Consider the following line element:
$ds^2=dr^2+r^2d\theta^2 \tag1 $
From which I obtained the following non-zero Christoffel symbols:
$\Gamma^r_{\theta\theta}=-r\ \mathrm{and}\ \Gamma^\theta_{r\theta}=1/r\tag2$
This is my problem: I am required to show that $r=\sec(\theta)$, where is a constant, is a geodesic on the above metric.
The method that I tried to solve it is to work out the derivative:
$\dfrac{dr}{ds}\ ,\ \dfrac{d\theta}{ds}\ ,\ \dfrac{d^2r}{ds^2}\ ,\ \dfrac{d^2\theta}{ds^2}\ ,\ \dfrac{drd\theta}{ds^2} $
and substitute them in the equation of the geodesic and get 0. Is this correct, and if not what should I do differently please?
Thanks in advance
It is convenient to stick to the independent variable given in first fundamental form metric 1) itself.
From curvature formula for geodesic straight line in the plane with $(r,\theta) $ polar coordinates, primed wrt $\theta$ we have
$$ \kappa_g= \dfrac{r^2+2r^{\prime^2}-rr^{\prime \prime}}{(r^2+r^{\prime^2})^{\frac32} }=0 $$
you need to only verify (plugging it in) that
$$ r^{\prime \prime}= \dfrac{2 r^{\prime}}{r}+r $$
for the general geodesic/straight line obtained by direct integration
$$ r = a \sec (\theta + \alpha) $$
where $ a,\alpha $ are two arbitrary constants. (They are pedal distance and initial rotation wrt pole respectively).