Sea [math]\alpha:\left(a,b\right)\rightarrow\mathbb{R}^2[/math] una curva regular. La curvatura [math]k\left[\alpha\right][/math] de [math]\alpha[/math] esta dada por la fórmula:[br][br][math]k\left[\alpha\right]\left(t\right)=\frac{a''\left(t\right)\cdot Ja'\left(t\right)}{\mid\mid a'\left(t\right)\mid\mid^3}[/math][br][br]Donde J es una transformación a el vector velocidad (rotación de 90 grados).

sea [math]\beta:\left(0,2\right)\rightarrow\mathbb{R}^2[/math], [math]\beta\left(t\right)=\left(t^2,3t^3\right)[/math] con [math]t=1[/math], nuestro vector es [math]\left(1,3\right)[/math][br]velocidad: [math]\beta'\left(t\right)=\left(2\left(1\right),9\left(1\right)^2\right)=\left(2,9\right)\rightarrow J\left(2,9\right)=\left(-9,2\right)[/math][br]aceleración: [math]\beta''\left(t\right)=\left(2,18\left(1\right)\right)=\left(2,18\right)[/math][br][br][math]k\left(t\right)=\frac{\left(2,18\right)\cdot\left(-9,2\right)}{\left(\sqrt{2^2+9^2}\right)^3}=\frac{-18+36}{\simeq721}=\frac{18}{721}=0.02[/math][br][br]luego, 0.02 es la curvatura cuanto [math]t=1[/math][br][br]Para términos generales hacemos:[br][br][math]k\left(t\right)=\frac{\alpha''\left(t\right)\cdot J\left(\alpha'\left(t\right)\right)}{\mid\mid\alpha´\left(t\right)\mid\mid^3}=\frac{\left(2,18t\right)\cdot J\left(2t,9t^2\right)}{\left(\sqrt{4t^2+81t^4}\right)^3}=\frac{\left(2,18t\right)\cdot\left(-9t^2,2t\right)}{\left(t\sqrt{4+81t^2}\right)^3}=\frac{-18t^2+36t^2}{t^3\left(\sqrt{4+81t^2}\right)^3}=\frac{18t^2}{t^3\left(\sqrt{4+81t^2}\right)^3}[/math][br][br][br][br]