Długość łuku cykloidy
-
MarekZGrabiny
- Użytkownik
- Posty: 15
- Rejestracja: 1 lut 2023, o 22:06
- Płeć: Mężczyzna
- wiek: 22
- Podziękował: 3 razy
Długość łuku cykloidy
Podstawiłem wszystko do wzoru ale nie wiem czy granice całkowania są dobrze określone, oraz jak ruszyć tę całkę.
-
Mariusz M
- Użytkownik
- Posty: 6909
- Rejestracja: 25 wrz 2007, o 01:03
- Płeć: Mężczyzna
- Lokalizacja: 53°02'N 18°35'E
- Podziękował: 2 razy
- Pomógł: 1246 razy
Re: Długość łuku cykloidy
Można też przez części
\(\displaystyle{ \int{ \sqrt{1-\cos{t}} \mbox{d}t} = \int{\frac{1-\cos{t}}{\sqrt{1-\cos{t}}}\mbox{d}t}\\
=\int{\frac{1}{\sqrt{1-\cos{t}}}\mbox{d}t} - \int{\frac{\cos{t}}{\sqrt{1-\cos{t}}}\mbox{d}t}\\
=\int{\frac{1}{\sqrt{1-\cos{t}}}\mbox{d}t} -\left( \frac{\sin{t}}{\sqrt{1-\cos{t}}} -\int{\sin{t} \cdot \left( -\frac{1}{2}\right) \cdot \frac{\sin{t}}{\left( 1-\cos{t}\right) \sqrt{1-\cos{t}} } } \right)\\
= -\frac{\sin{t}}{\sqrt{1-\cos{t}}}-\frac{1}{2} \int{\frac{\sin^2{t}}{\left( 1-\cos{t}\right) \sqrt{1-\cos{t}} }\mbox{d}t} +\int{\frac{1}{\sqrt{1-\cos{t}}}\mbox{d}t}\\
=-\frac{\sin{t}}{\sqrt{1-\cos{t}}}+\frac{1}{2}\int{\frac{2-2\cos{t}-\sin^{2}{t}}{\left( 1-\cos{t}\right) \sqrt{1-\cos{t}} }\mbox{d}t}\\
=-\frac{\sin{t}}{\sqrt{1-\cos{t}}}+\frac{1}{2}\int{\frac{1+1-\sin^{2}{t}-2\cos{t}}{\left( 1-\cos{t}\right) \sqrt{1-\cos{t}} }\mbox{d}t}\\
=-\frac{\sin{t}}{\sqrt{1-\cos{t}}}+\frac{1}{2}\int{\frac{1+\cos^{2}{t}-2\cos{t}}{\left( 1-\cos{t}\right) \sqrt{1-\cos{t}} }\mbox{d}t}\\
=-\frac{\sin{t}}{\sqrt{1-\cos{t}}}+\frac{1}{2}\int{\frac{\left( 1-\cos{t}\right)^2 }{\left( 1-\cos{t}\right) \sqrt{1-\cos{t}} }\mbox{d}t}\\
=-\frac{\sin{t}}{\sqrt{1-\cos{t}}}+\frac{1}{2}\int{\frac{\left( 1-\cos{t}\right) }{\sqrt{1-\cos{t}} }\mbox{d}t}\\
\int{ \sqrt{1-\cos{t}} \mbox{d}t} = -\frac{\sin{t}}{\sqrt{1-\cos{t}}}+\frac{1}{2}\int{ \sqrt{1-\cos{t}} \mbox{d}t}\\
\left( 1-\frac{1}{2}\right) \int{ \sqrt{1-\cos{t}} \mbox{d}t} = -\frac{\sin{t}}{\sqrt{1-\cos{t}}}+C_{1}\\
\frac{1}{2} \int{ \sqrt{1-\cos{t}} \mbox{d}t} = -\frac{\sin{t}}{\sqrt{1-\cos{t}}}+C_{1}\\
\int{ \sqrt{1-\cos{t}} \mbox{d}t} = -2\frac{\sin{t}}{\sqrt{1-\cos{t}}} + C\\
}\)
\(\displaystyle{ \int{ \sqrt{1-\cos{t}} \mbox{d}t} = \int{\frac{1-\cos{t}}{\sqrt{1-\cos{t}}}\mbox{d}t}\\
=\int{\frac{1}{\sqrt{1-\cos{t}}}\mbox{d}t} - \int{\frac{\cos{t}}{\sqrt{1-\cos{t}}}\mbox{d}t}\\
=\int{\frac{1}{\sqrt{1-\cos{t}}}\mbox{d}t} -\left( \frac{\sin{t}}{\sqrt{1-\cos{t}}} -\int{\sin{t} \cdot \left( -\frac{1}{2}\right) \cdot \frac{\sin{t}}{\left( 1-\cos{t}\right) \sqrt{1-\cos{t}} } } \right)\\
= -\frac{\sin{t}}{\sqrt{1-\cos{t}}}-\frac{1}{2} \int{\frac{\sin^2{t}}{\left( 1-\cos{t}\right) \sqrt{1-\cos{t}} }\mbox{d}t} +\int{\frac{1}{\sqrt{1-\cos{t}}}\mbox{d}t}\\
