Question

Find the value(s) of x where the curvature is greatest on the graph of $y=x^3.$

Short answer

When the value of xis $\pm \frac{1}{{\left(45\right)}^{{\displaystyle \frac{1}{4}}}}$the curvature is greatest on the graph of$y=x^3.$

Step-by-step solution

Step 1. Given Information.

The given graph is$y=x^3.$

Step 2. Find the value.

To find the value(s) of x where the curvature is greatest on the graph of$y=x^3,$ 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, $y=f\left(x\right)=x^3.$

So,

$$\begin{gathered} f\left(x\right)=x^3 \\ f\text{'}\left(x\right)=3x^2 \\ f\text{'}\text{'}\left(x\right)=6x \\ \mathrm{Now},\mathrm{the}\mathrm{curvature}\mathrm{of}\mathrm{the}\mathrm{graph}\mathrm{of}f\mathrm{is} \\ \mathrm{k}=\frac{\left|\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)\right|}{{\left(1+{\left(\mathrm{f}\text{'}\left(\mathrm{x}\right)\right)}^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ \mathrm{k}=\frac{\left|6x\right|}{{\left(1+{\left(3{\mathrm{x}}^2\right)}^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ \mathrm{k}=\frac{6x}{{\left(1+9{\mathrm{x}}^4\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \end{gathered}$$

Step 3. Use the first derivative test. 

Now, to find the value of x on the graph of$y=x^3$ where the curvature is greatest, we will apply the derivative test.

So,

$$\begin{gathered} \mathrm{Let}g\left(x\right)=\frac{6x}{{\left(1+9x^4\right)}^{{\displaystyle \frac{3}{2}}}} \\ g\text{'}\left(x\right)=\frac{{\left(1+9x^4\right)}^{\frac{3}{2}}\left(6\right)-\left(6x\right){\displaystyle \frac{3}{2}}{\left(1+9x^4\right)}^{{\displaystyle \frac{1}{2}}}\left(36x^3\right)}{{\left(1+9x^4\right)}^3} \\ g\text{'}\left(x\right)=\frac{6{\left(1+9x^4\right)}^{\frac{3}{2}}-324x^4\sqrt{\left(1+9x^4\right)}}{{\left(1+9x^4\right)}^3} \\ \mathrm{Put}g\text{'}\left(x\right)=0 \\ 0=\frac{6{\left(1+9x^4\right)}^{\frac{3}{2}}-324x^4\sqrt{\left(1+9x^4\right)}}{{\left(1+9x^4\right)}^3} \\ 0=6{\left(1+9x^4\right)}^{\frac{3}{2}}-324x^4\sqrt{\left(1+9x^4\right)} \\ 0=6\sqrt{1+9x^4}\left[\left(1+9x^4\right)-54x^4\right] \\ 0=1+9x^4-54x^4 \\ 0=1-45x^4 \\ \frac{1}{45}=x^4 \\ x=\pm \frac{1}{{\left(45\right)}^{{\displaystyle \frac{1}{4}}}} \end{gathered}$$

Thus, when$x=\pm \frac{1}{{\left(45\right)}^{{\displaystyle \frac{1}{4}}}}$the curvature is greatest on the graph of$y=x^3.$