geodesic curvature about boundary of geodesic disc

Question

Suppose $M^2$ is a surface in $\mathbb{R}^3$, choose a point $x\in M^2$, consider the geodesic disc $D_r(x)$. Why do we have following equality? $$\frac{d}{dt}(\int_{\partial D_t}1 \ ds)=\int_{\partial D_t}k_g\ ds$$ where $k_g$ is the geodesic curvature of $\partial D_t$.