# Confused about a few things

1. Apr 20, 2004

### Ebolamonk3y

Hello... I am new to the forum...

I just took part in the TN math comptetion today at University of Memphis and I had a few questions that I could not solve...

In particular...

Let $$f(x)=\sqrt{x+\sqrt{0+\sqrt{x+\sqrt{0+\sqrt{x...}}}}}$$... ad inf...
If $$f(a)=4$$, what is $$f'(a)$$?

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Just a general problem

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A dog sees a rabbit. Position of the dog is somewhere on the x-axis, the rabbit, on the y-axis... At this time, the dog moves directly towards the rabbit as the rabbit runs up the y-axis... if both the dog and the rabbit travel at some constant velocity V, does the dog catch the rabbit? If so, where? If not, why not? And how does one model this function?

Thats all for now...

Chang

Last edited: Apr 20, 2004
2. Apr 20, 2004

### matt grime

for the first one note that f^2 = x + stuff, and then after manipulation, that (f^2-x)^2 = f you can now differentiate put x=a in and you're done.

3. Apr 20, 2004

### jdavel

Change,

The rabbit gets away! The dog will never cross the y axis, so he will always have an x component of velocity, so the y component of his velocity will have to be less than 1, so the y component of the distance between him and the rabbit will always be increasing.

To model this?
Define the location of the dog as (Xd, Yd)
Define the location of the rabbit as (Xr, Yr)
Define a finite increment of time, t, for the model.

start>>In one time increment the dog's x and y coordinates will change by delta_x and delta_y which are calculated from sqrt((delta_x)^2 + (delta_y)^2) = vt and delta_y/delta_x = (Yr-Yd)/(Xr-Xd).
Yd = Yd +delta_y
Xd = Xd + delta_x
Yr = Yr+vt
print sqrt((Yr-Yd)^2 + (Xr-Xd)^2).
n=n+1
if n= a billion then stop
goto start

4. Apr 20, 2004

### Ebolamonk3y

Is there a more elegant solution?

5. Apr 20, 2004

### master_coda

What if the rabbit starts below the x-axis?

6. Apr 20, 2004

### Ebolamonk3y

how does that affect the situation any? he crosses the x but never the y...

Is there some exceptions to this case?

like what if...

the dog is on x and the rabbit is in the negative y-axis running to positive y infinity... And the Dog is mad close to the x=0... the the dog points to the rabbit and its almost like a headon collision but it cannot be because the angle dog nose makes with y-axis as dog goes to rabbit approaches 0... On the other hand... if they start the same distance away from the origion on their respective axises and rabit goes to positive inf and dog goes to rabbit... then its alright....

Another thing to ponder... Are things reversable?

Like...

$$\int\frac{1}{x^2}dx$$ from 1 to infinity... is just 1...

Say is some work function were the same... you keep walking infinite distance and you still get only 1 joule?! arg... :(

7. Apr 20, 2004

### jdavel

Ebolamonk asked: "What if the rabbit starts below the x-axis?"

Hadn't thought of that!

This is looking like a more interesting problem than I had originally thought. Unless the dog and rabbit start on the same line going toward eachother, I don't think the dog can ever catch the rabbit by always running toward the rabbit. But I'm not sure!

8. Apr 20, 2004

### master_coda

Yes, but the remark I was replying to said that the y component of the distance between them was always increasing...that isn't true if the rabbit starts below the y axis.

The proper way to solve this would probably be to find the position functions for the rabbit and dog (where time is a parameter). Then just find the time where the functions are equal (if such a time exists).

9. Apr 20, 2004

### Ebolamonk3y

This belongs on the darn Diffy Q forum :(

But anyways...

Does the angle of the between the Dog and Rabbit resemble the angle of the ladder problem if the ladder were to "slip" up? Like you push the ladder towards the Y axis from some point on x... $$\frac{dx}{dt}$$ is negative something and decreases like the graph of say... 1/x or something like that... arg... maybe I am confusing someone out there.. maybe not!