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 \(x\). $$ y=e^{x}, \quad x=0 $$

Short answer

The graph of the function \(y=e^x\) is an increasing curve. The circle of curvature for this function at \(x=0\) has a radius of \(2\sqrt{2}\) and is centered at \((0, 1+2\sqrt{2})\). This circle is tangent to the curve at the point (0,1).

Step-by-step solution

Graph the function

Use a graphing utility and plot the function \(y=e^x\). This is an exponential function, and its graph represents an increasing curve.

Find the curvature at x=0

The curvature \(k\) of a curve at a certain point is found using the formula \[k=\frac{|y''|}{(1+(y')^2)^{3/2}}\]. For \(y=e^x\), the first derivative \(y'\) is \(e^x\) and the second derivative \(y''\) is also \(e^x\). At \(x=0\), both \(y'\) and \(y''\) are 1. Substituting these into the formula gives \(k=\frac{|1|}{(1+1^2)^{3/2}}\), which simplifies to \(\frac{1}{2\sqrt{2}}\).

Find the circle of curvature

The radius \(R\) of the circle of curvature is the reciprocal of the curvature, so \(R=\frac{1}{k}=2\sqrt{2}\). The center of the circle of curvature is located at \((x-R\sinφ, y+R\cosφ)\), where φ is the angle of the tangent to the curve at the point of interest. For \(y=e^x\), φ equals \(y'(0)=1\), so the center is at (0-\(2\sqrt{2}\)*0, 1+ \(2\sqrt{2}\)*1), simplifies to \((0, 1+2\sqrt{2})\).

Graph the circle of curvature

Use the centre and the radius found in the previous step to graph the circle of curvature in the same viewing window as the function \(y=e^x\). The circle will be tangent to the curve at the point \((0,1)\), reflecting the curvature of the function at this point.