Calculus
posted by Sarah
Suppose you are in a room that is a*10 meters. You can walk normally along the "a" meter long wall at "s" meters per second and crab walk in any direction at 1 meter per second. You start at one corner of the room, walk normally along the wall of length "a" meters for some distance, and then crab walk straight to the corner opposite from where you started.
Suppose that the fastest route to the opposite corner of the room is to directly crab walk from one corner of the room to the other (i.e. you walk normally for 0 distance).
What can be said about the relationship between a and s?

Steve
If you walk a distance x, then the time t to reach the opposite corner is
t(x) = x/s + √((ax)^2 + 10^2)
Now, just find dt/dx. The fastest time is when dt/dx=0, so plug in the fact that it is when x=0. Then you will get an equation is a and s.
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