Calculus: Given parabola and vx in terms of x and c, prove constant velocity

[derelictee]

Here an AP Physics problem that's really bugging me.

Homework Statement

A particle moves along the parabola with equation y = .5x^2

part a) I believe I did this correct.

part b) Suppose that the particle moves with a velocity whose x-component is given by vx = c / (1 + x^2)^.5 Show that the particle's speed is constant.

Below I have images of the question and my attempted work. I think maybe for the first half of my work I was in the right direction; I got y in terms of t, and I was going to find the derivative to show that there is no acceleration, but I couldn't get the equation to equal y, and I ultimately became confused and went off track.

The Attempt at a Solution

http://img363.imageshack.us/img363/7313/scanqa9.th.jpg http://g.imageshack.us/thpix.php
http://img218.imageshack.us/img218/2346/scan0001fu2.th.jpg http://g.imageshack.us/thpix.php
I know; my work is a mess.

 

[LowlyPion]

Here an AP Physics problem that's really bugging me.

Homework Statement

A particle moves along the parabola with equation y = .5x^2

part a) I believe I did this correct.

part b) Suppose that the particle moves with a velocity whose x-component is given by vx = c / (1 + x^2)^.5 Show that the particle's speed is constant.

Below I have images of the question and my attempted work. I think maybe for the first half of my work I was in the right direction; I got y in terms of t, and I was going to find the derivative to show that there is no acceleration, but I couldn't get the equation to equal y, and I ultimately became confused and went off track.

I know; my work is a mess.

Have you considered that if Vx = c/(1+x²)^1/2 that x would be given by the integral?

In this regard isn't the x position as a function of time given by the integral of Vx(t)

\int \frac{c*dx}{(x^2 + 1)^{1/2}} = c*arcsinh(x) + C = c*arcsinh(x) + c

C is c because V=c at x=0