Thread: impossible problem =P

Ok, here we go. This is my first post on this forum, and its a fun one. I have a problem that is pretty dang hard. I have spent a lot of time thinking about it, but i have yet to come up with anything. Read the problem carefully because there are no numbers involved. Ok, here it is.

Ok, so there is a pond. The pond is a perfect circle and the radius of the circle/pond does not matter. In the middle of this pond there is a fish. The fish wants to escape the circle. However, at the edge of the circle there is a shark. The Shark CAN ONLY move on the outside of the circle. The fish, can move in any direction it wants to. The shark is trying to prevent the fish from escaping the circle. Both the shark and the fish move at constant speeds and can change direction instantly.

Ok, so here is what we are supposed to find. We have to find How many times faster does the shark have to swim (the shark swims x times faster than the fish) IN ORDER TO PREVENT the fish from escaping the circle.

The person who gave us this problem said you must use calculus to solve this problem. If the fish decides to swim in a straight line, then the shark has to swim Pi times faster than the fish. However, the fish is not dumb, it will try to move in a more complicated path than just a straight line. So we know the answer has to be greater than Pi.

The first step of the problem is finding the ideal path for the fish to swim. The second step is finding an equation to model this path. And the third step is probably to integrate that equation. (im guessing).

Sorry if this is a little long but I have to be sure to explain it. Have fun trying the problem but i will understand if you don't even want to try it at all XD. I know that there are some smart people on this forum so hopefully this problem can be a fun challenge for them

Originally Posted by WastedWizard

Ok, here we go. This is my first post on this forum, and its a fun one. I have a problem that is pretty dang hard. I have spent a lot of time thinking about it, but i have yet to come up with anything. Read the problem carefully because there are no numbers involved. Ok, here it is.

Ok, so there is a pond. The pond is a perfect circle and the radius of the circle/pond does not matter. In the middle of this pond there is a fish. The fish wants to escape the circle. However, at the edge of the circle there is a shark. The Shark CAN ONLY move on the outside of the circle. The fish, can move in any direction it wants to. The shark is trying to prevent the fish from escaping the circle. Both the shark and the fish move at constant speeds and can change direction instantly.

Ok, so here is what we are supposed to find. We have to find How many times faster does the shark have to swim (the shark swims x times faster than the fish) IN ORDER TO PREVENT the fish from escaping the circle.

The person who gave us this problem said you must use calculus to solve this problem. If the fish decides to swim in a straight line, then the shark has to swim Pi times faster than the fish. However, the fish is not dumb, it will try to move in a more complicated path than just a straight line. So we know the answer has to be greater than Pi.

The first step of the problem is finding the ideal path for the fish to swim. The second step is finding an equation to model this path. And the third step is probably to integrate that equation. (im guessing).

Sorry if this is a little long but I have to be sure to explain it. Have fun trying the problem but i will understand if you don't even want to try it at all XD. I know that there are some smart people on this forum so hopefully this problem can be a fun challenge for them

This is a reasonably complicated problem which I'm sure your teacher wants you to do yourself. Nevertheless, here's some reading for you which will help: Chases and escapes: the mathematics ... - Google Book Search

Here is a mathematical translation of the problem, and a few basic assumptions I believe are evident without the need for calculation.

Def 1: The fish follows a path expressible by $\displaystyle f(t)=(r\cos\theta,r\sin\theta)$ where t is time, r his distance from the center of the lake, and $\displaystyle \theta$ his direction. The shark follows the path $\displaystyle \alpha(t)=(\cos\alpha,\sin\alpha)$

Def 2: The fish's starting position is the center of the lake, $\displaystyle f(0)=(0,0)$. Without loss of generality, the shark begins at due west on the unit circle at the point $\displaystyle (-1,0)$.

Ass 1: The shark can move at least $\displaystyle \pi$ times faster than the fish, otherwise the fish could easily swim due east at max speed and escape.

Ass 2: At any point in time, when the fish is at a point $\displaystyle (r\cos\theta,r\sin\theta)$, the shark's optimal location is the closest point on the unit circle, $\displaystyle (\cos\theta,\sin\theta)$. Therefore $\displaystyle \frac{d\alpha}{dt}=\pm v_s$, i.e. the shark is swimming at max speed towards this optimal point along a minor arc.

Ass 3: The fish is always moving at max speed, $\displaystyle v_f$. He gains no advantage by slowing down his linear speed.

Ass 4: The fish gains no advantage in changing direction. If the fish instantaneously changes direction, then the shark can instantaneously change direction, resulting in a mirror image of the exact same situation. Therefore, WLOG, $\displaystyle \frac{d\theta}{dt}\geq 0$ where $\displaystyle \theta$ is the fish's direction of swimming.

