Curvature of a graph

Question

How to calculate the formula of the curvature of $\alpha= (t, f(t)), f$ is $C^{\infty}$ ?

I know that can be done easily using previous curvature formulas, but I couldn't do using jus the curvature definition $k(t)= \langle t'(t),n(t)\rangle$, where $t(t)$ is the unit tangent vector.

Answer 1

This is a reasonable question. But you have an error. Your formula for curvature is assuming that $t$ is the arclength parameter, which, of course it isn't. So when you working with a non-arclength parametrized curve you must use the chain rule to correct the computation. [See my differential geometry text linked in my profile for lots of examples.]

Since the independent variable is $t$, I'm going to use capital letters for the unit tangent and normal. Note that $$T(t)=\lambda(t) \big(1,f'(t)\big), \quad\text{where } \lambda(t) = (1+f'(t)^2)^{-1/2}.$$ When we differentiate by the product rule, we get two terms, one a multiple of $\alpha'(t)=(1,f'(t))$. This term will disappear when we dot with the unit normal. The term that remains is $\lambda(t) \big(0,f''(t)\big)$. When we dot with the unit normal $N(t)=\pm \lambda(t) \big(-f'(t),1\big)$, we get $$\pm\lambda(t) \big(0,f''(t)\big)\cdot \lambda(t) \big(-f'(t),1\big) = \pm \frac{f''(t)}{1+f'(t)^2}=\frac{|f''(t)|}{1+f'(t)^2}.$$ Ah, but the exponent is wrong on the denominator. This is because we need to correct by the chain rule: $$\frac{dT}{ds}= \frac{dT/dt}{ds/dt},$$ where, of course, $ds/dt = \|\alpha'(t)\| = \big(1+f'(t)^2\big)^{1/2}$. This gives us an extra factor of $\lambda$ and we have the correct answer, $\kappa(t)=\dfrac{|f''(t)|}{\big(1+f'(t)^2\big)^{3/2}}$.

Answer 2

If $\gamma: \mathbb{R} \to \mathbb{R}^3$ is a curve then the curvature at $t$ is given by;

$$\kappa(t) = \frac{\|\gamma'(t) \times \gamma''(t)\|}{\|\gamma'(t)\|^3}$$

If you wish to find the curvature of a graph for $f: \mathbb{R} \to \mathbb{R}$ then you parametrize by $\alpha(x) = \langle x, f(x), 0 \rangle$. Now if you follow that above formula you get;

$$\kappa(x) = \frac{\left|f''(x)\right|}{\left(1+f'(x)^2\right)^{\frac{3}{2}}}$$

Answer 3

Given the curve

$\alpha(t) = (t, f(t)), \tag 1$

we have its tangent vector

$\alpha'(t) = (1, f'(t)) \ne 0; \tag 2$

thus the curve $\alpha(t)$ is regular, and the typical differential geometric analysis may proceed, to wit: the magnitude of said tangent vector is

$\vert \alpha'(t) \vert = \sqrt{1 + (f'(t))^2} = (1 + (f'(t))^2)^{1/2}, \tag 3$

and hence the speed, or rate of change of arc-length with respect to $t$, is

$\dfrac{ds}{dt} = \vert \alpha'(t) \vert = (1 + (f'(t))^2)^{1/2}; \tag 4$

we shall also have occasion to make use of the quantity reciprocal to speed,

$\dfrac{dt}{ds} = \vert \alpha'(t) \vert^{-1} = (1 + (f'(t))^2)^{-1/2} = \dfrac{1}{(1 + (f'(t))^2)^{1/2}}. \tag 5$

The unit tangent vector to $\alpha(t)$ is

$T(t) = \dfrac{\alpha'(t)}{\vert \alpha'(t) \vert}$ $= \left (\dfrac{1}{(1 + (f'(t))^2)^{1/2}}, \dfrac{f'(t)}{(1 + (f'(t))^2)^{1/2}} \right )$ $= ((1 + (f'(t))^2)^{-1/2}, (1 + (f'(t))^2)^{-1/2}f'(t)), \tag 6$

and thus the curvature $\kappa(t)$ is defined in terms of the $s$-derivative of $T(t)$ via the first Frenet-Serret equation

$\kappa(t)N(t) = \dfrac{dT(t)}{ds}; \tag 7$

here we take $N(t)$ to be a unit vector,

$N(t) \cdot N(t) = 1, \tag 8$

which is orthogonal to $T(t)$ since $T(t)$ itself is of magnitude $1$; that is,

$T(t) \cdot T(t) = 1 \tag 9$

implies

$T(t) \cdot \dfrac{dT(t)}{ds} = \dfrac{1}{2} \dfrac{d(T(t) \cdot T(t))}{ds} = \dfrac{d(1)}{ds} = 0, \tag{10}$

and since $N(t)$ is collinear with $dT(t)/ds$ in accord with (7), we see that it too is orthogonal to $T(t)$. In light of these considerations, we call $N(t)$ the unit normal vector field to the curve $\alpha(t)$.

