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I have the following rational model equation which fits distance versus time data very well:

$$y = \frac{bt}{1+ct+dt^2}$$

The derivative of this equation is :

$$\frac{dy}{dt} = \frac{(b-bdt^2)}{(1+ct+dt^2)^2}$$

Now at $t = 0, y = 0$ and $$\frac{dy}{dt} =u$$

where $u$ is the velocity at $t=0$

At $t = t_1, y = r$ and $$\frac{dy}{dt} =v$$

what I would like to do is solve for the constants $b$, $c$ and $d$ in terms of $u$, $v$ and $r$:

I can easily get:

$$b = u$$

but I'm having trouble solving for the other coefficients. Is it possible to solve for $c$ and $d$ in terms of $u$, $v$ and $r$?

cs89
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rdemyan
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1 Answers1

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Let $\Delta := 1+ct_1+dt_1^2$.

At time $t_1$ you have $y = r$ so $\frac{b t_1}{ \Delta} = r$ and $\dot{y} = v$ so $\frac{b(1-dt_1^2)}{\Delta^2} = v$.

You already solved $b = u$. Hence, the first equation yields $\Delta = \frac{ut_1}{r}$. Now $\Delta$ is thus known.

Substituting in the second leads to $b(1-dt_1^2) = v \Delta^2$, from which you can determine $d$.

Eventually, going back to the definition of $\Delta$ and knowing $d$, you can compute $c$.

cs89
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  • In solving for d, I'm able to get $1-dt_1^2 = \frac{vut_1^2 }{r^2}$. However, I can't figure out how to get rid of $t_1$, so that I can solve for d without $t_1$. – rdemyan Mar 21 '23 at 20:02
  • As I understood your question, $t_1$ is known. Otherwise, you have two many unknowns for your number of equations. – cs89 Mar 21 '23 at 20:13
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    Since $d$ is determined by a condition involving $t_1$, isn't it reasonable to think that $d$ may actually depend on $t_1$? In other words, you can't get rid of it. – MPW Mar 21 '23 at 20:13
  • @MPW: Since v is the velocity and r is the distance at $t_1$, I was hoping they might be able to serve as a "proxy" for $t_1$. $v=ucos\beta$ where $\beta$ is a known angle. Therefore $v$ does not depend on $t_1$. But $r$ does except for specialized cases where I can assume it does not – rdemyan Mar 21 '23 at 23:00
  • Well, if you know that $v t_1 = u \cos \beta$, where $u$, $v$ and $\beta$ are known, then you can compute $t_1$ to "get rid of it" in the other equations. – cs89 Mar 22 '23 at 07:20
  • @cs89: I know that $v=ucos\beta$ not that $v t_1 = u \cos \beta$. However, I've tested all of my data and have found that the following approximation holds to within less than 5%: $t_1 = \frac{2r}{u+v}$. In other words, I can use the linear average of the starting velocity, u, and ending velocity, v, to estimate $t_1$. I'm still checking to see what the velocity profile looks like with this assumption. – rdemyan Mar 22 '23 at 14:31