Problem 69 - Calculus Early Transcendentals

Question

The curvature at a point \(P\) of a curve is defined as $$\kappa=\left|\frac{d \phi}{d s}\right|$$ where \(\phi\) is the angle of inclination of the tangent line at \(P\) as shown in the figure. Thus the curvature is the absolute value of the rate of change of \(\phi\) with respect to arc length. It can be regarded as a measure of the rate of change of direc- tion of the curve at \(P\) and will be studied in greater detail in Chapter \(13 .\) (a) For a parametric curve \(x=x(t), y=y(t),\) derive the formula $$\kappa=\frac{|\dot{x} y-x y|}{\left[\dot{x}^{2}+\dot{y}^{2}\right]^{3 / 2}}$$ where the dots indicate derivatives with respect to \(t,\) so \(\dot{x}=d x / d t .\left[\)Hint. Use \(\phi=\tan ^{-1}(d y / d x)\) and Formula 2\right. to find \(d \phi / d t\) . Then use the Chain Rule to find \(d \phi / d s\) . (b) By regarding a curve \(y=f(x)\) as the parametric curve \(x=x, y=f(x),\) with parameter \(x,\) show that the formula in part (a) becomes $$\kappa=\frac{\left|d^{2} y / d x^{2}\right|}{\left[1+(d y / d x)^{2}\right]^{3 / 2}}$$

Step-by-Step Solution

Answer

(a) \( \kappa = \frac{|\dot{x} \ddot{y} - \dot{y} \ddot{x}|}{(\dot{x}^2 + \dot{y}^2)^{3/2}} \), (b) \( \kappa = \frac{|f''(x)|}{(1+(f'(x))^2)^{3/2}} \).

Step 1: Compute the derivative of \( \phi \) with respect to \( t \)

Given \( \phi = \tan^{-1} \left( \frac{dy}{dx} \right) \), we need to find \( \frac{d\phi}{dt} \). Using the chain rule for derivatives and the identity \( \frac{d}{dx} \tan^{-1}(u) = \frac{1}{1+u^2} \cdot \frac{du}{dx} \),we have:\[ \frac{d\phi}{dt} = \frac{1}{1+(\frac{dy}{dx})^2} \cdot \frac{d}{dt}\left( \frac{dy}{dx} \right) \] Simplifying further with: \[ \frac{dy}{dx} = \frac{\dot{y}}{\dot{x}} \Rightarrow \frac{d}{dt}\left( \frac{dy}{dx} \right) = \frac{\dot{x}\ddot{y} - \dot{y}\ddot{x}}{\dot{x}^2} \].

Step 2: Compute \( \frac{d\phi}{ds} \) using the Chain Rule

Given \( ds = \sqrt{(dx)^2 + (dy)^2} = \sqrt{\dot{x}^2 + \dot{y}^2}dt \), we use the Chain Rule to convert \( \frac{d\phi}{dt} \) into \( \frac{d\phi}{ds} \):\[ \frac{d\phi}{ds} = \frac{d\phi}{dt} \cdot \frac{1}{\sqrt{\dot{x}^2 + \dot{y}^2}} \]Substituting from Step 1:\[ \frac{d\phi}{ds} = \frac{\dot{x}\ddot{y} - \dot{y}\ddot{x}}{\dot{x}^2 (1+(\frac{dy}{dx})^2) \cdot \sqrt{\dot{x}^2 + \dot{y}^2}} \].

Step 3: Simplify the expression for curvature \( \kappa \)

From the definition \( \kappa = \left| \frac{d\phi}{ds} \right| \), replace \( \sin^2 \phi + \cos^2 \phi \) with \( 1 \) and rearrange:\[\kappa = \left| \frac{\dot{x}\ddot{y} - \dot{y}\ddot{x}}{(\dot{x}^2 + \dot{y}^2)^{3/2}} \right| \]This confirms the desired form for parametric equations.

