Persamaan Polar Daun Descartes

Hendra Gunawan

Barangkali ada yang penasaran dengan persamaan kurva tepi daun Descartes dalam koordinat polar.

Sebagaimana telah dibahas dalam artikel sebelumnya, kita mempunyai

r^2 = \frac{(3at)^2(1+t^2)}{(1+t^3)^2};

\theta = \arctan t.

Nah, dari kedua persamaan di atas, kita peroleh

r = \frac{3at \sqrt{1+t^2}}{1+t^3} = \frac{3a\tan \theta \sec \theta}{1+\tan^3 \theta}.

*

Bandung, 31-03-2018

Published by Hendra Gunawan

I'm a professor of mathematics at the Faculty of Mathematics and Natural Sciences, Bandung Institute of Technology, Bandung, Indonesia.

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