Check if avoiding 2nd quad with 1/2 speed is faster via A-O-D:
Crawling times along:
AOD : 26.849/s = less time. (be double checked)
ABCD: 33.616/s
Hi there I need some help with this problem
sorry if this is in the wrong place, Im a new member here
A bug is crawling on the coordinate plane from (5,9) to (-15,-7) at constant
speed one unit per second except in the second quadrant where it travels at 1/2
units per second. What path should the bug take to complete its journey in as
short a time as possible?
So far I drew a picture of it and I was assume the shortest amount of time would be to draw a diagonal line connecting the points together and then calculating the time, but it doesn't feel correct to me. I hope someone can help/guide me through this problem, so I can understand how to do this, thanks.
Hey there Jasper thanks for helping me out with this problem, I looked over your solution and doubled checked it. What I've gotten for the total time for ABCD was 21.612 seconds and for AOD it was 26.849 seconds. I've included the 1/2 speed from B to C and discover that avoiding the 2nd quadrant still takes longer than going straight to the destination
I was wondering if there was a proof to determine if going a straight line was the shortest time to the destination or avoiding it.
I find, using differential calculus and help from the computer finding the critical point, that the quickest route is from (5,9) to (0,0) to (-15,-7), giving a time of about 26.85 s. If the bug crawls along the straight line between the two given points, then the time is about 42.08 s. We must in effect, double the distance in the 2nd quadrant, since the speed there is 1/2.