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jeff1evesque
Jul2-09, 12:42 PM
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
Jul2-09, 01:30 PM
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
Jul2-09, 01:35 PM
That's great, thanks a lot :smile:.


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}.