Question

In Exercises 31–35 find the curvature of the given function at the indicated value of x. Then sketch the curve and the osculating circle at the indicated point.

$y=e^x,x=0$

Short answer

The curvature at$x=0\text{is}k=\frac{\sqrt{2}}{4}$.

The graph is

Step-by-step solution

Step 1. Given information.

The given function and point is$y=e^x,x=0$.

Step 2. Curvature at the point.

The curvature is given by,

$$\begin{gathered} k=\frac{\left|f^{\text{'}}\left(x\right)\right|}{{\left(1+{\left(f^{\text{'}}\left(x\right)\right)}^2\right)}^{\frac{3}{2}}} \\ {f}^{\text{'}}\left(x\right)=e^x \\ {f}^{\text{'}\text{'}}\left(x\right)=e^x \\ k=\frac{\left|e^x\right|}{{\left(1+{\left(e^x\right)}^2\right)}^{\frac{3}{2}}} \\ k=\frac{e^x}{{\left(1+e^{2x}\right)}^{\frac{3}{2}}} \\ \text{At}x=0,k=\frac{e^0}{{\left(1+e^{2\left(0\right)}\right)}^{\frac{3}{2}}} \\ k=\frac{1}{2^{\frac{3}{2}}} \\ k=\frac{2^{\frac{1}{2}}}{2^{\frac{3}{2}}·2^{\frac{1}{2}}} \\ k=\frac{\sqrt{2}}{2^2} \\ k=\frac{\sqrt{2}}{4} \end{gathered}$$

Step 3. Graph.

The graph will be,