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Given a parametric function $\begin{align*} x &= x_o + v_x t + \frac{1}{2} a_x t^2\\ y &= y_o + v_y t + \frac{1}{2} a_y t^2\end{align*}$

and point $P$ at $(c, d)$ , find all point(s) on the function that are distance $R$ from point $P$, assuming that $R > 0$ and none of the constants equal $0$.

This is part of a programming project (that I'm doing on my own, not for homework), so the answer needs to be obtainable using only expressions. Graphically solving the problem won't be possible.

Thanks in advance.

1 Answers 1

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I'm note sure if (homework).

Hint:

I'm assuming that we're given $R, c, d, x_0, v_x, a_x, y_0, v_y, a_y.$

The squared distance between $P = (c,d)$ and $(x,y)$ is given by $ R^2 = (x - c)^2 + (y - d)^2 \tag{1}$ Substitute the definition of $x = x_o + v_x * t + \frac{1}{2} * a_x * t^2 \tag{2} \\ y = y_o + v_y * t + \frac{1}{2} * a_y * t^2$ in $(1) $

You're left with a polynomial in $t.$ Solve for $t,$ this will give you different values of $t = \{t_1, \ldots, t_4\}.$ Substitute each $t_i$ in $(2)$ to get different $(x,y)$ points.


Update # 1:

I'm too lazy to LaTeX the following equation:

        2          2   4                            3 (0.25 ay  + 0.25 ax ) t  + (1.0 vy ay + 1.0 vx ax) t        2                                       2               2  + (vx  + 1.0 xo ax - 1.0 ax c - 1.0 ay c + vy  + 1.0 yo ay) t                                                      2       2       2  + (-2. vy c + 2. yo vy + 2. xo vx - 2. vx c) t + xo  + 2. c  - 1. R                    2  - 2. xo c - 2. yo   = 0 

Anyways, that's the degree $4$ polynomial in $t.$ To find a closed-form expressions for the roots $\{t_1 = \cdots, t_2 = \cdots, t_3 = \cdots, t_4 = \cdots\},$ you will need to do a lot of algebra on this polynomial. For example this.

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    @JonW Indeed. The big expression above gives you $A, B, C, D, E.$ You're on the right path.2012-03-30