How to Define Differential Length Vectors

**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.
$LaTeX Code: \\frac{dy}{dx} = \\frac{1}{2}\\frac{-4x}{\\sqrt{a^2-x^2}} \\Rightarrow dy = \\frac{-2x}{y}dx \\Rightarrow \\vec{dl} = dx\\hat{x} - \\frac{2x}{y}dx\\hat{y}$

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

Thanks,

JL

**HallsofIvy** · Jul2-09, 01:30 PM

Originally Posted by 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.
$LaTeX Code: \\frac{dy}{dx} = \\frac{1}{2}\\frac{-4x}{\\sqrt{a^2-x^2}} \\Rightarrow dy = \\frac{-2x}{y}dx \\Rightarrow \\vec{dl} = dx\\hat{x} - \\frac{2x}{y}dx\\hat{y}$

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

Thanks,

JL

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

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

**jeff1evesque** · Jul2-09, 01:35 PM

That's great, thanks a lot

.

Jeffrey

Originally Posted by HallsofIvy

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

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