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A fleeing object leaves the origin and moves up the $y$-axis. At the same time, a pursuer leaves the point $(1,0)$ and moves always toward the fleeing object. If the pursuer's speed is twice that of the fleeing object, the equation of the path is $$y = \frac{1}{3}(x^\frac{3}{2} -3x^\frac{1}{2} + 2)$$

How far has the fleeing object traveled when it is caught? Show that the pursuer has traveled twice as far.

Would I take the derivative of the equation to get the speed? Then double it to get the speed of the pursuer?


Second thought would I just be able to take the integral of the arc length from 0 to 1 of the equation as it is and then compare it to the same thing multiple by $\frac{2}{3}$ ?

Edit #2

Using the arc length formula

$$ \int^1_0 \sqrt{1 + (f'(x))^2}dx$$

Leaves me with

$$ \int^1_0 \sqrt{(\frac{x + 2 + x^{-1}}{4})} dx $$

Then getting rid of the square root

$$ \int^1_0 \frac{x^{1/2} + x^{-1/2}}{2} dx $$

Then plug that into my Simpson's formula

$$ \frac{b - a}{6} [f(a) + 4f(x_0) + f(a)] $$

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up vote 0 down vote accepted

If you can believe the path you are given, the object is caught when the pursuer is at $x=0$, so $y=\frac 23$. Now you need to show the path length of the pursuer is $\frac 43$.

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@JoseBruchez: You are right. I started taking a derivative to set to zero and stripped off the $\frac 13$ because it didn't matter. Forgot to restore it for this purpose. – Ross Millikan Mar 10 '13 at 15:47
@JoseBruchez: yes, I agree with your edit. – Ross Millikan Mar 10 '13 at 15:49
Thank you Ross, I appreciate your help :) – JoseBruchez Mar 10 '13 at 21:56

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