Separating vector differential equation into components

1. Sep 16, 2016

Dustgil

1. The problem statement, all variables and given/known data
Write down the component form of the differential equations of motion of a projectile if the air resistance is proportional to the square of the speed. Are the equations seperated? Show that the x component of the velocity is given by

$$\dot{x}=\dot{x}_0e^{^-\gamma s}$$

where s is the distance the projectile has traveled along the path of motion and $$\gamma = c_2 / m$$

2. Relevant equations

3. The attempt at a solution

So, the differential equation in vector form is

$$m \frac {d^2r} {dt^2} = -c_2\vec{v}|v| -gk$$
$$\frac {d^2r} {dt^2} = -\gamma\sqrt{V_x^2+V_y^2+V_z^2}(V_xi+V_yj+V_zk) -gk$$

so x in particular is:

$$\ddot{x}=-\gamma\sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2}\dot{x}$$

But this isn't separable, making things very difficult. I do think that

$$s = \int_a^b|r'(t)|dt = \sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2}$$

or something very close to that. Yet I'm still not sure how I'm supposed to proceed. Maybe I'm getting something fundamentally wrong in the setup? Something else that's obvious? I'm fairly new to differential equations..

2. Sep 17, 2016

Hamal_Arietis

Along x-axis, we have: $m\ddot{x}=-\mu \dot{x}^2$
Then $-\mu(\frac{dx}{dt})^2=m\frac{dv}{dt}$
Then $-\mu v.dx=m. dv$
Solve this equation you will have this result

3. Sep 18, 2016

Dustgil

Thanks, this helped a lot!