Question

Prove that sqrt((y^2/a^2)+1) is a catenary

Answer 1

\[\sqrt{(y^2/a^2)+1}=x\]

Answer 2

i was thinking of doing dx/dy

Answer 3

What do you have about the mathematical definition of a catenary?

Answer 4

I'd say that's a step in the right direction.

Answer 5

\[x^2=(y^2/a^2)+1\]
\[y^2=a^2(x^2+1)\]

catenary is cosh(u)

Answer 6

\[2y * y \prime =2a^2x\]
\[y \prime = 2a^2x / 2y\]
\[y \prime = 2a^2x/ \sqrt{a^2(x^2+1)}\]

i'm not too sure if i can take the a^2 out of the sqrt to cancel it.

Answer 7

I think you'll want to use this definition
\[y = \frac{e^{ax} +e^{-ax}}{2a}\]
Where a is a constant.

Answer 8

@SmoothMath , ill check that out. thanks!

Answer 9

This explains it well, and I'm feeling pretty sure that's how to do it. I'll see if I can formalize it.
http://www.proofwiki.org/wiki/Catenary

Answer 10

\(x = \sqrt{(y^2/a^2)+1} \)
\(x^2 =y^2/a^2 +1 \)
\(x^2 - 1 = y^2/a^2 \)
\(y^2 = (x^2-1)*a^2\)
\( y = \sqrt{(x^2-1)*a^2} = \sqrt{x^2-1}\sqrt{a^2} = \sqrt{x^2-1}*a\)

Answer 11

hmph...

Answer 12

or maybe this is the route to go...
\(\sqrt{x^2a^2-a^2}\)

Answer 13

\[y \prime = 2a/\sqrt{x^2+1} \]

\[\int\limits_{}^{} y \prime = 2a \int\limits_{}^{}1/ \sqrt{x^2+1} \]

\[y = 2a*sinh^{-1} x \]

wtf. now it's in sine. and sine is not a catenary. hahahhaa

i'll try to continue.