Answer on Question #70824 – Math – Geometry
Question
1. Find the length of astroid x = a cos 3 t , y = a sin 3 t , [ 0 , 2 π ] x = a\cos^3 t, y = a\sin^3 t, [0, 2\pi] x = a cos 3 t , y = a sin 3 t , [ 0 , 2 π ]
In view of the symmetry of the curve, it's enough to find one-fourth l l l L L L L = 4 l L = 4l L = 4 l π 2 \frac{\pi}{2} 2 π
Find the differentials:
d x = − 3 a cos 2 t sin t ; d y = 3 a sin 2 t cos t ; dx = -3a\cos^2 t \sin t; \quad dy = 3a\sin^2 t \cos t; d x = − 3 a cos 2 t sin t ; d y = 3 a sin 2 t cos t ;
Hence we find
d x = ( d x ) 2 + ( d y ) 2 = 9 a 2 cos 4 t sin 2 t + 9 a 2 sin 4 t cos 2 t = = 9 a 2 cos 2 t sin 2 t ( cos 2 t + sin 2 t ) = 9 a 2 cos 2 t sin 2 t = 3 a cos t sin t = 3 2 a sin 2 t \begin{aligned}
dx &= \sqrt{(dx)^2 + (dy)^2} = \sqrt{9a^2\cos^4 t\sin^2 t + 9a^2\sin^4 t\cos^2 t} = \\
&= \sqrt{9a^2\cos^2 t\sin^2 t(\cos^2 t + \sin^2 t)} = \sqrt{9a^2\cos^2 t\sin^2 t} = 3a\cos t \sin t \\
&= \frac{3}{2}a \sin 2t
\end{aligned} d x = ( d x ) 2 + ( d y ) 2 = 9 a 2 cos 4 t sin 2 t + 9 a 2 sin 4 t cos 2 t = = 9 a 2 cos 2 t sin 2 t ( cos 2 t + sin 2 t ) = 9 a 2 cos 2 t sin 2 t = 3 a cos t sin t = 2 3 a sin 2 t
Integrating the resulting expression for d x dx d x π 2 \frac{\pi}{2} 2 π
l = ∫ 0 π / 2 3 2 a sin 2 t d t = 3 2 a ∫ 0 π / 2 sin 2 t d t = − 3 4 a cos 2 t ∣ π 2 0 ∣ = 3 4 a + 3 4 a = 3 2 a ⇒ L = 4 × 3 2 a = 6 a \begin{aligned}
l &= \int_0^{\pi/2} \frac{3}{2} a \sin 2t \, dt = \frac{3}{2} a \int_0^{\pi/2} \sin 2t \, dt = -\frac{3}{4} a \cos 2t \left| \begin{array}{c} \frac{\pi}{2} \\ 0 \end{array} \right| = \frac{3}{4} a + \frac{3}{4} a = \frac{3}{2} a \\
&\Rightarrow L = 4 \times \frac{3}{2} a = 6a
\end{aligned} l = ∫ 0 π /2 2 3 a sin 2 t d t = 2 3 a ∫ 0 π /2 sin 2 t d t = − 4 3 a cos 2 t ∣ ∣ 2 π 0 ∣ ∣ = 4 3 a + 4 3 a = 2 3 a ⇒ L = 4 × 2 3 a = 6 a
**Answer**: the length of the astroid is 6 a 6a 6 a
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