sketch the astroid in curves also calculate its tangent vector at each point .at which point is the tangent vector zero:
(i) γ(t) = (cos^2 t, sin^2 t)
(ii) γ(t) = (e^t, t^2)
x(t)=cos2t,y(t)=sin2(t)⇒x+y=1,x,y∈[0,1]x\left( t \right) =\cos ^2t,y\left( t \right) =\sin ^2\left( t \right) \Rightarrow x+y=1,x,y\in \left[ 0,1 \right]x(t)=cos2t,y(t)=sin2(t)⇒x+y=1,x,y∈[0,1]
The tangent vector is 0 if {x′(t)=0y′(t)=0⇒{2sintcost=02sintcost=0⇒t=πn2\left\{ \begin{array}{c} x'\left( t \right) =0\\ y'\left( t \right) =0\\\end{array} \right. \Rightarrow \left\{ \begin{array}{c} 2\sin t\cos t=0\\ 2\sin t\cos t=0\\\end{array} \right. \Rightarrow t=\frac{\pi n}{2}{x′(t)=0y′(t)=0⇒{2sintcost=02sintcost=0⇒t=2πn
x(t)=et,y(t)=t2⇒y(x)=ln2xx\left( t \right) =e^t,y\left( t \right) =t^2\Rightarrow y\left( x \right) =\ln ^2xx(t)=et,y(t)=t2⇒y(x)=ln2x
The tangent vector is 0 if {x′(t)=0y′(t)=0⇒{et=02t=0⇒t∈∅\left\{ \begin{array}{c} x'\left( t \right) =0\\ y'\left( t \right) =0\\\end{array} \right. \Rightarrow \left\{ \begin{array}{c} e^t=0\\ 2t=0\\\end{array} \right. \Rightarrow t\in \emptyset{x′(t)=0y′(t)=0⇒{et=02t=0⇒t∈∅
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