To find the curvature of a parametric curve given by \( r(t) = \langle x(t), y(t) \rangle \), we use the formula for curvature \( \kappa \) of a plane curve:
\[
\kappa = \frac{|x'(t)y''(t) - y'(t)x''(t)|}{(x'(t)^2 + y'(t)^2)^{3/2}}
\]
- Compute the first derivatives \( x'(t) \) and \( y'(t) \).
- Compute the second derivatives \( x''(t) \) and \( y''(t) \).
- Substitute these into the curvature formula.
- Evaluate the curvature at \( t = \frac{\pi}{3} \).
Dado \( r(t) = \langle t, \sin t \rangle \), calculamos las derivadas primera y segunda de \( x(t) = t \) y \( y(t) = \sin t \):
- \( x'(t) = 1 \)
- \( y'(t) = \cos t \)
- \( x''(t) = 0 \)
- \( y''(t) = -\sin t \)
La fórmula para la curvatura \( \kappa \) de una curva en el plano es:
\[
\kappa = \frac{|x'(t)y''(t) - y'(t)x''(t)|}{(x'(t)^2 + y'(t)^2)^{3/2}}
\]
Sustituimos las derivadas calculadas:
\[
\kappa = \frac{|1 \cdot (-\sin t) - \cos t \cdot 0|}{(1^2 + (\cos t)^2)^{3/2}} = \frac{|\sin t|}{(1 + \cos^2 t)^{3/2}}
\]
Sustituimos \( t = \frac{\pi}{3} \) en la expresión de la curvatura:
\[
\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, \quad \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}
\]
\[
\kappa = \frac{\left|\frac{\sqrt{3}}{2}\right|}{\left(1 + \left(\frac{1}{2}\right)^2\right)^{3/2}} = \frac{\frac{\sqrt{3}}{2}}{\left(1 + \frac{1}{4}\right)^{3/2}} = \frac{\frac{\sqrt{3}}{2}}{\left(\frac{5}{4}\right)^{3/2}}
\]
Calculamos el denominador:
\[
\left(\frac{5}{4}\right)^{3/2} = \left(\frac{5}{4}\right)^{1.5} \approx 1.6818
\]
Por lo tanto, la curvatura es:
\[
\kappa \approx \frac{0.8660}{1.6818} \approx 0.6197
\]
La curvatura de la curva en \( t = \frac{\pi}{3} \) es \(\boxed{0.6197}\).
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