so.. hmm notice the picture below
from the west of the radar station, 12.5 away
is the boat, traveling south at 57.3 degrees
so.. we have sides "b" the length it has traveled,
and side "y", the distance ever changing as it
travels south-eastward
at what point, is "y" the smallest
well... we have two sides, and one angle encroached by them,
namely, the law of cosines
we have a side that's constant, 12.5 miles,
and one that's ever changing with db/dt or the rate of
the boat, 11mph
so [tex]\bf \textit{Law of Cosines}\\ \quad \\
c^2 = {{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)\implies
c = \sqrt{{{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)}\\
----------------------------\\
y=\sqrt{b^2+12.5^2-2(b)(12.5)cos(57.3^o)}
\\ \quad \\
y=\sqrt{b^2+156.25-25b\cdot cos(57.3^o)}\impliedby cos(57.3^o)\textit{ is a constant}[/tex]
so.. all you need to do, is to find the "minumum" point for that equation of "y"
of course, you'll need to get dy/dx first,
then set it to 0, to get the value of "b", or the boat distance,
when the tangent is a horizontal line, thus an extrema
at that point "b", you found, do a first derivate test to see if it's a minimum,
so, "y" is at a minimum at that point "b"
we know the boat has a db/dt of 11, is going 11mph
or 11 miles in 60minutes
then do the conversion for "b" distance
