Shortest distance of line $y=x$ to the curve $y=\cosh x$

Question

I used the line point distance formula and tried to minimize that

$$\frac{Ax+By+C}{\sqrt {A^2 + B^2 }}$$ Putting $y=\cosh x$ $$\frac{Ax+B\cosh x+C}{\sqrt {A^2 + B^2 }} = G(x)$$ Using values of A and B from $y=x$

$$\frac{x-\cosh x}{\sqrt {2 }} = G(x)$$ and I minimized $G(x)$ using derivatives

$$\frac{1-\sinh x}{\sqrt {2 }} = G'(x)\implies \sinh x=1$$ but then the $x$ value for that doesn't seem to be the $x$ value which minimizes the distance.

Answer 1

You are correct, the shortest distance between the line $y=x$ and the curve $y=\cosh(x)$ is attained when $\sinh(x)=1$, that is at $$x_0=\text{arcsinh}(1)=\ln(1+\sqrt{2}).$$ which yields that the desired shortest distance is $$d=\frac{|x_0-\cosh(x_0)|}{\sqrt {2 }}=\frac{|\ln(1+\sqrt{2})-\sqrt{2}|}{\sqrt {2 }}=1-\frac{\ln(1+\sqrt2)}{\sqrt2}\approx 0.37677.$$

Answer 2

Let $y=x+n$ is a tangent line to the $y=\cosh x$, which is a parallel to $y=x$.

Thus, $$\left(\frac{e^x+e^{-x}}{2}\right)'=1.$$ Can you end it now?

I got $1-\frac{1}{\sqrt2}\ln(1+\sqrt2).$

Answer 3

Try to think geometrically: since $\cosh x$ is a convex function, its graph lies above any tangent line, and there are tangent lines with any slope since $\frac{d}{dx}(\cosh x)=\sinh x$ is bijective. Consider now the tangent line with unit slope, going through $(\text{arcsinh} 1,\cosh\text{arcsinh} 1)=(\log(1+\sqrt{2}),\sqrt{2})$. Its distance from the line $y=x$ is $1-\frac{1}{\sqrt{2}}\log(1+\sqrt{2})$, and this is also the answer to your problem. Can you figure out why?

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Let $\ds{\mathcal{D}^{2}\pars{x,\alpha} \equiv \pars{x - \alpha}^{2} + \bracks{x - \cosh\pars{\alpha}}^{\, 2}}$

\begin{align} \left\{\begin{array}{rcrcl} \ds{2\pars{x - \alpha}} & \ds{+} & \ds{2\bracks{x - \cosh\pars{\alpha}}} & \ds{=} & \ds{0} \\ \ds{2\pars{x - \alpha}\pars{-1}} & \ds{+} & \ds{2\bracks{x - \cosh\pars{\alpha}}\bracks{-\sinh\pars{\alpha}}} & \ds{=} & \ds{0} \end{array}\right. \end{align} In "adding" both equations, I found $\ds{\braces{\sinh\pars{\alpha} = 1,\cosh\pars{\alpha} = \root{2}} \implies \alpha = \operatorname{arcsinh}\pars{1} = \ln\pars{1 + \root{2}}}$. In addition, $\ds{x = {\alpha + \cosh\pars{\alpha} \over 2} = {\ln\pars{1 + \root{2}} + \root{2} \over 2}}$.

Then, \begin{align} &\bbox[15px,#ffd,border:1px solid navy]{\mathcal{D}^{2}\pars{x,\alpha}} \\[5mm] = &\ \bracks{{\ln\pars{1 + \root{2}} + \root{2} \over 2} - \ln\pars{1 + \root{2}}}^{2} \\ + & \,\,\bracks{{\ln\pars{1 + \root{2}} + \root{2} \over 2} - \root{2}}^{2} \\[5mm] = &\ {\bracks{\root{2} - \ln\pars{1 + \root{2}}}^{2} \over 2} \\[5mm] \implies &\ \bbox[15px,#ffd,border:1px solid navy]{\mathcal{D}\pars{x,\alpha} = 1 - {\root{2} \over 2}\,\ln\pars{1 + \root{2}}}\ \approx\ 0.3768 \end{align}