Not sure if I'm in the correct subforum as it is only part calculus, moderators feel free to move...
Basically, I do not know where to start. If somebody could guide me through the steps it would be greatly appreciated.
Not sure if I'm in the correct subforum as it is only part calculus, moderators feel free to move...
Basically, I do not know where to start. If somebody could guide me through the steps it would be greatly appreciated.
I have a lot of problems with the way this is worded! First, the parameterization of what? I presume you are talking about the parameterization of the surface you get by rotating the catenary.
You get the catenoid by rotating, not the parameterization!
But the original parameterization, (cosh(t), t) is in two dimensions. Which is the "x3-axis"? The only interpretation that makes sense (and gives the correct result) is that we really have (cosh(t), 0, t) in three dimensions.
Okay, rotating the curve around an axis gives a surface which requires two parameters. Since we already have "t", use that as one parameter and, since we are rotating, use the angle of rotation, $\displaystyle \theta$, as the other. In polar (or cyindrical) coordinates, $\displaystyle x= r cos(\theta)$, $\displaystyle y= r sin(\theta)$. Since we are rotating around the x3 (z) axis, it is the first, x= cosh(t) that is our "r", or distance from the axis. Then $\displaystyle x= r cos(\theta)= cosh(t)cos(\theta)$ and $\displaystyle y= r sin(\theta)= cosh(t)sin(\theta)$. Of course, z, measured along the axis of rotation, is not changed: z= t. That gives $\displaystyle (cosh(t)cos(\theta), cosh(t)sin(\theta), t)$.
yes, sorry.
I should have made it more clear that:
- the parametrisation for the catenoid is σ(t, θ)
so that rotating the catenary γ = (cosht, t) about the x3 axis yields a parametrisation for the catenoid, σ(t, θ).
Can you clarify if what you said before still applies, because I have a feeling it doesn't!
Thx!