Calculus help?

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

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

Rewrite as a function:
f(x) = (1/3)[x^(3/2)-3x^(1/2)+2]

Set x to 0 (since that's where the pursuer will meet the fleeing object)
f(0) = (1/3)[x^(3/2)-3x^(1/2)+2]
f(0) = 1/3[0 - 0 + 2)
f(0) = 2/3

The fleeing object has travelled 2/3.

The arc length of a function between (a,f(a)) and (b,f(b)) is ∫ab1+(f'(x)) 2dx.

So, for this arc from 1,f(1) to 0,(f(0), The arc length is:
∫(0*1)+(f'(x)) 2dx.
2∫f'(x) dx.

Find f'(x):
f(x) = (1/3)[x^(3/2)-3x^(1/2)+2]
f'(x) = 1/3 * 3/2 * x^(1/2) - 1/3 * 3 * 1/2 * x^(-1/2)
f'(x) = 1/2x^(1/2) - 1/2 x^(-1/2)

Back to the arclength:
2∫f'(x) dx
2∫(1/2x^(1/2) - 1/2 x^(-1/2)) dx
2 * (1/3x^(3/2) - x^1/2) from x = 0 to 1.

1/3(1)^(3/2) - (1)^1/2 - 1/3(0)^(3/2) - (0)^1/2
2 * ((1/3 - 1) - (0 - 0))
2 * -2/3
distance = 4/3 (twice the fleeing object's distance - solution!)