Understanding Velocity: ds/dt and dr/dt in Particle Motion

[Adjoint]

This is one very basic question. But I just need to confirm if I understood it right.

Suppose a particle moves along a curve and crosses Δs path in Δt time. Then we can say the velocity of the particle is \vec{v} = ds/dt \hat{u}
Where \hat{u} is tangent to the curve.

Also if the same particle, as it crosses Δs, goes through a displacement Δ\vec{r} in the same time interval Δt we say \vec{v} = d\vec{r}/dt

Is the V's calculated above are same (ie equal)?

I know the question is silly, but at present this forum is the only place for me to get help.

 

[verty]

This is one very basic question. But I just need to confirm if I understood it right.

Suppose a particle moves along a curve and crosses Δs path in Δt time. Then we can say the velocity of the particle is \vec{v} = ds/dt \hat{u}
Where \hat{u} is tangent to the curve.

Also if the same particle, as it crosses Δs, goes through a displacement Δ\vec{r} in the same time interval Δt we say \vec{v} = d\vec{r}/dt

Is the V's calculated above are same (ie equal)?

I know the question is silly, but at present this forum is the only place for me to get help.

The are the same. I think I know what you are thinking, that perhaps the second one is slightly smaller because the path may be curved, but any difference between $\Delta{s}$ and $|\Delta\vec{r}|$ goes to 0 in the limit, and any angle between $\Delta\vec{r}$ and the path goes to 0 in the limit as well. So in the limit they are perfectly the same, $d\vec{r} \over dt$ is exactly the vector rate of change of position which is velocity.

 

[Adjoint]

any difference between $\Delta{s}$ and $|\Delta\vec{r}|$ goes to 0 in the limit, and any angle between $\Delta\vec{r}$ and the path goes to 0 in the limit as well. So in the limit they are perfectly the same

Thanks.