(adsbygoogle = window.adsbygoogle || []).push({}); parabolic path at constant speed? (having issues)

I found a brief mention of this problem at a few web sites. An object follows a parabolic path at constant speed. I'm having some issues with the math involved:

Assume the path is defined by y = 1/2 x^{2}, and that the constant speed is c.

velocity vector can be defined in terms of x:

vx = dx/dt = c / sqrt(x^{2}+ 1)

vy = dy/dt = c x / sqrt(x^{2}+ 1)

speed = sqrt(vx^{2}+ vy^{2})

... = sqrt( c^{2}/(x^{2}+ 1) + (c x)^{2}/(x^{2}+ 1) )

... = sqrt( c^{2}(1+x^{2})/(x^{2}+ 1) )

... = sqrt( c^{2})

... = c

so that checks.

Since y = 1/2 x^{2}, dy = x dx, dy/dt = x dx/dt = c x / sqrt(x^{2}+ 1) which matches vy as defined above.

time can be defined as a function of x:

dx/dt = c / sqrt(x^{2}+ 1)

sqrt(x^{2}+ 1) dx = c dt

(1/2)(x sqrt(x^{2}+1) + ln(x + sqrt(x^{2}+1)) = c t + C

assuming t=0 when x=0, the integration constant will be zero

(1/2)(x sqrt(x^{2}+1) + ln(x + sqrt(x^{2}+1)) = c t

t = ( x sqrt(x^{2}+1) + ln(x + sqrt(x^{2}+1)) ) / (2 c)

I then plugged this info into a spreadsheet, using a range of x values as input. I confirmed that speed was c and that the slope vy/vx = x as expected (since dy = x dx).

The issue I ran into was trying to determine acceleration. Using a crude numerical method (ax = Δvx / Δt, ay = Δvy / Δt) resulted in calculating acceleration that wasn't perpendicular to velocity, but that conflicts with the fact that the speed is constant. I'm not sure how to take the second derivative for dx/dt and dy/dt when they are funcions of x and not t. I can substitute x = sqrt(2 y) for dy/dt:

dx/dt = c / sqrt(x^{2}+ 1)

dy/dt = c x / sqrt(x^{2}+ 1) = c sqrt(2 y) / (sqrt(2 y + 1), for positive x

dy/dt = -c sqrt(2 y) / (sqrt(2 y + 1), for negative x

but I'm not sure how to take derivates to get accelerations for ax and ay.

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# Parabolic path at constant speed

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