# Consider a mobile moving along this path that starts at the point (0,0). What are the coordinates of the point of arrival of the mobile ? (See image below)

## A "Suits and Series" problem.

Jul 30, 2018

Point of arrival is: $\left(\frac{32}{7} , \frac{10}{3}\right)$

#### Explanation:

In the x-direction, defined as horizontal direction with positive direction being left to right:

$x = 8 - 6 + \frac{9}{2} - \frac{27}{8} + \ldots$

This is an alternating geometric series.

$= 8 - 8 \left(\frac{3}{4}\right) + 8 {\left(\frac{3}{4}\right)}^{2} - 8 {\left(\frac{3}{4}\right)}^{3} + \ldots$

$= 8 \left({\underbrace{\left(1 + {\left(\frac{3}{4}\right)}^{2} + \ldots\right)}}_{a = 1 q \quad r = {\left(\frac{3}{4}\right)}^{2}} - \underbrace{\left(\frac{3}{4} + \frac{3}{4} {\left(\frac{3}{4}\right)}^{2} + \ldots\right)} {\setminus}_{a = \frac{3}{4} q \quad r = {\left(\frac{3}{4}\right)}^{2}}\right)$

The sum to infinity for a geometric series with $\left\mid r \right\mid < 1$ is:

• ${S}_{\infty} = \frac{a}{1 - r}$

""_xS_oo= 8 ( 1/(1- (3/4)^2 ) - (3/4)/(1- (3/4)^2 ))

:. ""_xS_oo= 32/7 qquad qquad qquad [= 3.375]

In the y-direction, defined as vertical direction with upward positive:

$y = 6 - \frac{25}{4} + \frac{96}{25} - \frac{384}{125} + \ldots$

$= 6 - 6 \left(\frac{4}{5}\right) + 6 {\left(\frac{4}{5}\right)}^{2} - 6 {\left(\frac{4}{5}\right)}^{3} + \ldots$

$= 6 \left(\left(1 + {\left(\frac{4}{5}\right)}^{2} + \ldots\right) - \left(\left(\frac{4}{5}\right) - {\left(\frac{4}{5}\right)}^{3} - \ldots\right)\right)$

With the same reasoning:

""_yS_oo= 6 ( 1/(1- (4/5)^2 ) - (4/5)/(1- (4/5)^2 )) = 10/3

Point of arrival is: $\left(\frac{32}{7} , \frac{10}{3}\right)$