**Unformatted text preview: **v , w , a , and b , then it is easy to determine this. With a little cleverness, however, we can determine the minimum in general. We have seen that f ±± ( x ) is always positive, so the derivative f ± ( x ) is always increasing. We know that at wb/ √ v 2-w 2 the derivative is zero, so for values of x less than that critical value, the derivative is negative. This means that f (0) > f ( a ), so the minimum occurs when x = a . So the upshot is this: If you start farther away from C than wb/ √ v 2-w 2 then you always want to cut across the sand when you are a distance wb/ √ v 2-w 2 from point C . If you start closer than this to C , you should cut directly across the sand....

View
Full Document

- Fall '07
- JonathanRogawski
- Math, Calculus, Critical Point, Derivative