=-\frac{\sin{t}}{\sqrt{1-\cos{t}}}+\frac{1}{2}\int{\frac{2-2\cos{t}-\sin^{2}{t}}{\left( 1-\cos{t}\right) \sqrt{1-\cos{t}} }\mbox{d}t}\\
=-\frac{\sin{t}}{\sqrt{1-\cos{t}}}+\frac{1}{2}\int{\frac{1+1-\sin^{2}{t}-2\cos{t}}{\left( 1-\cos{t}\right) \sqrt{1-\cos{t}} }\mbox{d}t}\\
=-\frac{\sin{t}}{\sqrt{1-\cos{t}}}+\frac{1}{2}\int{\frac{1+\cos^{2}{t}-2\cos{t}}{\left( 1-\cos{t}\right) \sqrt{1-\cos{t}} }\mbox{d}t}\\
=-\frac{\sin{t}}{\sqrt{1-\cos{t}}}+\frac{1}{2}\int{\frac{\left( 1-\cos{t}\right)^2 }{\left( 1-\cos{t}\right) \sqrt{1-\cos{t}} }\mbox{d}t}\\
=-\frac{\sin{t}}{\sqrt{1-\cos{t}}}+\frac{1}{2}\int{\frac{\left( 1-\cos{t}\right) }{\sqrt{1-\cos{t}} }\mbox{d}t}\\
\int{ \sqrt{1-\cos{t}} \mbox{d}t} = -\frac{\sin{t}}{\sqrt{1-\cos{t}}}+\frac{1}{2}\int{ \sqrt{1-\cos{t}} \mbox{d}t}\\
\left( 1-\frac{1}{2}\right) \int{ \sqrt{1-\cos{t}} \mbox{d}t} = -\frac{\sin{t}}{\sqrt{1-\cos{t}}}+C_{1}\\
\frac{1}{2} \int{ \sqrt{1-\cos{t}} \mbox{d}t} = -\frac{\sin{t}}{\sqrt{1-\cos{t}}}+C_{1}\\
\int{ \sqrt{1-\cos{t}} \mbox{d}t} = -2\frac{\sin{t}}{\sqrt{1-\cos{t}}} + C\\
}\)
-
janusz47
- Użytkownik
- Posty: 7917
- Rejestracja: 18 mar 2009, o 16:24
- Płeć: Mężczyzna
- Podziękował: 30 razy
- Pomógł: 1671 razy
Re: Długość łuku cykloidy
\(\displaystyle{ \gamma(t): \begin{cases} x = a[(\cos(t) +t\sin(t)] \\ y = a[1 -\cos(t)], \ \ a >0 \ \ 0 \leq t \leq 2\pi. \end{cases} }\)
\(\displaystyle{ x'(t) = a(1-\cos(t), \ \ y'(t) = a\sin(t),}\)
\(\displaystyle{ x'^2 (t) + y'^2(t) = ( a^2[1 -2\cos(t)+\cos^2(t)] + a^2\sin^2(t) = a^2 -2a^2 \cos(t) +a^2\cos^2(t) +a^2\sin^2(t) = }\)
\(\displaystyle{ a^2 -2a^2\cos(t) + a^2[\cos^2(t) + \sin^2(t)] = a^2 -2a^2\cos(t) +a^2 = 2a^2[1 -\cos(t)] = 4a^2\sin^2\left(\frac{t}{2}\right).}\)
\(\displaystyle{ |\mathcal{L}| = \int_{0}^{2\pi} \sqrt{4a^2\sin^2\left(\frac{t}{2}\right)} dt = 2a\int_{0}^{2\pi} \sin\left(\frac{t}{2}\right)dt = 4a \int_{0}^{\pi}sin(z)dz = 4a \cos(z)\mid_{\pi}^{0} = 4a[\cos(0) - \cos(\pi)] = 4a[1 -(-1)] = 4a(1+1) = }\)
\(\displaystyle{ = 4a\cdot 2 = 8a. }\)
\(\displaystyle{ x'(t) = a(1-\cos(t), \ \ y'(t) = a\sin(t),}\)
\(\displaystyle{ x'^2 (t) + y'^2(t) = ( a^2[1 -2\cos(t)+\cos^2(t)] + a^2\sin^2(t) = a^2 -2a^2 \cos(t) +a^2\cos^2(t) +a^2\sin^2(t) = }\)
\(\displaystyle{ a^2 -2a^2\cos(t) + a^2[\cos^2(t) + \sin^2(t)] = a^2 -2a^2\cos(t) +a^2 = 2a^2[1 -\cos(t)] = 4a^2\sin^2\left(\frac{t}{2}\right).}\)
\(\displaystyle{ |\mathcal{L}| = \int_{0}^{2\pi} \sqrt{4a^2\sin^2\left(\frac{t}{2}\right)} dt = 2a\int_{0}^{2\pi} \sin\left(\frac{t}{2}\right)dt = 4a \int_{0}^{\pi}sin(z)dz = 4a \cos(z)\mid_{\pi}^{0} = 4a[\cos(0) - \cos(\pi)] = 4a[1 -(-1)] = 4a(1+1) = }\)
\(\displaystyle{ = 4a\cdot 2 = 8a. }\)