Particle undergoes a displacement

[sherlockjones]

Lets say we have \vec{r} = x\vec{i} + y\vec{j} + z\vec{k}. If the particle undergoes a displacement \Delta \vec{r} in time \Delta t then we know that \Delta r \doteq \frac{d\vec{r}}{dt} \Delta t = \vec{v}\Delta t.

How is \vec{v} \Delta t tangent to the particles trajectory, when we take \Delta t \rightarrow 0? Wouldn't the expression become 0? I can see how \Delta \vec{r} \rightarrow 0 as \Delta t \rightarrow 0.

Also let's say we have the following:

\vect{r} = A(e^{\alpha t}\vec{i} + e^{-\alpha t}\vec{j})

I know that \vec{v} = A(\alpha e^{\alpha t}\vec{i} + e^{-\alpha t}\vec{j})

As t\rightarrow \infty, e^{\alpha t} \rightarrow \infty and e^{-\alpha t} \rightarrow 0. From this how do we come to the conclusion that \vec{r} \rightarrow Ae^{\alpha t}\vec{i}? I thought that \vec{r} \rightarrow \infty.

Similarily, \vec{v} \rightarrow \alpha Ae^{\alpha t}\vec{i} when it seems like it should approach \infty. Perhaps it has to do something with the x-components?

\alpha is a constant.

Thanks for your help

 

[Galileo]

How is \vect{v} \Delta t tangent to the particles trajectory, when we take \Delta t \rightarrow 0? Wouldn't the expression become 0? I can see how \Delta \vect{r} \rightarrow 0 as \Delta t \rightarrow 0.

The expression v\Delta t would approach zero. It's the vector v=\lim_{t\to 0}\Delta r/Delta t that is tangential to the particle's trajectory.

Also let's say we have the following:

\vect{r} = A(e^{\alpha t}\vect{i} + e^{-\alpha t}\vect{j})

I know that \vect{v} = A(\alpha e^{\alpha t}\vect{i} + e^{-\alpha t}\vect{j})

As t\rightarrow \infty, e^{\alpha t} \rightarrow \infty and e^{-\alpha t} \rightarrow 0. From this how do we come to the conclusion that \vect{r} \rightarrow Ae^{\alpha t}\vect{i}? I thought that \vect{r} \rightarrow \infty.

Similarily, \vect{v} \rightarrow \alpha Ae^{\alpha t}\vect{i} when it seems like it should approach \infty. Perhaps it has to do something with the x-components?

Thanks for your help

I`m not sure what you want to do here. There's an error in the expression for v (not sure if that's a typo). Then (assuming alpha>0) you take the limit as t-> infinity (what for?).

It's true that r-> infinity if t-> infinity, but If you want to approximate the trajectory for large t, you can drop the second term since it is vanishingly small in comparison with the first term.

 

[sherlockjones]

They wanted me to sketch the trajectory. But \vec{r} = Ae^{\alpha t }\vec{i}} + Ae^{-\alpha t}\vec{j}?
Then the second term approaches 0 and the first term approaches \infty. So then doesn't \vec{r} \rightarrow \infty?

 

[Galileo]

Yes, but that doesn't help in sketching the trajectory. You want the asymptotic behaviour of r, not the limit as t-> infinity.