Integrating with Velocity: A Question of Correctness

[Fallen Seraph]

Sorry for the undescriptive title, but I couldn't think of a better one.

My question is essentially this: is the following procedure correct?


a=-kx and we want v.

Multiply both sides by v to get:

$$\frac {d^2x} {dt^2} * \frac {dx} {dt} = -kx *\frac {dx} {dt}$$

Now, by dt, and some rearranging :

$$(\frac {d^2x} {dt^2} dt) * \frac {dx} {dt} = -kx *dx$$

And the step which I'm not convinced can be done:


$$( \int \frac {d^2x} {dt^2} dt) * \frac {dx} {dt} = -k \int x *dx$$


and so we end up with v^2 = -(kx^2)/2 +c which is but an errant 1/2 away from what I need.

I just don't trust integrating a wrt t, while having a v just sitting there as if it were a constant. The only reason that I haven't already disregarded it is that it checks out if you analyse the dimensions...

 

[StatusX]

No, that's not right. As you mention, you're treating v as a constant, and it isn't.

You started out right, you just need to rewrite:

$$\frac {d^2x} {dt^2} * \frac {dx} {dt}$$

as:

$$\frac{1}{2} \frac{d}{dt} \left( \left(\frac{dx}{dt} \right)^2 \right)$$,

then integrate.

 

[Fallen Seraph]

Wow, quite the nifty identity. Thanks a lot for that help.