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Arc length question

  • #1

Homework Statement


1okol3.jpg



Homework Equations


l = int( sqrt( 1 + (dy/dx)^2) dx) from a to b


The Attempt at a Solution


So far I'm stuck at the R^2 thing. I know if it was just R it would mean the set of all real numbers, but I'm not sure as to what R^2 means and I don't know how to google that. Normally I would ignore it and just try to do the arc length of the 9x^2 = 4y^2 from 0 to 1 part, but that equation is just two straight opposite sloped lines right? So I know the R^2 has to mean something important, but I can't figure it out.
 

Answers and Replies

  • #2
Dick
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R^2 is just the xy plane. And it is a pretty easy problem. The curve is just two line segments as you said.
 
  • #3
Oh...okay I guess that makes sense, it means like the set of real numbers in two dimensions?

So if I just treat the curve as two line segments..

9x^2 - 4y^2 = 0
18x - 8y(y') = 0
-8y(y') = -18x
y' = 9x/4y
(y')^2 = 81x^2 / 16y^2
1 + (y')^2 = 1 + 81x^2 / 16y^2 = (16y^2 + 81x^2)/16y^2
= (16y^2 + 36y^2)/16y^2 = 52y^2 / 16y^2 = 13/4

l = integral sqrt(13)/2 dx from 0 to 1
= [sqrt(13)x/2] evaluated from 0 to 1
= sqrt(13)/2

So the arc length is sqrt(13)/2?
 
  • #4
Times 2 since in the second quadrant 0 <= y <= 1 is still true...so sqrt(13)?
 
  • #5
Dick
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618
Careful. You drew a picture of the curve, right? It isn't x that goes from 0 to 1. It's y. What's the x range? You've also got a second segment with negative x values.
 
  • #6
Oh so I do the integral from 0 to 2/3? And then get 2(sqrt(13)/3) because the "curve" is in quadrant 1 and 2?
 
  • #7
Dick
Science Advisor
Homework Helper
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618
Oh so I do the integral from 0 to 2/3? And then get 2(sqrt(13)/3) because the "curve" is in quadrant 1 and 2?
That's what I get. You can also check it by drawing the triangles and using the Pythagorean Theorem, right?
 
  • #8
Yeah I get the same answer, thank you!
 

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