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?
- JimLv 61 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
edges equal the circle's radius.參考資料： my puny brain