Question

Finding radii of curvature Find the radius of curvature (see Exercise 70 ) of the following curves at the given point. Then write the equation of the circle of curvature at the point. $$y=\ln x \text { at } x=1$$

Short answer

Question: Determine the radius of curvature and the equation of the circle of curvature for the curve $$y=\ln x$$ at the point $$x=1$$. Answer: The radius of curvature for the given curve at $$x=1$$ is $$2^\frac{3}{2}$$. The equation of the circle of curvature at this point is $$(x-1)^2+(y-2^\frac{3}{2})^2=4$$.

Step-by-step solution

Find the first and second derivatives of the curve with respect to x.

To find the derivatives, we will differentiate $$y=\ln x$$ with respect to $$x$$: First derivative: $$\frac{dy}{dx}=\frac{1}{x}$$ Second derivative: $$\frac{d^2y}{dx^2}=-\frac{1}{x^2}$$

Evaluate the derivatives at $$x=1$$

To find the value of the derivatives at $$x=1$$, we will plug $$x=1$$ into the derivatives: $$\frac{dy}{dx}\Big|_{x=1}=\frac{1}{1}=1$$ $$\frac{d^2y}{dx^2}\Big|_{x=1}=-\frac{1}{1^2}=-1$$

Find the radius of curvature using the formula

The formula for radius of curvature $$\rho$$ is given by: $$\rho=\frac{(1+(\frac{dy}{dx})^2)^\frac{3}{2}}{\left|\frac{d^2y}{dx^2}\right|}$$ Plug in the values we found in step 2: $$\rho=\frac{(1+(1)^2)^\frac{3}{2}}{\left|-1\right|}=\frac{2^\frac{3}{2}}{1}=2^\frac{3}{2}$$

Find the equation of the circle of curvature

The circle of curvature is centered at the point $$\left(x_0, y_0+\rho\right)$$, where $$(x_0, y_0)$$ is the given point on the curve. In this case, we have $$x_0=1$$ and $$y_0 = \ln(1) = 0$$. Therefore, the center of the circle of curvature is at $$\left(1,2^\frac{3}{2}\right)$$. The equation of the circle of curvature will be: $$(x-1)^2+(y-2^\frac{3}{2})^2=\rho^2$$ Plug in the radius we found in step 3: $$(x-1)^2+(y-2^\frac{3}{2})^2=\left(2^\frac{3}{2}\right)^2$$ So, the final equation of the circle of curvature is: $$(x-1)^2+(y-2^\frac{3}{2})^2=4$$

Key concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus. They measure how a function changes as its input changes. Essentially, the derivative at a certain point tells us the slope of the tangent line to the curve at that point.

In our specific problem, we differentiate the function \(y = \ln x\) with respect to \(x\). The first derivative, \(\frac{dy}{dx} = \frac{1}{x}\), tells us how the natural logarithm function is changing at any point \(x\).

The second derivative, \(\frac{d^2y}{dx^2} = -\frac{1}{x^2}\), provides information about the curvature of the function. Specifically, whether it is bending upwards or downwards. Here, it shows the concavity of the curve by indicating if it's convex or concave at a particular point.

Evaluating these derivatives at \(x=1\) provides us with specific rates of change for the curve at that point, which are necessary for finding the radius of curvature.

Equation of a Circle

The equation of a circle in the Cartesian plane is a simple yet powerful concept. It accounts for all the points that maintain a consistent distance, known as the radius, from a central point. For a circle with center \((h, k)\) and radius \(r\), the equation is:

• \((x - h)^2 + (y - k)^2 = r^2\)

In the context of our problem, we determined the center of the circle of curvature to be at \((1, 2^{\frac{3}{2}})\). This required reasoning through the fact that the point of tangency on the curve is \((1, 0)\), and by extending vertically by the radius of curvature (\(2^{\frac{3}{2}} = \sqrt{8}\)), we establish the center point.

Plugging these values into our circle equation, we establish:

• \((x-1)^2+(y-2^{\frac{3}{2}})^2 = 4\)

This formula allows us to graphically represent the circle that wraps the curve at \(x = 1\) very closely.

Solving Calculus Problems

Tackling calculus problems requires a balance of understanding theory and applying computational skills. The exercise at hand involves multiple layers. You are asked to find the radius of curvature and derive the corresponding circle's equation.

Begin with analyzing the curve through its derivatives. This insight is essential as it involves:

• Calculating the slope and the concavity at given points (using derivatives).
• Identifying how these mathematical properties connect to curvature.

Utilize the radius of curvature formula:

• \( \rho = \frac{(1 + (\frac{dy}{dx})^2)^{\frac{3}{2}}}{|\frac{d^2y}{dx^2}|} \)

This formula combines insights from both the first and second derivatives. Understanding its components is key to comprehend how circles snugly fit a curve at a precise point \(x = 1\).

Finally, the original equation links circle properties from the curvature results. Each of these steps highlights core calculus problem-solving strategies, entwining differentiation with geometric representation.