Question

Use a graphing utility to graph the function. In the same viewing window, graph the circle of curvature to the graph at the given value of \(x\). $$ y=x+\frac{1}{x}, \quad x=1 $$

Short answer

The function is a hyperbola and its graph at \(x=1\) includes a identifiable curvature. The circle of curvature at \(x=1\) has a radius of 0.5, and its center is at the point (1, 2.5).

Step-by-step solution

Graph the Function

Input the function \(y = x + \frac{1}{x}\) into the graphing utility and plot the graph. Observe the features of the graph.

Determine the Point on the Curve

Find the point on the curve corresponding to \(x=1\). Substitute \(x=1\) into the function to get \(y = 1 + \frac{1}{1} = 2\). Hence, the point on the curve is (1, 2).

Calculate the Circle of Curvature

The formula for the radius of curvature is \(R = \frac{(1+(dy/dx)^2)^\frac{3/2}}{|d^2y/dx^2|}\). First, find the derivative of the function, \(dy/dx = 1 - \frac{1}{x^2}\). Second, plug \(x=1\) into the derivative to get \(dy/dx = 1 - 1 = 0\). Then, find the second derivative \(d^2y/dx^2 = \frac{2}{x^3}\), and substitute \(x=1\) to get \(d^2y/dx^2 = 2\). Lastly, substitute these results into the radius of curvature formula to get \(R = \frac{(1+0)^{3/2}}{2} = 0.5\).

Graph Circle of Curvature

The center of the circle of curvature will be at the point (1, 2.5) - (0.5 units above the point on the curve due to 0 slope). Add the circle of radius 0.5 and centre at (1, 2.5) to the graph.