Second Derivative of Circle Not a Constant?

In summary: The curvature of a circle is constant (1 over its radius) and the curvature is related to the second derivative but not equal to it.
  • #1
Joseph Nechleba
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Using the standard equation of a circle x^2 + y^2 = r^2, I took the first and second derivatives and obtained -x/y and -r^2/y^3 , respectively. I understand that the slope is going to be different at each point along the circle, but what does not make sense to me is that the rate of change of the slope is dependent on the y value of a point along the circle. For some reason, I want to believe that, conceptually, the second derivative of a circle is a constant, which produces the "circle" shape. Can someone please clear my misunderstanding?
 
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  • #2
Only polynomials of degree 2 or lower can have a constant second derivative.
EDIT: I'm talking about polynomials which only consist of the x variable.
 
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  • #3
Joseph Nechleba said:
Using the standard equation of a circle x^2 + y^2 = r^2, I took the first and second derivatives and obtained -x/y and -r^2/y^3 , respectively.

You take the derivative of what with respect to what? What function are you trying to find the derivative of?

the second derivative of a circle is a constant

What does "derivative of a circle" mean? You can only take derivatives of functions.
 
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  • #4
A circle is easiest given by x = r⋅cos(t), y = r⋅sin(t), t∈[0, 2π). Then [itex]\dot{x}= -r\cdot \sin(t) = -y [/itex] and [itex]\dot{y}= r\cdot \cos(t) = x [/itex]. Continuing, we get [itex] \ddot{x}=-r\cdot\cos(t)=-x[/itex] and [itex] \ddot{y}=-r\cdot\sin(t)=-y[/itex].
 
  • #5
Joseph Nechleba said:
Using the standard equation of a circle x^2 + y^2 = r^2, I took the first and second derivatives and obtained -x/y and -r^2/y^3 , respectively. I understand that the slope is going to be different at each point along the circle, but what does not make sense to me is that the rate of change of the slope is dependent on the y value of a point along the circle. For some reason, I want to believe that, conceptually, the second derivative of a circle is a constant, which produces the "circle" shape. Can someone please clear my misunderstanding?

The curvature of a circle is constant.
 
  • #6
Joseph Nechleba said:
Using the standard equation of a circle x^2 + y^2 = r^2, I took the first and second derivatives and obtained -x/y and -r^2/y^3 , respectively. I understand that the slope is going to be different at each point along the circle, but what does not make sense to me is that the rate of change of the slope is dependent on the y value of a point along the circle. For some reason, I want to believe that, conceptually, the second derivative of a circle is a constant, which produces the "circle" shape. Can someone please clear my misunderstanding?

A vertical parabola has constant second derivative (of its y coordinate as a function of its x coordinate), and only intersects a circle in at most 4 points. Conversely, if we start from the assumption that the second derivative of y is a constant function of x, we may use the Fundamental Theorem of Calculus to imply that the graph of y as a function of x is a parabola. I'm not sure why you believe a circle should have the same second derivative as a parabola.
The expressions you gave are the second derivatives of the y coordinate of a circle over intervals where that coordinate is a function of the x coordinate. The largest interval of x coordinates on which this occurs is [-1, 1], assuming we restrict ourselves to values of y that satisfy either [itex]y\geq 0[/itex] or [itex]y\leq 0[/itex], since otherwise we do not have y as a function of x. These semi-circles are not parabolas, as you can also see by solving for y algebraically and comparing the expression to the standard Cartesian coordinate equation of a parabola.
More geometrically, we note that the second derivative is supposed to tell us the slopes of tangent lines to the graph of the first derivative. The first derivative in turn tells us the slopes of tangent lines to the curve. However, at the endpoints of our semi-circles, the tangent line is vertical, which yields undefined, or infinite, first derivatives. The claim that the second derivative is a constant essentially implies that the graph of the first derivative function is a straight line. This contradicts the geometry of the semi-circle, since straight lines do not approach infinities near -1 or 1 (we are assuming the derivative is continuous, but this is easy enough to show separately).
 
