Value of 0 on X-Axis for Circle Problem (r=2)

[eddybob123]

if you have a quadrant sitting on top of the x-axis and on the right of the y axis, when you draw a line perpendicular to the x-axis that splits the circle into two equal parts, then what is the value of 0 on the x-axis to the line mark when r=2? too bad i don't have a diagram to show you

 

[Delong]

I can't find an exact value but I approximated it to be about negative .027. I used integration to try to find the value needed for the left part to be equal in area to the right part and I arrived at this equation where the value cannot be calculated explicitly. So I graphed the equation instead to try to find the intersection of the graphs and use the x-value of that graph

 

[Delong]

Which is the wrong answer when I checked so I don't know I'm sorry.

 

[Mentallic]

Sorry I'm not understanding

then what is the value of 0 on the x-axis to the line mark

 

[uart]

I'm pretty sure he's either asking for the value of "x" of the vertical line that splits the area of a quadrant in two, or he's asking for the "y" value where the that vertical line intersects the circle.

Best to do the calculation for a unit circle and scale the results by the radius as needed.

To find "x" solve,

\int_0^x \, \sqrt{1-u^2} \, du = \pi/8

Which gives,

x \, \sqrt{1-x^2} + \sin^{-1}(x) = \pi/4

Solving numerically (to 3 dp) gives x \simeq 0.404 and the corresponding "y" value is y = \sqrt{(1-x^2)} \simeq 0.915.

Multiply the results by 2 for a radius of 2.

 

[Delong]

I think you may have integrated incorrectly. I got a one half multiplied with the first term in that equation.

 

[Delong]

So I checked my calculations and I found out that I accidently used degrees rather than radians. If I used radians and solved it I find an answer around .8318 which is similar to what uart obtained.

 

[uart]

I think you may have integrated incorrectly. I got a one half multiplied with the first term in that equation.

And one half multiplied by the second term, hence the reason that I multiplied both sides of the equation by two. Look at the RHS of those two equations!

 

[HallsofIvy]

if you have a quadrant sitting on top of the x-axis and on the right of the y axis, when you draw a line perpendicular to the x-axis that splits the circle into two equal parts, then what is the value of 0 on the x-axis to the line mark when r=2? too bad i don't have a diagram to show you

I assume you mean a circle "sitting on top of the x-axis and on the right of the y axis". You don't say how far to the right but I am going to assume you mean that the circle is tangent to both axes. That is, the circle is given by (x- r)^2+ (y- r)^2= r^2. If r= 2 then that is (x- 2)^2+ (y- 2)^2= 4. The "line perpendicular to the x-axis that splits the circle into two equal parts" would be the diameter that passes through the center, (2, 2) to the point (2, 0) on the x-axis.

If I have not correctly understood your question, please clarify.

 

[Delong]

He might have mis-stated. If it were just a circle it would be too easy.

Uart- well I did it slightly different from you in that I started with a circle radius 2 and solved for pi/4 instead of solving for the unit circle and multiplying by two. But I'm thinking we should get the same results anyway and since I got something like .8318 I'm guessing we did.

 

[uart]

He might have mis-stated. If it were just a circle it would be too easy.

Uart- well I did it slightly different from you in that I started with a circle radius 2 and solved for pi/4 instead of solving for the unit circle and multiplying by two. But I'm thinking we should get the same results anyway and since I got something like .8318 I'm guessing we did.

You haven't shown your working, but for a radius of two it's x=0.808 to three decimal places. So something looks wrong with your answer.

For the unit circle the answer to 14 significant figures is x = 0.40397275329952, y = 0.91477101757304. Double these for the R=2 circle.

 

[Delong]

arg I don't know how to show my work online but I will found out how later. After I get done some work.

 

[Mentallic]

Just as a heads up, when you show your work, a lot of beginners make the mistake of thinking we know exactly what you're talking about. We don't. You need to use brackets whenever applicable. For example, if you write a/b+c we would interpret this as \frac{a}{b}+c but you could mean \frac{a}{b+c} in which case you should be writing a/(b+c).

 

[Delong]

find* out how to do it later.

 

[eddybob123]

i am going to say this one more time. the semicircle is plotted by sqrt(4-x^2). we are only looking at the positive section, which is in the shape of a quadrant.the 'line' is parallel to the y axis, and splits the quadrant into two equal parts. i wish i had a diagram to show you.

