Question

Show that the curvature on the parabola defined by$y=x^2$ is greatest at the origin.

Short answer

It is proved that the curvature on the parabola defined by $y=x^2$is greatest at the origin.

Step-by-step solution

Step 1. Given Information. 

The given parabola is$y=x^2.$

Step 2. Showing.

To show that the curvature on the parabola defined by $y=x^2$is greatest at the origin, we will use the formula for the curvature in the plane.

The curvature in the plane is $k=\frac{\left|f\text{'}\text{'}\left(x\right)\right|}{{\left(1+{\left(f\text{'}\left(x\right)\right)}^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}.$

Now, $y=f\left(x\right)=x^2.$

So,

$$\begin{gathered} f\left(x\right)=x^2 \\ f\text{'}\left(x\right)=2x \\ f\text{'}\text{'}\left(x\right)=2 \\ \mathrm{Now},\mathrm{the}\mathrm{curvature}\mathrm{of}\mathrm{the}\mathrm{graph}\mathrm{of}f\mathrm{is} \\ \mathrm{k}=\frac{\left|\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)\right|}{{\left(1+{\left(\mathrm{f}\text{'}\left(\mathrm{x}\right)\right)}^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ \mathrm{k}=\frac{\left|2\right|}{{\left(1+{\left(2\mathrm{x}\right)}^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ \mathrm{k}=\frac{2}{{\left(1+4{\mathrm{x}}^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \end{gathered}$$

Step 3. Use the first derivative test.

Now, to find the point on the parabola where the curvature is greatest, we will apply the derivative test.

So,

$$\begin{gathered} \mathrm{Let}g\left(x\right)=\frac{2}{{\left(1+4x^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ g\text{'}\left(x\right)=2\left[-\frac{3}{2}{\left(1+4x^2\right)}^{-\frac{5}{2}}\left(8x\right)\right] \\ g\text{'}\left(x\right)=\frac{-24x}{{\left(1+4x^2\right)}^{{\displaystyle \frac{5}{2}}}} \\ \mathrm{Put}g\text{'}\left(x\right)=0 \\ 0=\frac{-24x}{{\left(1+4x^2\right)}^{{\displaystyle \frac{5}{2}}}} \\ 0=-24x \\ x=0 \end{gathered}$$

Substitute $x=0\mathrm{in}\mathrm{the}\mathrm{equation}y=x^2$to find the value,

$y=0$

Thus, at the origin$\left(0,0\right),$ the curvature is greatest.