# Finding radius of curvature of a catenary with length of string given

I had the following question, asking about the radius of curvature (in$$\text{ cm}$$) of the string, at the bottom most point, fixed at two ends, with the angle with the vertical being $$\pi/4$$ at the fixed points, and length of string $$40 \text{ cm}$$.

My Attempt:

Let the catenary be $$y=a\cosh(x)=\mathrm d^2y/\mathrm dx^2$$. Then $$\mathrm dy=a\sinh(x)\mathrm dx$$. As per constraints, $$a\sinh(x)=1$$ and the length of the half string is $$20\text{ cm}$$. So that can be used by writing the arc length integral as follows.

$$\int_{0}^{x}\sqrt{1+a^2\sinh^2(x)}\mathrm dx=20$$

$$\boxed{r=\frac{\left(1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^{2}\right)^{3/2}}{\left|\frac{\mathrm d^2y}{\mathrm dx^2}\right|}}$$

How to go about solving this? The $$a$$ is the cause of trouble inside the integrand. For $$a=1$$, it evaluates to $$\sinh(x)$$, but that does not satisfy the slope condition at $$x$$. How to proceed? Any hints are appreciated. Thanks

Edit $$1$$:

I would like to put forth a general version of the problem. Say we have a hanging chain of length $$l$$ making angle $$\theta$$ with the fixed supports. Knowing that it is a catenary, we can say the equation of the curve will be of the form $$y=a\cosh(bx)$$ with the coordinate system origin at the mid-point of the hanging chain and $$x$$-axis in the horizontal direction and $$y$$ in the vertical. If we let the $$x$$-coordinate at the point from the which the chain is hanging to be $$x_0$$, then we have the following system of equations.

\begin{aligned}ab\sinh(bx_0)&=\tan(\theta)\\ 2\int_{0}^{x_0}\sqrt{1+a^2b^2\sinh^2(x)}\mathrm dx&=l\end{aligned}

Any ideas on how to solve $$a,b$$ in terms of $$l,\theta$$. Thanks

Hoping that I properly understood, we have $$y(x)=\int_{0}^{x}\sqrt{1+a^2\sinh^2(t)}\,dt=20$$
Making $$t=i u$$ $$\int\sqrt{1+a^2\sinh^2(t)}\,dt=i \int\sqrt{1-a^2\sin^2(u)}\,du=i E\left(u\left|a^2\right.\right)=-i E\left(i t\left|a^2\right.\right)$$ $$y(x)=-i E\left(i x\left|a^2\right.\right)=20$$ $$\frac {dy(x)}{dx}=\sqrt{1+a^2 \sinh ^2(x)} \qquad \text{and} \qquad \frac {d^2y(x)}{dx^2}=\frac{a^2 \sinh (x) \cosh (x)}{\sqrt{1+a^2 \sinh ^2(x)}}$$
$$r^2=\frac{\left(1+\left(\frac{ dy}{ dx}\right)^{2}\right)^{3}}{\left(\frac{d^2y}{ dx^2}\right)^2}=\frac{\left(a^2 \sinh ^2(x)+2\right)^3 } {\frac{a^4 \sinh ^2(x) \cosh ^2(x)}{1+a^2 \sinh ^2(x)} }$$ $$r^2=\frac 1 {a^4}\left(\text{csch}^2(x)\, \text{sech}^2(x) \left(1+a^2 \sinh ^2(x)\right) \left(2+a^2 \sinh ^2(x)\right) \right)$$ You gave as constraint $$a\sinh(x)=1 \implies x=\sinh ^{-1}\left(\frac{1}{a}\right)\implies r^2=\frac{6}{1+a^2}$$ but we also have $$-i E\left(i\ \text{csch}^{-1}(a)|a^2\right)=20$$
I am probably mistaken since this would lead to something close to $$a=10^{-9}$$