# 1st variational form of surface area in the tangent direction, directional derivative of surface along the local tangent direction

This question would be a follow-up to the question here, where Yuri and Xipan describes the variation in the normal direction (how about in the local tangent direction?).

My question is more like finding the directional derivative of the area in the local tangent direction. For example, the surface I have looks like a typical car engine hood described as (u,w,f(u,w)). At a given point $$p_{1}(u_{1},v_{1})$$ and a closed boundary $$\gamma(u,v)$$ given as $$(u-u_{1})^{2}+(v-v_{1})^{2} = 0.01$$ (for example). I can find the surface area of $$\gamma$$ as $$s_{1}$$. I can also find a tangent plane at $$p_{1}$$. Let's say there is a very close point $$p_{2}(u_{2},v_{2})$$ around $$p_{1}$$ in the $$f_{u}$$ direction. With a similar closed boundary $$\gamma$$ and the surface area is $$s_{2}$$, how do I know if $$s_{2}$$ is going to be greater or smaller or approximately equal to $$s_{1}$$?

I apologize first for not writing this as a proper math question since I'm a MS student in engineering and my native language is not English. I would appreciate any help and feedback.

One observation is that your proposed definition of "surface area around a point" will depend on the way your surface is parametrized. This is because you draw a small circle around $$(u_1,v_1)$$ in the parameter space, and then compute the area of the corresponding patch on the surface. Depending on whether the parameter lines on the surface around this point are far apart or close together, this will give you a smaller or larger surface area:
In this image, picking a point on the left handle will yield a smaller surface area than on the right handle (assuming you keep your squared radius $$r^2=0.01$$ constant). The local area measure is given by $$dA=\|f_u\times f_v\|$$. This tells you roughly the size of one the white/black parallelograms in the image. If you want to know whether this measure decreases or increases in direction $$f_u$$, you can look at $$\frac{d}{du}\|f_u\times f_v\|$$.
• thank you for the response. But I believe my question is how to find $\frac{d}{du}||f_{u}\times f_{v}||$ (I believe this is the directional derivative of the 1st fundamental form). Is there a analytical expression for the derivation in a way similar to the 1st variational form in the normal direction?
• It is not exactly the first fundamental form, but we can compute the area measure $\|f_u\times f_v\|$ from the first fundamental form: if $g$ is the matrix representation of the fundamental form, then the area measure is given by $\sqrt{\det g}$. I am not sure one can give anything more concrete than the formula I have already given. If you write out $\frac{d}{du} \sqrt{\det g}$, you will end up with a formula just in terms of derivatives of $g$, which is maybe more desirable? May 1, 2021 at 21:26