Conditions of a quarter-circle running through the origin

[CSA]

Homework Statement

and

Homework Equations

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I was working on a math project involving certain simulations and a quarter circle. The equation of the quarter circle I used is y = r - SQRT(r^2 - x^2). Where r is the radius of a circle and x is greater than 0.

The purpose of this is to model is to get a concave up, quarter-circle that runs through both the origin and a chosen point (x_final, y_final) where both x and y are again positive. This point must not necessarily be at the end of the quarter circle (i.e. when the gradient is vertical).

When I was running through my simulations I noticed that this model broke down if x_final was smaller than y_final. Hence the question is that is x_final greater than y_final a necessary condition of the curve.

The Attempt at a Solution

It makes intuitive sense to me that this condition must be true because if I imagine all the possible circles of different radii that rest with their bottom on the origin and x-axis, a point where x_final is smaller than y_final can only lie on the upper left quarter of the circle. It almost feels like x_final = y_final is the limit to this model. However, I can't think of a proper mathematical way to show this.

 

[Mark44]

Homework Statement

and

Homework Equations

[/B]
I was working on a math project involving certain simulations and a quarter circle. The equation of the quarter circle I used is y = r - SQRT(r^2 - x^2). Where r is the radius of a circle and x is greater than 0.

The purpose of this is to model is to get a concave up, quarter-circle that runs through both the origin and a chosen point (x_final, y_final) where both x and y are again positive. This point must not necessarily be at the end of the quarter circle (i.e. when the gradient is vertical).

When I was running through my simulations I noticed that this model broke down if x_final was smaller than y_final. Hence the question is that is x_final greater than y_final a necessary condition of the curve.

The Attempt at a Solution

It makes intuitive sense to me that this condition must be true because if I imagine all the possible circles of different radii that rest with their bottom on the origin and x-axis, a point where x_final is smaller than y_final can only lie on the upper left quarter of the circle. It almost feels like x_final = y_final is the limit to this model. However, I can't think of a proper mathematical way to show this.

I'm assuming that your quarter circle looks like this:

Suppose that (x_f, y_f) is some point on the quarter circle other than (0, 0) or (r, r). A line segement that connects the endpoints of this figure is part of a line whose equation is y = x. Every point below and to the right of this line segment satisfies the inequality $y_f < x_f$, so what you're concerned about won't happen, again assuming that we're both talking about the same quarter circle.

 

[tnich]

A couple of things puzzle me about this problem. You say that

The purpose of this is to model is to get a concave up, quarter-circle that runs through both the origin and a chosen point (x_final, y_final) where both x and y are again positive. This point must not necessarily be at the end of the quarter circle (i.e. when the gradient is vertical).

But who choses the point $(x_{final},y_{final})$? You? Your teacher?
One thing that is clear is that you cannot choose $(x_{final},y_{final})$ after you have already defined the function $y=f(x)$ that represents your quarter circle. You can only choose $x_{final}$, then $y_{final}$ must equal $f(x_{final})$.
To read your post, it sounds like you first need to choose the point $(x_{final},y_{final})$ first and then construct the equation of the quarter circle that contains that point.
Are your supposed to look at different quarter circles that contain the point?
I don't understand what problem you are trying to solve. Maybe the text of the original problem statement as given to you would help.

 

[CSA]

I'm assuming that your quarter circle looks like this:
View attachment 222041
Suppose that (x_f, y_f) is some point on the quarter circle other than (0, 0) or (r, r). A line segement that connects the endpoints of this figure is part of a line whose equation is y = x. Every point below and to the right of this line segment satisfies the inequality $y_f < x_f$, so what you're concerned about won't happen, again assuming that we're both talking about the same quarter circle.

Yes, that makes sense. Thank you!

 

[CSA]

A couple of things puzzle me about this problem. You say that

To read your post, it sounds like you first need to choose the point $(x_{final},y_{final})$ first and then construct the equation of the quarter circle that contains that point.