Nota Bene: The astute reader may have observed that the signs of $\kappa(t)$ and $N(t)$ are not in fact fixed by (7), but that the sign of each may be reversed whilst still preserving the validity of this equation. Thus (7) admits the possibility of negative curvature; we observe, however, that taking $\kappa(t) < 0$ forces $N(t)$ to point in a direction opposite to that of the turning of of $T(t)$ along $\alpha(t)$; therefore we along with many others will take $\kappa(t) \ge 0$, noting that when $\kappa(t) = 0$, $N(t)$ cannot be defined via (7). $\kappa(t) = 0$ clearly occurs precisely when (that is, is forced by) the condition $dT(s)/ds = 0$. End of Note.

We may compute $dT(t)/ds$ by means of (5) and the chain rule

$\dfrac{dT(t)}{ds} = \dfrac{dT(t)}{dt}\dfrac{dt}{ds}; \tag 8$

now we have

$\dfrac{dT(t)}{dt} = \left (-\dfrac{1}{2}(1 + (f'(t))^2)^{-3/2}(2f'(t)f''(t)), -\dfrac{1}{2}(1 + (f'(t))^2)^{-3/2}(2f'(t)f''(t))f'(t) + (1 + (f'(t))^2)^{-1/2}f''(t) \right )$ $= \left (-(1 + (f'(t))^2)^{-3/2}f'(t)f''(t), -(1 + (f'(t))^2)^{-3/2}(f'(t))^2(f''(t)) + (1 + (f'(t))^2)^{-1/2}f''(t) \right )$ $= \left ( -\dfrac{f'(t)f''(t)}{(1 + (f'(t))^2)^{3/2}}, -\dfrac{(f'(t))^2f''(t)}{(1 + (f'(t))^2)^{3/2}} + \dfrac{f''(t)}{(1 + (f'(t))^2)^{1/2}} \right ), \tag 9$

and thus

$\dfrac{dT(t)}{ds} = \dfrac{dT(t)}{dt} \dfrac{dt}{ds}$ $= \dfrac{1}{ (1 + (f'(t))^2)^{1/2}}\left ( -\dfrac{f'(t)f''(t)}{(1 + (f'(t))^2)^{3/2}}, -\dfrac{(f'(t))^2(f''(t)}{(1 + (f'(t))^2)^{3/2}} + \dfrac{f''(t)}{(1 + (f'(t))^2)^{1/2}} \right )$ $= \left ( -\dfrac{f'(t)f''(t)}{(1 + (f'(t))^2)^2}, -\dfrac{(f'(t))^2(f''(t)}{(1 + (f'(t))^2)^2} + \dfrac{f''(t)}{1 + (f'(t))^2} \right )$ $= \left ( -\dfrac{f'(t)f''(t)}{(1 + (f'(t))^2)^2}, -\dfrac{(f'(t))^2(f''(t)}{(1 + (f'(t))^2)^2} + \dfrac{(1 + (f'(x))^2)f''(t)}{(1 + (f'(t))^2)^2} \right )$ $= \dfrac{f''(t)}{(1 + (f'(t))^2)^{3/2}}\left ( -\dfrac{f'(t)}{(1 + (f'(t))^2)^{1/2}}, -\dfrac{f'(t))^2(f''(t)}{(1 + (f'(t))^2)^{1/2}} + \dfrac{(1 + (f'(x))^2)f''(t)}{(1 + (f'(t))^2)^{1/2}} \right )$ $= \dfrac{f''(t)}{(1 + (f'(t))^2)^{3/2}}\left ( -\dfrac{f'(t)}{(1 + (f'(t))^2)^{1/2}}, \dfrac{1}{(1 + (f'(t))^2)^{1/2}} \right ); \tag{10}$

scrutiny of this equation reveals that the vector

$\left ( -\dfrac{f'(t)}{(1 + (f'(t))^2)^{1/2}}, \dfrac{1}{(1 + (f'(t))^2)^{1/2}} \right ) \tag{11}$

is both of unit magnitude and normal to $T(t)$; thus in keeping with the above paragraph titled Nota Bene, the vector $dT(t)/ds$ is aligned along (11) when $f''(t) > 0$; when $f''(t) < 0$, $dT(t)/ds$ and (11) point in opposite directions. However, by reversing if necessary the signs of both (11) and the curvature factor

$\dfrac{f''(t)}{(1 + (f'(t))^2)^{3/2}}, \tag{12}$

we attain a situation in which both $\kappa(t)$ positive and (11) is in the direction of $dT(t)/ds$ and not against it. Thus we take the curvature to be

$\kappa(t) = \dfrac{\vert f''(t) \vert}{(1 + (f'(t))^2)^{3/2}}, \tag{12}$

in accord with answers of both Faraad Armwood and Ted Shifting given here.