Page 71 - Calc for the Clueless

Yeeeow!!!

The second solution is not much nicer. We will do this with trig functions in terms of the angle A.

cot A = AB/8 (always try to get the unknown on top). AB = 8 cot A. AC = 8 cot A + 1.

sec A = AE/AC. AE = L. L = AC sec A = (8 cot A + 1) sec A = 8 csc A + sec A (using trig identities).

8 cot A csc A = tan A sec A

This says the ratio of DB to AB is 2:1. DB = 8 feet; AB = 4 feet. Therefore AC = 4 + 1 = 5. By the same
reasoning, EC must be 10 feet.

We finish the max and min problems with what I consider the ultimate in distance problems—not the hardest,
but one of the slickest.

Example 15—

A man is on an island which is 4 miles from the nearest point on a straight shoreline. He wishes to go to a
house which is 12 miles from this nearest point. If he rows at 3 miles per hour and runs at 5 miles per hour, find
the shortest time to reach the house.

Insight: If he rowed faster than he ran, he would have rowed straight for the house. If he ran muuuch faster than
he rowed, he would row straight for the shore and then run. So we are looking for the point on the shore where
he must land. The picture is below.