How to Define Differential Length Vectors

[jeff1evesque]

Hello, I am looking at some notes and cannot understand the following example:


Suppose x² + y² = a²
Note: (Optional) use the equation of the curve to convert all vector components to the same differential, e.g.
$$\frac{dy}{dx} = \frac{1}{2}\frac{-4x}{\sqrt{a^2-x^2}} \Rightarrow dy = \frac{-2x}{y}dx \Rightarrow \vec{dl} = dx\hat{x} - \frac{2x}{y}dx\hat{y}$$

Question: I don't understand the second arrow which leads to the following conclusion:
$$\vec{dl} = dx\hat{x} - \frac{2x}{y}dx\hat{y}$$
Why wouldn't there be a dy?

Thanks,

JL

 

[HallsofIvy]

Hello, I am looking at some notes and cannot understand the following example:


Suppose x² + y² = a²
Note: (Optional) use the equation of the curve to convert all vector components to the same differential, e.g.
$$\frac{dy}{dx} = \frac{1}{2}\frac{-4x}{\sqrt{a^2-x^2}} \Rightarrow dy = \frac{-2x}{y}dx \Rightarrow \vec{dl} = dx\hat{x} - \frac{2x}{y}dx\hat{y}$$

Question: I don't understand the second arrow which leads to the following conclusion:
$$\vec{dl} = dx\hat{x} - \frac{2x}{y}dx\hat{y}$$
Why wouldn't there be a dy?

Thanks,

JL

There is! That is simply the standard equation for a tangent vector,
$$\vec{dl}= dx\hat{x}+ dy\hat{y}$$
with "dy" replaced using the equation just before the second arrow:
$$dy= \frac{-2x}{y}dx$$

$$\vec{dl}= dx\hat{x}+ dy\hat{y}= dx\hat{x}- \frac{2x}{y}dx\hat{y}$$.

 

[jeff1evesque]

That's great, thanks a lot .


Jeffrey

There is! That is simply the standard equation for a tangent vector,
$$\vec{dl}= dx\hat{x}+ dy\hat{y}$$
with "dy" replaced using the equation just before the second arrow:
$$dy= \frac{-2x}{y}dx$$

$$\vec{dl}= dx\hat{x}+ dy\hat{y}= dx\hat{x}- \frac{2x}{y}dx\hat{y}$$.