- #1

Taufik

Now, what will the trajectory be near that point? Since dy/dx = 0/0, which is not defined, therefore the slope at that point is not defined. So, will it be something like a sharp edge or a smooth curve?

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- #1

Taufik

Now, what will the trajectory be near that point? Since dy/dx = 0/0, which is not defined, therefore the slope at that point is not defined. So, will it be something like a sharp edge or a smooth curve?

- #2

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That is not right. The rule that ##\frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}## only applies at points where ##\frac{dx}{dt}\neq 0##, and hence not at that point.Since dy/dx = 0/0

In fact, the particle's path may have

In a real physical system, there will always be a specific velocity, and hence direction, at any point in time, but not necessarily at a point in space.

Another way to see the problems of asking for direction at a point in space is to consider a particle mapping out a figure eight. What is the direction at the crossing point?

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Taufik

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Then it's stopped.Then, dx/dt =0 & dy/dt =0 at that point.

It's still stopped.Now, what will the trajectory be near that point?

The fact that it's stopped doesn't tell you how it was moving before it stopped, nor how it will move at some future time.

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$$\ddot{\vec{x}}=-\omega^2 \vec{x}.$$

Take the initial condition ##\vec{v}(0)=0## and ##\vec{x}(0)=\vec{x}_0 \neq 0## then the solution obviously is

$$\vec{x}(t)=\vec{x}_0 \cos(\omega t).$$

Everything is smooth as a function of ##t##. As a function ##y=y(x)## there's of course a singularity since ##t=0## (by construction) is the turning point of the linear motion along the direction ##\vec{x}_0##.

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