Ass 5: The fish's opening strategy at $\displaystyle t=0$ is to swim at max speed due east. By Ass 2, WLOG, the shark's opening strategy is to swim at max speed counterclockwise (south).

I believe this is a safe place to start. If memory serves, the fish's optimal strategy is a spiral, $\displaystyle \theta(r)=kr$ and the most efficient spiral is the "trivial" one, $\displaystyle k=0$. In other words, the fish has no better strategy than simply swimming at max speed due east until he reaches shore. That's my conjecture, at any rate.

Interesting problem.

One question. Why does the shark have to travel $\displaystyle \pi$ times faster if the fish travels in a straight line? Wouldn't the shark only need to be $\displaystyle \frac {\pi}{2}$ times faster because $\displaystyle \frac {\pi}{2}$ is the longest distance the shark has to travel. i.e. if the fish swam in the other direction of the shark, the shark needs to travel half a circle, and therefore $\displaystyle \frac {\pi}{2}$. You don't need $\displaystyle \pi$ because the shark can swim in the other direction, and therefore half a circle is the longest distance needed to travel.

Originally Posted by chengbin

Interesting problem.

One question. Why does the shark have to travel $\displaystyle \pi$ times faster if the fish travels in a straight line? Wouldn't the shark only need to be $\displaystyle \frac {\pi}{2}$ times faster because $\displaystyle \frac {\pi}{2}$ is the longest distance the shark has to travel. i.e. if the fish swam in the other direction of the shark, the shark needs to travel half a circle, and therefore $\displaystyle \frac {\pi}{2}$. You don't need $\displaystyle \pi$ because the shark can swim in the other direction, and therefore half a circle is the longest distance needed to travel.

If the fish were to swim in a straight line, then the distance it has to travel is equal to the radius of the circle. The shark, on the other hand, has to swim $\displaystyle \pi$ times the radius of the circle. Therefore, in order for the shark to get there before the fish, it must swim $\displaystyle \pi$ times faster than the fish.

I don't think you get what I'm saying.

Let's suppose the outer rim is described in degrees from 0 to 360. If the shark is at 0, the maximum distance the shark has to go is at 180 degrees, which is the other side of the circle, which is only $\displaystyle \frac {\pi}{2}$.

If the fish decides to go to 270 degree, the shark doesn't swim 270, or $\displaystyle \frac {3\pi}{4}$degrees to get to 270 degree, but it travels 90 degrees in the other direction, or $\displaystyle -\frac{\pi}{4}$to get to 270 degree.

Originally Posted by chengbin

I don't think you get what I'm saying.

Let's suppose the outer rim is described in degrees from 0 to 360. If the shark is at 0, the maximum distance the shark has to go is at 180 degrees, which is the other side of the circle, which is only $\displaystyle \frac {\pi}{2}$.

If the fish decides to go to 270 degree, the shark doesn't swim 270, or $\displaystyle \frac {3\pi}{4}$degrees to get to 270 degree, but it travels 90 degrees in the other direction, or $\displaystyle -\frac{\pi}{4}$to get to 270 degree.

I sort of get what you are saying but the way I see it is that the fish has to swim a distance of r (the radius of the circle). The equation for the circumference of a circle is 2$\displaystyle \pi$r. However, the shark only has to travel HALF of the circle, so the shark travels a distance of $\displaystyle \pi$r. So the shark has to swim $\displaystyle \pi$ times longer than the fish in less time than the fish. Therefore the shark must swim $\displaystyle \pi$ times faster than the fish to catch it if the fish choose to escape in a straight line.

Originally Posted by chengbin

I don't think you get what I'm saying.

Let's suppose the outer rim is described in degrees from 0 to 360. If the shark is at 0, the maximum distance the shark has to go is at 180 degrees, which is the other side of the circle, which is only $\displaystyle \frac {\pi}{2}$.

If the fish decides to go to 270 degree, the shark doesn't swim 270, or $\displaystyle \frac {3\pi}{4}$degrees to get to 270 degree, but it travels 90 degrees in the other direction, or $\displaystyle -\frac{\pi}{4}$to get to 270 degree.

We get what you are saying, but you are forgetting your unit circle. $\displaystyle 360^\circ$ is $\displaystyle 2\pi$, $\displaystyle 270^\circ$ is $\displaystyle \frac{3\pi}2$, $\displaystyle 180^\circ$ is $\displaystyle \pi$, and $\displaystyle 90^\circ$ is $\displaystyle \frac\pi2$ . If the fish swims in a straight line to the shore, it's escape distance is r. The shark must swim at most half the perimeter of the circle, which is$\displaystyle \frac12 2\pi r$, $\displaystyle \pi$ times the fish's distance. Your logic is correct, your recollection of the unit circle incorrect.