Answer:
Step-by-step explanation:
[tex]\frac{dx}{dt} =-sin~t\\\frac{dy}{dt} =2 cos~t\\\frac{dy}{dx} =\frac{\frac{dy}{dt} }{\frac{dx}{dt} } =\frac{2 cos t}{-sin t} =-2 cot t\\when ~t=\frac{-\pi }{4} \\\frac{dy}{dx} =-2~cot (\frac{-\pi }{4} )\\\frac{dy}{dx} =-2(- cot \frac{\pi }{4} )=2(1)=2\\at ~t=-\frac{\pi }{4} \\x=cos (\frac{-\pi }{4} )=cos \frac{\pi }{4} =\frac{1}{\sqrt{2} } \\at~t=-\frac{\pi }{4} \\y=2 sin (\frac{-\pi }{4} )\\y=-2 ~sin (\frac{\pi }{4} )=-2 \times \frac{1}{\sqrt{2} }=\frac{-2}{\sqrt{2} } \\[/tex]
[tex]eq. ~of~tangent~is \\y-(\frac{-2}{\sqrt{2} } )=2(x-\frac{1}{\sqrt{2} } )\\y-y_{1}=m(x-x_{1})[/tex]
[tex]\frac{d^2y}{dx^2} =\frac{d}{dx} (\frac{dy}{dx} )=\frac{d}{dx} (-2~cot~t)\\=\frac{d}{dt} (-2~cot~t)\frac{dt}{dx} \\=-2(-cosec^2t) \times \frac{1}{\frac{dx}{dt} } \\=\frac{2 ~cosec^2 t}{-sin~t} \\=-2cosec^2~t (cosec~t)=-2 cosec ^3t[/tex]
=-2/(sin³t)
at t=-π/4
[tex]\frac{d^2y}{dx^2} =\frac{-2}{sin^3(\frac{-\pi }{4} )} \\\frac{d^2y}{dx^2} =\frac{-2}{-sin^3(\frac{\pi }{4} )} \\=\frac{2}{\frac{1}{\sqrt{2} } } \\=2\sqrt{2}[/tex]
