The interval of α for which a function increases

[Oiler]

Given a projectile, P is launched at a velocity [MATH]u[/MATH] on a plane at an angle to the horizontal, [MATH]\alpha[/MATH], [MATH] \left[0<\alpha <\frac{1}{2}\pi\right][/MATH] from point O (the origin)
I am trying to prove that the length of OP is increasing throughout P's entire flight, if [MATH]\alpha [/MATH] is such that [MATH]\sin(\alpha)<\frac{2\sqrt{2}}{3}[/MATH].

I have not made much progress on it, so far I have only been able to conclude that if OP is to keep increasing for the duration of P's flight, then the rate of change of OP's length( i call it x) with respect to time must be greater than zero for the duration of its flight time, so I tried to form an [MATH]x(t)[/MATH] function with the aim of differentiating it, setting [MATH]x'(t) \geq 0[/MATH] and hopefully re-arranging for an expression of [MATH]\alpha[/MATH][MATH] x(t)=\vec i \left(ut\cos(\alpha)\right)+\vec j \left(ut\sin(\alpha)-\frac{g}{2}t^2)\right)[/MATH][MATH]\longrightarrow \left(u^2t^2-ut^3g\sin(\alpha)+\frac{g^2}{4}t^4\right)^\frac{1}{2}[/MATH]
But this is definitely not the correct way of approaching it, since differentiating x(t) will give me a cubic to rearrange for [MATH]\alpha[/MATH]Basically, I've not a clue lol.

 

[Dr.Peterson]

The expression before the arrow a vector, and what follows is its length, right? And it's the latter that you are calling x(t).

To keep things simple, you can differentiate the square of the distance, since if that is always increasing, so will the distance itself (because it is positive).

This does yield your desired result. The cubic equation is not too hard to work with, because you can partially factor it. Don't solve for alpha; just work with the discriminant. You'll see what I mean when you do it!

 

[JeffM]

The horizontal component of the initial velocity is

[MATH]u * cos ( \alpha ).[/MATH]
The vertical component of the initial velocity is

[MATH]u * sin ( \alpha ).[/MATH]
Both those are positive numbers if [MATH]0 < \alpha < \dfrac{\pi}{2}.[/MATH]
Ignoring air resistance, there is no force acting on the horizontal component. What can you conclude?

 

[Oiler]

The expression before the arrow a vector, and what follows is its length, right? And it's the latter that you are calling x(t).

To keep things simple, you can differentiate the square of the distance, since if that is always increasing, so will the distance itself (because it is positive).

This does yield your desired result. The cubic equation is not too hard to work with, because you can partially factor it. Don't solve for alpha; just work with the discriminant. You'll see what I mean when you do it!

firstly, thanks.
I have tried to do as you recommended and power-rule differentiated the square of the displacement function:
[MATH]x(t)=\sqrt{u^2t^2-ug\sin(\alpha)t^3+\frac{g^4}{4}t^4}[/MATH][MATH]\rightarrow x(t)^2=u^2t^2-ug\sin(\alpha)t^3+\frac{g^4}{4}t^4[/MATH][MATH]\rightarrow x'(t)^2=2u^2t-3ug\sin(\alpha)t^2+g^4t^3[/MATH] from where I partially factorised t, just like you said it could be done!
[MATH]x'(t)^2=t\left(g^4t^2+t(-3ug\sin(\alpha))+2u^2\right)[/MATH]but if I were to compute the discriminant of the quadratic part of[MATH]x'(t)^2[/MATH] then unfortunately my confusion with basic algebra rears its ugly head
because I can't seem to figure out how many real solutions [MATH]x'(t)^2[/MATH] would have.
One of them is going to be zero, but i already factored out [MATH]t[/MATH].
Not really sure how to figure out the number of times [MATH]g^4t^2+t(-3ug\sin(\alpha))+2u^2[/MATH] intersects with the x axis honestly.
All i can figure out is that since the question is asking the reader to prove OP is increasing for the interval of [MATH]\alpha[/MATH] given, then that must be true, so the ball's maximum displacement from O happens when it lands; and its minimum displacement is when its just launched at [MATH]t=0[/MATH]. Thats two stationary points on [MATH]x(t)[/MATH] hence, these are two real solutions for [MATH]x'(t)[/MATH]i dont really know where to go from there.

 

[Oiler]

The horizontal component of the initial velocity is

[MATH]u * cos ( \alpha ).[/MATH]
The vertical component of the initial velocity is

[MATH]u * sin ( \alpha ).[/MATH]
Both those are positive numbers if [MATH]0 < \alpha < \dfrac{\pi}{2}.[/MATH]
Ignoring air resistance, there is no force acting on the horizontal component. What can you conclude?

The horizontal velocity will remain a constant?

