Question

Q.Find the length of the Asteroid x=acos3t, y=asin3t.

Answer 1

Answer on Question #70464, Math / Statistics and Probability

Q. Find the length of the astroid $x = \text{acos}^3 t$ , $y = \text{asin}^3 t$ , $0 \leq t \leq 2\pi$ .

Solution.

Fig. 1. Astroid.

The length of the curve, given parametrically in interval $t_1 \leq t \leq t_2$ is calculated by the formula

$$L = \int_ {t _ {1}} ^ {t _ {2}} \sqrt {\left(x _ {t} ^ {\prime}\right) ^ {2} + \left(y _ {t} ^ {\prime}\right) ^ {2}} d t.$$

So, the length of the astroid is calculated by the formula

$$L = \int_ {0} ^ {2 \pi} 3 a \sin t \cos t d t.$$

Given the symmetry of astroid (fig. 1), it is enough to find the fourth part of the length of the arc $L$ , which corresponds to a change in the parameter $t$ from 0 to $\frac{\pi}{2}$ :

$$\frac {1}{4} L = \int_ {0} ^ {\frac {\pi}{2}} 3 a \sin t \cos t d t = \frac {3 a}{2} \int_ {0} ^ {\frac {\pi}{2}} \sin 2 t d t = \frac {3 a}{4} (- \cos 2 t) \Big | _ {0} ^ {\frac {\pi}{2}} = \frac {3 a}{4} (1 + 1) = \frac {3}{2} a.$$

Hence $L = 6a$

Answer. The length of the astroid equals $6a$ .

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