Step 4: Deriving curvature formula for \( y = f(x) \)

For \( y = f(x) \), consider the parametric form \( x = x, y = f(x) \) and let \( \dot{x} = 1 \), \( \dot{y} = f'(x) \). The second derivative \( \ddot{y} = f''(x) \) since we differentiate \( f'(x) \) w.r.t. \( x \).Substitute into the result:\[ \kappa = \frac{|\ddot{y}|}{(1 + \dot{y}^2)^{3/2}} = \frac{|f''(x)|}{(1+(f'(x))^2)^{3/2}} \]This confirms the formula for \( y = f(x) \).

Key Concepts

Parametric Equations

Parametric equations are a way to express a set of related quantities as explicit functions of an independent parameter, often denoted as \( t \). In the context of curves, these equations provide a method to describe the position of a point in the plane. Instead of using a single equation like \( y = f(x) \), which relates \( y \) directly to \( x \), parametric equations use two separate equations:

• \( x = x(t) \)
• \( y = y(t) \)

Here, \( x \) and \( y \) are both expressed as functions of \( t \), allowing the tracing of a curve in a more flexible way. This is particularly useful for curves that cannot be represented as functions of \( x \) alone. By varying \( t \), one can generate various coordinate pairs \((x, y)\) that comprise the curve. This format is convenient for curves like circles, ellipses, and more complex shapes.
Understanding parametric equations allows for more creative and comprehensive representation of geometrical shapes in coordinate geometry.

Derivative

In calculus, a derivative represents the rate at which a function changes as its input changes. For parametric equations, derivatives are crucial in computing how one parameter affects the output functions \( x(t) \) and \( y(t) \). The derivative with respect to \( t \) is expressed as \( \dot{x} = \frac{dx}{dt} \) and \( \dot{y} = \frac{dy}{dt} \).The concept of the derivative helps us understand the tangent line to a curve at any point, as well as how the curve's shape changes. For curvature calculations, knowing derivatives is essential because they are used to derive the formula for curvature:

• \( \ddot{x} = \frac{d^2x}{dt^2} \) is the second derivative of \( x(t) \)
• \( \ddot{y} = \frac{d^2y}{dt^2} \) is the second derivative of \( y(t) \)

These higher-order derivatives help in determining the behavior of the curvature of a curve by indicating changes in the directions of the tangent lines as the parameter \( t \) varies.

Arc Length

The arc length of a curve is the distance along the curve between two points and is a fundamental concept in calculus and differential geometry. It involves integrating to find the total length of the curve segment as parameter \( t \) changes. For a parametric curve defined by \( x = x(t) \) and \( y = y(t) \), we compute the arc length \( s \) using the formula:\[ s = \int \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]This formula results from the Pythagorean theorem, considering the infinitesimal distances along the curve, and gives a precise measure of the curve's length, enhancing comprehension of the curve's geometric properties. By understanding arc length, we also gain insights into measuring how wrinkles or bends in the curve contribute to its total length, which ties directly into the concept of curvature.

Chain Rule

The chain rule is a fundamental derivative rule in calculus that allows us to differentiate composite functions. When dealing with parametric equations, the chain rule becomes indispensable to express derivatives in terms of an independent variable like \( t \). For example, if we need to find \( \frac{d\phi}{ds} \), where \( \phi \) is a function of \( x \) and \( x \) is a function of \( s \), the chain rule facilitates this by stating:\[ \frac{d\phi}{ds} = \frac{d\phi}{dt} \cdot \frac{dt}{ds} \]Using the chain rule allows us to break complex derivative computations into manageable steps by treating each part of the function individually. In our context, it was used to convert \( \frac{d\phi}{dt} \) into \( \frac{d\phi}{ds} \), crucial for determining curvature since curvature involves the rate of change of the tangent angle with respect to the arc length \( s \). This conversion highlights how smoothly the angle of inclination changes as one moves along the curve.