Question #265916

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

1
Expert's answer
2021-11-15T16:38:53-0500
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)×r(t)=ijk(sint)/2cost(3sint)/2(cost)/2sint(3cost)/2r'(t)\times r''(t)=\begin{vmatrix} i & j & k \\ -(\sin t)/2 & -\cos t & (3\sin t)/2\\ -(\cos t)/2 & \sin t & (3\cos t)/2 \end{vmatrix}


=icost(3sint)/2sint(3cost)/2=i\begin{vmatrix} -\cos t & (3\sin t)/2 \\ \sin t & (3\cos t)/2 \end{vmatrix}

j(sint)/2(3sint)/2(cost)/2(3cost)/2-j\begin{vmatrix} -(\sin t)/2 & (3\sin t)/2 \\ -(\cos t)/2 & (3\cos t)/2 \end{vmatrix}


+k(sint)/2cost(cost)/2sint+k\begin{vmatrix} -(\sin t)/2 &-\cos t \\ -(\cos t)/2 & \sin t \end{vmatrix}

=32i+34sintcostj12k=-\dfrac{3}{2}i+\dfrac{3}{4}\sin t\cos tj-\dfrac{1}{2}k


r(t)×r(t)=94+916sin2tcos2t+14|r'(t) \times r''(t)|=\sqrt{\dfrac{9}{4}+\dfrac{9}{16}\sin^2t\cos^2t+\dfrac{1}{4}}

=1440+9sin2tcos2t=\dfrac{1}{4}\sqrt{40+9\sin^2t\cos^2t}

r(t)=14sin2t+cos2t+94sin2t|r'(t)|=\sqrt{\dfrac{1}{4}\sin^2t+\cos^2t+\dfrac{9}{4}\sin^2t}

=124+6sin2t=\dfrac{1}{2}\sqrt{4+6\sin^2t}

k=r(t)×r(t)r(t)3=1440+9sin2tcos2t18(4+6sin2t)4+6sin2tk=\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}=\dfrac{\dfrac{1}{4}\sqrt{40+9\sin^2t\cos^2t}}{\dfrac{1}{8}(4+6\sin^2t)\sqrt{4+6\sin^2t}}

=40+9sin2tcos2t(2+3sin2t)4+6sin2t=\dfrac{\sqrt{40+9\sin^2t\cos^2t}}{(2+3\sin^2t)\sqrt{4+6\sin^2t}}



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