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 \(y=\frac{1}{x^{2}+1} ;\) points given at \(x=0\) and \(x=2\).

Short answer

Curvature is greater at \(x=0\) with \(\kappa=2\). At \(x=2\), \(\kappa\approx 0.028\).

Step-by-step solution

Understand the Concept of Curvature

Curvature (\(\kappa\)) is a measure of how quickly a curve changes direction. For a function\(y = f(x)\), curvature can be calculated using the formula:\[\kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}}\]where\(y'\)and\(y''\)are the first and second derivatives of the function respectively.

Find the Derivatives

Given\(y = \frac{1}{x^2 + 1}\), first, find the first derivative\(y'\) using the quotient rule:\[y' = \frac{d}{dx} \left(\frac{1}{x^2 + 1}\right) = \frac{-2x}{(x^2 + 1)^2}\]Next, find the second derivative\(y''\):\[y'' = \frac{d}{dx} \left(\frac{-2x}{(x^2 + 1)^2}\right) = \frac{2(3x^2 - 1)}{(x^2 + 1)^3}\]

Calculate Curvature Formula Components

Now substitute\(y'\)and\(y''\)into the curvature formula. Compute:\(|y''|\) and\((1 + (y')^2)^{3/2}\).For\(x = 0\):\[y' = 0,\quad y'' = -2\Rightarrow |y''| = 2\]\[(1+(y')^2)^{3/2} = 1\]For\(x = 2\):\[y' = -\frac{4}{25},\quad y'' = \frac{10}{343}\Rightarrow |y''| = \frac{10}{343}\]\[(1 + (\frac{4}{25})^2)^{3/2} \approx 1.038\]

Evaluate Curvature at Given Points

Substitute these values into the curvature formula: For\(x = 0\):\[\kappa = \frac{2}{1} = 2\]For\(x = 2\):\[\kappa = \frac{\frac{10}{343}}{1.038} \approx 0.028\]

Sketch the Curve and Analyze

Sketch the curve\(y = \frac{1}{x^2 + 1}\). Notice that at\(x = 0\), the graph is at its peak, indicating a sharper curve (greater curvature). At\(x = 2\), the curve flattens out, indicating lesser curvature.

Key concepts

First Derivative

When we talk about the first derivative of a function, we're really exploring how a function changes as the input changes. Mathematically, the first derivative, denoted as \(y'\) or \(\frac{dy}{dx}\), gives us the slope of the tangent line to the curve at any given point. This tells us whether the curve is rising or falling as you move along it.
For the function given in the exercise, \(y = \frac{1}{x^2 + 1}\), the first derivative was calculated using a method called the quotient rule. The first derivative is represented by the formula:

• For \(x = 0\), the first derivative \(y'\) results in \(0\), indicating a horizontal tangent line.
• For \(x = 2\), the first derivative \(y'\) is \(-\frac{4}{25}\), indicating that the curve is slightly sloping downwards at this point.

Understanding this gives a good idea of how the curve behaves near these points.

Second Derivative

The second derivative, often symbolized as \(y''\), gives us insights into the curvature of a function. While the first derivative tells us about the slope, the second derivative tells us about the change of that slope—that is, how the rate of change itself is changing. This can indicate the concavity of the curve: whether it is curving upwards or downwards.
For the function \(y = \frac{1}{x^2 + 1}\), the second derivative was found to be:

• For \(x = 0\), \(y'' = -2\), which signifies that the curve is concave downward at this point.
• For \(x = 2\), \(y'' = \frac{10}{343}\), suggesting a very mild concave upward shape.

By understanding the second derivative, we can understand how tightly or loosely the curve bends at a given point.

Quotient Rule

The quotient rule is a crucial tool in calculus for taking derivatives of functions that are ratios of two differentiable functions. The formula for the quotient rule is: \[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \] where \(u\) and \(v\) are functions of \(x\), and \(u'\) and \(v'\) are their respective derivatives.
In our problem, to find the first derivative \(y'\) of \(y = \frac{1}{x^2 + 1}\), we treat \(u(x) = 1\) and \(v(x) = x^2 + 1\). Using the quotient rule enabled us to derive \(y' = \frac{-2x}{(x^2 + 1)^2}\). This calculation is consistent with the algebra involved in checking how rapidly and in what direction our function \(y\) is changing as \(x\) shifts.

Function Evaluation

The concept of function evaluation is pretty straightforward. It involves substituting specific values into the function to determine outcomes for particular inputs. In the context of curvature, function evaluation is applied in two crucial ways:
First, we compute the values of the first and second derivatives at specific points to determine the nature of the curve at those points.

• At \(x = 0\), the function behavior is evaluated to provide \(y' = 0\) and \(y'' = -2\).
• At \(x = 2\), evaluation delivers \(y' = -\frac{4}{25}\) and \(y'' = \frac{10}{343}\).

Function evaluation helps us apply these values directly into the curvature formula \(\kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}}\). This leads to calculate the curvature for a clearer illustration of the curve's geometry at given points.