r(t) is the position function of a particle moving in 2 or 3 space.

**I don't know how to do integral symbol, so I'll just denote that symbol as integ(f(x))**

integ( ||dr/dt|| ) from t0 to t1

I know dr/dt = v(t) which is the velocity function, and the norm of v(t) is the magnitude of the vector, aka, the numerical speed of the particle at time t, but when we integrate the speed, wouldn't we just end up with the position function again?

That is what my intuition is telling me, but my intuition has failed me before, so I would rather be shown why my thinking is wrong, if it even is. Thanks.

**I don't know how to do integral symbol, so I'll just denote that symbol as integ(f(x))**

integ( ||dr/dt|| ) from t0 to t1

I know dr/dt = v(t) which is the velocity function, and the norm of v(t) is the magnitude of the vector, aka, the numerical speed of the particle at time t, but when we integrate the speed, wouldn't we just end up with the position function again?

That is what my intuition is telling me, but my intuition has failed me before, so I would rather be shown why my thinking is wrong, if it even is. Thanks.

If I were to assume some random value for || v(t) || to be 10 m/s, upon integration I would end up with 10t, where t is in seconds, and 10 is still in meters per second. So by simple plug and chug, where t = 1...

10 m/s * 1 s = 10m therefore proving that the above statement is the displacement of the particle after t seconds.

Right?

The integral of dr/dt is of course (dr/dt)dt and so is just dr.

dr is the change in position so you are correct.

The indefinite integral of the velocity is the position function (of course were you to compute it, you would get that + C which would be the initial position).

The definite integral of the velocity is the displacement (distance from start to end).

Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.