Yes, the point is chosen first and then the radius of the circle is found based on that point

 

[Mark44]

Yes, the point is chosen first and then the radius of the circle is found based on that point

Since the "bottom" of the circle is at the origin, the center is necessarily at (0, r). The equation of the full circle is $x^2 + (y - r)^2 = r^2$.

Substitute the point $(x_f, y_f)$ into this equation and solve for r. Again, since this is the lower right quarter of the circle, it must be true that for any point (x, y) on the arc, $y \le x$, with equality only at the origin and the upper point on the arc.

 

[CSA]

Since the "bottom" of the circle is at the origin, the center is necessarily at (0, r). The equation of the full circle is $x^2 + (y - r)^2 = r^2$.

Substitute the point $(x_f, y_f)$ into this equation and solve for r. Again, since this is the lower right quarter of the circle, it must be true that for any point (x, y) on the arc, $y \le x$, with equality only at the origin and the upper point on the arc.

Would this make sense then? Using the equation for the bottom quarter circle. Assuming,
x_f < y_f

x_f < r - sqrt(r² - x_f²)

r - x_f > sqrt(r² - x_f²)

square both sides, since both sides are positive and also assuming x_f is not zero, it eventually reduces to

x_f > r

Therefore, by contradiction, we can say that y_f ≤ x_f.

 

[Mark44]

Would this make sense then? Using the equation for the bottom quarter circle. Assuming,
x_f < y_f

No it doesn't make sense. On the lower right quarter circle, $x_f > y_f$, not the other way around as you have it.
Think about it -- the points (0, 0) and (r, r) are on the line y = x. All points above and to the left of this line satisfy y > x. All points below and to the right of this line satisfy y < x, or equivalently, x > y. I said essentially this in posts #2 and #6, but you didn't seem to catch on.

Since you started with an invalid assumption, I didn't read any further.

x_f < r - sqrt(r² - x_f²)

r - x_f > sqrt(r² - x_f²)

square both sides, since both sides are positive and also assuming x_f is not zero, it eventually reduces to

x_f > r

Therefore, by contradiction, we can say that y_f ≤ x_f.

 

[CSA]

No it doesn't make sense. On the lower right quarter circle, $x_f > y_f$, not the other way around as you have it.
Think about it -- the points (0, 0) and (r, r) are on the line y = x. All points above and to the left of this line satisfy y > x. All points below and to the right of this line satisfy y < x, or equivalently, x > y. I said essentially this in posts #2 and #6, but you didn't seem to catch on.

Since you started with an invalid assumption, I didn't read any further.

Oh, I think I did understand what you meant. But, I just meant to prove by contradiction that the initial assumption, in my first line, is invalid by manipulating the inequalities. But, I guess I will just stick to your way of explaining it then. Thanks so much for helping!

 

[SammyS]

... But, I just meant to prove by contradiction that the initial assumption, in my first line, is invalid by manipulating the inequalities.
...

Now that you point out that you were doing a proof by contradiction, what you did in post #7 makes sense.

An appropriate additional statement would have made it a valid argument. (I have added these in red below.)

Would this make sense then? Using the equation for the bottom quarter circle.
Proof by contradiction that y_f ≤ x_f.
Assume,
x_f < y_f .

x_f < r - sqrt(r² - x_f²)

r - x_f > sqrt(r² - x_f²)

square both sides, since both sides are positive and also assuming x_f is not zero, it eventually reduces to

x_f > r, which is a contradiction.

Therefore, by contradiction, we can say that y_f ≤ x_f.

 

[Mark44]

Now that you point out that you were doing a proof by contradiction, what you did in post #7 makes sense.

Yes, that would have been helpful, but a proof by contradiction is more work than just a direct argument. Since the circle was defined as having its lowest point at (0, 0) and with radius r, it's clear that the circle also passes through the point (r, r). This point and the origin determine the line whose equation is y = x. All points below and to the right of this line satisfy the inequality y < x, and this includes all of the points on the lower right quarter circle.

 

[SammyS]

Yes, that would have been helpful, but a proof by contradiction is more work than just a direct argument.

Yes!
I agree.

 

[CSA]

Alright, thanks again!