A curve is defined parametrically by
x=t2+1,y=t3−3tx=t^2+1,\qquad y=t^3-3tx=t2+1,y=t3−3t, where t>0t>0t>0.
Find dydx\dfrac{dy}{dx}dxdy in terms of ttt.
Show that
d2ydx2=3(t2+1)4t3\dfrac{d^2y}{dx^2}=\dfrac{3(t^2+1)}{4t^3}dx2d2y=4t33(t2+1).
Hence classify the stationary point corresponding to t=1t=1t=1.