 

[JeffM]

Assuming no other force acts horizontally on the projectile, the horizontal velocity remains constant until the projectile hits the ground. The fact that the constant is not zero means that it will travel some non-zero distance.

The exact distance depends primarily on the force of gravity (a negative vertical force), the angle of elevation, and air resistance. Frequently the latter is ignored as immaterial.

 

[Dr.Peterson]

firstly, thanks.
I have tried to do as you recommended and power-rule differentiated the square of the displacement function:
[MATH]x(t)=\sqrt{u^2t^2-ug\sin(\alpha)t^3+\frac{g^4}{4}t^4}[/MATH] <--- You had g^2 here, not g^4, right?
[MATH]\rightarrow x(t)^2=u^2t^2-ug\sin(\alpha)t^3+\frac{g^4}{4}t^4[/MATH][MATH]\rightarrow x'(t)^2=2u^2t-3ug\sin(\alpha)t^2+g^4t^3[/MATH] from where I partially factorised t, just like you said it could be done!
[MATH]x'(t)^2=t\left(g^4t^2+t(-3ug\sin(\alpha))+2u^2\right)[/MATH]but if I were to compute the discriminant of the quadratic part of[MATH]x'(t)^2[/MATH] then unfortunately my confusion with basic algebra rears its ugly head
because I can't seem to figure out how many real solutions [MATH]x'(t)^2[/MATH] would have.
One of them is going to be zero, but i already factored out [MATH]t[/MATH].
Not really sure how to figure out the number of times [MATH]g^4t^2+t(-3ug\sin(\alpha))+2u^2[/MATH] intersects with the x axis honestly.
All i can figure out is that since the question is asking the reader to prove OP is increasing for the interval of [MATH]\alpha[/MATH] given, then that must be true, so the ball's maximum displacement from O happens when it lands; and its minimum displacement is when its just launched at [MATH]t=0[/MATH]. Thats two stationary points on [MATH]x(t)[/MATH] hence, these are two real solutions for [MATH]x'(t)[/MATH]i dont really know where to go from there.

You should really give a new name to the function [MATH](x(t))^2[/MATH], as your [MATH]x'(t)^2[/MATH] would properly mean the square of the derivative, not the derivative of the square. So either write [MATH]((x(t))^2)'[/MATH] or let [MATH]q(x) = (x(t))^2[/MATH] and write [MATH]q'(x) = \dots[/MATH] .

But I know what you meant.

Your goal is to find the conditions under which [MATH]q'(t)=t\left(g^4t^2+t(-3ug\sin(\alpha))+2u^2\right)[/MATH] stays positive; that is, it does not change sign except at x=0. So you want [MATH]g^2t^2+(-3ug\sin(\alpha))t+2u^2[/MATH] to have no zeros. Find the discriminant of that quadratic, namely [MATH]b^2 - 4ac[/MATH]. Set that to be less than zero, and solve.

 

[Dr.Peterson]

Assuming no other force acts horizontally on the projectile, the horizontal velocity remains constant until the projectile hits the ground. The fact that the constant is not zero means that it will travel some non-zero distance.

The exact distance depends primarily on the force of gravity (a negative vertical force), the angle of elevation, and air resistance. Frequently the latter is ignored as immaterial.

How is this related to the problem we're working on, which is about the distance from O to the projectile throughout the flight? And isn't it already shown in the OP (no pun intended)?

 

[JeffM]

How is this related to the problem we're working on, which is about the distance from O to the projectile throughout the flight? And isn't it already shown in the OP (no pun intended)?

I goofed.

 

[Oiler]

You should really give a new name to the function [MATH](x(t))^2[/MATH], as your [MATH]x'(t)^2[/MATH] would properly mean the square of the derivative, not the derivative of the square. So either write [MATH]((x(t))^2)'[/MATH] or let [MATH]q(x) = (x(t))^2[/MATH] and write [MATH]q'(x) = \dots[/MATH] .

But I know what you meant.

Your goal is to find the conditions under which [MATH]q'(t)=t\left(g^4t^2+t(-3ug\sin(\alpha))+2u^2\right)[/MATH] stays positive; that is, it does not change sign except at x=0. So you want [MATH]g^2t^2+(-3ug\sin(\alpha))t+2u^2[/MATH] to have no zeros. Find the discriminant of that quadratic, namely [MATH]b^2 - 4ac[/MATH]. Set that to be less than zero, and solve.

Thanks again, and yes I did mistype the power of g. sorry about that and other notation mistakes.

