Problem 52 - Calculus III

Question

Find the arc length of the curve on the given interval.Find the radius of curvature of \(y=\ln (x+1)\) at point \((2, \ln 3)\).

Step-by-Step Solution

Answer

The radius of curvature of the curve at point \((2, \ln 3)\) is approximately 18.26.

Step 1: Recall the Formula for Radius of Curvature

The formula for the radius of curvature, \( R \), of a curve \( y = f(x) \) is given by:\[ R = \frac{{[1 + (f'(x))^2]^{3/2}}}{|f''(x)|} \] where \( f'(x) \) is the first derivative of \( y \) with respect to \( x \) and \( f''(x) \) is the second derivative.

Step 2: Calculate the First Derivative

Find \( f'(x) \) for \( y = \ln(x+1) \). Use the derivative formula for natural logarithms:\[ f'(x) = \frac{1}{x+1}\]So, \( f'(x) = \frac{1}{x+1} \).

Step 3: Calculate the Second Derivative

Find the second derivative, \( f''(x) \), by differentiating \( f'(x) \):\[ f''(x) = \frac{d}{dx}\left(\frac{1}{x+1}\right) = -\frac{1}{(x+1)^2}\]So, \( f''(x) = -\frac{1}{(x+1)^2} \).

Step 4: Evaluate the Derivatives at \( x = 2 \)

Substitute \( x = 2 \) into \( f'(x) \) and \( f''(x) \):\[ f'(2) = \frac{1}{3} \quad \text{and} \quad f''(2) = -\frac{1}{9}\]

Step 5: Calculate the Radius of Curvature

Substitute \( f'(2) \) and \( f''(2) \) into the formula for the radius of curvature:\[ R = \frac{[1 + (\frac{1}{3})^2]^{3/2}}{|-\frac{1}{9}|}\]Simplify the expression:\[ R = \frac{[1 + \frac{1}{9}]^{3/2}}{\frac{1}{9}} = \frac{[\frac{10}{9}]^{3/2} \times 9}{1}\]Calculate the final result:\[ R = \frac{10^{3/2}}{3^{1/2}} \approx \frac{31.62}{1.73} \approx 18.26\]

Step 6: Conclusion

The radius of curvature at \( x = 2 \) for the curve \( y = \ln(x+1) \) is approximately \( 18.26 \).

Key Concepts

Arc Length

The concept of arc length is all about measuring the distance along a curve. Imagine you have a piece of string lying along a curvy path. If you straighten it out, the length of that string represents the arc length. In calculus, the arc length of a function over a certain interval can be calculated using an integral. For a curve defined by a function \( y = f(x) \), the formula for arc length \( S \) on the interval \([a, b]\) is:

• \[ S = \int_a^b \sqrt{1 + (f'(x))^2} \, dx \]

This nifty formula combines the simple length of a line segment with the change in direction of the curve, given by the derivative. It helps us find out how long the curve is even if it bends and twists.

Derivative

The derivative is a fundamental concept in calculus that tells us about the rate of change of a function. When you see a function like \( y = \ln(x+1) \), taking its derivative with respect to \( x \) helps us understand how \( y \) changes as \( x \) increases or decreases.For example, if \( f'(x) = \frac{1}{x+1} \), it means that for every small change in \( x \), \( y \) changes by this derivative value. Derivatives have many important applications such as:

• Finding the slope of a curve at any point.
• Understanding how quantities change over time, such as speed or temperature.

Derivatives give us a 'sneak peek' into the behavior of functions at any given point.

Second Derivative

The second derivative tells us about the curvature or concavity of a function. If the first derivative gives the slope, the second derivative informs us whether the slope is increasing or decreasing. In our example, where the first derivative of \( y = \ln(x+1) \) is \( \frac{1}{x+1} \), the second derivative becomes \( -\frac{1}{(x+1)^2} \).This negative second derivative suggests that the function is concave down, meaning it bends in a 'n' shape rather than a 'u'. Curvature is vital for understanding how the graph of a function behaves. Knowing the second derivative, we can:

• Determine points of inflection where the graph changes concavity.
• Identify maximum or minimum points for optimization problems.

Understanding second derivatives helps us gain deeper insights into the nature of functions beyond the immediate rate of change.