is given by the formula }} \\ {\kappa=\frac{|\ddot{x} \ddot{y}-\dot{y} \ddot{x}|}{\left(\dot{x}^{2}+\dot{y}^{2}\right)^{3 / 2}}}\end{array} $$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 \( \kappa \) for (b) is \( \csc(t) \) and for (c) is \( \frac{1}{\cosh t} \).

Step 1: Understanding the Curvature Formula

The curvature \( \kappa \) of a smooth parametrized plane curve \( \mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} \) is given by \( \kappa = \frac{|g''(t) f'(t) - f''(t) g'(t)|}{((f'(t))^2 + (g'(t))^2)^{3/2}} \). Here, \( f'(t) \) and \( g'(t) \) denote the first derivatives, and \( f''(t) \) and \( g''(t) \) denote the second derivatives with respect to \( t \).

Step 2: Apply the Formula to \( \mathbf{r}(t)=t \mathbf{i}+\ln(\sin t) \mathbf{j} \)

For \( f(t) = t \), we have \( f'(t) = 1 \) and \( f''(t) = 0 \). For \( g(t) = \ln(\sin t) \), we have \( g'(t) = \cot(t) \) and \( g''(t) = -\csc^2(t) \). Substitute these into the curvature formula to get \[ \kappa = \frac{|0 \cdot \cot(t) - (-\csc^2(t)) \cdot 1|}{(1^2 + \cot^2(t))^{3/2}} = \frac{\csc^2(t)}{(1 + \cot^2(t))^{3/2}}. \] As \( 1 + \cot^2(t) = \csc^2(t) \), the curvature simplifies to \( \kappa = \frac{\csc^2(t)}{\csc^3(t)} = \csc(t). \)

Step 3: Apply the Formula to \( \mathbf{r}(t)=\tan^{-1}(\sinh t) \mathbf{i}+\ln(\cosh t) \mathbf{j} \)

For \( f(t) = \tan^{-1}(\sinh t) \), we have \( f'(t) = \frac{1}{\cosh t} \) and \( f''(t) = -\frac{\sinh t}{\cosh^2 t} \). For \( g(t) = \ln(\cosh t) \), we have \( g'(t) = \tanh t \) and \( g''(t) = \text{sech}^2 t \). Substitute these into the curvature formula to get \[ \kappa = \frac{|\text{sech}^2 t \cdot \frac{1}{\cosh t} - (-\frac{\sinh t}{\cosh^2 t}) \cdot \tanh t|}{\left(\left(\frac{1}{\cosh t}\right)^2 + \tanh^2 t\right)^{3/2}}. \] Simplifying gives \( \kappa = \frac{1}{\cosh t} \).

Key Concepts

Parametrized Plane Curve

A parametrized plane curve is a path traced out by points defined by a pair of equations. These points depend on a parameter, usually denoted as \( t \). In the exercise you're tackling, the curve is expressed with position vector \( \mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} \), where \( f(t) \) and \( g(t) \) are functions of the parameter \( t \). Each function gives the position along the \( x \)-axis and \( y \)-axis respectively.

Think about a vector coming from the origin to a point \( (f(t), g(t)) \) on the plane. As \( t \) changes, this point moves, creating a curve. This is how we trace paths like circles, ellipses, or even more complex shapes.

In this framework, it's easier to study properties like curvature and tangents of curves since the parameter \( t \) can provide a smooth progression from one point on the curve to another.

Differentiable Functions

Differentiable functions are those smooth, continuous functions where derivatives exist at all points. Differentiation provides the rate of change or the slope of the function at any point.

In the curvature formula, we use the first derivatives \( f'(t) \) and \( g'(t) \) to determine the slope of the tangent to the curve at any point. The second derivatives \( f''(t) \) and \( g''(t) \) involve the curvature and describe how the curve turns or bends. These derivatives ensure that the curve is sufficiently smooth so that curvature can be properly defined.

Without differentiable functions, analyzing curves would be tricky since abrupt changes would occur. For the curves in calculations, knowing these derivatives allows you to substitute into the curvature formula without hassle.

Trigonometric Functions

Trigonometric functions like sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)) are essential in dealing with angles and periodic phenomena. They often feature in the parametrization of curves, especially when describing circular or wave-like shapes.

In your particular exercise, consider the function \( g(t) = \ln(\sin t) \). Understanding both the differentiation process and how to manipulate trigonometric identities can be crucial.

When calculating derivatives like \( g'(t) \) and \( g''(t) \), the knowledge of how to manipulate \( \sin t \), \( \cos t \), and related expressions like the co-secant (\( \csc t \)) and cotangent (\( \cot t \)) helps us simplify expressions. These functions let us grasp complex mathematical landscapes with just a few specialized tools.

Hyperbolic Functions

Hyperbolic functions, such as sine hyperbolic (\( \sinh \)) and cosine hyperbolic (\( \cosh \)), resemble trigonometric functions but relate more to the geometry of hyperbolas. They grow exponentially and are often used in models of hyperbolic geometry or for solving real-world problems involving rapid growth or decay.

In the problem, \( f(t) = \tan^{-1}(\sinh t) \) and \( g(t) = \ln(\cosh t) \) highlight functions involving hyperbolic sine and cosine. Understanding their behavior and derivatives is vital since they differ slightly from their circular counterparts.

The derivatives reveal important properties; \( \sinh t \) and \( \cosh t \) often appear in physics and engineering because they describe natural motions like the shape of a hanging cable, known as a catenary. Recognizing that \( \cosh t \) is always greater than or equal to 1 helps identify solution boundaries and optimize the expression for curvature.