How does multiplying by delta theta relate to spherical coordinates?

[tmbrwlf730]

Hi,

I'm reading through one of my books and it's explaining how a vector is eqaul to multiplying sin$\phi$ and $\Delta\vartheta$. the way it's written in the text is as followed,

|$\Delta$i| $\approx$ (sin$\phi$)$\Delta$$\vartheta$

I have never understood how things like this work. Could someone please explain to me why this is true and how it works? I included a picture of the figure it uses. Thank you.

 

[HallsofIvy]

You are talking about spherical coordinates, right? Also you are using "physics" notation which switches $\theta$ and $\phi$ from "mathematics" notation. In your notation, $\phi$ is the "longitude" and $\theta$ is the "co-latitude".

In these coordinates, for a fixed [itex]\theta[/tex] and r, sweeping $\phi$through 0 to $2\pi$, the point $(r, \phi, \theta)$ sweeps through a circle, but not a circle of radius r. If we draw the vertical axis, the line from (0, 0, 0) to the point $(r, \phi, \theta)$, and the line from that point perpendicular to the vertical axis, we get a right triangle with hypotenuse of length r and base angle of $\phi$. If we call the opposite side to that angle "x" then $sin(\phi)= x/r$ so $x= r sin(\phi)$. That will be the radius of the circle swept out.[/itex]