# 英文微積分數學問題?

A rectangular vessel is divided into two equal compartments by a vertical permeable membrane. Liquid in one compartment, initially at a depth of 20 cm, passes into the other compartment, initially empty at a rate proportional to the difference in the levels.

(a)If the depth of liquid in one of the vessels at any time, t minutes, is x cm, derive the differential equation

(b)If the level in the second compartment rises 2 cm in the first 5 minutes, after what time will the difference in levels be 2 cm?

### 1 個解答

• Jim
Lv 6
1 0 年前
最佳解答

To establish a lower bound on the area of a circle of radius r,

suppose we inscribe a regular n-sided polygon within the circle.

Notice that this polygon can be decomposed into n identical triangular

slices meeting at the center of the circle, as illustrated below for

the case n=6.

The outer edges of these triangles approximate the circumference of

the circle. Since a straight line is the shortest path connecting two

given points, the sum of these n linear segments is necessarily (slightly)

less than the circumference. The "height" of these individual triangles

(taking the outer edge as the "base") is slightly less than the radius

of the circle. Now, by increasing n, the sum of the outer edges can be

made arbitrarily close to the circumference C of the circle, and the

"heights" of the triangles can be made arbitrarily close to the radius

r. Also, since the area of each triangle is half its "height" times its

base, and the sum of their areas is half their height times the sum of

their bases, we see that it's possible to inscribe within the circle

a regular n-gon whose area is arbitrarily close to rC/2. It follows

that the area of the circle cannot be less than rC/2.

Similarly, by circumscribing a regular n-sided polygon around a circle,

as shown below for n=6, we can easily see that the area of the polygon

cannot be greater than rC/2.

Therefore, since the circle's area cannot be either less than or greater

than rC/2, it must be rC/2. This is the form of the argument as presented

by Archimedes. Of course, since C = 2 pi r, Archimedes' result is

equivalent to the modern formula A = pi r^2 for the area of a circle.

This establishes the fact that the number pi, which we originally

defined as the ratio of the circumference to the diameter of a circle,

is also the ratio of the circle's area to the area of a square whose