I'm still enjoying Amanda's calculus course. She uses her book, so I read up the topics on the syllabus in my old textbook, or in Martin Gardner's *Calculus Made Easy*, and we then work the problems from her homework together. The couple that calculates together, sticks together.

So here's a recent problem in "related rates." A ten foot ladder stands straight against a wall. Then someone starts moving its bottom end away from the wall at the rate of one foot per second. How fast is the top of the ladder going down the side of the wall at the moment the bottom of the ladder is six feet out from the wall? Eight feet out from the wall?

Maybe your intuition is that the top of the ladder moves, if not at the same steady rate as the bottom, at least at some constant rate. But it is relatively easy to verify that that is not so. After six seconds, for example, the base of the ladder will be six feet from the wall. How far will the top of the ladder have gone down? Since the ladder is always ten feet long, we are considering a right triangle with a base of six feet and a hypotenuse of ten feet. From the Pythagorean Theorem, it follows that the height is eight feet (8^{2} + 6^{2} = 10^{2}), and that in the first six seconds the ladder has gone down two feet. Two seconds later, however, the base is eight feet and the height is only six feet, so the ladder went down as far in those two seconds as it did in the first six. Determining instantaneous velocity at certain points is a problem in calculus.

The key to solving the problem is to analyze this constantly changing right triangle. If we call the height of the ladder on the wall h, and the distance from the bottom of the ladder to the wall b, then

h

^{2}+ b^{2}= 10^{2}.

What we are looking for is the ratio of the change in height to the change in time--dh/dt, in the language of calculus. Differentiating implicitly, we get

2h (dh/dt) + 2b (db/dt) = 0.

Now, let us recall that we are looking for dh/dt when b is six and eight, respectively; also, that db/dt, the ratio of the change in the length of the base to the change in time, is a constant one foot per second. So all that really remains is to make the necessary substitutions and then do the algebra:

2h (dh/dt) + 2b = 0

h (dh/dt) = -b

dh/dt = -b/h

When b is six, h is eight; and when b is eight, h is six--remember our right triangle with the hypotenuse a constant ten. So at the instant the bottom of the ladder is six feet from the wall, the top of the ladder is going down the wall at the rate of 9 inches per second. When the bottom is eight feet from the wall, the top is going down at the rate of 16 inches per second.

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