Mathematics, 09.07.2020 02:01 DnakMeme6e7272727272
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)
Answers: 2
.![\kappa = \frac{|\dot {x}\cdot \ddot {y}-\dot{y}\cdot \ddot{x}|}{[\dot{x}^{2}+\dot{y}^{2}]^{\frac{3}{2} }}](/tpl/images/0703/3969/72543.png)
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![\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}}} }](/tpl/images/0703/3969/4ebef.png)
![\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} }}](/tpl/images/0703/3969/60f6a.png)
![\kappa = \frac{72\cdot e^{2\cdot t}}{[72\cdot e^{2\cdot t}]^{\frac{3}{2} } }](/tpl/images/0703/3969/cae4a.png)
![\kappa = [72\cdot e^{2\cdot t}]^{-\frac{1}{2} }](/tpl/images/0703/3969/4a04f.png)
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The curvature of a plane parametric curve x = f(t), y = g(t) is $ \kappa = \dfrac{|\dot{x} \ddot{y}...
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