How to find the solution for this equation?

Question

I wanted to find the value that makes the derivative of the following function equal to $0$ $$X(e)=v_i \sqrt{1-e^2}\left(\frac{v_ie}{g}+\sqrt{\frac{v_i^2e^2}{g^2}+\frac{2y_i}{g}}\right)$$ After deriving and simplifying i reached the following $$\frac{v_i[g \left(v_i(1-2e^2) \sqrt{\frac{v_i^2e^2}{g^2}+\frac{2y_i}{g} }-2ey_i \right)+v_i^2e(1-2e^2)]}{g^2\sqrt{\frac{v_i^2e^2}{g^2}+\frac{2y_i}{g} }\sqrt{1-e^2}}$$ But I am having a hard time trying to find the values of $e$ that make the derivative equal to $0$.

$$g[v_i(1-2e^2) \sqrt{\frac{v_i^2e^2}{g^2}+\frac{2y_i}{g} }-2ey_i]+v_i^2e(1-2e^2)=0$$

Answer 1

You can start by rearranging it to get rid of the square root (I'm dropping the subscripts because I'm lazy):

$$\begin{eqnarray} g\left[v(1-2e^2)\sqrt{\frac{v^2 e^2}{g^2} + \frac{2y}{g}} - 2ey \right] + v^2 e(1 - 2e^2) & = & 0 \\ gv(1-2e^2)\sqrt{\frac{v^2 e^2}{g^2} + \frac{2y}{g}} & = & 2egy - v^2 e(1 - 2e^2) \\ g^2v^2(1-2e^2)^2\left[\frac{v^2 e^2}{g^2} + \frac{2y}{g}\right] & = & 4e^2g^2y^2 - 4e^2gyv^2(1-2e^2) + v^4e^2(1-2e^2)^2 \end{eqnarray}$$

Then notice that the first term on the left cancels the last term on the right. With some more simplification, you can reduce the whole thing to a quadratic expression in $2e^2$ that you can then solve via normal means.

Answer 2

EDIT1:

Removing constants $(y_i,g, v)$ and defining new non-dimensional constant

$$ p^2=\frac{2y_i g}{v^2} $$

$$ Y(e)= \sqrt{1-e^2}(e+ \sqrt{e^2+p^2})$$ Applying Chain Rule

$$ \frac{e+\sqrt{e^2+p^2}}{\sqrt{1-e^2}} = \frac{(1+\dfrac{e}{\sqrt{e^2+p^2}})\sqrt{1-e^2}}{e} $$

Cross multiplying, the $f(e)$ to be solved is

$$e({e+\sqrt{e^2+p^2}})= (1-e^2)(1+\dfrac{e}{\sqrt{e^2+p^2}})$$

Resulting in a simpler equation

$$ \frac{1-e^2}{e}=\sqrt{e^2+p^2}$$

EDIT2

To continue: Squaring and simplifying

$$ p^2e^2 +2 e^2-1=0 $$

has a solution of the quadratic

$$ e= \frac{-1\pm \sqrt{1+p^2}}{2p^2}$$

from which the negative value is discarded.

Answer 3

I am going to share the the answer to the question with detailed steps and the things I had trouble with , in case someone encounters the same problems hope this will be helpful ,

so first the equation was in fact $$X(\theta)=v_icos(\theta)\left(\frac{v_isin(\theta)}{g}+\sqrt{\frac{v_i^2sin(\theta)^2}{g^2}+\frac{2y_i}{g}}\right)+x_i$$ where 0<\theta<90

I went and substituted $ sin(\theta)$ with $e$ and used the chain rule
so we have $$\frac{\delta x}{\delta e}\frac{\delta e}{\delta \theta}$$ now $sin(\theta)$ can't be equal to zero in the period given so we know that if we want $$\frac{\delta x}{\delta \theta}=0$$ then $$\frac{\delta x}{\delta e} \text{ would =0 }$$

the derivation is a bit long took me a page from here but it would be equal to the following $$\frac{v_i[g \left(v_i(1-2e^2) \sqrt{\frac{v_i^2e^2}{g^2}+\frac{2y_i}{g} }-2ey_i \right)+v_i^2e(1-2e^2)]}{g^2\sqrt{\frac{v_i^2e^2}{g^2}+\frac{2y_i}{g} }\sqrt{1-e^2}}$$ now here $$g \left(v_i(1-2e^2) \sqrt{\frac{v_i^2e^2}{g^2}+\frac{2y_i}{g} }-2ey_i \right)+v_i^2e(1-2e^2)=0$$
$$\begin{eqnarray} gv(1-2e^2)\sqrt{\frac{v^2 e^2}{g^2} + \frac{2y}{g}} & = & 2egy - v^2 e(1 - 2e^2) \\ g^2v^2(1-2e^2)^2\left[\frac{v^2 e^2}{g^2} + \frac{2y}{g}\right] & = & 4e^2g^2y^2 - 4e^2gyv^2(1-2e^2) + v^4e^2(1-2e^2)^2 \end{eqnarray}$$

$$v^4e^2(1-2e^2) + 2ygv^2(1-2e^2)^2 = 4e^2g^2y^2 - 4e^2gyv^2(1-2e^2) + v^4e^2(1-2e^2)^2$$

$$ 2ygv^2(1-2e^2)^2 = 4e^2g^2y^2 - 4e^2gyv^2(1-2e^2) + $$ now my second mistake was continuing to simplify from here , the oversimplifying caused more miscalculations I was left with terms like $e^4$ and they weren't getting deleted and even when substituting $2e^2=z$ I mistyped some signs and went on solving equations with sqrt of sqrt and the results were wrong

now we divide by $2gy$ and substituting $z=2e^2$

$$ v^2(1-2e^2)^2 = 2e^2gy - 2e^2v^2(1-2e^2) + $$ $$ v^2(1-z)^2 = zgy - zv^2(1-z) + $$ $$ 0 =(-v^2z^2+v^2z^2)+2zv^2+ zgy - zv^2-v^2 $$ $$ 2zv^2+ zgy - zv^2-v^2=0 $$ $$ z(gy+2v^2-v^2)-v^2 \\ z(gy+v^2)-v^2 \\ 2e^2(gy+v^2)-v^2 \\ e = \frac{0 \pm \sqrt{0- 4*2(gy+v^2)*-v^2}}{2*2(gy+v^2)} $$ $$ e = \pm \sqrt{\frac{ 8(gy+v^2)*-v^2}{16(gy+v^2)^2}} \\e = \pm \sqrt{\frac{ v^2}{2(gy+v^2)}}$$ we divide both numerator and denominator by $v^2$ $$ e = \pm \sqrt{\frac{ 1}{2(\frac{gy}{v^2}+1)}}$$ finally e is equal to sin in the period ]0,90[ so we take the positive value only .