# Geodesic problem

If $f:[a,b] \to R$ is a positive real function and$\gamma(u,v) = ( f(u)\cos (v), f(u) \sin (v), u)$ then show that

$\gamma(t) = \sigma(u(t), c)$ is a geodesic in $M$where $c$ is a constant between 0 and$2\pi$ and
$M=\sigma(U)$ where $U= \{ (u,v)| a<u<b and 0<v< 2\pi \}$

Actually , I tried to calculate the second derivative of
$\sigma(t)$ but that did not work and also I still have u in the first derivative

which means it is not constant

any suggestion? :\

Thanx no help?

lavinia
Gold Member no help?

i don't know what u(t) is and M seems incorrectly defined

lavinia
Gold Member
If $f:[a,b] \to R$ is a positive real function and$\gamma(u,v) = ( f(u)\cos (v), f(u) \sin (v), u)$ then show that

$\gamma(t) = \sigma(u(t), c)$ is a geodesic in $M$where $c$ is a constant between 0 and$2\pi$ and
$M=\sigma(U)$ where $U= \{ (u,v)| a<u<b and 0<v< 2\pi \}$

Actually , I tried to calculate the second derivative of
$\sigma(t)$ but that did not work and also I still have u in the first derivative

which means it is not constant

any suggestion? :\

Thanx

OK. Now I think I understand your question.

Choose u(t) so that the curve $\sigma(u(t), c)$ is parameterized by arclength.

This curve may be written as (u(t),f(u(t))cos(c),f(u(t))sin(c))

Its tangent has length one which is expressed in the equation

u'(t)^2 + (f'(u)u'(t))^2 = 1

Differentiate this equation with respect to t. The left hand side is the inner product of the tangent vector with the second derivative. The right hand side is zero. Thus the acceleration of the curve is perpendicular to its tangent. Now you need to show that it is perpendicular to the surface, U.

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