Day 11: Escape the Field

We only need to consider the case where $0 \leq \theta < \frac{\pi}{4}$ , as the other cases are identical to this one.

So what we want is: $\frac{\int_0^\frac{\pi}{4} \sec{x} \, \mathrm{d}x }{(\frac{\pi}{4})} = \frac{\left[ \ln (\sec x + \tan x) \right] ^{\frac{\pi}{4}} _0 }{(\frac{\pi}{4})} = 1.1221...$

But our answer needs to be in metres, therefore our answer is $\boxed{1122}$ as required.

I would rather take ln tan [Pi/ 4 + (Pi/ 4)/ 2] to evaluate.

Given that for angle $\theta$ the distance is $1/\cos(\theta)$ then for a distance $x$ we have that the angle is $F(x)=\arccos(1/x)$ . The probability that he travels of a distance between [a,b] is equal to $4(F(b)-F(a))/\pi$ . Therefore $f(x)= F'(x) 4/\pi$ is the distribution of probability. The expectaction can be calculated with the following formula $\int_1^{\sqrt{2}}f(x)xdx$ where 1 and $\sqrt{2}$ are the minimum and maximum distance he can travel.

So $f(x)=\frac{4}{\pi x \sqrt{x^2-1}}$ and $\int_1^{\sqrt{2}} \frac{4}{\pi \sqrt{x^2-1}}dx =1.1222$ . Therefore the answer in meters is 1222.

The integral is calculated by replacing $z=\sqrt{x^2-1}$ to have $\int \frac{4}{\pi \sqrt{z^2+1}}dz$ which is equal to $\sinh^{-1}(z)4/\pi=\ln(z+\sqrt{z^2+1})4/\pi$ . which results in $\ln(x+\sqrt{x^2-1})4/\pi$ .