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  • #7
Joseph Nechleba said:
For some reason, I want to believe that, conceptually, the second derivative of a circle is a constant, which produces the "circle" shape. Can someone please clear my misunderstanding?
The curvature of a circle is constant (1 over its radius) and the curvature is related to the second derivative but not equal to it. For y= f(x), the curvature is [tex]\frac{f''(x)}{(1+f'(x)^2)^{3/2}}[/tex]
 
  • #8
Svein said:
A circle is easiest given by x = r⋅cos(t), y = r⋅sin(t), t∈[0, 2π). Then [itex]\dot{x}= -r\cdot \sin(t) = -y [/itex] and [itex]\dot{y}= r\cdot \cos(t) = x [/itex]. Continuing, we get [itex] \ddot{x}=-r\cdot\cos(t)=-x[/itex] and [itex] \ddot{y}=-r\cdot\sin(t)=-y[/itex].
In differential geometry, the tangent vector at (x, y) is given by [itex] \vec{t}=\frac{\vec{\dot{r(t)}}}{\lvert \vec{\dot{r(t)}} \rvert}=\frac{(-r\sin(t), r \cos(t))}{\sqrt{r^{2}(\sin^{2}(t)+\cos^{2}(t)}}=((-\sin(t), \cos(t))[/itex]. The absolute value of the curvature is given by [itex]\lvert \kappa\rvert=\lvert \vec{t} '\rvert = \lvert \frac{1}{r} (-\cos(t), -\sin(t))\rvert=\frac{1}{r}[/itex].
 
  • #9
PWiz said:
Only polynomials of degree 2 or lower can have a constant second derivative.
EDIT: I'm talking about polynomials which only consist of the x variable.
##|x|^2## is not a polynomial (I think), and it has a constant second derivative.
 
  • #10
certainly said:
##|x|^2## is not a polynomial (I think), and it has a constant second derivative.

##|x|^2 = x^2## for all ##x##.
 
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  • #11
micromass said:
##|x|^2 = x^2## for all ##x##.
But analytically speaking aren't we first taking the absolute value and then squaring it, so it shouldn't be a polynomial right ? (I wasn't sure of this, that is why I said "I think")
 
  • #12
certainly said:
But analytically speaking aren't we first taking the absolute value and then squaring it, so it shouldn't be a polynomial right ? (I wasn't sure of this, that is why I said "I think")

The two functions ##f(x) = x^2## and ##g(x) = |x|^2## are equal. Since ##f## is a polynomial, so is ##g##. The function concept does not address which operation is done first, it only cares whether the inputs and the outputs correspond.
 
  • #13
Oh! I see, thanks!
 
  • #14
But there are also the floor and ceiling functions which have constant 2nd derivatives, where they are continuous and also the sawtooth function (the fractional part of ##x##) which will have a constant 2nd derivative when ##x## is not an integer.
 
  • #15
Every function defined and twice differentiable on an interval whose second derivative is constant is a polynomial.
 
  • #16
Hmmmm...I see your point of view.
[Edit:- I am, more or less self taught. So sometimes there exist these small gaps in my knowledge, that make me look like a complete beginner :-)]
 
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  • #17
micromass said:
Every function defined and twice differentiable on an interval whose second derivative is constant is a polynomial.
Proving that statement...
(D^2)y= k
( D^1) y = kx + C
y(x)= (1/2)k(x^2 ) + C x + K
 
  • #18
K,k,c are all constants of integration.
 
  • #19
At first glance your proposal seems logical, but does stand up against analysis.

You can disprove your 'intuitive' hypothesis here:

See "Tangent Line" slope here [item 4.3] :
https://en.wikipedia.org/wiki/Circle

Further, maybe think 'intuitively' of the velocity and acceleration [second derivative] perspective:
The tangent line 'slope' is constantly changing direction...so it MUST be a function, not a constant.
 
  • #20
Thank you to everyone who replied; everything makes a lot more sense!
 
  • #21
certainly said:
##|x|^2## is not a polynomial (I think), and it has a constant second derivative.

When ## x<0 , |x|= -x ; |x|^2= (-x)^2= x^2##. Similar for when ## x \geq 0 ##
 
  • #22
For unknown reasons, I can't edit my post post above...#19...
It SHOULD read "...but doesn't stand up against analysis..."
 

Related to Second Derivative of Circle Not a Constant?

1. What is the second derivative of a circle?

The second derivative of a circle is a constant value of 0. This means that the rate of change of the slope of a circle remains constant, and the curvature of a circle remains the same at all points.

2. How is the second derivative of a circle calculated?

The second derivative of a circle can be calculated by taking the derivative of the first derivative of the circle equation. Since the first derivative of a circle is a constant value of 0, the second derivative will also be 0.

3. What does the second derivative of a circle represent?

The second derivative of a circle represents the rate of change of the slope of a circle. This can be thought of as the acceleration of the curve at any given point on the circle.

4. Is the second derivative of a circle always 0?

Yes, the second derivative of a circle is always 0. This is because the slope of a circle is a constant value, and taking the derivative of a constant value results in 0.

5. Why is the second derivative of a circle important?

The second derivative of a circle is important because it helps us understand the behavior of a circle. It tells us that the curvature of a circle remains constant, and there is no change in the slope of the circle at any point. This can be useful in various applications such as designing curves and analyzing circular motion.

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