The curvature of a plane parametric curve x = f(t), y = g(t) is $ \kappa = \dfrac{|\dot{x} \ddot{y} - \dot{y} \ddot{x}|}{[\dot{x}^2 + \dot{y}^2]^{3/2}}$ where the dots indicate derivatives with respect to t. Use the above formula to find the curvature. x = 6et cos(t), y = 6et sin(t)

Respuesta :

Answer:

The curvature is modelled by [tex]\kappa = \frac{e^{-t}}{6\sqrt{2}}[/tex].

Step-by-step explanation:

The equation of the curvature is:

[tex]\kappa = \frac{|\dot {x}\cdot \ddot {y}-\dot{y}\cdot \ddot{x}|}{[\dot{x}^{2}+\dot{y}^{2}]^{\frac{3}{2} }}[/tex]

The parametric componentes of the curve are:

[tex]x = 6\cdot e^{t} \cdot \cos t[/tex] and [tex]y = 6\cdot e^{t}\cdot \sin t[/tex]

The first and second derivative associated to each component are determined by differentiation rules:

First derivative

[tex]\dot{x} = 6\cdot e^{t}\cdot \cos t - 6\cdot e^{t}\cdot \sin t[/tex] and [tex]\dot {y} = 6\cdot e^{t}\cdot \sin t + 6\cdot e^{t} \cdot \cos t[/tex]

[tex]\dot x = 6\cdot e^{t} \cdot (\cos t - \sin t)[/tex] and [tex]\dot {y} = 6\cdot e^{t}\cdot (\sin t + \cos t)[/tex]

Second derivative

[tex]\ddot{x} = 6\cdot e^{t}\cdot (\cos t-\sin t)+6\cdot e^{t} \cdot (-\sin t -\cos t)[/tex]

[tex]\ddot x = -12\cdot e^{t}\cdot \sin t[/tex]

[tex]\ddot {y} = 6\cdot e^{t}\cdot (\sin t + \cos t) + 6\cdot e^{t}\cdot (\cos t - \sin t)[/tex]

[tex]\ddot{y} = 12\cdot e^{t}\cdot \cos t[/tex]

Now, each term is replaced in the the curvature equation:

[tex]\kappa = \frac{|6\cdot e^{t}\cdot (\cos t - \sin t)\cdot 12\cdot e^{t}\cdot \cos t-6\cdot e^{t}\cdot (\sin t + \cos t)\cdot (-12\cdot e^{t}\cdot \sin t)|}{\left\{\left[6\cdot e^{t}\cdot (\cos t - \sin t)\right]^{2}+\right[6\cdot e^{t}\cdot (\sin t + \cos t)\left]^{2}\right\}^{\frac{3}{2}}} }[/tex]

And the resulting expression is simplified by algebraic and trigonometric means:

[tex]\kappa = \frac{72\cdot e^{2\cdot t}\cdot \cos^{2}t-72\cdot e^{2\cdot t}\cdot \sin t\cdot \cos t + 72\cdot e^{2\cdot t}\cdot \sin^{2}t+72\cdot e^{2\cdot t}\cdot \sin t \cdot \cos t}{[36\cdot e^{2\cdot t}\cdot (\cos^{2}t -2\cdot \cos t \cdot \sin t +\sin^{2}t)+36\cdot e^{2\cdot t}\cdot (\sin^{2}t+2\cdot \cos t \cdot \sin t +\cos^{2} t)]^{\frac{3}{2} }}[/tex]

[tex]\kappa = \frac{72\cdot e^{2\cdot t}}{[72\cdot e^{2\cdot t}]^{\frac{3}{2} } }[/tex]

[tex]\kappa = [72\cdot e^{2\cdot t}]^{-\frac{1}{2} }[/tex]

[tex]\kappa = 72^{-\frac{1}{2} }\cdot e^{-t}[/tex]

[tex]\kappa = \frac{e^{-t}}{6\sqrt{2}}[/tex]

The curvature is modelled by [tex]\kappa = \frac{e^{-t}}{6\sqrt{2}}[/tex].