Let's solve the problem step by step: ### **Step 1: Find the derivatives** Given: \[ y = \ln x \] We need the first and second derivatives: \[ y' = \frac{1}{x} \] \[ y'' = -\frac{1}{x^2} \] ### **Step 2: Find the curvature at \( x = 1 \)** The formula for curvature (\(\kappa\)) at a point for \( y = f(x) \) is: \[ \kappa = \frac{|y''|}{\left[1 + (y')^2\right]^{3/2}} \] At \( x = 1 \): - \( y' = 1 \) - \( y'' = -1 \) So, \[ \kappa = \frac{|-1|}{\left[1 + (1)^2\right]^{3/2}} = \frac{1}{(2)^{3/2}} = \frac{1}{2\sqrt{2}} \] ### **Step 3: Find the radius of curvature** The radius of curvature \( R = \frac{1}{\kappa} \): \[ R = \frac{1}{\frac{1}{2\sqrt{2}}} = 2\sqrt{2} \] --- ### **Step 4: Find the center of the circle of curvature** The center \((h, k)\) is given by moving a distance \( R \) from the point in the direction of the unit normal vector. - Point on the curve at \( x = 1 \): \( (1, 0) \) - Slope of the tangent: \( y' = 1 \Rightarrow \) Slope of normal: \( -1 \) - Unit tangent vector: \( \frac{(1, y')}{\sqrt{1 + (y')^2}} = \frac{(1, 1)}{\sqrt{2}} \) - Unit normal vector (perpendicular, pointing to the concave side): \( \frac{(-1, 1)}{\sqrt{2}} \) So, the center: \[ (h, k) = (1, 0) + 2\sqrt{2} \cdot \frac{(-1, 1)}{\sqrt{2}} = (1, 0) + 2(-1, 1) = (1 - 2, 0 + 2) = (-1, 2) \] --- ### **Step 5: Equation of the circle of curvature** The circle: \[ (x - h)^2 + (y - k)^2 = R^2 \] So, \[ (x + 1)^2 + (y - 2)^2 = (2\sqrt{2})^2 = 8 \] --- ## **Final Answers** ### **1. The radius of curvature at \( x = 1 \) is:** \[ \boxed{2\sqrt{2}} \] ### **2. The equation of the circle of curvature at \( x = 1 \) is:** \[ \boxed{(x + 1)^2 + (y - 2)^2 = 8} \]

Let's go through the detailed step-by-step solution to find the radius of curvature at \( x = 1 \) for the curve \( y = \ln x \), and then derive the equation of the circle of curvature at that point. --- ### **Step 1: Find the derivatives of \( y = \ln x \)** Given: \[ y = \ln x \] First derivative: \[ y' = \frac{dy}{dx} = \frac{1}{x} \] Second derivative: \[ y'' = \frac{d^2 y}{dx^2} = -\frac{1}{x^2} \] --- ### **Step 2: Calculate the curvature \( \kappa \) at \( x=1 \)** The formula for curvature for a function \( y = f(x) \): \[ \kappa = \frac{|y''|}{\left[1 + (y')^2\right]^{3/2}} \] At \( x = 1 \): \[ y' = \frac{1}{1} = 1 \] \[ y'' = -\frac{1}{(1)^2} = -1 \] Plugging into the curvature formula: \[ \kappa = \frac{|-1|}{\left[1 + (1)^2\right]^{3/2}} = \frac{1}{(1 + 1)^{3/2}} = \frac{1}{(2)^{3/2}} \] Recall: \[ (2)^{3/2} = 2^{1} \times 2^{1/2} = 2 \sqrt{2} \] Thus: \[ \kappa = \frac{1}{2 \sqrt{2}} \] --- ### **Step 3: Find the radius of curvature \( R \)** \[ R = \frac{1}{\kappa} = 2 \sqrt{2} \] **This is the radius of curvature at \( x=1 \).** --- ### **Step 4: Find the coordinates of the point \( P \) on the curve at \( x=1 \)** \[ P = (x, y) = (1, \ln 1) = (1, 0) \] --- ### **Step 5: Determine the direction of the normal vector at \( P \)** - Slope of tangent at \( P \): \[ m_t = y' = 1 \] - Slope of the normal: \[ m_n = -\frac{1}{m_t} = -1 \] - The tangent vector at \( P \): \[ \vec{T} = (1, y') = (1, 1) \] - The unit tangent vector: \[ \hat{T} = \frac{1}{\sqrt{2}} (1, 1) \] - The unit normal vector (perpendicular to tangent): \[ \hat{N} = \frac{1}{\sqrt{2}} (-1, 1) \] --- ### **Step 6: Find the center of the circle of curvature** The center \( C = (h, k) \) is located at a distance \( R \) along the normal from \( P \): \[ C = P + R \times \hat{N} \] Calculating: \[ h = 1 + R \times \left(-\frac{1}{\sqrt{2}}\right) = 1 - \frac{2 \sqrt{2}}{\sqrt{2}} = 1 - 2 \] \[ h = -1 \] Similarly: \[ k = 0 + R \times \frac{1}{\sqrt{2}} = 0 + \frac{2 \sqrt{2}}{\sqrt{2}} = 2 \] **So, the center of curvature is at \( (-1, 2) \).** --- ### **Step 7: Write the equation of the circle of curvature** The circle has center \( (-1, 2) \) and radius \( R = 2 \sqrt{2} \): \[ (x - h)^2 + (y - k)^2 = R^2 \] \[ (x + 1)^2 + (y - 2)^2 = (2 \sqrt{2})^2 = 8 \] --- ## **Final Results:** ### **1. Radius of curvature at \( x=1 \):** \[ \boxed{2 \sqrt{2}} \] ### **2. Equation of the circle of curvature at \( x=1 \):** \[ \boxed{(x + 1)^2 + (y - 2)^2 = 8} \] --- **Summary:** - The radius of curvature at the point \( (1, 0) \) on \( y = \ln x \) is \( 2 \sqrt{2} \). - The corresponding circle of curvature has center at \( (-1, 2) \) and radius \( 2 \sqrt{2} \), with the equation: \[ (x + 1)^2 + (y - 2)^2 = 8 \] Let me know if you'd like further clarifications!

The radius of curvature at a point on a curve quantifies the sharpness of the bend, representing the radius of the osculating circle that best approximates the curve at that point. The equation of the circle of curvature (osculating circle) is derived using the center and radius, providing a local geometric representation of the curve's behavior near the point.