Solving Arc Length of 9x^2 = 4y^2 from 0 to 1

[downwithsocks]

Homework Statement

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.

 

[Dick]

R^2 is just the xy plane. And it is a pretty easy problem. The curve is just two line segments as you said.

 

[downwithsocks]

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?

 

[downwithsocks]

Times 2 since in the second quadrant 0 <= y <= 1 is still true...so sqrt(13)?

 

[Dick]

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.

 

[downwithsocks]

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?

 

[Dick]

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?

 

[downwithsocks]

Yeah I get the same answer, thank you!