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Find the unit tangent vector T(t) and find a set of parametric equations for the line tangent to the space curve at point P. r(t) = ⟨2 sin t, 2 cos t, 4 sin² t⟩ P(1, √3, 1)


Answered 9 months ago
Answered 9 months ago
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r(t)=2sint,2cost,4sin2t,P(1,3,1)\color{#4257b2}{\mathbf{r}(t)=\left\langle 2 \sin t, 2 \cos t, 4 \sin ^{2} t\right\rangle, \quad P(1, \sqrt{3}, 1)}

\hspace*{5mm}Now is

r(t)=4+64sin2tcos2tT(t)=r(t)r(t)=2cost,2sint,8sintcost4+64sin2tcos2t\begin{align*} \left\|\vec{r}^{\prime}(t)\right\|&=\sqrt{4+64 \sin ^{2} t \cos ^{2} t} \\ \vec{T}(t)&=\frac{\vec{r}^{\prime}(t)}{\left\|\vec{r}^{\prime}(t)\right\|}\\ &=\frac{\langle 2 \cos t,-2 \sin t, 8 \sin t \cos t\rangle}{\sqrt{4+64 \sin ^{2} t \cos ^{2} t}} \end{align*}

\hspace*{5mm}For t=π6t=\frac{\pi}{6} is

T(π6)=2cos(π6),2sin(π6),8sin(π6)cos(π6)4+64sin2(π6)cos2(π6)=3,1,2316\begin{align*} T\left(\frac{\pi}{6}\right)&=\frac{\left\langle 2 \cos \left(\frac{\pi}{6}\right),-2 \sin \left(\frac{\pi}{6}\right), 8 \sin \left(\frac{\pi}{6}\right) \cos \left(\frac{\pi}{6}\right)\right\rangle}{\sqrt{4+64 \sin ^{2}\left(\frac{\pi}{6}\right) \cos ^{2}\left(\frac{\pi}{6}\right)}} \\ &=\frac{\langle\sqrt{3},-1,2 \sqrt{3}\rangle}{\sqrt{16}} \end{align*}

\hspace*{5mm}The direction number of the tangent vector are

a=3,b=1,c=23a=\sqrt{3}, b=-1, c=2 \sqrt{3}

\hspace*{5mm}So then is

x=1+3ty=3tz=1+23t\color{#c34632}{\begin{matrix}x=1+\sqrt{3} t \\ y=\sqrt{3}-t \\ z=1+2 \sqrt{3} t \end{matrix}}

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