Como calcular o comprimento da curva?

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Nesse exercício vamos calcular o comprimento da curva em coordenadas polares.

Assim como em coordenadas cartesianas, temos:

$$ds=d\theta\sqrt{\left(\dfrac{dr}{d\theta}\right)^2+r^2}$$

Integrando o diferencial, temos o comprimento:

$$l=\int_0^{\pi/2}ds=\int_0^{\pi/2}\sqrt{\left(\dfrac{dr}{d\theta}\right)^2+r^2}d\theta$$

Substituindo a função dada, temos:

$$l=\int_0^{\pi/2}\sqrt{\left(-3\sin\frac{\theta}{2}\cos\frac{\theta}{2}\right)^2+\left(3\cos^2\frac{\theta}{2}\right)^2}d\theta$$

$$l=\int_0^{\pi/2}\sqrt{9\sin^2\frac{\theta}{2}\cos^2\frac{\theta}{2}+9\cos^2\frac{\theta}{2}\cos^2\frac{\theta}{2}}d\theta$$

$$l=\int_0^{\pi/2}\sqrt{\left(3\cos\frac{\theta}{2}\right)^2\left(\sin^2\frac{\theta}{2}+\cos^2\frac{\theta}{2}\right)}d\theta=\int_0^{\pi/2}3\cos\frac{\theta}2d\theta$$

Integrando:

$$l=\left[6\sin\frac{\theta}2\right]_0^{\pi/2}=6\sin\frac\pi4$$

Logo:

$$\boxed{l=3\sqrt2}$$

Nesse exercício vamos calcular o comprimento da curva em coordenadas polares.

Assim como em coordenadas cartesianas, temos:

$$ds=d\theta\sqrt{\left(\dfrac{dr}{d\theta}\right)^2+r^2}$$

Integrando o diferencial, temos o comprimento:

$$l=\int_0^{\pi/2}ds=\int_0^{\pi/2}\sqrt{\left(\dfrac{dr}{d\theta}\right)^2+r^2}d\theta$$

Substituindo a função dada, temos:

$$l=\int_0^{\pi/2}\sqrt{\left(-3\sin\frac{\theta}{2}\cos\frac{\theta}{2}\right)^2+\left(3\cos^2\frac{\theta}{2}\right)^2}d\theta$$

$$l=\int_0^{\pi/2}\sqrt{9\sin^2\frac{\theta}{2}\cos^2\frac{\theta}{2}+9\cos^2\frac{\theta}{2}\cos^2\frac{\theta}{2}}d\theta$$

$$l=\int_0^{\pi/2}\sqrt{\left(3\cos\frac{\theta}{2}\right)^2\left(\sin^2\frac{\theta}{2}+\cos^2\frac{\theta}{2}\right)}d\theta=\int_0^{\pi/2}3\cos\frac{\theta}2d\theta$$

Integrando:

$$l=\left[6\sin\frac{\theta}2\right]_0^{\pi/2}=6\sin\frac\pi4$$

Logo:

$$\boxed{l=3\sqrt2}$$