1. Problem: Farmer Tim is lost in the densely-forested Cartesian plane. Starting from the origin he walks a sinusoidal path in search of home; that is, after $t$ minutes he is at position $(t, \sin t)$.

Five minutes after he sets out, Alex enters the forest at the origin and sets out in search of Tim. He walks in such a way that after he has been in the forest for $m$ minutes, his position is $(m, \cos t)$.
What is the greatest distance between Alex and Farmer Tim while they are walking in these paths?
Solution: At arbitrary time $t$, Farmer Tim is at position $(t, \sin t)$ and Alex is at position $(t-5, \cos t)$. Hence at time $t$, the distance, $d$, between Tim and Alex is $d=\sqrt{(\sin t-\cos t)^{2}+25}$. To find the maximum value of $d$, we solve for $t$ such that $\frac{d d}{d t}=0$.
$\frac{d d}{d t}=\frac{(\sin t-\cos t)(\cos t+\sin t)}{\sqrt{(\sin t-\cos t)^{2}+25}}$. Then $\frac{d d}{d t}=0 \Rightarrow \sin ^{2} t-\cos ^{2} t=0 \Rightarrow \sin ^{2} t=\cos ^{2} t$. Equality happens if $t$ is any constant multiple of $\frac{\pi}{4}$.

Notice that to maximize $d$, we need to maximize $(\sin t-\cos t)^{2}$. This is achieved when $\cos t=-\sin t$. Because we determined earlier that $t$ is a constant multiple of $\frac{\pi}{4}$, then with this new condition, we see that $t$ must be a constant multiple of $\frac{3 \pi}{4}$.
Then $(\sin t-\cos t)^{2}=2 \Rightarrow d=\sqrt{29}$.