Projectile Equation - Parabola

sankalpmittal

Here's a equation of a projectile thrown in a parabolic path :

y=xtanθ - gx²/2u²cos²θ

where x is the corresponding x - coordinate or the range of a projectile at a point ,
y is the corresponding y - coordinate or the height of projectile at point,
θ is the angle made with the horizontal arbitrary x-axis through which projectile is thrown.
u is magnitude of initial velocity vector making angle θ with horizontal.

Note: Point from where the projectile began is assumed to be origin , i.e. (0,0) coordinate.

I was once told that in order to solve difficult problems of projectile , calculus is a very powerful tool.

We can consider y in equation as a function of θ and then can differentiate y with respect to θ {find y'(θ)}.

dy/dθ = .........

Then we can put dy/dθ =0 and also dx/dθ =0. Hence we can easily solve for optimal angle θ_o for maximum range. Sometimes we just put dy/dθ =0 when we are given "x" and we can solve for θ.

Also if we imagine the equation to be plotted on a graph then we can obviously consider y as a function of x and then differentiate y with respect to x {find y'(x)}.

Then we put dy/dx=0 to find point of maxima or minima , I think. We also find d²y/dx².

So here are my questions :

What's the use or when do we find dy/dθ or dy/dx in the projectile equation? I cannot understand. Also I cannot fathom this all theoretically. Why are we doing such thing ? Can someone explain ? Is this really a powerful tool to solve projectile problems ? When do we put dy/dθ =0 and also dx/dθ =0 or even dy/dx =0 and why , while differentiating w.r.t θ or x in y=xtanθ - gx²/2u²cos²θ ?

haruspex

Since x and θ are both being considered as variables here, in each case, it should really be the partial derivative: ∂y/∂θ, ∂y/∂x etc. E.g. ∂y/∂θ says how y changes when θ is varied but x is held constant.
∂y/∂x = tanθ - gx/u²cos²θ:
For a given θ, ∂y/∂x tells you the slope of the trajectory as you look along it. When this becomes zero, the projectile is moving horizontally, so is at the peak of the parabola:
tanθ = gx/u²cos²θ
u² sin(2θ) = 2gx
I.e. the distance at which the arc peaks is u² sin(2θ) /2g.

For max range, we would first find what the range is. This is the nonzero x which makes y zero:
0 = x_rangetanθ - gx_range²/2u²cos²θ
x_range = u² sin(2θ)/g
(twice the distance to the peak, unsurprisingly). To maximise wrt θ:
0 = dx_range/dθ = 2u² cos(2θ)/g
So θ = π/4.

It's not obvious to me what use would be made of setting ∂x/∂θ = 0 in the original equation. ∂y/∂θ will be 0 when y is at the maximum value it can reach by varying θ only. This value may depend on x:

∂(x tanθ - gx²/2u²cos²θ)/∂θ =
x sec²θ - (gx²/2u²)2 sec²θ tanθ
Setting this zero:
1 = (gx/u²) tanθ
tan θ = u²/gx
That is, if I want y to be as great as possible at distance x_crit then I should set θ = atan( u²/gx_crit)

sankalpmittal

What if y changes when both x and θ are being varied. What will we get when we find ∂y/∂θ or ∂y/∂x or dy/du even in equation ? :

y=xtanθ - gx²/2u²cos²θ

Also what if u is varied when y changes but θ and x both are held constant. Logically , I guess this does not even make sense.

Well ok , but in this case , it'll mean that how y changes with respect to x but θ is held constant. This does not make sense to me.

Well ok...

So here this means that y changes w.r.t to θ but x is held constant. What if we do not set ∂y/∂θ =0 ? Also what if we do not set ∂x/∂θ =0 in projectile equation ? What new can we get ?

Also if in equation :

y=xtanθ - gx²/2u²cos²θ .... (i)

I put x = u²sin2θ/g

I get y = (2u²sin²θ - u⁴tanθ)/g

If I put y=u²sin²θ/2g in (i)

I obtain x= u²sin2θ/2g

If I put y=u²sin²θ/2g and x = u²sin2θ/g
in (i) I get :

u² = sinθcosθ
2u² = sin2θ

These equations I obtain , are they correct ?

