I am trying to find the curvature of the parabola $f(x) = x^2$ at the point $x=0$.
For this I decided to use this simple definition of the curvature: it's the limit of the tangent angle change when the distance travelled (on the curve) goes to zero.
I did some calculations by hand and I think this approach leads me to this limit.
$$\lim_{a \to 0} \frac{arctan(2a)}{ (0.5 \cdot a \cdot \sqrt{4a^2+1}) + 0.25 \cdot \ln{(\sqrt{4a^2+1} + 2a)} } $$
Is my limit correct?
I think it should be, the denominator is the length of the curve from $(0,0)$ to $(a,a^2)$, right?
Can I compute this limit in a simple way? How?
Here $arctan$ is the inverse of the $\tan$ function.
Btw, WA is giving me that the answer is 2 for the limit above.
limit ( (arctan(2x) / ((0.5 * x * sqrt(4x^2+1)) + (0.25 * ln(sqrt(4x^2+1) + 2x)) ) ) ) as x goes to 0
Is 2 the correct curvature of $y=x^2$ at the point $(0,0)$ ?
For the following parametrization $(x,f(x))=(x,x^2)$ the curvature is given by
$$\kappa =\frac{|f''(t)|}{\left(1+f'(t)^2\right)^\frac32}=\frac{2}{\left(1+4x^2\right)^\frac32}$$
which is equal to $2$ at $x=0$.
Following your idea and more simply avoiding integration, by l’Hospital we have
$$\kappa =\lim_{a\to 0}\frac{\arctan(2a)}{\int_0^a \sqrt{1+4x^2}dx}=\lim_{a\to 0}\frac{\frac{2}{1+4a^2}}{\sqrt{1+4a^2}}=2$$