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Second Derivative of Circle Not a Constant?

  1. Jun 8, 2015 #1
    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?
  2. jcsd
  3. Jun 8, 2015 #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.
    Last edited: Jun 8, 2015
  4. Jun 8, 2015 #3
    You take the derivative of what with respect to what? What function are you trying to find the derivative of?

    What does "derivative of a circle" mean? You can only take derivatives of functions.
  5. Jun 8, 2015 #4


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    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].
  6. Jun 8, 2015 #5


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    The curvature of a circle is constant.
  7. Jun 8, 2015 #6
    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).
    Last edited: Jun 8, 2015
  8. Jun 9, 2015 #7


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    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]
  9. Jun 9, 2015 #8


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    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].
  10. Jun 9, 2015 #9
    ##|x|^2## is not a polynomial (I think), and it has a constant second derivative.
  11. Jun 9, 2015 #10
    ##|x|^2 = x^2## for all ##x##.
  12. Jun 9, 2015 #11
    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")
  13. Jun 9, 2015 #12
    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.
  14. Jun 9, 2015 #13
    Oh! I see, thanks!
  15. Jun 9, 2015 #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.
  16. Jun 9, 2015 #15
    Every function defined and twice differentiable on an interval whose second derivative is constant is a polynomial.
  17. Jun 9, 2015 #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 :-)]
    Last edited: Jun 9, 2015
  18. Jun 12, 2015 #17
    Proving that statement.....
    (D^2)y= k
    ( D^1) y = kx + C
    y(x)= (1/2)k(x^2 ) + C x + K
  19. Jun 12, 2015 #18
    K,k,c are all constants of integration.
  20. Jun 12, 2015 #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] :

    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.
  21. Jun 21, 2015 #20
    Thank you to everyone who replied; everything makes a lot more sense!
  22. Jun 21, 2015 #21


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    When ## x<0 , |x|= -x ; |x|^2= (-x)^2= x^2##. Similar for when ## x \geq 0 ##
  23. Jun 30, 2015 #22
    For unknown reasons, I can't edit my post post above.....#19...
    It SHOULD read "....but doesn't stand up against analysis....."
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