How to graph parametric equations?

[Meowy]

Hi, so after doing some calculus for projectile motion with air resistance, I obtained two equations of vx(t) and vy(t) that describes the vertical and horizontal motion of the projectile.

Please tell me if I'm wrong, but I believe since both vx and vy are functions of t, can't they be parametrized? This meaning I can graph them as parametric equations and for each moment in time, there's a specific (x,y) coordinate that describes where the projectile is?

I want to know that if this method is valid and whether I can somehow show the (x,y) values given t=? on a graph. If anyone is kind enough to tell me if there's another way to connect them, that would be great as well! Thanks.

 

[Simon Bridge]

Hi, so after doing some calculus for projectile motion with air resistance, I obtained two equations of vx(t) and vy(t) that describes the vertical and horizontal motion of the projectile.

Please tell me if I'm wrong, but I believe since both vx and vy are functions of t, can't they be parametrized?

All equations can be parametrized - in this case they are parametrized wrt time t. But it does mean you can plot vy vs vx.

This meaning I can graph them as parametric equations and for each moment in time, there's a specific (x,y) coordinate that describes where the projectile is?

You'd have to solve the vx and vy equations to get x(t) and y(t) to do that. But, having done that, you can certainly plot y vs x if you want to.

I want to know that if this method is valid and whether I can somehow show the (x,y) values given t=? on a graph. If anyone is kind enough to tell me if there's another way to connect them, that would be great as well! Thanks.

Whether the method is "valid" depends on what you want to know.

The path traced out by (x(t),y(t)) is called the "trajectory", and it is a reasonable thing to want to find out about. You have probably already seen this done for ballistic motion without air resistance - the trajectory is a parabola.

 

[ellipsis]

Note: I tried doing this with Newtonian gravity not too long ago (such that acceleration is a function of height according to inverse square law), and it turned into an intractable second-order differential equation that could only be numerically solved.

I think your case is simple enough to do though; if you already have vx(t) and vy(t) solved, you can integrate each to find x(t) and y(t) respectively. Once that's done, use your favorite graphing software: MATLAB if you want something professional, or maybe Winplot if you want to easily animate the trajectory.

EDIT: For the last three hours, I've been trying to find a closed solution for height based on constant gravitational acceleration (9.81) and a simplified drag model where acceleration due to drag is (1/10)v^2 opposite the current direction of velocity.

The trouble is 'opposite the current direction'... The square factor deletes all knowledge of velocity direction from the equation, and trying to use sign(v) in the differential equation leads to being unable to find a closed solution for v.

Only method I see are two solutions: One equation for the ascending part of the trajectory, and another for the descending part.

Meowy, based on this nastiness I've run into, I deduce that you're using an even simpler drag model - probably where the deceleration due to drag is directly proportional to velocity. Am I right?