If I have a parabola as seen below, and I know Vmax, Vi, and the area, 'd' under the curve from x = n to x = t, is it possible to find the equation of the parabola? Or do I need more information?

n and t are not known.

I've been trying this for a while but always seem to run into a problem somewhere. Not sure if I need more givens or if I'm overlooking something.



  • $\begingroup$ 3 unknowns -> need 3 constraints (rule of thumb) $\endgroup$ – MichaelChirico Mar 17 '15 at 2:37

Edited / redone:

The general equation belonging to such a graph, orientated as shown, is $$y=-c(x-a)^2+b.$$

where $a,b,c$ are positive numbers. Note that during the argumentation below only the positive square roots will be considered. If one took into account the negative square roots then the alternative results would belong to the same graph but shifted to the left by $t>0$.

First, divide both sides of the equation above by $c$ and introduce the following notations: $$v=\frac{\ \ V_{max}}{c}, \ \ v_i=\frac{\ \ V_i}{c}, \ \ \delta=\frac{d}{c}, \ \ y'=\frac{\ y}{c}.$$

Note that the location of the maximum and the zero crossings of the graph have not been changed by the manipulation above. Now, one can easily conclude that


To see this consider that the graph goes through the origin, so $a=\sqrt{b}$. Also, the graph takes its maximum at $\frac{t}{2}$, so $0=-2(\frac{t}{2}-a)$, that is, $a=\frac{t}{2}$, where the maximum is taken. Substituting $a$ back to the equation for $x$, one gets that $b=v$.

Since the points $(n,v_i),\ (t,0)$ are also on the graph, one has $$t=2\sqrt{v}, \text{ and }n=\sqrt{v-v_i}+\sqrt{v}.$$

Finally, use the following information: $$\int_{\sqrt{v-v_i}+\sqrt{v}}^{2\sqrt{v}} ( -(x-\sqrt{v})^2+v ) \ \ dx=\delta.$$

Behind the parameters of the equation above there is only one unknown quantity: $c$. Finding the right $c$ may be tiresome but it is possible.

  • $\begingroup$ This is flawed from the start. A parabola actually has three parameters, $y=a(x-h)^2+k$. You have assumed the polynomial is monic, i.e. that its leading coefficient is 1, i.e. that $a=1$. $\endgroup$ – symplectomorphic Mar 17 '15 at 2:57
  • $\begingroup$ While writing up the previous solution I just forgot about the division by $c$. Sorry for that... $\endgroup$ – zoli Mar 17 '15 at 21:03

You don't need to know the area.

If you know that the roots are at $x=0$ and $x=T$, then the parabola must have the form $$v(t) = -a(t-0)(t-T)=at(T-t)$$ for some positive real number a. So \begin{align} v\left(\frac T2 \right) &= v_{max}\\ a \frac T2 \left( t - \frac T2 \right) &= v_{max}\\ a \frac{T^2}{4} &= v_{max}\\ a = \frac{4 v_{max}}{T^2} \end{align}

And the equation of the parabola is

$$v(t) = \frac{4 v_{max}}{T^2}t(T-t)$$

Just for the record,

$$\int_0^T v(t) dt = \frac 23 v_{max}$$


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