Question #265924

Find curvature r(t) = ( ½ Cost, 1-sint, -3/2 cost)

1
Expert's answer
2022-01-18T06:15:55-0500
r(t)=12cost,1sint,32costr(t)=\langle\dfrac{1}{2}\cos t, 1-\sin t, -\dfrac{3}{2}\cos t\rangle

r(t)=12sint,cost,32sintr'(t)=\langle-\dfrac{1}{2}\sin t, -\cos t, \dfrac{3}{2}\sin t\rangle

r(t)=12cost,sint,32costr''(t)=\langle-\dfrac{1}{2}\cos t, \sin t, \dfrac{3}{2}\cos t\rangle

r(t)=(12sint)2+(cost)2+(32sint)2|r'(t)|=\sqrt{(-\dfrac{1}{2}\sin t)^2+( -\cos t)^2+(\dfrac{3}{2}\sin t)^2}

=106cos2t2=\dfrac{\sqrt{10-6\cos ^2t}}{2}

r(t)×r(t)=ijk12sintcost32sint12costsint32costr'(t)\times r''(t)=\begin{vmatrix} i & j & k \\ \\ -\dfrac{1}{2}\sin t & -\cos t &\dfrac{3}{2}\sin t \\ \\ -\dfrac{1}{2}\cos t & \sin t &\dfrac{3}{2}\cos t \end{vmatrix}

=icost32sintsint32costj12sint32sint12cost32cost=i\begin{vmatrix} -\cos t & \dfrac{3}{2}\sin t \\ \\ \sin t & \dfrac{3}{2}\cos t \end{vmatrix}-j\begin{vmatrix} -\dfrac{1}{2}\sin t & \dfrac{3}{2}\sin t \\ \\ -\dfrac{1}{2}\cos t& \dfrac{3}{2}\cos t \end{vmatrix}

+k12sintcost12costsint=32i12k+k\begin{vmatrix} -\dfrac{1}{2}\sin t & -\cos t \\ \\ -\dfrac{1}{2}\cos t & \sin t \end{vmatrix}=-\dfrac{3}{2}i-\dfrac{1}{2}k

r(t)×r(t)=(32)2+(0)2+(12)2=102|r'(t)\times r''(t)|=\sqrt{(-\dfrac{3}{2})^2+(0)^2+(-\dfrac{1}{2})^2}=\dfrac{\sqrt{10}}{2}

Find curvature


κ(t)=r(t)×r(t)(r(t))3\kappa(t)=\dfrac{|r'(t)\times r''(t)|}{(|r'(t)|)^{3}}

=102(106cos2t2)3=\dfrac{\dfrac{\sqrt{10}}{2}}{(\dfrac{\sqrt{10-6\cos ^2t}}{2})^{3}}

=410(106cos2t)3/2=\dfrac{4\sqrt{10}}{(10-6\cos ^2t)^{3/2}}

=25(53cos2t)3/2=\dfrac{2\sqrt{5}}{(5-3\cos ^2t)^{3/2}}

κ(t)=25(53cos2t)3/2\kappa(t)=\dfrac{2\sqrt{5}}{(5-3\cos ^2t)^{3/2}}


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