# A snail starts at the origin of an argand diagram and walks along the real axis for an hour, covering a distance of 8 metres...at the end of each hour it changes its direction by Pi/2...

A snail starts at the origin of an argand diagram and walks along the real axis for an hour, covering a distance of 8 metres...

at the end of each hour it changes its direction by Pi/2 anticlockwise; and in each hour it walks half as far as it did in the previous hour. Find where it is after 4 hours,

Answers 6+3i

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### 2 Answers

Let each unit on the axes represent 1meter.

(1) After 1 hour the snail is at (8+0`i` )

(2) During hour two, the snail travels half as far (4 units) parallel to the `i` axis, thus it arrives at (8+4`i` )

(3) During hour three, the snail moves parallel to the real axis in the negative direction 2 units thus ending up at (6+4`i` )

(4) During hour four, the snail moves parallel to the `i` axis in the negative direction 1unit, thus arriving at (6+3`i` ).

To calculate the position the snail has on the real axis x, after 4 hours, you need to use the following formula:

`((2^(4-4))*(2^(4+1) - 2cos(4pi/2) + sin(4pi/2)))/(4+1) = ((2^0)*(2^5 - 2cos2pi + sin2pi))/5`

Replacing sin 2pi = 0 and cos 2pi = 1 yields:

`((2^(4-4))*(2^(4+1) - 2cos(4pi/2) + sin(4pi/2)))/(4+1) = (32-2)/5`

`` `((2^(4-4))*(2^(4+1) - 2cos(4pi/2) + sin(4pi/2)))/(4+1) = 30/5`

`((2^(4-4))*(2^(4+1) - 2cos(4pi/2) + sin(4pi/2)))/(4+1) = 6`

x = 6

To calculate the position the snail has on the imaginary axis y, after 4 hours, you need to use the following formula:

`y = 2^4*(2^4 - 1)(cos 2pi + 2sin 2pi)/(2^4*5)`

Reducing by `2^4` yields:

`y = (15*1)/5 =gt y = 3`

**After 4 hours, the snail is at the position z = 6 + 3i on Argand's Diagram.**