от Anubis » 23 Юни 2012, 09:35
Зад. 1. [tex]L: \, \begin{array}x = e^t (\sin t + \cos t) \\ y = e^t (\sin t - \cos t) \end{array}, \, t \in [0; \, 1][/tex]
Дължината на кривата е равна на [tex]l=\int_{0}^{1} \sqrt{\dot{x}^2+\dot{y}^2} \operatorname{d}t[/tex].
[tex]\dot{x}=2e^t \cos t, \, \dot{y}=2e^t \sin t \Rightarrow \dot{x}^2+\dot{y}^2=(2e^t)^2 \Rightarrow l = \int_{0}^{1} 2e^t \operatorname{d}t = 2(e^1-e^0)=2(e-1)[/tex]
Зад. 3. [tex]\int_{L} y \operatorname{d}x + x \operatorname{d}y, \quad L: \, \begin{array}x = \sin t \\ y = \cos t\end{array}, \quad t \in \left [ 0; \, \frac{\pi}{4} \right ][/tex]
[tex]\operatorname{d}x=\cos t \operatorname{d}t, \quad \operatorname{d}y=-\sin t \operatorname{d}t \Rightarrow \int_{0}^{\frac{\pi}{4}} (\cos^2 t - \sin^2 t) \operatorname{d}t = \int_{0}^{\frac{\pi}{4}} \cos 2t \operatorname{d}t = \frac{1}{2} \left ( \sin\frac{\pi}{2}-\sin 0\right ) = \frac{1}{2}[/tex]
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