<s>[INST] Parametric equations - stating values. A curve is defined by the parametric equations: $x=\cos2t, y=\sin2t, 0<t<π.$
a) Use parametric equations to find $\frac{dy}{dx}$. Hence find the equation of the tangent when $t=\frac{π}{8}$.
b) Obtain an expression for $\frac{d^2y}{dx^2}$ and hence show that: $\sin2t\left(\frac{d^2y}{dx^2}\right)+\left(\frac{dy}{dx}\right)^2=k$, where $k$ is an integer. State the value of $k$.
 [/INST] Using $$\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
So here $x=\cos 2t\;,$ Then $$\displaystyle \frac{dx}{dt} = -2\sin 2t$$ and $y=\sin 2t\;,$ Then $$\displaystyle \frac{dy}{dt} = 2\cos 2t$$
So $$\displaystyle \frac{dy}{dx} = -\frac{2\cos 2t}{2\sin 2t} = -\cot 2t\;,$$ Now $$\displaystyle \left(\frac{dy}{dx}\right)_{t=\frac{\pi}{8}} = -\left[\cot 2t \right]_{t=\frac{\pi}{8}} = -1$$
Now $$\displaystyle \frac{d}{dx}\left(\frac{dy}{dx}\right) = -\frac{d}{dx}(\cot 2t) =-\frac{d}{dt}(\cot 2t)\cdot \frac{dt}{dx} = 2\csc^2 2t\cdot \frac{1}{-2\sin 2t} $$
So $$\displaystyle \frac{d^2y}{dx^2} = -\frac{1}{\sin^3 2t}$$
Now $$\displaystyle \sin 2t\cdot \frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2=-\frac{1}{\sin^2 2t}+\frac{\cos^2 2t}{\sin^2 2t} = -\frac{1}{\sin^2 2t}(1-\cos^2 2t)$$
So we get $$\displaystyle \sin 2t\cdot \frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2 = -\frac{\sin^2 2t}{\sin^2 2t} = -1=k$$(Given)
So we get $$k=-1$$
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