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]

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.

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.

∂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.

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.

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.

Well ok...

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)

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]

What if y changes when both x and θ are being varied.

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

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.

Yes, it makes sense. Holding θ constant is obvious; holding x constant at $\hat{x}$ means you're focusing on the point in the trajectory distance $\hat{x}$ from the start. Then you can vary the launch velocity and observe the effect on the height at $\hat{x}$.

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.

It means watching what happens to the height at a given distance as you vary θ.

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 ?

Maybe you don't understand the significance of the places where a derivative becomes zero.
If we fix x = $\hat{x}$ and vary θ, we see how the height of the trajectory varies at distance $\hat{x}$. 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 $\hat{x}$, 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.

Also what if we do not set ∂x/∂θ =0 in projectile equation ? What new can we get ?

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.

Also if in equation :

y=xtanθ - gx²/2u²cos²θ ... (i)

I put x = u²sin2θ/g

I get y = (2u²sin²θ - u⁴tanθ)/g

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.)

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 ?

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 $\hat{x}$:
y_tgt = $\hat{x}$tanθ - g$\hat{x}$²/2u²cos²θ (1)
We can now consider this to be an equation for $\hat{x}$ as a function of y_tgt, u and θ. We want to maximise $\hat{x}$ wrt θ, so differentiate wrt θ and set ∂$\hat{x}$/∂θ = 0:
0 = $\hat{x}$∂tanθ/∂θ - (g$\hat{x}$²/2u²)∂sec²θ/∂θ
= sec²θ - (g$\hat{x}$/2u²)2sec²θ tanθ
tan θ = u²/g$\hat{x}$ (2)
Using 2 we can substitute for $\hat{x}$ in (1) to obtain an equation relating θ to y_tgt.

 

[PJC01]

The projectile equation - parabola is a mathematical representation of the motion of a projectile in a parabolic path. It is derived from the equations of motion in classical mechanics, and it takes into account the initial velocity, angle of projection, and gravitational acceleration.

The equation can be used to determine the trajectory of a projectile and its maximum height and range. In order to solve for these quantities, we need to differentiate the equation with respect to either the angle of projection (θ) or the horizontal displacement (x). This is where calculus comes in as a powerful tool.

When we differentiate the equation with respect to θ, we are finding the slope of the trajectory at different angles. Setting dy/dθ = 0 allows us to find the angle (θo) at which the projectile will have the maximum range. This is useful in determining the optimal angle for a projectile to be launched in order to achieve a desired range.

Similarly, when we differentiate the equation with respect to x, we are finding the slope of the trajectory at different horizontal positions. Setting dy/dx = 0 allows us to find the point of maximum height or maximum range. This is useful in determining the maximum height or range that a projectile can achieve.

In summary, differentiating the projectile equation with respect to θ or x allows us to find important quantities such as the optimal angle, maximum height, and maximum range of a projectile. It is a powerful tool in solving projectile problems because it allows us to analyze the motion of a projectile in a mathematical and systematic way.