Формула нахождение длины дуги кривой на отрезке [a;b]
![L=\int\limits^b_a {\sqrt{1+(y')^2}} \, dx](https://tex.z-dn.net/?f=L%3D%5Cint%5Climits%5Eb_a%20%7B%5Csqrt%7B1%2B%28y%27%29%5E2%7D%7D%20%5C%2C%20dx)
Для данной задачи:
![a=\frac{1}{4} =0,25\\b=\frac{1}{2}=0,5\\y=\ln(1-x^2)+4\\y'=\frac{1}{1-x^2} *(1-x^2)'=\frac{2x}{x^2-1}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B1%7D%7B4%7D%20%3D0%2C25%5C%5Cb%3D%5Cfrac%7B1%7D%7B2%7D%3D0%2C5%5C%5Cy%3D%5Cln%281-x%5E2%29%2B4%5C%5Cy%27%3D%5Cfrac%7B1%7D%7B1-x%5E2%7D%20%2A%281-x%5E2%29%27%3D%5Cfrac%7B2x%7D%7Bx%5E2-1%7D)
![L=\int\limits^{0,5}_{0,25} {\sqrt{1+(\frac{2x}{x^2-1})^2}} \, dx=\int\limits^{0,5}_{0,25} {\sqrt{\frac{(x^2-1)^2+4x^2}{(x^2-1)^2} }} \, dx=\int\limits^{0,5}_{0,25} {\sqrt{\frac{x^4+2x^2+1}{(x^2-1)^2}}} \, dx=\\=\int\limits^{0,5}_{0,25} {\frac{x^2+1}{x^2-1}} \, dx=\int\limits^{0,5}_{0,25} {\frac{(x^2-1)+2}{x^2-1}} \, dx=\int\limits^{0,5}_{0,25} (1+\frac{2}{x^2-1} )\, dx=](https://tex.z-dn.net/?f=L%3D%5Cint%5Climits%5E%7B0%2C5%7D_%7B0%2C25%7D%20%7B%5Csqrt%7B1%2B%28%5Cfrac%7B2x%7D%7Bx%5E2-1%7D%29%5E2%7D%7D%20%5C%2C%20dx%3D%5Cint%5Climits%5E%7B0%2C5%7D_%7B0%2C25%7D%20%7B%5Csqrt%7B%5Cfrac%7B%28x%5E2-1%29%5E2%2B4x%5E2%7D%7B%28x%5E2-1%29%5E2%7D%20%7D%7D%20%5C%2C%20dx%3D%5Cint%5Climits%5E%7B0%2C5%7D_%7B0%2C25%7D%20%7B%5Csqrt%7B%5Cfrac%7Bx%5E4%2B2x%5E2%2B1%7D%7B%28x%5E2-1%29%5E2%7D%7D%7D%20%5C%2C%20dx%3D%5C%5C%3D%5Cint%5Climits%5E%7B0%2C5%7D_%7B0%2C25%7D%20%7B%5Cfrac%7Bx%5E2%2B1%7D%7Bx%5E2-1%7D%7D%20%5C%2C%20dx%3D%5Cint%5Climits%5E%7B0%2C5%7D_%7B0%2C25%7D%20%7B%5Cfrac%7B%28x%5E2-1%29%2B2%7D%7Bx%5E2-1%7D%7D%20%5C%2C%20dx%3D%5Cint%5Climits%5E%7B0%2C5%7D_%7B0%2C25%7D%20%281%2B%5Cfrac%7B2%7D%7Bx%5E2-1%7D%20%29%5C%2C%20dx%3D)
![=(x+\ln|\frac{x-1}{x+1} |) \int\limits^{0,5}_{0,25}=0,5+\ln|\frac{-0,5}{1,5}|- (0,25+ln|\frac{-0,75}{1,25}|)=\\=0,5-\ln(3)-0,25-\ln\frac{3}{5}=0,25-\ln(\frac{9}{5})](https://tex.z-dn.net/?f=%3D%28x%2B%5Cln%7C%5Cfrac%7Bx-1%7D%7Bx%2B1%7D%20%7C%29%20%5Cint%5Climits%5E%7B0%2C5%7D_%7B0%2C25%7D%3D0%2C5%2B%5Cln%7C%5Cfrac%7B-0%2C5%7D%7B1%2C5%7D%7C-%20%280%2C25%2Bln%7C%5Cfrac%7B-0%2C75%7D%7B1%2C25%7D%7C%29%3D%5C%5C%3D0%2C5-%5Cln%283%29-0%2C25-%5Cln%5Cfrac%7B3%7D%7B5%7D%3D0%2C25-%5Cln%28%5Cfrac%7B9%7D%7B5%7D%29)
Поскольку длина не может быть отрицательной:
![|L|=|0,25-\ln(\frac{9}{5})|=\ln(\frac{9}{5})-0,25](https://tex.z-dn.net/?f=%7CL%7C%3D%7C0%2C25-%5Cln%28%5Cfrac%7B9%7D%7B5%7D%29%7C%3D%5Cln%28%5Cfrac%7B9%7D%7B5%7D%29-0%2C25)
Ответ: ![\ln(\frac{9}{5})-0,25](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B9%7D%7B5%7D%29-0%2C25)
Известный румынский математик прошлого века - Б. Угуртов очень просто через логарифмы нашёл решение этой задачи.
*cos2beta = 2cos2beta = psinbeta - 4 = 0
Подставляем логарифм E=log(2) со значением переменной - 2
Получаем 8cos2beta = log(2) 2 cos 2 beta + 9 sin beta (2) = 121.
Со вторым тоже самое.
1) (m^2 - 2m)^2 - 1=(m²-2m-1)(m²-2m+1)=(m²-2m-1)(m-1)(m-1)
2) 16 - (m^2 + 4m)^2=4²-(m²+4m)²=(4+m²+4m)(4-m²-4m)=(m+2)(m+2)(4-m²-4m)
<span>3) x^2 - 18xy + 81y^2 - z^2=(x-9y)</span>²-z²=(x-9y-z)(x-9y+z)
Окупаемость 20.4= 5 лет
амортизация 500000.5=100000 реблей в год