Find the arc length?

Question

A fleeing object leaves the origin and moves up the y-axis. At the same time, a pursuer leaves the point (1,0) and always moves toward the fleeing object. If the pursuer's speed is twice that of the fleeing object, the equation of the path is

y = (1/3)[x^(3/2)-3x^(1/2)+2]

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

Answer 1

The distance traveled by the fleeing object is y(0) = 2/3.

The distance traveled by the pursuer is:
integral from 1to 0 (sqrt(1+ (y')^2)dx
y' = 1/2 x^1/2 - 1/2x^-1/2 = 1/2(x^1/2 - x^-1/2)
y'^2 = 1/4(x +1/x -2)
So integ 1 to 0 [sqrt(1 + x/4 +1/(4x) -1/2)]dx = pursuer distance
= integ 1 to 0 1/2[x^1/2 + x^-1/2)
=1/2(2/3x^3/2 +2x^1/2) from 1 to 0
1/2(2/3 +2*1) = 1/3 +1 = 4/3
So pursuer's distance is twice fleeing objects distance.
I did not use calculator. Would you be allowed to in a test?

Answer 2

When the pursuer is caught the value of x is 0. Enter that into your equation to solve for y and that is how far the fleeing object traveled seeing as how it never leaves the y-axis.

Answer 3

First you find out how far the object has travelled up the y axis. It is the line x=0 so plugging in 0 to the equation you get y=2/3. This is when they hit.

To find arc length you use the arc length formula. My calculator says the length from 0 to 1 is 1.33333 which is 2/3 * 2