Задача 70652 Найти длину дуги кривой y=ln(x^2–1),...
Условие
Найти длину дуги кривой y=ln(x^2–1), отсеченной прямыми х=2, х=5
Решение
[m]y`=\frac{2x}{x^2-1}[/m]
[m]1+(y`)^2=1+(\frac{2x}{x^2-1})^2[/m]
[m]1+(y`)^2=1+\frac{(2x)^2}{(x^2-1)^2}[/m]
[m]1+(y`)^2=\frac{(x^2-1)^2+(2x)^2}{(x^2-1)^2}[/m]
[m]1+(y`)^2=\frac{x^4-2x^2+1+4x^2}{(x^2-1)^2}[/m]
[m]1+(y`)^2=\frac{x^4+2x^2+1}{(x^2-1)^2}[/m]
[m]1+(y`)^2=\frac{(x^2+1)^2}{(x^2-1)^2}[/m]
[m]\sqrt{1+(y`)^2}=\frac{x^2+1}{x^2-1}[/m]
[m]L= ∫ ^{5}_{2}\frac{x^2+1}{x^2-1}dx=∫ ^{5}_{2}\frac{x^2-1+2}{x^2-1}dx=∫ ^{5}_{2}(1+\frac{2}{x^2-1})dx=[/m]
[m]=(x+2\cdot \frac{1}{2}ln|\frac{x-1}{x+1}|)|^{5}_{2}=(5+ln|\frac{5-1}{5+1}|)-(2+ln|\frac{2-1}{2+1}|)=[/m]
[m]=3+ln\frac{4}{6}-ln\frac{1}{3}=3+ln(\frac{4}{6}\cdot 3)=3+ln8=3+3ln2[/m]