estimate the perimeter of the island I'm assigned a task involving solving a problem that can be described as follows: Suppose I'm driving a car around a lake. In the lake there is an island of irregular shape. I have a GPS with me in the car so I know how far I've driven and every turns I've made. Now suppose I also have a camera that takes picture of the island 30 times a second, so I know how long sidewise the island appears to me all the time. Also assume I know the straight line distance between me and the island all the time. Given these conditions, if I drive around the lake for one full circle, will I be able to estimate the perimeter of the island? If yes, how? Thanks.
Assume the island $\Omega$ is convex. Parametrize the observed width $w$ of $\Omega$ with the direction from which it is observed. So we have a function $\phi\to w(\phi)$ of period $\pi$. The average observed width is $$\bar w:={1\over 2\pi}\int_0^{2\pi }w(\phi)\ d\phi\ .$$
Consider a line element $ds$ of $\partial\Omega$. Its contribution to $\bar w$ is half its average projection length over all directions $\phi$ (here we use the convexity of $\partial\Omega$), so it is $\ \lambda\thinspace\thinspace ds\ $ for some universal constant $\lambda&gt;0$ yet to be determined. In any case we have  a formula of the following kind:

$$\bar w = \lambda \int_{\partial\Omega} ds =\lambda\ L(\partial\Omega)\ .$$

When $\Omega$ is the unit disk then $\bar w=2$ and $L(\partial\Omega)=2\pi$. It follows that $\lambda={1\over\pi}$, and solving for $L(\partial\Omega)$ we get

$$L(\partial\Omega)=\pi\ \bar w={1\over 2}\int_0^{2\pi} w(\phi)\ d\phi\ .$$
I am going to take a stab at this, although I would like to hear a flaw in my argument. 

If you have a reference point on the island where your camera is always pointed to then, since you know your exact travel path and distance from your path to the edge of the island, you can plot the shape of the island, then, finding the area can be done in multiple ways (integration being one of the ways). 

If you do not have a reference point that the camera(your eyes) can always point to, then I think the problem is tricky, and I believe there is no solution.