Calculus/Arc Length

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Question
I am confused by this question. I know it has to do with arc length but how would you perform the calculations for arc length if they don't give you an interval on which to evaluate it? Thanks for any help.

A hawk flying at 19 m/s at an altitude of 165 m accidentally drops its prey. The parabolic trajectory of the falling prey is described by the equation y = 165- (x^2)/57 until it hits the ground, where y is its height above the ground and x is the horizontal distance traveled in meters. Calculate the distance traveled by the prey from the time it is dropped until the time it hits the ground.

Answer
I'm found a article on finding the arc length of a curve. It is http://tutorial.math.lamar.edu/Classes/CalcII/ArcLength.aspx What it comes down to is
L = ∫(squareroot(1 + (dy/dx)) dx, which is a few pages down in the paper.

Here, y = 165 - x/57, so dy/dx = -2x/57.
This make dy/dx = 4x/3249.
We would then get the length L = ∫(squareroot(1 + 4x/3249) dx

This looks like a triangle probelm with the near leg as 1,
the far leg as 2x/57, so the hypotenuse would be squareroot(1 + 4x/3249).
Take the angle as u.

Using this, we could say that tan u = 2x/57, so du = 2/57 dx.
The value in the integral is the length of the hypotenuse
and the near side is 1, so the function turns into sec(u).

This changes the problem to ∫57*sec(u)/2 du.

This is the same as (57/2)∫sec(u) du.

To remember how to do this if you don't know, see
http://math2.org/math/integrals/more/sec.htm

Calculus

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