Question

Q.Find the length of the astroid x=acos3t, y=asin3t, [0, 2π].

Answer 1

Answer on Question #70498 - Math - Geometry

Q. Find the length of the astroid $x = a\cos 3t, y = a\sin 3t, t \in [0,2\pi]$.

Solution:

$$l = \int_{0}^{2\pi} \sqrt{[x'(t)]^2 + [y'(t)]^2} dt = \int_{0}^{2\pi} \sqrt{(-3a\sin 3t)^2 + (3a\cos 3t)^2} dt = \\ = 3a \int_{0}^{2\pi} \sqrt{\sin^2 3t + \cos^2 3t} dt = 3a \int_{0}^{2\pi} dt = 6\pi a$$

Answer:

Length of the asteroid equals $6\pi a$.

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