Parabolic path at constant speed

[rcgldr]

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 x², and that the constant speed is c.

velocity vector can be defined in terms of x:

vx = dx/dt = c / sqrt(x² + 1)
vy = dy/dt = c x / sqrt(x² + 1)

speed = sqrt(vx² + vy²)
... = sqrt( c²/(x² + 1) + (c x)²/(x² + 1) )
... = sqrt( c² (1+x²)/(x² + 1) )
... = sqrt( c² )
... = c

so that checks.

Since y = 1/2 x², dy = x dx, dy/dt = x dx/dt = c x / sqrt(x² + 1) which matches vy as defined above.

time can be defined as a function of x:

dx/dt = c / sqrt(x² + 1)

sqrt(x² + 1) dx = c dt

(1/2)(x sqrt(x²+1) + ln(x + sqrt(x²+1)) = c t + C

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

(1/2)(x sqrt(x²+1) + ln(x + sqrt(x²+1)) = c t
t = ( x sqrt(x²+1) + ln(x + sqrt(x²+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(x² + 1)

dy/dt = c x / sqrt(x² + 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.

 

[Redbelly98]

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(x² + 1)

dy/dt = c x / sqrt(x² + 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.

Interesting problem. My initial thoughts are to use the chain rule to get dv_x/dt:

$$\begin{align} a_x = \frac{dv_x}{dt} \\ = \frac{dv_x}{dx} \frac{dx}{dt} \\ = v_x \frac{dv_x}{dx} \end{align}$$

Similarly, $a_y = v_x \frac{dv_y}{dx}$

Hopefully that helps.

 

[rcgldr]

Interesting problem. My initial thoughts are to use the chain rule to get dv_x/dt:

The orignial problem was due to my crude numerical derivative method. If acceleration is perpendicular to velocity, then inverting the slope for acceleration should match the slope for velocity: -ax/ay == vy/vx. My original estimated acceleration was based on

a_i = (v_i - v_i-1) / (t_i - t_i-1).

but -ax/ay was low by a near constant value, ~(t₁ - t₀). I changed this to

a_i = (v_i+1 - v_i-1) / (t_i+1 - t_i-1)

which solved the original problem.

Using the chain rule as suggested provided the actual solution:

ax = -c² x / (x²+1)²
ay = c² / (2y+1)²
ay = c² / (x²+1)²

As expected, acceleration is perpendicular to velocity, -ax/ay = vy/vx = x.

 

[Redbelly98]

Glad it worked out.

My original estimated acceleration was based on

a_i = (v_i - v_i-1) / (t_i - t_i-1).

but -ax/ay was low by a near constant value, ~(t₁ - t₀). I changed this to

a_i = (v_i+1 - v_i-1) / (t_i+1 - t_i-1)

which solved the original problem.

Yes. If you must evaluate a derivative numerically, your second method is just as simple and a much, much more accurate way than the first one.

 

[rcgldr]

I had actually tried chain method before, just messed up on the math and didn't follow up on checking for my mistake when comparing the chain method to my estimated accelerations on the spread sheet. Seeing you thought that chain method should also work, I did it again, this time paying more attention and getting the correct answer as posted above.

On a side note, I forgot to include radius of curvature for a general function {x, y(x)}:

radius(x) = (1 + (y'(x))²)^3/2 / | y''(x) |

in this case, y = 1/2 x², y' = x, y'' = 1, so

radius(x) = (1 + x²)^3/2 / |1|

I could have used this to confirm the magnitude of acceleration, a = v² / r, in this case:

a = c² / (x²+1)^3/2

checking

ax = -c² x / (x²+1)²
ay = c² / (x²+1)²

a = ( ax² + ay² )^1/2
a = ( (-c² x / (x²+1)²)² + (c² / (x²+1)²) )^1/2
a = ( c⁴ (x²+1) / (x²+1)⁴) )^1/2
a = ( c⁴ / (x²+1)³) )^1/2
a = c² / (x²+1)^3/2