Seja $\rho(\theta)$ uma função diferenciável no intervalo $(a, b)$, chamando de $C$ seu comprimento quando $\theta$ varia de $a$ a $b$:
${\small C = \displaystyle\lim_{N \rightarrow 0} \displaystyle\sum \sqrt{\left\{\left[\rho(\theta_{i+1})\right]\left[\cos(\theta_{i+1})\right] - \left[\rho(\theta_i)\right]\left[\cos(\theta_i)\right]\right\}^2 + \left\{\left[\rho(\theta_{i+1})\right]\left[\sin(\theta_{i+1})\right] - \left[\rho(\theta_i)\right]\left[\sin(\theta_i)\right]\right\}^2}}$
Sejam $\theta_{k_i}$ tais que $\theta_i \le \theta_{k_i} \le \theta_{i+1}$, pelo TVM (Teorema do Valor Médio):
${\scriptsize C = \displaystyle\lim_{N \rightarrow 0} \displaystyle\sum \sqrt{\left\{(\theta_{i+1} - \theta_i)\left[\rho'(\theta_{k_i})\cos \theta_{k_i} - \rho(\theta_{k_i})\sin \theta_{k_i}\right]\right\}^2 + \left\{(\theta_{i+1} - \theta_i)\left[\rho'(\theta_{k_i})\sin \theta_{k_i} + \rho(\theta_{k_i})\cos \theta_{k_i}\right]\right\}^2} =}$
${\scriptsize = \displaystyle\lim_{N \rightarrow 0} \displaystyle\sum \sqrt{\left[\rho'(\theta_{k_i})\cos \theta_{k_i} - \rho(\theta_{k_i})\sin \theta_{k_i}\right]^2 + \left[\rho'(\theta_{k_i})\sin \theta_{k_i} + \rho(\theta_{k_i})\cos \theta_{k_i}\right]^2} (\theta_{i+1} - \theta_i)}$
Logo, pela definição de integral:
$\fbox{$C = \displaystyle\int_a^b \sqrt{\left[\rho'(\theta)\cos \theta - \rho(\theta)\sin \theta\right]^2 + \left[\rho'(\theta)\sin \theta + \rho(\theta)\cos \theta\right]^2}\ d\theta$}$
Exemplo: seja $\rho(\theta) = 1$, $a = 0$ e $b = 2\pi$ (o ciclo trigonométrico):
$C = \displaystyle\int_0^{2\pi} \sqrt{\sin^2 \theta + \cos^2 \theta}\ d\theta = \left.\theta\right|_0^{2\pi} = 2\pi$.
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