Question

Вычислить длину дуги кривой

Answer 1

Вычислить длину дуги кривой, заданную в полярной системе координат:

$\rho(\phi)=3e^{3\phi/4},\ \ \ \ 0\le\phi\le\dfrac{\pi}3$

Воспользуемся формулой:

$\displaystyle L=\int\limits^\beta _\alpha \sqrt{\rho^2+\left(\rho'\right)^2} \, d\phi$

$\rho^2=\left(3e^{3\phi/4}\right)^2=9\cdot e^{3\phi/2}\\\\\left(\rho'\right)^2=\left(\left(3e^{3\phi/4}\right)'\right)^2=\left(3e^{3\phi/4}\cdot \dfrac 34\right)^2=\dfrac{81}{16}\cdot e^{3\phi/2}$

$\displaystyle L=\int\limits^{\pi/3} _0 \sqrt{\rho^2+\left(\rho'\right)^2} \, d\phi=\\\\=\int\limits^{\pi/3} _0 \sqrt{9\cdot e^{3\phi/2}+\dfrac {81}{16}\cdot e^{3\phi/2}} \, d\phi=\\\\=\int\limits^{\pi/3} _0 \sqrt{\dfrac {225}{16}\cdot e^{3\phi/2}} \, d\phi=\int\limits^{\pi/3} _0 \dfrac {15}4\cdot e^{3\phi/4} \, d\phi=\\\\=\dfrac {15}4\cdot \int\limits^{\pi/3} _0 e^{3\phi/4} \, d\phi=\dfrac {15}4\cdot e^{3\phi/4}\cdot \dfrac43\ \ \bigg|^{\pi/3}_0=$

$=5\cdot e^{3\phi/4}\ \ \bigg|^{\pi/3}_0=5\cdot\bigg(e^{3\cdot \frac{\pi}3/4}-e^{3\cdot0/4}\bigg)=\\\\\\\boldsymbol{=5\Big(e^{\pi/4}-1\Big)}$

Ответ: $\boldsymbol{5\Big(e^{\pi/4}-1\Big)}\approx5{,}97$