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.