A Sliding Ladder Problem...Thanks to Doug Shaw
It is a beautiful Spring evening. You and your friends are sitting in your room, reading your Calculus books. At one point, one of them gets to a related rates problem and starts to laugh:
A 10 foot long ladder rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 foot/second, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is six feet from the wall.
"Why do you laugh so?" you ask.
"I laugh out of delight," says your friend. "I love mathematics so. Observe as I do this problem."
Your friend draws the following picture:
"We want to find dy/dt. And we can set up: x2 + y2 = 100. Now, we want dy/dt when dx/dt = 1 and x = 6. Substituting x = 6 into the equation gives 36 + y2 = 100. Taking derivatives yields 2y dy/dt = 0, so the answer is dy/dt = 0."
The problem is, of course, that this answer does not make any sense. Why doesn't this answer make sense? What error did your friend make, and how can it be corrected?
Note: For a corresponding problem, see the Golden Oldies section.
Source: http://www.dougshaw.com/findtheerror/FTErelated.html
Hint: Sorry...no hint....except to follow Doug Shaw's advice.
Solution Commentary: Again, out of respect for Doug Shaw's web site, no solution will be provided. Nonetheless, think about the order of things performed in the problem...and review the key "generic" steps in setting up a related rates problem.