 

[eddybob123]

I assume you mean a circle "sitting on top of the x-axis and on the right of the y axis". You don't say how far to the right but I am going to assume you mean that the circle is tangent to both axes. That is, the circle is given by (x- r)^2+ (y- r)^2= r^2. If r= 2 then that is (x- 2)^2+ (y- 2)^2= 4. The "line perpendicular to the x-axis that splits the circle into two equal parts" would be the diameter that passes through the center, (2, 2) to the point (2, 0) on the x-axis.

If I have not correctly understood your question, please clarify.

the quadrant is touching the y-axis to the right and touching the x-axis on top. (sqrt(4-x^2))

 

[Mentallic]

Is it like this?

[PLAIN]http://img580.imageshack.us/img580/3204/graphcircle.png

Where area A = area B and we want to find the coordinates at which that horizontal line touches the circle?

edit: and the graph is instead y=\sqrt{4-x^2} than what is in the picture (typo).

 

[Delong]

No dude the line is parallel to the y-axis.

Okay here I will try to show my work. So I'm trying to find an x-value such that the area under the curve to the left of that x value and to the right of zero is equal to the area under the curve to the right of that x value and to the left of where the curve touches the x-axis. The curve, as Eddybob has stated, is modeled by the equation:

(4-x^2)^(1/2) from x=0 to x=2 for the top right quadrant. To find the x-value I integrate the equation from 0 to x and set it equal to the definite integral from x to 2.

(integrate, 0..n): [(4-x^2)^(1/2)] dx = (integrate, n..2): [(4-x^2)^(1/2)] dx

the integral of the equation requires integration by parts where du is dx, u is x, and v is (4-x^2)^(1/2). That means dv is -xdx/(4-x^2)^(1/2). So the equation on the left side now looks like
x(4-x^2)^(1/2)- (integrate): [-(x^2)/(4-x^2)^(1/2)]dx
with the limits as 0 to n on the left side and n to 2 on the right side.

Everything in bold is for the second integral

Now to compute the other integral I use trigonometric substitution where I set x equal to 2 sin\theta, and dx=2cos\thetad\theta.

That now makes the integrand: -[4(sin\theta)^2)/(4-4[sin\theta]^2)^[1/2])* 2cos\thetad\theta.

the denominator is equal to 2cos\theta and so I cancel that on the top and bottom as well. That leaves me with
-4(sin\theta)^2 which equals to
-4/2*(1-cos2\theta). When I integrate that I have
-2\theta+sin2\theta. I turn sin(2\theta) into its double angle identity to get

-2\theta+2sin\theta*cos\theta.

I remember that 2sin\theta=x so using a triangle I calculate that \theta= arcsin(x/2) and cos\theta= (4-x^2)^(1/2)/2.

That expression above becomes: -2arcsin(x/2)+x(4-x^2)^(1/2)/2
OKAY NOW that was all just the second integrand.

Putting that into the equation I got from above with integration by parts I get:

x(4-x^2)^(1/2) - [-2arcsin(x/2)+x(4-x^2)^(1/2)/2]
with limits 0 to n on the left side and limits n to 2 on the right side.

Simplifying the expression I get:

x(4-x^2)^(1/2)/2 +2arcsin(x/2). Now if I just plug in the limits and solve I get:

n(4-n^2)^(1/2)/2 +2arcsin(n/2) - 0 = 2arcsin(1)-n(4-n^2)^(1/2)/2-2arcsin(n/2)

when I simplify I get:

n(4-n^2)^(1/2)+4arcsin(n/2)=2*(\pi/2)

Now to solve for n I graph the expression on the left and see where it intersects with y=\pi and then make it negative to answer the question. Using my calculator I get the value of .80794551 (by using the intersection program thingy). The answer should be -.80794551. So I hope that was clear thanks.

Edit: So I think I got the same answer as you uart I might not have computed closely enough on my calculator. Also I think Eddybob may be asking for the x-value if we take n to be the y-axis but I could be wrong.

 

[uart]

Yeah that answer is now correct Delong. Notice that I made use of the fact that we already know what half the area of a quadrant is, so you don't really need two integrals.

 

[eddybob123]

i used another method that doesn't involve pi, but i still don't understand it.

 

[eddybob123]

can you please explain how you got your answer?

 

[Delong]

I'd like to but I kind of just did. I don't really know how to explain it more thoroughly. I can summarize by saying that I just did integration by parts twice and stuff. THen I got this equation with the variable and using the graph to find the intersection to find the answer and stuff and then making it negative. bla bbla bla

I don't know how you can solve it without pi. Any particular part you have trouble understanding?

 

[eddybob123]

well, i think it would be possible, because the curve is defined as sqrt(4-x^2).