I am reading Hamilton's An isoperimetric estimate for the Ricci Flow on the Two-Sphere. It is the 9th paper of Collected papers on Ricci flow. For explain my problem, I use the three parts of this paper, which are the first three pictures below.
I want to know how to show $$ \frac{dL}{dr}=\int k ds $$ namely, the first red line. But I feel for general curve $\Lambda$, there should be not the first red line. I think I should use the characteristic of $\overline\Lambda$ to get the red line. But I don't know how to start.
Besides, I afraid that I misunderstand it. Therefore, I draw a picture in bottom to describe my understand. Thanks for any help.
What I try(2023-4-9): Consider the unit round sphere $$ \hat S(\varphi,\theta) = (\cos \varphi\cos\theta, \cos\varphi\sin\theta,\sin\varphi) $$ where $\varphi\in[-\pi/2,\pi/2], \theta\in [0, 2\pi)$. Let $\gamma(t,s)$ are the curves on $\hat S$,namely $$ \gamma(t,s)=\hat S(\varphi(t,s),\theta(t,s)) = (\cos \varphi(t,s)\cos\theta(t,s), \cos\varphi(t,s)\sin\theta(t,s),\sin\varphi(t,s)) $$ and $\gamma(0,s)=\gamma(1,s)$ for any $s\in (-\delta,\delta)$. Then I have $$ |\gamma_t|^2=\varphi_t^2 + \theta_t^2\cos^2\varphi $$ Let $$ L(s)=\int_0^1|\gamma_t| dt $$ So $$ \frac{dL}{ds}=\int_0^1 \partial_s|\gamma_t| dt $$ where $$ \partial_s|\gamma_t|=\frac{1}{|\gamma_t|}[ \varphi_t\varphi_{ts} + \theta_t\theta_{ts}\cos^2\varphi-\varphi_s\theta_t^2\cos\varphi\sin\varphi ] \tag{2} $$ Since $\langle \hat S_\varphi, \hat S_\theta\rangle =0$ and $|\hat S_\varphi|=1, |\hat S_\theta|=\cos^2\varphi$, the first fundamental form is $$ I=d\varphi^2 + |\cos\varphi|^2d\theta^2 $$
I want to use the geodesic curvature of curve on sphere is $$ k_g=\frac{d(\gamma_t,\varphi_{curve})}{dt}-\sin \varphi \frac{d\theta}{dt} \tag{3} $$ where $(\gamma_t,\varphi_{curve})$ is the angle of $\gamma$ and longitude.
In fact, I calculate $$ (\gamma_t,\varphi_{curve})= \arccos\frac{\varphi_t}{|\gamma_t|} $$ As (3), I have (The process is a little complicated, I omit it) $$ k_g=\frac{ \varphi_t\theta_t\theta_{tt}\cos^2\varphi -\varphi_t^2\theta_t^2\cos\varphi\sin\varphi -\varphi_{tt}\theta_t^2\cos^2\varphi }{|\gamma_t|^2 |\theta_t \cos\varphi|} -\theta_t\sin\varphi \tag{4} $$ But seemly, it is not equal to $\partial_s|\gamma_t|$.
Notation: Given tangent vectors $V,W$, denote $V\cdot W = g(V,W)$ and $|V|^2 =g(V,V)$.
First, there is an ambiguity in the sign of the curvature. The standard definition of the curvature of a curve in a surface is: $$ \kappa = N\cdot\nabla_TT, $$ where $T$ is the unit tangent vector and $N$ is the unit normal to a curve. On the other hand, the curvature of the curve can be viewed as being the second fundamental form of the curve and therefore the covariant of the normal, i.e., $$ \kappa = T\cdot\nabla_TN. $$ These two definitions differ only in sign. Below, I will use the latter, because that appears to be what Hamilton uses.
Now let $\gamma: [0,L] \times (-\delta,\delta) \rightarrow S$, where $\gamma(\cdot,0)$ is the curve $\Lambda_0$ parameterized by arclength and, for each $s \in (0,L)$, $\gamma(s,\cdot)$ is a unit speed geodesic normal to $\Lambda_0$ at $\gamma(s,0)$.
The length of the curve $\gamma(\cdot,r)$ is $$ L(r) = \int_{s=0}^{s=L} |\partial_s\gamma(s,r)|\,ds, $$ and therefore \begin{align*} \frac{dL}{dr} &= \frac{d}{dr} \int_{s=0}^{s=L} (\partial_s\gamma\cdot\partial_s\gamma)^{1/2}\,ds\\ &= \int_{s=0}^{s=L} \frac{\partial_s\gamma\cdot\nabla_{\partial_r\gamma}\partial_s\gamma}{|\partial_s\gamma|}\,ds\\ &= \int_{s=0}^{s=L} \frac{\partial_s\gamma\cdot\nabla_{\partial_s\gamma}\partial_r\gamma}{|\partial_s\gamma|}\,ds\\ \end{align*} It follows that if $T(s) = \partial_s\gamma(s,0)$ and $N(s) = \partial_r\gamma(s,0)$, then, since $|T|=|N|=1$, \begin{align*} \left.\frac{dL}{dr}\right|_{r=0} &= \int_{s=0}^{s=L} T\cdot\nabla_TN\,ds\\ &= \int_{s=0}^{s=L} \kappa(s)\,ds. \end{align*}