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Sketch the curve represented by the vector-valued function and give the orientation of the curve. r(t) = ⟨cos t + t sin t, sin t - t cos t, t⟩


Answered 9 months ago
Answered 9 months ago
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We have

x(t)=cost+tsint,y(t)=sinttcost,andz(t)=t\begin{align*} x(t)=\cos t+t\sin t,&&y(t)=\sin t-t\cos t,&&\text{and}&&z(t)=t \end{align*}


x2+y2=(cost+tsint)2+(sinttcost)2=cos2(t)+2tcos(t)sin(t)+t2sin2(t)+sin2(t)2tsin(t)cos(t)+t2cos2(t)=t2sin2(t)+t2cos2(t)+cos2(t)+sin2(t)=t2(sin2(t)+cos2(t))=1+1=t2+1\begin{align*} x^2+y^2&=\left(\cos t+t\sin t\right)^2+\left(\sin t-t\cos t\right)^2\\ &=\cos ^2\left(t\right)+2t\cos \left(t\right)\sin \left(t\right)+t^2\sin ^2\left(t\right)+\sin ^2\left(t\right)-2t\sin \left(t\right)\cos \left(t\right)+t^2\cos ^2\left(t\right)\\ &=t^2\sin ^2\left(t\right)+t^2\cos ^2\left(t\right)+\cos ^2\left(t\right)+\sin ^2\left(t\right)\\ &=t^2\underbrace{\left(\sin ^2\left(t\right)+\cos ^2\left(t\right)\right)}_{\color{#c34632}=1}+1\\ &=t^2+1 \end{align*}

Substitute z=tz=t in the above equation.

x2+y2=z2+1x2+y2z2=1\begin{align*} x^2+y^2&=z^2+1\\ x^2+y^2-z^2&=1\\ \end{align*}

The required curve and the orientation of the curve is shown below:

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