Find the parametric equation for a curve with given curvature

Question

Find the parametric equations for a curve satisfying $k_{g}=\frac{1}{1+s^{2}}$.Show that this curve corresponds to the catenary $y=coshx$.

If we parametric this curve by arc length, we will have the unit tangent vectors at every point $\overrightarrow{T}=(cos(\theta(s)),sin(\theta(s)))$.And $\theta'(s)=k_{g}$. By integrating ,I get the curve should be $\overrightarrow{X}=(\int cos(arctans+\theta_{0})ds+e,\int sin(arctans+\theta_{0})ds+f).$ But I don't know how to relate this equation with $coshx$.

Answer 1

Assume for simplicity that the curve is $y=y(x)$. Then $$ k=k(x)=\frac{y''(x)}{(1+(y')^2)^{3/2}}. $$ Set $y'(x)=g(x)$, then $$ k=k(x)=\frac{g'(x)}{(1+g(x)^2)^{3/2}}. $$ Integrating this last equation we arrive to $$ \int^{x}_{0} k(t)dt=\int^{g(x)}_{c_1}\frac{dt}{(1+t^2)^{3/2}}\textrm{, }g(0)=c_1 $$ Hence $$ K(x)=\int^{x}_{0}k(t)dt=\frac{g(x)}{\sqrt{1+g(x)^2}}+C $$ and $$ K(x)-C=\frac{g(x)}{\sqrt{1+g(x)^2}} $$ Solving wuth respect to $g(x)$ we arive to $$ g(x)=\pm\frac{K(x)-C}{\sqrt{1-(K(x)-C)^2}} $$ and lastly $$ y(x)=\pm\int g(x)dx=\pm\int\left(\frac{\int^{x}_{0}k(t)dt-C}{\sqrt{1-\left(\int^{x}_{0}k(t)dt-C\right)^2}}\right)dx. $$