A point that is equidistant from all three sides of a triangle is the center of the triangle

A point that is equidistant from all three sides of a triangle is the center of the triangle

Intersection of three bisectors
A triangle has three angles, that is, three bisectors can be made, and the three bisectors intersect with a point (that is, the heart), At the same time, the distance from the three angles to the corresponding sides is equal. So the point with equal distance to the triangle three times is the intersection point of the bisector

How to understand the concept of Green's formula for higher numbers
In the concept of Green's formula: what is the meaning of piecewise smooth curve l?
Is green's formula also true for complex connected regions? How to calculate it,

Curve integral condition: piecewise smooth. Smooth: with tangent, please refer to the calculation process of two kinds of curve integral, and think about why it is smooth, not differentiable. Piecewise: (finite multi segment) please teach one variable integral (including generalized integral) condition: finite discontinuities, and piecewise integrable, please think about why it is finite

The problem of Green's formula of higher number!
Why should we use (0,0) point to discuss p146 case 4 in Volume 2 of the fifth edition of advanced mathematics, whether it belongs to region D, and other points are not?
It's a book from Tongji University

Green's formula requires that the integrand function and its first partial derivative exist in the region D. if there is a point (0,0) in the region bounded by the curve given in its title, the integrand function and its first partial derivative do not exist at this point, so we need to find a very small circle (radius tends to 0)

The application of Green's formula in higher numbers
Now I'm a freshman and I've just finished advanced mathematics
1. Suddenly, there is a definition of Green's formula that should be applied to smooth curves
However, there is no special treatment for broken line, and there is no mistake. The teacher never mentioned this one. Why?
2. A simple problem of Green's formula depressed me: line L goes from (0,0) to (2,0) along y = 1 - | X-1 | broken line
If I want to find ∫ L (- YDX + XDY), I use the common method to make it - 2, and the answer is - 2, but if I use Green's formula, the result is 2,
The student said thank you

1. The boundary conditions required by Green's formula need not be smooth curve, but simple curve
To put it simply, it's a common curve that doesn't intersect itself, that is, the starting point and the starting point are shown on the curve
The end point is allowed to coincide, other points are not allowed to coincide, so the curve can be
2. You've used the Green's formula wrong. The Green's formula requires that the boundary be a closed curve, which is not in this problem, so you need to fill the line
In addition, it is also required that the curve should be counter clockwise. The line segment s from (0,0) to (2,0) should not be counter clockwise,
So you need to add a minus sign
The specific method is as follows: the direction of S is from (0,0) to (2,0), so l and S ^ (-) are clockwise, where s ^ (-) is from (0,0) to (2,0)
(2,0) to (0,0)
Original integral = integral of L and S ^ (-) + integral on S
Integral of = - 2 triangle area + s (*)
The area of triangle is 1, and the parameter of S is y = 0,0

Green's formula of higher number and its application

Then, where ∫ d2dxdy = the area of D2

Advanced mathematics page 205 Green's formula example 4
How can there be two answers

There are two cases. One is that when the area enclosed by L does not contain the origin, the integral is 0 according to Green's formula. The other is that when the area enclosed by L contains the origin, Green's formula can not be used directly, and the auxiliary line x ^ 2 + y ^ 2 = R ^ 2 is considered according to the denominator of integrand. The specific method is to read a book

If Green's formula holds, what conditions should be satisfied?

The curve is a closed space, but the product of Green's formula is partial differential, so there may be a missing term, but it is usually not considered. In the proof, it is assumed that it can be integrated back to the original function. There are many factors, so we can study it by ourselves

The application condition of Green's formula

A:
1) Region D must be simply connected, that is to say, region D is continuous. Generally speaking, there is no "hole" in region D;
2) The curve of region D must be continuous;
3) The curve L (which may be composed of segments) has positive regulation;
4) The integrand has continuous first order partial derivatives in D

X ^ 2 + y ^ 2 = e ^ y for dy / DX

The two sides are derived from X
2x+2yy'=e^y*y

X ^ Y-Y ^ x = 2, find dy / DX

Let x ^ y = u, then: ylnx = LNU, ∧ lnxdy / DX + Y / x = (1 / U) Du / DX, ∧ U (lnxdy / DX + Y / x) = Du / DX,
∴du/dx＝x^y·lnx·dy/dx＋（y/x）x^y.
Let y ^ x = t, then: xlny = LNT, LNY + (x / y) dy / DX = (1 / T) DT / DX, t [LNY + (x / y) dy / DX] = DT / DX,
∴dt/dx＝y^x·lny＋（x/y）y^x·dy/dx.
∵x^y＋y^x＝2,∴u＋t＝2,∴du/dx＋dt/dx＝0,
∴［x^y·lnx·dy/dx＋（y/x）x^y］＋［y^x·lny＋（x/y）y^x·dy/dx］＝0,
∴［x^y·lnx＋（x/y）y^x］dy/dx＝－［（y/x）x^y＋y^x·lny］,
∴dy/dx＝－［（y/x）x^y＋y^x·lny］/［x^y·lnx＋（x/y）y^x］.