precalc question?

Question

Hi, it'd be great if someone could answer this and explain this problem to me:

Two bugs start at the origin, one travelling along the positive half of the x-axis and the other travelling along the positive half of the y-axis. Each bug moves 1 unit on the first move, 1/2 unit on the second move, and so on, where on any subsequent move, the bug moves only HALF as far as the immediately preceding move. How far will the bugs be apart, assuming their moves are frequent and over an infinite amount of time?

I think this has something to do with limits? But I don't know how to set it up. Thanks in advance to anyone who'll help me! :)

Answer 1

First, let's look at the x bug.

After an infinite amount of time, the x bug is at:
1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32....
As the number of moves goes to infinity, this sum goes to 2.

The x-coordinate of the x-bug is:
x = 2 - (1/2)^t
where t is the number of moves made.

Similarly, the y-coordinate of the y-bug is:
y = 2 - (1/2)^t

And the distance between the two is given by Pythagoras. In a 45-45-90 triangle, the hypotenuse is sqrt(2) * the length of one of the other sides.

Hence, the distance between the two bugs is given by:
sqrt(2) * (2 - (1/2)^t)

As t goes to infinity, this limit goes to 2*sqrt(2).

Hope that helps!

Answer 2

hello!

with each step their distance apart is the hypotenuse of a right triangle.

after first step, their distance apart is Sqrt[2].

After second step, their distance is 3 Sqrt[2] / 2

After third, distance is 7 Sqrt[2] / 4

After fourth, 15 Sqrt[2] / 8...

see a pattern...

pattern is Sqrt[2] + Sum[(2^n - 1)/(2^(n-1))][Sqrt[2]]

Sum is from 2 to infinity (or change if you wish).

now, simplify the summation and it's done. fairly easy, actually. looks like answer might be 2Sqrt[2]. i'll check...

hint: 2^n / 2^n-1 = 2^n-(n-1) = 2^n-n+1 = 2^1 = 2.

check Sum[2^1/n] from 1 to infinity.

Answer 3

x-direction: 1+1/2+1/4+1/8+... = 2
y-direction: 1+1/2+1/4+1/8+... = 2
d = √(2^2+2^2) = 2√2 units

Answer 4

Plz post your question on http://www.tutorbuddy.org I will answer as soon as u post