Setting the discriminant of [MATH]g^2t^2+t(-3ug\sin(\alpha))+2u^2 [/MATH] to less than zero does indeed yield the correct range of [MATH]\sin(\alpha)[/MATH]
But I'm not sure I understand why [MATH]g^2t^2+t(-3ug\sin(\alpha))+2u^2[/MATH] has to have a discriminant less than zero, since the quadratic
can be positive for the time interval of P's flight, even if it had two or one solutions in the negative part of the x-axis. If that were the case, then it would still have a positive slope for [MATH]t\geq 0[/MATH] satisfying the condition of [MATH]x(t)[/MATH] being an increasing function for the duration of P's flight, (if [MATH]\sin(\alpha)<\frac{2\sqrt{2}}{3}[/MATH])

 

[Dr.Peterson]

Very perceptive. I didn't claim that was all you had to do for a full proof. This was the next issue I might have suggested after doing what I said; I held back to avoid making things too complicated too soon. But you're able to handle it.

Either before or after using the discriminant, look at the whole quadratic formula and see whether real roots can be negative when they exist.

Another approach might be to complete the square (equivalent to using the formula), so you can see what this parabola looks like in general.

 

[JeffM]

Consider the curve described by

[MATH]x = ut \text { and } y = w * t - \dfrac{g}{2} * t^2, \ u,\ w, \text { and } g \text { all positive, and}[/MATH]
[MATH]0 \le t \le \dfrac{2w}{g}.[/MATH]
We can restate this as

[MATH]0 \le x \le \dfrac{2uw}{g},\ t = \dfrac{x}{u}, \text { and }[/MATH]
[MATH]y = w * \dfrac{x}{u} - \dfrac{g}{2} * \left ( \dfrac{x}{u} \right)^2 \ = - \dfrac{g}{2u^2} * x^2 + \dfrac{w}{u} * x.[/MATH]
To simplify our lives

[MATH]\text {Let } \dfrac{g}{2u^2} = p < 0 \text { and } \dfrac{w}{u} = q > 0 \implies y = - px^2 + qx.[/MATH] Obviously

[MATH]y = 0 = - px^2 + qx = x(-px + q) \iff x = 0 \text { or } x = \dfrac{q}{p} \iff[/MATH]
[MATH]x = 0 \text { or } \dfrac{w}{u} \div \dfrac{g}{2u^2} = \dfrac{2uw}{g}.[/MATH]
The distance from the origin to a point on the curve is

[MATH]d = \sqrt{x^2 + y^2} = \sqrt{x^2 + (-px^2 + qx)^2} =[/MATH]
[MATH]\sqrt{p^2x^4 -2pqx^3 + (q^2 + 1)x^2} = r^{1/2}.[/MATH]
Obviously d exceeds 0 if x is not zero. So the derivative if x > 0 is

[MATH]d' = \dfrac{4p^2x^3 - 6pqx^2 + 2(q^2 + 1)x}{2r^{1/2}} = \dfrac{x(2p^2x^2 - 3pqx + q^2 + 1)}{d}.[/MATH]
If x > 0, the sign of d' depends on the sign of [MATH]2p^2x^2 - 3pqx + q^2 + 1.[/MATH]
But that is just a parabola with a positive leading coefficient. Its derivative is

[MATH]4p^2x - 3pq.[/MATH]
The minimum will occur at [MATH]x = \dfrac{3q}{4p} > 0.[/MATH]
The value there will be

[MATH]2p^2 * \left ( \dfrac{3q}{4p} \right )^2 - 3pq * \dfrac{3q}{4p} + q^2 + 1 =[/MATH]
[MATH]\dfrac{9q^2}{8} - \dfrac{9q^2}{4} + q^2 + 1 = \dfrac{9q^2 - 18q^2 + 8q^2 + 8}{8} = 1 - q^2.[/MATH]
The minimum will be non-negative if and only if

[MATH]1 - q^2 \ge 0 \implies 1 \ge q.[/MATH]
But [MATH]q = \dfrac{w}{u}.[/MATH]
Therefore the sign of the derivative of distance will be non-negative if and only if

[MATH]u \ge w.[/MATH]
And u and w are respectively the horizontal and vertical components the initial velocity v. If the angle of elevation is alpha, then

[MATH]cos(\alpha) = \dfrac{u}{v} \text { and } sin (\alpha) = \dfrac{w}{v}.[/MATH]
[MATH]u \ge w \implies \dfrac{u}{v} \ge \dfrac{w}{v} \implies \\ cos( \alpha ) \ge sin ( \alpha ) \implies \alpha \le \dfrac{\pi}{4}.[/MATH]
The only question left is

[MATH]\dfrac{\dfrac{\pi}{4}}{\dfrac{2\sqrt{2}}{3}} = \dfrac{\pi}{4} * \dfrac{3}{2\sqrt{2}} = \dfrac{3 \pi}{8 \sqrt{2}} < 1.[/MATH]
Thus, unless I screwed up the algebra somewhere, you cannot prove what you want to because it is not true.