Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the curve in the xy-plane. Then, for the given point, find the curvature and the radius of curvature. Finally, \(y=e^{-x^{2}},(1,1 / e)\)

Step-by-Step Solution

Answer

Curvature at (1, 1/e) is \( \kappa = \frac{2}{e(1 + 4/e^2)^{3/2}}\). Radius of curvature is \( R = \frac{1}{\kappa} \).

Step 1: Understand the Function

We begin by understanding the given function, which is the curve: \[ y = e^{-x^2}. \]This is the curve we need to sketch and analyze for curvature.

Step 2: Sketch the Curve

To sketch the curve \( y = e^{-x^2} \), note that it is a Gaussian, bell-shaped curve symmetric about the y-axis. It has a maximum at \( x = 0 \) with \( y = 1 \), and as \( x \to \pm \infty \), \( y \to 0 \). It decays rapidly past \( x = 1 \) and \( x = -1 \). Plot this graph on the xy-plane to visualize it.

Step 3: Recall Curvature Formula

The curvature \( \kappa \) at any point on a curve is given by:\[ \kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}}. \] We need to find the first derivative \( y' \) and the second derivative \( y'' \) of the function to calculate curvature.

Step 4: Find the First Derivative

Compute the first derivative of \( y = e^{-x^2} \):\[ y' = \frac{d}{dx}(e^{-x^2}) = -2x e^{-x^2}. \]

Step 5: Find the Second Derivative

Compute the second derivative of \( y \), starting from the first derivative:\[ y'' = \frac{d}{dx}(-2x e^{-x^2}) = -2 e^{-x^2} + 4x^2 e^{-x^2}. \]Simplifying gives:\[ y'' = e^{-x^2}(-2 + 4x^2). \]

Step 6: Evaluate at Point (1, 1/e)

Plug \( x = 1 \) into \( y' \) and \( y'' \):\[ y'(1) = -2(1) e^{-1} = -\frac{2}{e}, \]\[ y''(1) = e^{-1}(-2 + 4 \cdot 1^2) = e^{-1}(-2 + 4) = \frac{2}{e}. \]

Step 7: Calculate the Curvature

Using the curvature formula at \( x = 1 \):\[ \kappa = \frac{|\frac{2}{e}|}{(1 + ( -\frac{2}{e} )^2)^{3/2}} = \frac{\frac{2}{e}}{(1 + \frac{4}{e^2})^{3/2}}. \]Simplify the expression to find the numerical value of \( \kappa \).

Step 8: Find the Radius of Curvature

The radius of curvature \( R \) is the reciprocal of the curvature:\[ R = \frac{1}{\kappa}. \]Calculate \( R \) using the found value of \( \kappa \).

Key Concepts

Understanding Curvature Calculation

Curvature is a measure of how sharply a curve bends at a given point. To calculate the curvature of a curve, we use the formula: \[ \kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}}. \] Here, \( y' \) is the first derivative of the function, and \( y'' \) is the second derivative. The numerator represents the absolute value of the second derivative, which shows how much the slope of the curve is changing at that point.

• First, we find the derivatives of the function \( y = e^{-x^2} \).
• The first derivative \( y' = -2x e^{-x^2} \) shows the rate of change of the function.
• The second derivative \( y'' = e^{-x^2}(-2 + 4x^2) \) gives us how this rate itself is changing.

By plugging these derivatives into the curvature formula, we can determine how the curve bends at the point of interest, namely \( (1, 1/e) \). More curvature implies a sharper bend.

Radius of Curvature Demystified

The radius of curvature is the reciprocal of curvature and provides the radius of the osculating circle at a particular point on the curve. Given the curvature \( \kappa \), the radius \( R \) is simply \[ R = \frac{1}{\kappa}. \] This relationship gives us a practical way to comprehend how tightly a curve turns at a given point.

• If the curve has a high curvature, the radius of curvature is small, indicating a sharper turn.
• Conversely, a lower curvature results in a larger radius, suggesting a gentler turn.

So for the point \( (1, 1/e) \) on our function, after calculating \( \kappa \), we find \( R \) to understand the bend's nature better. This simple reciprocal relationship allows us to visualize how the curve hugs around a point.

The Gaussian Function Explored

The Gaussian function, of the form \( y = e^{-x^2} \), is a classic bell-shaped curve commonly known in probability and statistics. It's symmetric about the y-axis with a peak at \( x = 0 \) where \( y = 1 \). As \( x \to \pm \infty \), \( y \to 0 \), emphasizing its rapid decay.

• This function is used extensively in the Gaussian distribution or normal distribution.
• The symmetry around the y-axis makes it particularly easy to analyze and compute for curvature and radius.
• Because it’s smooth and continuous, derivatives can be easily calculated to analyze properties like curvature.

Understanding this function's nature aids in comprehending its behavior on a graph, and preparing us for calculating other properties, such as those derived from derivatives.

Mastering Derivatives for Curve Analysis

Derivatives are essential in understanding the behavior of a curve. The first derivative \( y' \) of a function represents the slope or tangent at any given point on the curve. For the Gaussian function \( y = e^{-x^2} \), the first derivative is \[ y' = -2x e^{-x^2}, \] which tells us how steep the curve is at any point \( x \).

• If \( y' \) is positive, the curve is rising; if negative, it's falling.
• Zero implies a flat section, potentially a peak or trough.

The second derivative \( y'' \) checks the rate at which this slope changes, showing concavity or convexity of the graph. For our function, it is \[ y'' = e^{-x^2}(-2 + 4x^2). \]

• If \( y'' \) is positive, the curve is concave up, like a valley.
• If negative, it's concave down, like a hill.

By comprehending these derivatives, analyzing curves for shapes, curvatures, and turning points becomes more manageable.