Find the point(s) on the curve of xy=1 at which the curvature is largest.

Question

idealist10 · Answer

$$k(x)=\frac{ \left| y'' \right| }{ [1+(y')^2]^{3/2} }$$

idealist10 · Answer

$$y'=-\frac{ 1 }{ x^2 }$$

idealist10 · Answer

$$y''=\frac{ 2 }{ x^3 }$$

idealist10 · Answer

$$k(x)=\frac{ \left| \frac{ 2 }{ x^3 } \right| }{ [1+\frac{ 1 }{ x^4 }]^{3/2} }$$

idealist10 · Answer

@SithsAndGiggles

anonymous · Answer

$$\frac{\left|\dfrac{2}{x^3}\right|}{\left(1+\dfrac{1}{x^4}\right)^{3/2}}=\frac{x^6\left|\dfrac{2}{x^3}\right|}{\left(x^4+1\right)^{3/2}}=\frac{2\left|x^3\right|}{\left(x^4+1\right)^{3/2}}$$
where the second equality is true for $x\neq0$, which is the case since $xy=1$ is not defined there.

Since we have $|x^3|$ in the numerator, we should expect two answers, which makes sense because, graphically, $xy=1$ is symmetric about the origin. Let's assume $x>0$, so that $|x^3|=x^3$.
$$k(x)=\frac{2x^3}{(x^4+1)^{3/2}}$$
Take your derivative:
$$k'(x)=\frac{6x(x^4+1)^{3/2}-\dfrac{3}{2}(x^4+1)^{1/2}(4x^3)(2x^3)}{(x^4+1)^3}=\frac{6x^5+6x-12x^6}{(x^4+1)^{5/2}}$$
(assuming I haven't made a silly mistake).

Find your critical points:
$$\begin{align*}-12x^6+6x^5+6x&=-6x\left(2x^5-x^4-1\right)\\
&=-6x\bigg((x^5-x^4)+(x^5-1)\bigg)\\
&=-6x\bigg(x^4(x-1)+(x-1)(x^4+x^3+x^2+x+1)\bigg)\\
&=-6\color{blue}{x}(x-1)\color{red}{(2x^4+x^3+x^2+x+1)}
\end{align*}$$
You can ignore the root $x=0$ because it's not in our domain, and you can ignore the red factor as well, since it doesn't have any real roots.

idealist10 · Answer

That was very helpful! Thank you!