Problem 147 - Calculus Volume 3

Question

Find the radius of curvature of \(y=\ln (x+1)\) at point \((2, \ln 3)\).

Step-by-Step Solution

Answer

The radius of curvature is \( \frac{10\sqrt{10}}{3} \).

Step 1: Understand the Formula for Radius of Curvature

The formula for the radius of curvature \( R \) of a function \( y = f(x) \) is given by \[ R = \frac{(1 + (f'(x))^2)^{3/2}}{|f''(x)|} \]. We need to find the first and second derivatives of the given function to apply this formula.

Step 2: Find the First Derivative

The given function is \( y = \ln(x+1) \). The first derivative \( y' \) is found using the chain rule: \[ y' = \frac{d}{dx}(\ln(x+1)) = \frac{1}{x+1} \].

Step 3: Find the Second Derivative

To find the second derivative \( y'' \), we differentiate the first derivative: \[ y'' = \frac{d}{dx}\left(\frac{1}{x+1}\right) = -\frac{1}{(x+1)^2} \].

Step 4: Evaluate Derivatives at the Given Point

The given point is \((2, \ln 3)\). Substitute \( x = 2 \) into the first and second derivatives: \[ y'(2) = \frac{1}{2+1} = \frac{1}{3} \] and \[ y''(2) = -\frac{1}{(2+1)^2} = -\frac{1}{9} \].

Step 5: Apply the Radius of Curvature Formula

Substitute \( y'(2) \) and \( y''(2) \) into the radius of curvature formula: \[ R = \frac{(1 + (\frac{1}{3})^2)^{3/2}}{| -\frac{1}{9} |} = \frac{(1+\frac{1}{9})^{3/2}}{\frac{1}{9}} = \frac{(\frac{10}{9})^{3/2}}{\frac{1}{9}} \].

Step 6: Simplify the Expression for R

Calculate \( (\frac{10}{9})^{3/2} \) and simplify the radius of curvature: \[ R = 9 \times (\frac{10}{9})^{3/2} = 9 \cdot \left(\frac{\sqrt{10}^3}{9\sqrt{9}}\right) = 9 \times \frac{10\sqrt{10}}{27} = \frac{30\sqrt{10}}{9} = \frac{10\sqrt{10}}{3} \].

Key Concepts

First Derivative

The first derivative of a function tells us how the function is changing at any given point. It's like a speedometer for the function, showing the rate of change or the slope of the tangent line at a particular point. To find the first derivative of the function \( y = \ln(x+1) \), we apply differentiation techniques.To differentiate \( \ln(x+1) \), we use the derivative rule for logarithmic functions: the derivative of \( \ln(u) \) with respect to \( x \) is \( \frac{1}{u} \cdot \frac{du}{dx} \). Since \( u = x+1 \), and \( \frac{du}{dx} = 1 \), we get:

• \( y' = \frac{1}{x+1} \).

This result gives us the instantaneous rate of change of \( y \) with respect to \( x \), at any point \( x \).

Second Derivative

Once we have the first derivative, finding the second derivative is about determining how the rate of change itself is changing. This can indicate the curvature or concavity of the graph. The second derivative of a function is found by differentiating the first derivative.For the function \( y = \ln(x+1) \), with the first derivative \( y' = \frac{1}{x+1} \), the second derivative \( y'' \) is calculated as:

• \( y'' = \frac{d}{dx}\left(\frac{1}{x+1}\right) = -\frac{1}{(x+1)^2} \).

The negative sign in \( y'' \) indicates the function is concave down, showing a curve that bends downward at all points. This gives important insights when locating points of inflection or analyzing concavity in practical applications.

Chain Rule

The chain rule is a fundamental concept in calculus used when differentiating composite functions. It enables us to differentiate functions that are made up of other functions, linked together like a chain.For example, in the function \( y = \ln(x+1) \), we are dealing with the natural logarithm of \( x+1 \), where "\( x+1 \)" is the inner function and "\( \ln \)" is the outer function.

• According to the chain rule: \( \frac{d}{dx}[h(g(x))] = h'(g(x)) \cdot g'(x) \). Here, \( h(x) = \ln(x) \) and \( g(x) = x+1 \).
• Thus, \( y' = \frac{1}{x+1} \cdot 1 = \frac{1}{x+1} \).

The chain rule simplifies the process of finding the derivative quickly and is invaluable when working with complex functions that have layers of operations tied together.

Logarithmic Function

Logarithmic functions form an important class of functions and are the inverse of exponential functions. The natural logarithm \( \ln(x) \) specifically uses the base \( e \), where \( e \approx 2.718 \).The properties of logarithmic functions, such as converting products into sums or powers into products, arise from these being closely linked to exponentiation. For instance, the natural logarithm satisfies:

• \( \ln(ab) = \ln a + \ln b \)
• \( \ln(a^b) = b \ln a \)

In calculus, these properties enable the simplification of complex expressions.In our exercise with the function \( y = \ln(x+1) \), the property allowing differentiation is crucial. Navigating through the logarithmic framework helps unravel its behavior, showcasing how adjustments in \( x \) transform \( y \), thereby affecting curvature and growth dynamics smoothly.