Recognitions:
Gold Member

## Projectile Equation - Parabola

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 θo 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 ?
 Recognitions: Homework Help Science Advisor 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 = xrangetanθ - gxrange2/2u2cos2θ xrange = u2 sin(2θ)/g (twice the distance to the peak, unsurprisingly). To maximise wrt θ: 0 = dxrange/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 xcrit then I should set θ = atan( u2/gxcrit)

Recognitions:
Gold Member
 Quote by 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.
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 = xrangetanθ - gxrange2/2u2cos2θ xrange = u2 sin(2θ)/g (twice the distance to the peak, unsurprisingly). To maximise wrt θ: 0 = dxrange/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 xcrit then I should set θ = atan( u2/gxcrit)
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 ?

Recognitions:
Homework Help

## Projectile Equation - Parabola

 Quote by sankalpmittal 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 $\hat{x}$:
ytgt = $\hat{x}$tanθ - g$\hat{x}$2/2u2cos2θ (1)
We can now consider this to be an equation for $\hat{x}$ as a function of ytgt, 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 ytgt.