Question

A curve \(C\) is described along with 2 points on \(C\). (a) Using a sketch, determine at which of these points the curvature is greater. (b) Find the curvature \(\kappa\) of \(C\), and evaluate \(\kappa\) at each of the 2 given points. \(C\) is defined by \(y=\sqrt{1-x^{2}}\) on (-1,1)\(;\) points given \(a t\) \(x=0\) and \(x=1 / 2\).

Short answer

The curvature is greater at \( x = \frac{1}{2} \), with \( \kappa = \frac{4}{3} \).

Step-by-step solution

Understand the Definition of Curvature

The curvature of a curve at a given point quantifies how fast the curve is changing direction at that point. For a function \( y = f(x) \), the curvature \( \kappa \) is given by:\[\kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}}\]

Differentiate the Function

Given \( y = \sqrt{1-x^2} \), first find the first derivative, \( y' \), using differentiation:\[y' = \frac{d}{dx}(\sqrt{1-x^2}) = \frac{-x}{\sqrt{1-x^2}}\]

Find the Second Derivative

Calculate the second derivative, \( y'' \), of the function \( y = \sqrt{1-x^2} \) using the first derivative:\[y'' = \frac{d}{dx} \left( \frac{-x}{\sqrt{1-x^2}} \right) = \frac{-1}{(1-x^2)^{3/2}}\]

Substitute into the Curvature Formula

Substitute \( y' \) and \( y'' \) into the curvature formula:\[\kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}} = \frac{|\frac{-1}{(1-x^2)^{3/2}}|}{\left(1 + \left(\frac{-x}{\sqrt{1-x^2}}\right)^2\right)^{3/2}}\]Simplify the expression to obtain:\[\kappa = \frac{1}{(1-x^2)}\]

Evaluate the Curvature at Given Points

Evaluate \( \kappa \) at \( x = 0 \):\[\kappa(0) = \frac{1}{(1-0^2)} = 1\]Evaluate \( \kappa \) at \( x = \frac{1}{2} \):\[\kappa\left(\frac{1}{2}\right) = \frac{1}{\left(1-\left(\frac{1}{2}\right)^2\right)} = \frac{4}{3}\]

Compare Curvatures at Given Points

Compare the curvatures at \( x = 0 \) and \( x = \frac{1}{2} \):\( \kappa(0) = 1 \) and \( \kappa\left(\frac{1}{2}\right) = \frac{4}{3} \).Since \( \frac{4}{3} > 1 \), the curvature is greater at \( x = \frac{1}{2} \).

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus that focuses on finding how a function changes or the slope of the function at any point. It allows us to determine the rate at which one quantity changes with respect to another. In the context of curves, differentiation helps us understand how the curve behaves at different points.

For a function given as \( y = f(x) \), performing differentiation means finding the derivative, often denoted as \( y' \) or \( \frac{dy}{dx} \). The derivative can be interpreted as the slope of the tangent line to the curve at any point \( x \). It gives us the best linear approximation of the function near that point.

For instance, in the exercise provided, we begin with the function \( y = \sqrt{1-x^2} \). To find the first derivative, we apply the rules of differentiation to obtain:
\[ y' = \frac{-x}{\sqrt{1-x^2}} \]
This derivative tells us how the function \( y \) changes as \( x \) changes.

Second Derivative

The second derivative, denoted as \( y'' \) or \( \frac{d^2y}{dx^2} \), provides insight into the curvature of a function or curve. It essentially measures how the rate of change of a function's slope changes. In more simple terms, it tells you how the rate at which \( y \) is increasing or decreasing is itself growing or shrinking.

For curvature, the second derivative is crucial as it forms part of the formula that measures how "bendy" a curve is. In the example exercise, the second derivative of \( y = \sqrt{1-x^2} \) was calculated as:
\[ y'' = \frac{-1}{(1-x^2)^{3/2}} \]
This result helps indicate the nature of the curve's bending at each point. A positive second derivative suggests the curve is concave up, like a valley, while a negative one suggests it is concave down.

Curvature Formula

Curvature is a measure of how quickly a curve is changing direction at any given point. The formula for curvature \( \kappa \) for a function \( y = f(x) \) is:
\[ \kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}} \]
This formula incorporates both the first and second derivatives, making it a tool that combines differentiation's insights on linear change and the specific bending of the curve.

In the provided solution, the second derivative \( y'' \) already calculated is used along with the first derivative \( y' \) to determine \( \kappa \) for \( y = \sqrt{1-x^2} \). Through simplifying using algebraic skills, we find that the curvature \( \kappa \) becomes:
\[ \kappa = \frac{1}{(1-x^2)} \]
With this simplified formula, we can easily compute the curvature at any point \( x \). The simplification reflects the curve's geometric properties effectively.

Mathematical Analysis

Mathematical analysis involves studying the foundational concepts of calculus, such as limits, continuity, and differentiation. It provides tools for rigorously exploring the behavior of functions and drawing conclusions about them.

In relation to curvature, mathematical analysis lets us understand how functions change and stabilize, ultimately leading to conclusions about the shape and nature of curves. For example, after determining \( \kappa \), we evaluated it at specific points to see where the curve was more pronounced or "bent".

In the solved problem, the curvature was evaluated at \( x = 0 \) and \( x = \frac{1}{2} \), yielding values of 1 and \( \frac{4}{3} \) respectively. Comparing these values enabled a conclusion on where the curve \( C \) bends more sharply, which is an example of applying mathematical analysis to real-world problems. Generally, understanding analyses like these helps solve complex calculus problems efficiently.

• Curvature quantifies the bending at points.
• Second derivatives assist in revealing such properties.
• Analytical techniques streamline problem-solving in calculus.