MQ53 --- Chasing Problem

圖片參考：http://imgcld.yimg.com/8/n/HA05107138/o/20121117135804.jpg

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2012-11-17 14:54:14 補充：
Non calculus way:
Suppose the velocity vectors for bug 1 and bug 2 are v1 and v2 respectively
|v1|=|v2|=v
And they are at right angle to each other
The relative velocity of bug 1 with respect to bug 2 is vr=v1 – v2
Therefore |vr|=(√2)v

2012-11-17 14:54:23 補充：
And this vr is always at 45 degrees with the line joining bug 1 and bug 2
Therefore the component of relative velocity along this joining line is |vr| cos(45) = v
Hence for bug 1 to meet bug 2 time taken = a/v
The actual distance bug 1 travelled = vt = v(a/v) = a

2012-11-17 18:44:27 補充：
A simpler view on non calculus way. Let bug 2 be origin O & note the velocity of bug 1 wrt 2 by using a polar coordinate terms. The radial velocity is v towards O. The angular component is v clockwise. It does not change the radial but angular position=>it takes t=a/v to reduce radial distance to 0.

A lower-form way xD

The direction of running of a bug is always perpendicular to the bug it is chasing.
i.e. The directions of 4 bugs always form a square, which the square formed getting smaller and smaller until it becomes the center of original square ...(i)

Let bug A run, before it change direction, a small distance:
d1 = a/x

Bug B, which bug A is chasing, also run for a distance a/x.
The distance bug A do not walk at the square
= a - a/x
= a(x - 1)/x

ratio between d1 and the side of square:
d1/side of square
= (a/x) / a
= 1/x

ratio between the distance ant A didn't run (= a - d1) and the side of square:
(a - d1) / a
= [a - (a/x)] / a
= (x - 1)/x

ratio between the distance

i.e.
After it change its direction for the 1st time,
The distance bug A run after 1st direction change and before 2nd direction change
= d2
= (1/x)[(a/x)^2 + (a - a/x)^2]^(1/2)
= (1/x)(a/x)(x^2 - 2x + 2)^(1/2)

i.e. bug B also run (1/x)(a/x)(x^2 - 2x + 2)^(1/2) after 1st direction change and before 2nd direction change

The distance bug A run after 2nd direction change and before 3rd direction change
= d3
= (1/x)〔[(1/x)(a/x)(x^2 - 2x + 2)^(1/2)]^(2)
+ {[(x - 1)/x](a/x)(x^2 - 2x + 2)^(1/2)}^2〕^(1/2)
= (1/x^2)(a/x)(x^2 - 2x + 2)

d3 / d2
= (1/x^2)(a/x)(x^2 - 2x + 2) / (1/x)(a/x)(x^2 - 2x + 2)^(1/2)
= (1/x)(x^2 - 2x + 2)^(1/2)

d2 / d1
= (1/x)(a/x)(x^2 - 2x + 2)^(1/2) / (a/x)
= (1/x)(x^2 - 2x + 2)^(1/2)

i.e. d1, d2, d3 are geometric sequence
with d1 = a/x, ratio = (1/x)(x^2 - 2x + 2)^(1/2)
Easily seen by (i) that ratio < 1

d1 = A/x tends to 0
as x tends to infinity,
i.e.
Required answer
= lim(x -> ∞) (a/x) / [1 - (1/x)(x^2 - 2x + 2)^(1/2)]
= a lim(x -> ∞) (1/x){1/[1 - (1/x)(x^2 - 2x + 2)^(1/2)]}
= a lim(x -> ∞) (1/x){1 + (1/x)(x^2 - 2x + 2)^(1/2)]} / [1 - (1/x)^2(x^2 - 2x + 2)]
= a lim(x -> ∞) (1/x){1 + (1 - 2/x + 2/x^2)^(1/2)]} / (1 - 1 + 2/x - 2/x^2)
= a lim(x -> ∞) {1 + (1 - 2/x + 2/x^2)^(1/2)]} / (2 - 2/x)
= a{1 + (1 - 0 + 0)^(1/2)]} / (2 - 0)
= a(2/2)
= a