Moreover you got :

tan θ = u²/gx
x= u²/gtanθ
If I set
x = u²sin2θ/g

Then u²sin2θ/g = u²/gtanθ

I get after solving that θ = ∏/4

Moreover ,

These are for symmetric projectile. What if projectile is thrown from an elevation at an angle from horizontal or to an elevation ? What can we get in that case from projectile equation - i.e what will be optimal angle for maximum range ?

haruspex

If you make a small change Δx to x and Δθ to θ then the change in y will be Δy = Δx∂y/∂x + Δθ∂y/∂θ
Yes, it makes sense. Holding θ constant is obvious; holding x constant at [itex]\hat{x}[/itex] means you're focusing on the point in the trajectory distance [itex]\hat{x}[/itex] from the start. Then you can vary the launch velocity and observe the effect on the height at [itex]\hat{x}[/itex].
It means watching what happens to the height at a given distance as you vary θ.
Maybe you don't understand the significance of the places where a derivative becomes zero.
If we fix x = [itex]\hat{x}[/itex] and vary θ, we see how the height of the trajectory varies at distance [itex]\hat{x}[/itex]. If ∂y/∂θ > 0, increasing θ will increase y. If ∂y/∂θ < 0, increasing θ will decrease y. To get the maximum y, we want to increase θ so long as that makes y increase. As soon as it gets to the point where it would make y decrease we want to stop changing θ. Since these equations involve smooth functions, that change-over cannot happen suddenly. To get from ∂y/∂θ > 0 to ∂y/∂θ < 0 it must pass through the point ∂y/∂θ = 0. Likewise, for a transition from ∂y/∂θ < 0 to ∂y/∂θ > 0.
So when the value of θ, for the given [itex]\hat{x}[/itex], is such that ∂y/∂θ = 0, we must be at either a local maximum or local minimum of y wrt θ.
[OK, there is a third option: it could be an inflexion point. This happens when ∂y/∂θ hits zero but instead of crossing to the opposite sign it immediately turns around again: either >0, 0, > 0 or < 0, 0, < 0. So in general it is not enough to observe the derivative is zero at some point: you should look at whether it is positive or negative just either side of that point. In the present case however we are clearly dealing with local maxima.]
So, to answer your question, if you do not set ∂y/∂θ = 0 you will not be looking at a local extremum. You could set it to something else, but I struggle to think how you would usefully do that in the context of your set-up.
In the context we're discussing, y is the dependent variable. It is a function of the independent variables x, θ and u. The partial derivatives are only defined on that basis; e.g., it is only by knowing that u and x are the other independent variables that you know to interpret ∂y/∂θ as varying θ while holding x and u fixed. So it doesn't make sense to ask about ∂x/∂θ.
However, you can recast the equation, making x a function of y, u and θ. Provided you're clear that this is what you are doing, that now gives ∂x/∂θ a meaning. Namely, the recast has given you an equation that tells you at what distance the projectile will have a specified height, given its initial speed and angle. ∂x/∂θ now tells you how that will change as you vary the angle.
No, you should get y = u²2cos²θ/g - u²2sin²θ/g = 2u²cos(2θ)/g
I could tell you'd gone wrong somewhere because your equations became dimensionally invalid. (Always a useful check.)
The trajectory as a whole (x from -∞ to +∞) is symmetric about the peak. What you're describing here is simply looking at different sections of it.
If you want to maximise the range, but the target is at a different height from the start, we can do that. The equation is already set up for y to be 0 when x is zero. Set y to be the target height at [itex]\hat{x}[/itex]:
y_tgt = [itex]\hat{x}[/itex]tanθ - g[itex]\hat{x}[/itex]²/2u²cos²θ (1)
We can now consider this to be an equation for [itex]\hat{x}[/itex] as a function of y_tgt, u and θ. We want to maximise [itex]\hat{x}[/itex] wrt θ, so differentiate wrt θ and set ∂[itex]\hat{x}[/itex]/∂θ = 0:
0 = [itex]\hat{x}[/itex]∂tanθ/∂θ - (g[itex]\hat{x}[/itex]²/2u²)∂sec²θ/∂θ
= sec²θ - (g[itex]\hat{x}[/itex]/2u²)2sec²θ tanθ
tan θ = u²/g[itex]\hat{x}[/itex] (2)
Using 2 we can substitute for [itex]\hat{x}[/itex] in (1) to obtain an equation relating θ to y_tgt.