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.

$y=\frac{1}{3}{\left(x^2+2\right)}^{3/2},x=1$

Short answer

The value of curvature is$\frac{\sqrt{3}}{6}$

Step-by-step solution

Step 1. Given information.

The given function is$y=\frac{1}{3}{\left(x^2+2\right)}^{3/2},x=1$.

Step 2. Curvature.

Curvature is given by,

$$\begin{gathered} k=\frac{\left|f^{\text{'}\text{'}}\left(x\right)\right|}{{\left(1+{\left(f^{\text{'}}\left(x\right)\right)}^2\right)}^{\frac{3}{2}}} \\ \text{Let}y=f\left(x\right)=\frac{1}{3}{\left(x^2+2\right)}^{\frac{3}{2}} \\ {f}^{\text{'}}\left(x\right)=x\sqrt{x^2+2} \\ {f}^{\text{'}\text{'}}\left(x\right)=\frac{x^2}{\sqrt{x^2+2}}+\sqrt{x^2+2}=\frac{2x^2+2}{\sqrt{x^2+2}} \\ k=\frac{\left|\frac{2x^2+2}{\sqrt{x^2+2}}\right|}{{\left(1+{\left(x\sqrt{x^2+2}\right)}^2\right)}^{\frac{3}{2}}} \\ \text{At}x=1\text{,} \\ k=\frac{\frac{4}{\sqrt{3}}}{(1+3{)}^{\frac{3}{2}}} \\ k=\frac{1}{2\sqrt{3}} \\ k=\frac{\sqrt{3}}{6} \end{gathered}$$

Step 3. Graph.

The graph is