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

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In summary: Is the line like that and the radius is 2?If that is the case, the line would be x=2 and the value of 0 on the x-axis would be 2.
  • #1
eddybob123
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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
 
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  • #2
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
 
  • #3
Which is the wrong answer when I checked so I don't know I'm sorry.
 
  • #4
Sorry I'm not understanding

eddybob123 said:
then what is the value of 0 on the x-axis to the line mark
 
  • #5
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,

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

Which gives,

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

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

Multiply the results by 2 for a radius of 2.
 
  • #6
I think you may have integrated incorrectly. I got a one half multiplied with the first term in that equation.
 
  • #7
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.
 
  • #8
Delong said:
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! :smile:
 
  • #9
eddybob123 said:
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 [itex](x- r)^2+ (y- r)^2= r^2[/itex]. If r= 2 then that is [itex](x- 2)^2+ (y- 2)^2= 4[/itex]. 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.
 
  • #10
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.
 
  • #11
Delong said:
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 [itex]x=0.808[/itex] 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.
 
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  • #12
arg I don't know how to show my work online but I will found out how later. After I get done some work.
 
  • #13
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 [tex]\frac{a}{b}+c[/tex] but you could mean [tex]\frac{a}{b+c}[/tex] in which case you should be writing a/(b+c).
 
  • #14
find* out how to do it later.
 
  • #15
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.
 
  • #16
HallsofIvy said:
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 [itex](x- r)^2+ (y- r)^2= r^2[/itex]. If r= 2 then that is [itex](x- 2)^2+ (y- 2)^2= 4[/itex]. 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))
 
  • #17
Is it like this?

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

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 [tex]y=\sqrt{4-x^2}[/tex] than what is in the picture (typo).
 
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  • #18
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[tex]\theta[/tex], and dx=2cos[tex]\theta[/tex]d[tex]\theta[/tex].

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

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

-2[tex]\theta[/tex]+2sin[tex]\theta[/tex]*cos[tex]\theta[/tex].

I remember that 2sin[tex]\theta[/tex]=x so using a triangle I calculate that [tex]\theta[/tex]= arcsin(x/2) and cos[tex]\theta[/tex]= (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*([tex]\pi[/tex]/2)

Now to solve for n I graph the expression on the left and see where it intersects with y=[tex]\pi[/tex] 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.
 
  • #19
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.
 
  • #20
i used another method that doesn't involve pi, but i still don't understand it.
 
  • #21
can you please explain how you got your answer?
 
  • #22
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?
 
  • #23
well, i think it would be possible, because the curve is defined as sqrt(4-x^2).
 

1. What does the value of 0 represent on the X-Axis for the Circle Problem (r=2)?

The value of 0 on the X-Axis represents the center of the circle. In this case, with a radius of 2, the circle's center is located at (0,0) on the coordinate plane.

2. Why is the value of 0 important in the Circle Problem (r=2)?

The value of 0 is important because it helps determine the symmetry and shape of the circle. With a radius of 2, the circle is perfectly symmetrical around the origin (0,0).

3. How does the value of 0 affect the circumference of the circle in the Circle Problem (r=2)?

The value of 0 does not directly affect the circumference of the circle in the Circle Problem (r=2). The circumference is determined by the formula C = 2πr, where r is the radius. However, the center of the circle being located at (0,0) does affect how the circumference is measured from the center.

4. What is the significance of the value of 0 in relation to the circle's diameter in the Circle Problem (r=2)?

The value of 0 is significant in determining the circle's diameter. The diameter of a circle is equal to twice the radius, and with a radius of 2, the diameter of the circle is 4. This means that the diameter passes through the center of the circle at (0,0).

5. Can the value of 0 on the X-Axis change in the Circle Problem (r=2)?

No, the value of 0 on the X-Axis cannot change in the Circle Problem (r=2) as it represents the fixed center of the circle. However, the radius and circumference of the circle can change depending on the given parameters.

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