Question

find the length of the asteroid x2/3+y2/3=a2/3​

Answer 1

SOLUTION

TO DETERMINE

The length of the asteroid

$\displaystyle \sf{ {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } = {a}^{ \frac{2}{3} } }$

EVALUATION

Here the given equation of the curve is

$\displaystyle \sf{ {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } = {a}^{ \frac{2}{3} } }$

Above equation represents the equation of an asteroid

Let us consider the parametric form

$\displaystyle \sf{ x = a { \cos}^{3} \theta \: \: and \: \: y = a { \sin}^{3} \theta }$

Now

$\displaystyle \sf{ \frac{dx}{d\theta} = - 3 a { \cos}^{2} \theta \sin \theta }$

$\displaystyle \sf{ \frac{dy}{d\theta} = 3 a { \sin}^{2} \theta \cos \theta }$

Thus we get

$\displaystyle \sf{ { \bigg( \frac{dx}{d\theta} \bigg)}^{2} + { \bigg( \frac{dy}{d\theta} \bigg)}^{2} }$

$\displaystyle \sf{ = { \bigg( - 3 a { \cos}^{2} \theta \sin \theta \bigg)}^{2} + { \bigg( 3 a { \sin}^{2} \theta \cos \theta\bigg)}^{2} }$

$\displaystyle \sf{ = 9 {a}^{2}{ \sin}^{2} \theta { \cos}^{2} \theta { \bigg( { \cos}^{2} \theta + { \sin}^{2} \theta \bigg)}}$

$\displaystyle \sf{ = 9 {a}^{2}{ \sin}^{2} \theta { \cos}^{2} \theta }$

Hence the required length

$\displaystyle \sf{ = 4 \int\limits_{0}^{ \frac{\pi}{2} } \sqrt{ { \bigg( \frac{dx}{d\theta} \bigg)}^{2} + { \bigg( \frac{dy}{d\theta} \bigg)}^{2} } \, d \theta }$

$\displaystyle \sf{ = 4 \int\limits_{0}^{ \frac{\pi}{2} } \sqrt{ { 9 {a}^{2}{ \sin}^{2} \theta { \cos}^{2} \theta}} \, d \theta }$

$\displaystyle \sf{ = 4 \int\limits_{0}^{ \frac{\pi}{2} } 3a \sin \theta \cos \theta \, d \theta }$

$\displaystyle \sf{ = 12a \int\limits_{0}^{ \frac{\pi}{2} } \sin \theta \cos \theta \, d \theta }$

$\displaystyle \sf{ = 6a \int\limits_{0}^{ \frac{\pi}{2} } 2\sin \theta \cos \theta \, d \theta }$

$\displaystyle \sf{ = 6a \int\limits_{0}^{ \frac{\pi}{2} } \sin 2\theta \, d \theta }$

$\displaystyle \sf{ = 6a \times \frac{ - \cos 2 \theta}{2} \bigg|_{0}^{ \frac{\pi}{2}} }$

$\displaystyle \sf{ = 3a [ - \cos \pi + \cos 0] }$

$\displaystyle \sf{ = 3a [1 + 1] }$

$\displaystyle \sf{ = 3a \times 2 }$

$\displaystyle \sf{ = 6a }$

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