# Particle undergoes a displacement

1. Oct 31, 2006

### 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 lets 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.

Last edited: Oct 31, 2006
2. Oct 31, 2006

### Galileo

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.

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.

3. Oct 31, 2006

### 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 doesnt $$\vec{r} \rightarrow \infty$$?

Last edited: Oct 31, 2006
4. Oct 31, 2006

### Galileo

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