How to Calculate Maximum Error in Polar Coordinates

  • Thread starter Bryon
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In summary, the goal of the conversation was to find the maximum error in measuring the polar coordinates of a point (7.2, 2.5) given a possible error of 0.05 in each coordinate. The equations used were dx = deltax, dy = deltay, and dz = (partial diff of z with respect to x)dx + (partial diff of z with respect to y)dy. The conversation also touched on the possibility of using the equations x = rcos(theta), y = rsin(theta), and tan(theta) = y/x. Ultimately, the solution involved differentiating r = y/sin(theta) and theta = arcsin(y/r) in terms of the original Cartesian coordinates.
  • #1
Bryon
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Homework Statement



Rectangular to Polar Coordinates:
I am to find the approximate maximum error in measuring the polar coordinates of the point (7.2, 2.5). There is a possible error of 0.05 in each coordinate.

Homework Equations



dx = deltax
dy = deltay

dz = (partial diff of z with respect to x)dx + (partial diff of z with respect to y)dy

Not sure if I need the following:
x = rcos(theta)
y = rsin(theta)
tan(theta) = y/x


The Attempt at a Solution



I suspect that dx = dy = 0.05

I could set the rectangular coordinates to the polar conversions but I am not sure how that helps:

7.2 = rcos(theta)
2.5 = rsin(theta)

This is where I am held up at and not sure how to go about solving this one.
 
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  • #2
What are dr and [tex]d\theta[/tex] in terms of dx and dy?
 
  • #3
Would I be differentiating r = y/sin(theta) theta = arcin(y/r) and r = x/cos(theta) theta = arccos(x/r)?
 
  • #4
Bryon said:
Would I be differentiating r = y/sin(theta) theta = arcin(y/r) and r = x/cos(theta) theta = arccos(x/r)?

Close, but get the r's out of your [tex]\theta[/tex] formulas and the [tex]\theta[/tex]'s out of your r formulas.
 
  • #5
Ah, all I have to do is substitute r for r = y/sin(theta) in theta = arcsin(y/r) which would be
theta = arcsin(y/ysin(theta)) and so on! Thanks!
 
  • #6
No, you need the r's and [tex]\theta[/tex]'s entirely in terms of the original Cartesian coordinates.
 
  • #7
I figured it out! Thank you very much!
 

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