Parametric Equations Of a Point Moving on a Cubic Parabola with Constant Velocity

Question

A particle moves with constant speed $v$ along a cubic parabola of equation $y = \frac{x^3}{k}$, where $k$ is a constant. I would like to find the position of the particle at a specific time. I thought the best way would be to derive parametric equations $x(t)$ and $y(t)$ of the formula above. Therefore, I chose $x = ct$ and hence $y=\frac{c^3t^3}{k}$. The derivatives are $\frac{dx}{dt} = c$ and $\frac{dy}{dt} = \frac{3c^3t^2}{k}$. Then, from the velocity vector $v^2 = c^2 + \bigl(\frac{3c^3t^2}{k}\bigr)^2$, the $c$ parameter can be derived and substituted into the $x(t)$ and $y(t)$ expressions. However, this seems to be a higher order polynomial equation. Is it solvable analytically? Perhaps another parametric substitution would be better? Any help is appreciated.

Answer 1

I wasn't able to get a closed form solution, but you can use the following approach to get a solution:

First, solve for the arc length as a function of $x$: $$ s(x) = \int_{x_0}^{x}\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$$

or in this case:

$$ s(x) = \int_{x_0}^{x}\sqrt{1+\frac{9x^4}{k^2}}dx$$

Wolfram Alpha gives the indefinite integral as:

$$ I(x) = \frac{1}{3}x\left(2\ {}_2{F}_1\left(\frac{1}{4},\frac{1}{2}; \frac{5}{4};-\frac{9x^4}{k^2}\right)+\sqrt{1+\frac{9x^4}{k^2}}\right)$$

So: $$s(x)=I(x)-I(x_0)$$

Note that $2F_1$ is the hypergeometric function.

Since the speed is constant $s(t)=vt$. To solve for the position, you will need to find $x$ such that $vt=s(x)$. Newton's method can be used to solve for $x$ given a time $t$.