How Does Setting x=a Identify Minima or Maxima in Circular Equations?

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In summary: F(x,y)=-dy##.In summary, the equation for a circle is given, and its derivative is found. The max/min points are found when x=a.
  • #1
notgoodatphysics
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We have the equation for a circle, and its derivative:

$$(y-a)^2 + (x-a)^2 = r^2$$
$$\frac{dy}{dx} = \frac{a-x}{(r^{2}-(x-a)^2)^{1/2}} = 0$$

So ##x = a## then he subs it into the original equation to get the max/min.

Why does ##x = a## give the points of minima/ maxima if we didn’t know it was a circle in this case? Is there a specific rule? It’s not really explained well in the book.

It’s from calculus made easy and he says “Since no value whatever of x will make the denominator infinite, the only condition to give zero is x = a”

I’m pretty confused what is meant by that statement.
 
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  • #2
Isn't the slope at a max/min point 0? From there you can see that x must equal a to get a zero slope.
 
  • #3
jedishrfu said:
Isn't the slope at a max/min point 0? From there you can see that x must equal a to get a zero slope.
Right, but you’d know this without having to find the derivative. I guess if you didn’t know it was a circle there’s an explanation to “Since no value whatever of x will make the denominator infinite, the only condition to give zero is x = a”?
 
  • #4
notgoodatphysics said:
Right, but you’d know this without having to find the derivative. I guess if you didn’t know it was a circle there’s an explanation to “Since no value whatever of x will make the denominator infinite, the only condition to give zero is x = a”?
If the centre of the circle is the point ##(a,b)##, then ##y## has its maximum and minimum values above and below the centre. The maximum and minimum ##y## are the points ##(a, b \pm r)##. Where ##r## is the radius.

You don't need calculus to see that.
 
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  • #5
I understand that bit. But it’s presented in the book as an example, and he mentions specifically that you have no idea it’s circle, so suppose you were presented with the original equation and asked to find the minimum/ maximum.

You’d need find the derivative if I’m not mistaken? My confusion comes about when the derivative is turned into x = a. To paraphrase him - as there is no value of x that makes the denominator infinite. So the conclusion is x = a.
 
  • #6
notgoodatphysics said:
I understand that bit. But it’s presented in the book as an example, and he mentions specifically that you have no idea it’s circle, so suppose you were presented with the original equation and asked to find the minimum/ maximum.

You’d need find the derivative if I’m not mistaken?
Not necessarily. You know, for example, that ##(x-a)^2## has a minimum at ##x =a## without using calculus.
notgoodatphysics said:
My confusion comes about when the derivative is turned into x = a.
Do you mean when the derivative is evaluated at ##x =a##?
notgoodatphysics said:
To paraphrase him - as there is no value of x that makes the denominator infinite. So the conclusion is x = a.
With any curve you must do some work to establish what type of curve you are dealing with. Take ##y = x^3##. The derivative is zero at ##x =0## and the behaviour of the curve as ##x \to \pm \infty## tells you this must be an infection point.

The equation of a circle is not a function, so you need some care in working with its derivatives. Technically, you might want to confirm that ##y## is bounded, as are the allowable values of ##x##.

I think it's better to figure this out for yourself than to try to decrypt some obscure remarks from a professor.
 
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  • #7
The derivative vanishes if the numerator is 0 and the denominator isn't or if the denominator goes to infinity faster than the numerator does. The author is pointing out in this case that only the first case matters.
 
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  • #8
PS you do know that's a circle, because a circle of radius r is defined as the set of points precisely a distance ##r## from a given point. That, by inspection, is the equation you are given - with the centre at ##(a,a)##.
 
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  • #9
If you follow the logic of not knowing it's a circle, then it makes sense to take the derivative of the function and determine what values give you a zero slope.

In this case, we can see when x=a, the derivative will be zero.

The denominator comment reminds you that you're not dividing by zero, so things are good to say x=a is where the maximum/minimum is.
 
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  • #10
notgoodatphysics said:
... so suppose you were presented with the original equation and asked to find the minimum/ maximum.
I suppose you know how to do that if the functions is explicitly given as ##y=f(x)##. In your case it is implicitly given by an equation say ##F(x,y)=0##. To find the min/max of ##y## you do the same as in the explicit case except that the derivative now is calculated implicitly form the equation i.e. ##\frac{dy}{dx}=-\frac{F_x}{F_y}##.
 
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  • #11
notgoodatphysics said:
We have the equation for a circle, and its derivative:

$$(y-a)^2 + (x-a)^2 = r^2$$
$$\frac{dy}{dx} = \frac{a-x}{(r^{2}-(x-a)^2)^{1/2}} = 0$$
Your first equation is of a circle whose center is at the point (a, a) and whose radius is r. A more general equation would put the center at an arbitrary point (a, b), say.
I have a couple of problems with your second equation, though.
First, this says that the derivative is zero. That's not true in general and is true only for two points: (a, a + r) and (a, a - r).

Second, the derivative should have two values for a given value of x.

Starting with the equation ##(x - a)^2 + (y - a)^2 = r^2##, implicit differentiation gives ##2(x - a) + 2(y - a)\frac{dy}{dx} = 0##.

Solving for dy/dx gives $$\frac{dy}{dx} = \frac{a - x}{y - a} = \frac{a - x}{\pm\sqrt{r^2 - (x - a)^2}}$$

Note the plus/minus I've added.
 
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