Calculus: Analytic Geometry and Calculus, with Vectors

188 Functions, limits, derivatives

Our knowledge of the natures of graphs of

lines now shows that the graph of y = I(x)

over the interval x1 < x 5 x2 is a line seg-

ment having slope ml. Figure 3.795 shows

the graph of y = I(x) that can be constructed

from information given in a possible tax

1(x) table. Everybody says that tax rates are

different in different tax brackets, being for

01 x0 x1 xy x3 xS z example m1 (or 100m1 per cent) when

Figure 3.795 x1 < x < x2. Our fundamental problem is

the following. Show that if x1 < x < x2,

then our definition of a rate as a derivative is in agreement with the things that

have been said about tax rates. Show that when x is x1, the tax rate does not

exist but that the "right-hand rate" and "left-hand rate" do exist.

3.8 Related rates Useful information about derivatives and their

applications can be gained by solving problems more or less like the fol-

lowing one. Figure 3.81 represents a ladder which is 10 units (feet or

meters) long. The top of the ladder rests against a vertical wall and is

Figure 3.81 Figure 3.82

8 units above the horizontal floor upon which

the bottom of the ladder rests 6 units from

the wall. It is supposed that the bottom of

the ladder is moving away from the wall at

the rate of 2 units per second, and we are

required to find the rate at which the

invisible man at the top of the ladder is

plunging earthward. To solve this problem,

we begin by constructing the more propitious

Figure 3.82 in which the ladder still has length 10 but x and y are varia-

bles which (unlike 8 and 6) can have different values at different times i

when the ladder is skidding. The variables x and y are related by the

formula

(3.83) x2 + y2 = 100.

In order to obtain a relation involving dx/dt (the rate at which the bottom

of the ladder is moving away from the wall) and dy/dt (the rate at which

our man is moving upward), we need the fundamental idea that we should

consider x and y to be functions of t and differentiate with respect to t.

Equating the derivatives of the members of (3.83) and dividing by 2

gives the formula

(3.84) xdx+ydt=0

which relates the related rates dx/dt and dy/dt. Putting x = 6, y = 8,

and dx/dt = 2 shows that dy/dt = -J. This shows that our poor