A crazy ant is standing on the origin. It begins by walking 1 unit in

1.	A crazy ant is standing on the origin. It begins by walking 1 unit in the +x direction and then turns 60 degrees counterclockwise and walks 1/2 units in that direction. The ant then turns another 60 degrees and walks 1/3 units in that direction. The ant keeps doing this endlessly. What is the ant's final position?#Method by Which you can solve1) Cartesian Method2) Polar Coordinate method.#Hint:- Use formula for displacement, D = 1/n(cos∅+ isin∅)
Answer» (BASIC question of calculus) I had solved it using Polar coordinate method Solution:-Displacement of ant = $\frac{1}{n}( \cos \alpha + isin \alpha )$ where 1/n is the distance travelled. The FINAL position of the ant will be sum of all the consecutive displacements so we can write D= $1(cos0 + isin0) + \frac{1}{2}(sin \frac{\pi}{3} + isin \frac{\pi}{3}) + \frac{1}{3}(cos \frac{2\pi}{3} + isin \frac{2\pi}{3}) + .$ From Euler's rule,(which we had learnt in complex NUMBER) we know ${e}^{i \alpha } = cos \alpha + isin \alpha$ $on \: comparing \: we \: can \: write\\ d = \sum\limits_{n=0}^{n= \infty } \frac{ {e}^{ \frac{i(n - 1)\pi}{3} } }{n}$ now we know that $\frac{ln(1 - x)}{x } = - \sum\limits_{n=0}^{n= \infty } \frac{ {e}^{( n - 1)x} }{n} \\ comparing \: this \: with \: d \: we \: get \\ d = > \sum\limits_{n=0}^{n= \infty } \frac{ {e}^{( n - 1)x} }{n} = \sum\limits_{n=0}^{n= \infty } \frac{ {e}^{ \frac{i(n - 1)\pi}{3} } }{n} \\ from \: here \: we \: can \: say \: x = {e}^{i \frac{\pi}{3} } \\ now \\ d = \frac{ln(1 - {e}^{ \frac{i\pi}{3} }) }{ {e}^{ \frac{i\pi}{3} } } = \frac{i \frac{\pi}{3} }{cos \frac{\pi}{3} + isin \frac{\pi}{3} } = \frac{i \frac{\pi}{3} }{ \frac{1 + \sqrt{3}i }{2} } \\ = > \frac{2i \frac{\pi}{3} }{1 + \sqrt{3}i } = \frac{\pi}{2 \sqrt{3} } + \frac{i\pi}{6} \\ this \: shows \: that \: coordinates \: of \: final \: position \\ is \: ( \frac{\pi}{2 \sqrt{3} } \: \: \: \frac{\pi}{6} ) \\ distance \: from \: initial \: position = \sqrt{ {( \frac{\pi}{2 \sqrt{3} } })^{2} + ( { \frac{\pi}{6} })^{2} } \\ = > \sqrt{( { \frac{ \pi }{9} })^{2} } = \frac{\pi}{3} = 1.05 \\$ HENCE the ant will be 1.05m away from it's initial position Note:-Can also be solved by using Cartesian method,but will be a bit LENGTHY

A crazy ant is standing on the origin. It begins by walking 1 unit in the +x direction and then turns 60 degrees counterclockwise and walks 1/2 units in that direction. The ant then turns another 60 degrees and walks 1/3 units in that direction. The ant keeps doing this endlessly. What is the ant's final position?#Method by Which you can solve1) Cartesian Method2) Polar Coordinate method.#Hint:- Use formula for displacement, D = 1/n(cos∅+ isin∅)

Answer»

(BASIC question of calculus)

I had solved it using Polar coordinate method

Solution:-Displacement of ant =

$\frac{1}{n}( \cos \alpha + isin \alpha )$

where 1/n is the distance travelled.

The FINAL position of the ant will be sum of all the consecutive displacements

so we can write

D= $1(cos0 + isin0) + \frac{1}{2}(sin \frac{\pi}{3} + isin \frac{\pi}{3}) + \frac{1}{3}(cos \frac{2\pi}{3} + isin \frac{2\pi}{3}) + .$

From Euler's rule,(which we had learnt in complex NUMBER)

we know

${e}^{i \alpha } = cos \alpha + isin \alpha$

$on \: comparing \: we \: can \: write\\ d = \sum\limits_{n=0}^{n= \infty } \frac{ {e}^{ \frac{i(n - 1)\pi}{3} } }{n}$

now we know that

$\frac{ln(1 - x)}{x } = - \sum\limits_{n=0}^{n= \infty } \frac{ {e}^{( n - 1)x} }{n} \\ comparing \: this \: with \: d \: we \: get \\ d = > \sum\limits_{n=0}^{n= \infty } \frac{ {e}^{( n - 1)x} }{n} = \sum\limits_{n=0}^{n= \infty } \frac{ {e}^{ \frac{i(n - 1)\pi}{3} } }{n} \\ from \: here \: we \: can \: say \: x = {e}^{i \frac{\pi}{3} } \\ now \\ d = \frac{ln(1 - {e}^{ \frac{i\pi}{3} }) }{ {e}^{ \frac{i\pi}{3} } } = \frac{i \frac{\pi}{3} }{cos \frac{\pi}{3} + isin \frac{\pi}{3} } = \frac{i \frac{\pi}{3} }{ \frac{1 + \sqrt{3}i }{2} } \\ = > \frac{2i \frac{\pi}{3} }{1 + \sqrt{3}i } = \frac{\pi}{2 \sqrt{3} } + \frac{i\pi}{6} \\ this \: shows \: that \: coordinates \: of \: final \: position \\ is \: ( \frac{\pi}{2 \sqrt{3} } \: \: \: \frac{\pi}{6} ) \\ distance \: from \: initial \: position = \sqrt{ {( \frac{\pi}{2 \sqrt{3} } })^{2} + ( { \frac{\pi}{6} })^{2} } \\ = > \sqrt{( { \frac{ \pi }{9} })^{2} } = \frac{\pi}{3} = 1.05 \\$

HENCE the ant will be 1.05m away from it's initial position

Note:-Can also be solved by using Cartesian method,but will be a bit LENGTHY

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