Calculate the length of $\gamma(t)=(t,t), t \in [-1,-\frac{1}{2}]$ with the metric $g=\frac{dx^2+dy^2}{y^2}$ and compare with euclidean metric

Question

Consider the metric $g=\frac{dx^2+dy^2}{y^2}$ on $\mathbb{R}_+^2=\{(x,y) \in \mathbb{R}^2 : y>0\}$.

Calculate the length of the curve $\gamma(t)=(t,t), t \in [-1,-\frac{1}{2}]$ and compare with the length of the same curve with the euclidean metric on $\mathbb{R}^2$.

I know that, in theory, the length of the curve is given by $$\int_{-1}^{-\frac{1}{2}} \sqrt{g(\gamma',\gamma')}dt$$ where $\gamma'(t)=D_{\gamma_t}(\frac{d}{dt})$.

In practise I don't know how to proceed.

Also, I presume that the length of the given curve with respect to the euclidean metric is $$\int_{-1}^{-\frac{1}{2}}\|\gamma'(t)\|dt = \frac{\sqrt2}{2}.$$

Am I making some confusion here? Some help would be appreciated.

Thanks!

Answer 1

To say $g = \frac{dx^2+dy^2}{y^2}$ is to say $g = \frac{1}{y^2} (dx \otimes dx+dy \otimes dy)$. Then $\gamma'(t) = \frac{d\gamma^1}{dt}\frac{\partial}{\partial x}\bigg{|}_{(t,t)}+\frac{d\gamma^2}{dt}\frac{\partial}{\partial y}\bigg{|}_{(t,t)}$ hence $$g(\gamma'(t),\gamma'(t)) = \left(\frac{1}{y^2}(dx \otimes dx+dy \otimes dy)\right)\left(\frac{\partial}{\partial x}\bigg{|}_{(t,t)}+\frac{\partial}{\partial y}\bigg{|}_{(t,t)},\frac{\partial}{\partial x}\bigg{|}_{(t,t)}+\frac{\partial}{\partial y}\bigg{|}_{(t,t)}\right)$$ which gives $$ g(\gamma'(t),\gamma'(t)) = \frac{2}{t^2}$$ hence $\sqrt{g(\gamma'(t),\gamma'(t))}dt = \sqrt{2}dt/|t| = -\sqrt{2}dt/t$ (as $t<0$) finally, just integrate over $[-1,-1/2]$ to obtain $\sqrt{2}\ln(2)$. And, yes, I do think your Euclidean length is correct.