Question

Berechne den Krümmungsradius von \(f(x)=e^{x}(x \in \mathbb{R})\) für \(x=0\) und \(x=1\)

Short answer

The radius of curvature at \(x=0\) is \(\sqrt{2}\) and at \(x=1\) is approximately 2.65.

Step-by-step solution

Compute the First Derivative

The derivative of the given function \(f(x)=e^{x}\) (where \(x \in \mathbb{R}\)) is given by \(f'(x)=e^{x}\) since the derivative of \(e^{x}\) is itself.

Compute the Second Derivative

The second derivative of the function is obtained by taking the derivative of the first derivative. In this case, \(f''(x)=e^{x}\) since the derivative of \(e^{x}\) is itself.

Compute the Radius of Curvature for x=0

Substitute \(x=0\) into the derivatives obtained and the radius of curvature at \(x=0\) is given by the formula \[ R = \left| \frac{(1+(f'(x))^2)^{\frac{3}{2}}}{f''(x)} \right| = \left| \frac{(1+(1)^2)^{\frac{3}{2}}}{1} \right| =\sqrt{2}.\]

Compute the Radius of Curvature for x=1

Substitute \(x=1\) into the derivatives obtained and the radius of curvature at \(x=1\) is given by the formula \[ R = \left| \frac{(1+(f'(x))^2)^{\frac{3}{2}}}{f''(x)} \right| = \left| \frac{(1+(e)^2)^{\frac{3}{2}}}{e} \right| =\sqrt{e^2+1},\] which is approximately 2.65 when rounded to two decimal places.

Key concepts

Ableitung

When we talk about the "Ableitung" or "derivative" of a function, we are referring to the rate at which the function is changing at any given point. It's a fundamental concept in calculus, often used to find slopes of curves or rates of changes.
To find the derivative of an exponential function like \( f(x)=e^{x} \), we use basic derivative rules:

• The derivative of \( e^{x} \) with respect to \( x \) is simply \( e^{x} \) itself. This means the slope of the tangent line to the curve is the same as the value of the function at that point.

Finding a derivative helps us calculate the slope, which leads to better understanding how the function behaves at different points. In our particular exercise, this ability allows us to later find the radius of curvature at specific values of \( x \), such as 0 and 1. The first step is computing the first and then the second derivatives. For \( f(x)=e^{x} \), they are \( f'(x)=e^{x} \) and \( f''(x)=e^{x} \). This simplicity is due to the unique property of the exponential function.

Exponentielle Funktion

An "Exponentielle Funktion" or "exponential function" is a mathematical function of the form \( f(x) = a^{x} \), where \( a \) is a constant. However, the most common base is \( e \), Euler's number, approximately equal to 2.718.
Exponential functions are fundamental in mathematics because they model real-world phenomena:

• They describe growth or decay processes, such as population growth, radioactive decay, and interest calculations.

Our function, \( f(x) = e^{x} \), beautifully illustrates these principles since it is self-replicating when differentiated. That's why it plays a critical role in calculus, retaining its form through differentiation. This property allows easy calculation of higher order derivatives, all equal to \( e^{x} \). This behavior is pivotal when determining things like the curvature at various points of the function. Exponential functions have a unique growth trajectory: it starts slow but once they pass a certain point, the value accelerates rapidly.

Radius der Krümmung

The "Radius der Krümmung" or "radius of curvature" is a measure of how sharply a curve bends at a particular point. It is an important concept in differential geometry and calculus because it gives insight into the behavior of curves.
To calculate the radius of curvature \( R \) at a given point \( x \), we use the formula:

• \( R = \left| \frac{(1+(f'(x))^2)^{\frac{3}{2}}}{f''(x)} \right| \)

This formula involves both the first and second derivatives of the function, which describe the curve's slope and how that slope changes.
For our function \( f(x) = e^{x} \), substituting \( x = 0 \) yields the radius of curvature \( R = \sqrt{2} \). Here, \( f'(0) = e^{0} = 1 \) and \( f''(0) = e^{0} = 1 \), leading to this constant radius at the origin.
Likewise, at \( x = 1 \), substituting in \( f'(1) = e \) and \( f''(1) = e \), results in \( R = \sqrt{e^2 + 1} \), which gives us an intuitive measure of how the curve bends differently as \( x \) varies. Understanding the radius of curvature helps in analyzing how smooth or sharp a change is on a graph.